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15900	https://www.chegg.com/homework-help/questions-and-answers/2-relativistic-doppler-effect-describes-observed-frequency-relatively-moving-source-change-q91654895	Solved 2. The relativistic Doppler effect describes how the \| Chegg.com Skip to main content Books Rent/Buy Read Return Sell Study Tasks Homework help Understand a topic Writing & citations Tools Expert Q&A Math Solver Citations Plagiarism checker Grammar checker Expert proofreading Career For educators Help Sign in Paste Copy Cut Options Upload Image Math Mode ÷ ≤ ≥ o π ∞ ∩ ∪           √  ∫              Math Math Geometry Physics Greek Alphabet Science Advanced Physics Advanced Physics questions and answers 2. The relativistic Doppler effect describes how the observed frequency of a relatively moving source changes with respect to the proper frame of the source. In Fig. (1) a satellite is shown to be moving parallel with the ground at a height h with a velocity v toward an observer at 0. x x h 1 'L e Ground Figure 1: An approaching satellite moving parallel to Your solution’s ready to go! Our expert help has broken down your problem into an easy-to-learn solution you can count on. See Answer See Answer See Answer done loading Question: 2. The relativistic Doppler effect describes how the observed frequency of a relatively moving source changes with respect to the proper frame of the source. In Fig. (1) a satellite is shown to be moving parallel with the ground at a height h with a velocity v toward an observer at 0. x x h 1 'L e Ground Figure 1: An approaching satellite moving parallel to Show transcribed image text Here’s the best way to solve it.Solution Share Share Share done loading Copy link (a… View the full answer Previous questionNext question Transcribed image text: The relativistic Doppler effect describes how the observed frequency of a relatively moving source changes with respect to the proper frame of the source. In Fig. (1) a satellite is shown to be moving parallel with the ground at a height h with a velocity v toward an observer at 0. x x h 1 'L e Ground Figure 1: An approaching satellite moving parallel to the ground is emitting light toward a ground observer. For the approaching satellite, determine the angle for which no Doppler effect is measured. For what speed of particle will this point coincide with the observer ground position (i.e. x = 0)? (3,2 marks] If the satellite emission frequency is 1 Hz and B = 1/2, plot a graph of effective frequency against cos 0 for the approaching satellite up to when it is overhead for the observer. [4 marks] The satellite changes its emitter to produce visible light at 500 nm wavelength. Determine the effective frequency for both the approach- ing and receding parts of the trip when the satellite is 100 m away from the observer in ground coordinates. Take the satellite height to be 50 m. (2,2 marks) Not the question you’re looking for? Post any question and get expert help quickly. 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15901	https://math.answers.com/other-math/How_much_if_you_add_10_percent_of_1500_to_1500	How much if you add 10 percent of 1500 to 1500? - Answers Create 0 Log in Subjects>Math>Other Math How much if you add 10 percent of 1500 to 1500? Anonymous ∙ 13 y ago Updated: 4/28/2022 10% of 1500 is 150. So, 1500 + 150 = 1650. Wiki User ∙ 13 y ago Copy Add Your Answer What else can I help you with? Search Continue Learning about Other Math ### What is 10 percent from 1500? To find 10 percent of a number, you simply multiply the number by 0.10. In this case, 10 percent of 1500 is calculated as 1500 0.10 = 150. Therefore, 10 percent of 1500 is 150. ### What is ten percent of one thousand five hundred dollars? 10% of 1500 = 10% 1500 = 0.1 1500 = 150 ### How much is 17.18 plus 10 percent tax? Just multiply 17.18 by .10 and you get 1.781 and add that to 17.18 and you get the answer ### How much is ten percent of fifteen hundred dollars? $150Quick tip - you can always find 10% of something by shifting the decimal place to the left.So - 1500 > 150.0 > 150 ### If i have a 100 dallors and i need to take 10 percent out of it how much will the 10 percent be? 10 dallors Related Questions Trending Questions Does 1 square yard equal 9 square feet?What polygon has two sets of parallel sides?If 4 apples and 2 oranges equal 1 and 2 apples and 3 oranges equals 0.70 how much does each apple and each orange cost?How many zeros in 35 lakhs?How many weeks in 1000000 days?What is the LCM of 11 3 and 2?Is the word surprisingly an adverb?What is 180 divided by 22?What is 89 times 50?What is the name of 20000000000000000000000?Why do some people sit with one leg curled under them?What does a vertical straight up and down line on a graph mean?What is 3.01 as a mixed number?What is bigger a quarter of two fifths?What does 4MgO stands for?typhoon yolanda took the lives of over 6,300 filipinos and displaced thousand others. write a five paragraph that explains why there were so many casualities even when they were warned of a storm surge use cause and effect pattern of development?How would you write 0.000056 in scientific notation?What is 14.4 round to nearest tenth?What are the next 4 multiples of one fifth?Is math an exact science? Resources LeaderboardAll TagsUnanswered Top Categories AlgebraChemistryBiologyWorld HistoryEnglish Language ArtsPsychologyComputer ScienceEconomics Product Community GuidelinesHonor CodeFlashcard MakerStudy GuidesMath SolverFAQ Company About UsContact UsTerms of ServicePrivacy PolicyDisclaimerCookie PolicyIP Issues Copyright ©2025 Answers.com. All Rights Reserved. The material on this site can not be reproduced, distributed, transmitted, cached or otherwise used, except with prior written permission of Answers.
15902	https://www.frontiersin.org/journals/environmental-science/articles/10.3389/fenvs.2022.929558/full	Frontiers \| Urban Sprawl and Haze Pollution: Based on Raster Data of Haze PM2.5 Concentrations in 283 Cities in Mainland China Frontiers in Environmental Science About us About us Who we are Mission and values History Leadership Awards Impact and progress Frontiers' impact Our annual reports Publishing model How we publish Open access Peer review Research integrity Research Topics FAIR² Data Management Fee policy Services Societies National consortia Institutional partnerships Collaborators More from Frontiers Frontiers Forum Frontiers Planet Prize Press office Sustainability Career opportunities Contact us All journalsAll articlesSubmit your researchSearchLogin Frontiers in Environmental Science Sections Sections Atmosphere and Climate Biogeochemical Dynamics Drylands Ecosystem Restoration Environmental Citizen Science Environmental Economics and Management Environmental Informatics and Remote Sensing Environmental Policy and Governance Environmental Systems Engineering Freshwater Science Interdisciplinary Climate Studies Land Use Dynamics Social-Ecological Urban Systems Soil Processes Toxicology, Pollution and the Environment Water and Wastewater Management ArticlesResearch TopicsEditorial board About journal About journal Scope Field chief editors Mission & scope Facts Journal sections Open access statement Copyright statement Quality For authors Why submit? 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Article types Author guidelines Editor guidelines Publishing fees Submission checklist Contact editorial office Frontiers in Environmental Science Sections Sections Atmosphere and Climate Biogeochemical Dynamics Drylands Ecosystem Restoration Environmental Citizen Science Environmental Economics and Management Environmental Informatics and Remote Sensing Environmental Policy and Governance Environmental Systems Engineering Freshwater Science Interdisciplinary Climate Studies Land Use Dynamics Social-Ecological Urban Systems Soil Processes Toxicology, Pollution and the Environment Water and Wastewater Management ArticlesResearch TopicsEditorial board About journal About journal Scope Field chief editors Mission & scope Facts Journal sections Open access statement Copyright statement Quality For authors Why submit? Article types Author guidelines Editor guidelines Publishing fees Submission checklist Contact editorial office Submit your researchSearchLogin Your new experience awaits. Try the new design now and help us make it even better Switch to the new experience ORIGINAL RESEARCH article Front. Environ. Sci., 30 June 2022 Sec. Environmental Economics and Management Volume 10 - 2022 \| Urban Sprawl and Haze Pollution: Based on Raster Data of Haze PM2.5 Concentrations in 283 Cities in Mainland China Zhenhua Wang1Jian Yang 1Jinqi Jiang 2 1 College of Economics and Management, Shenyang Agricultural University, Shenyang, China 2 Department of Agricultural and Resources Economics, College of Economics and Management, Shenyang Agricultural University, Shenyang, China Incorporating the urban sprawl and its quadratic term into the analytical framework of the environmental Kuznets curve and considers the spatial and threshold effects of pollution, this paper used the raster data of haze PM2.5 concentrations in 283 cities in mainland China to verify the non-linear effects of urban sprawl on urban haze pollution. It finds that: the inter-city spillover effect of haze pollution is significant, and the environmental Kuznets curve holds on haze pollution; there is an inversed “U” relationship between urban population size and haze pollution; the enlarge of the urban built-up area of city would increase haze pollution significantly; the impact of urban population size on haze pollution has a threshold effect that it would decline with the urban built-up area expansion; the coordination between population urbanization and land urbanization has an notable effect on haze pollution that its incoordination in China’s urbanization has aggravated haze pollution in city and this impact would lagged 1–2 period in time. 1 Introduction In recent years, haze pollution has been a prominent environmental problem in mainland China (Li et al., 2021). Haze pollution is not only the greatest public health risk, but also has serious negative impacts on economic and social development (Huang and Chew, 2021). Therefore, it is worthy to deeply discuss the economic and social determinants of haze pollution. Haze pollution is coupled with human economic activities (Liao et al., 2020). A major economic phenomenon in mainland China in recent years is continued urbanization, mainly in the form of expanding urban population size and urbanized built-up area. (Li et al., 2022). So, the relationship of urbanization and haze pollution has triggered many theoretical and empirical discussions. Some studies have shown that this is a complex causal relationship between urbanization and haze pollution (Yang and Yan, 2021), that it would vary by region, economic stage and scale (Feng and Wang, 2020; Yang et al., 2020). For this, we consider that there may be a threshold effect of urban sprawl on hazy pollution, and then we deduce a nonlinear causality of urbanization and hazy pollution. However, this nonlinear relationship is hardly investigated in a linear analysis framework in the existing literature. Consequently, we introduce a quadratic term of urbanization to explore it. In addition, the Chinses urbanization is characterized by lagging population urbanization, over-urbanization of land, and an incoordination between population and land in urbanization (Zhang et al., 2022). And this incoordination would even affect the relationship of urbanization and pollution. Therefore, the questions concerned in this paper are whether there is a significant causal relationship between urban expansion and haze pollution? Is there a threshold effect of urban sprawl on haze pollution? Does the incoordination of population urbanization and land urbanization significantly affect the relationship between urbanization and haze pollution? The possible contributions of this paper are: firstly, using a prefectural city-level raster data of PM2.5 concentrations in 283 cities in mainland China and establishing a spatial econometric model based on the environmental Kuznets curve to discuss the influence of urban expansion on haze pollution from the two perspectives of population and land, and also analyzing the spatial effects of haze pollution; secondly, constructing threshold models with population and urban built-up area as threshold variable respectively to discuss the non-linear effect of urbanization on haze pollution; Third, a greater contribution of this article is to include coordination between population urbanization and land urbanization in the study to investigate whether the incompatibility between these two has an impact on haze pollution, and analyze its impact mechanism. 2 Literature Review and Theoretical Analysis 2.1 Literature Review The experience of economic development and environmental pollution in various countries shows that developed economies have experienced serious environmental pollution, including haze pollution, in the early stages of industrialization, such as the United Kingdom, Germany, the US and Japan (Jungwook K. and Jinkyeong K., 2021). Scholars have summarized this experience as a general economic law between economic development and environmental pollution: the environmental Kuznets curve (Maddison D, 2006; Müller-Fürstenberger and Wagner, 2007). The environmental Kuznets curve points to an inverted “U" relationship between economic development and environmental pollution: in the early stages of rapid economic development, environmental pollution in an economy will gradually increase as resource inputs increase; as the economy develops further, the level of technology improves and the quality of economic growth increases, environmental pollution will gradually decrease (Song et al., 2021). Under this rule, other economic variables will also affect environmental pollution, including government regulation, foreign investment, urbanization development, etc. The impact of these variables on environmental pollution will eventually manifest as changes in the slope and inflection point of the inverted “U” curve (Wang and Padmanabhan P, 2021; Song and Ye, 2021). Among the existing empirical studies, scholars have also explored other factors influencing environmental problems. Some studies have found that renewable energy, biomass energy and per capita income are closely related to environmental pollution, and that the use of renewable energy is not only beneficial to environmental protection but also promotes economic recovery and sustainable development (C. Magazzino et al., 2022 and M. Mele et al., 2021; H. Altintas and Kassouri, 2020; M. Mele et al., 2021), in addition, C. Magazzino and M. Mele et al. (2021) suggest that fossil fuel combustion and economic growth have a significant impact on atmospheric pollutants such as PM2.5, PM10 and CO 2. It has also been found that the penetration of information and communication technologies accelerates electricity consumption while promoting economic growth, which in turn leads to higher CO 2 emissions and environmental pollution, in which urbanization development has played an important role (C. Magazzino, M. Mele et al., 2021). S. Katircioglu et al. (2018) suggest that the environmental Kuznets curve is characterised by an inverted ''U'' shape with increasing urbanisation and population. In addition, Y. Kassouri (2021) argues that urbanisation has significant direct or indirect spatial spillover effects on the ecological and water footprints. In the face of the serious situation of environmental pollution, it is also crucial to identify the factors influencing haze as one of the primary pollutants. Scholars have explained the influencing factors of haze pollution in terms of economic development, industrial upgrading, foreign investment, government regulation, and transportation (Lan and Pan, 2019). Yang and Yan (2021) measured the impact of urban sprawl on haze pollution in terms of urban built-up area and urban utility occupancy and found a non-linear relationship between the two. Scholars have also found that population concentration has a spatial impact on haze pollution through scale and intensification effects, and that the expansion of urban population size exacerbates haze pollution (Li and Zhou et al.,2012), including both direct and indirect effects (Feng and Wang, 2020), at the same time, China’s rapid urbanisation has brought not only economic and population agglomeration, but also serious haze pollution, but the haze also had a significant impact on the level of urbanisation ((Fan et al., 2019; Liu et al., 2021a). In addition, the development of land urbanization also increases haze pollution, but this effect is only manifested as an indirect effect (Yu X. et al., 2020). Further research by scholars found inter-regional and inter-city differences in the relationship between city size and haze pollution, with the expansion of city size helping to suppress haze pollution in megacities and first-tier cities (Fan et al., 2019), and scale heterogeneity being significantly present in the results of tests of the effect of urbanization on haze pollution (Liu and Ragusa, 2019). There are several points to add to the existing literature: firstly, the spatial effects of haze pollution are ignored in the literature discussing urban expansion-related elements on haze pollution, which may lead to unreliable conclusions of existing studies. The spatial mobility of haze pollution determines its possible spatial correlation, as confirmed by many scholars (Hao and Liu, 2016; Fan et al., 2019). Secondly, the findings of existing studies imply that there may be a threshold effect of variables such as city size on haze pollution, i.e. the relevant variables may have a non-linear effect on haze pollution (Yu X. et al., 2020), but the existing literature only artificially divides the sample for testing and does not use a threshold model to discuss this issue objectively. Thirdly, haze pollution is a matter of economic development stage (Zhang et al., 2021), and the theory behind it lies in the environmental Kuznets curve, which is confirmed by scholars to exist in haze pollution (Maddison D, 2006; Müller-Fürstenberger and Wagner, 2007). Relevant studies therefore need to build models in this theoretical framework, which corroborates existing studies that have found different effects of urbanization development-related indicators on haze pollution in different cities, regions and stages. Finally, the urbanization in mainland China is characterized by a lag in population urbanization and a lack of coordination between the two, which has led to many problems in the economic development of mainland China (Liu et al., 2021b), and therefore the impact of the lack of coordination between population urbanization and land urbanization on haze pollution needs to be discussed, which has not been analyzed in depth in the literature. This paper precisely attempts to start from the shortcomings of the above-mentioned several existing studies, based on the theoretical framework of environmental Kuznets curve, the impact of urban expansion on haze pollution is systematically discussed. 2.2 Theoretical Analysis The mechanism of haze formation is complicated, but briefly three conditions are required: a certain concentration of pollutants, static atmospheric conditions, and a certain humidity of air. Among these, human economic activity mostly influences the concentration of the pollutants which mainly are aerosols. The secondary aerosols are formed largely by human economic activity, and particles include S O X S O X, N O X N O X, S V O C X S V O C X, C O X C O X, organic compounds containing halogens, etc. And these particles can attached toxic heavy metals and other toxic, which then form haze in stationary weather and high humidity (Yu J. et al., 2020). Urban population expansion will affect haze pollution in three ways: the first is that population expansion would increase material and resource consumption and then result in increased pollutants emission and bring about environmental pollution such as haze. And meanwhile, population urbanization could also change the consumption structure, resulting in more domestic and production waste emissions and deteriorating environmental quality (Feng and Wang, 2020). For example, urban population expansion increases housing demand. And this boosting housing demand would raise construction dust and then promote primary aerosol particle concentrations. It also increases the organic matter volatilized from painting materials, which increases secondary aerosol particle concentrations. All of these would increase haze pollution. The second is that urbanized mode of production and living would increase energy and consumer goods consumption per capita and then leads more industrial and domestic waste, which would increase environmental and haze pollution. In here, the urban waste disposal and incineration and the winter heating all would increase sulphur oxide and nitrogen oxide emission, leading to an increase in secondary aerosol particle concentrations and causing haze pollution (Wang et al., 2021). The third is the expansion of urban population will increase the government’s ability to govern environmental pollution and the productivity of the urban economy and then reduce pollution levels per unit of output and haze pollution (Sun et al., 2020; Jiang et al., 2021). Moreover, urban population expansion will increase the people’s opportunity for better education, increase the utilization of public transport and then reduce environmental pollution. In addition, the concentration of population in cities would promote the development of service industries, squeeze out industries with high pollution emission intensity, and even could reduce haze pollution. At the early stage of urbanization, the first two ways play the major role, and urban population expansion will increase haze pollution; as urbanization further developed, the third way will dominate and haze pollution gradually be reduced (Sun et al., 2020; Jiang et al., 2021; Wang et al., 2021). Therefore, Hypothesis 1 proposed as: Hypothesis 1.There is a non-linear effect of urban population expansion on haze pollution that it will increase haze pollution at first and then decrease haze pollution.Previous studies have confirmed that urban land expansion increases environmental pollution (Zhang et al., 2020), and this even true for haze pollution (Yang and Yan, 2021). Firstly, the process of urban built-up area expansion, initial building construction, later infrastructure support and continued construction of residential areas will increase dust emissions, which in turn increase the concentration of primary aerosol particles and secondary aerosol particles and then haze pollution. In addition, the renovation materials used in the construction process will increase VOCs, which also increase the concentration of secondary aerosol particles and increase haze pollution (Yang and Yan, 2021). Secondly, the expansion of urban areas will lift the amount of motor vehicles and traffic distances, which in turn increase the nitrogen oxides caused by motor vehicle emissions, and therefore also increase haze pollution by the higher concentration of secondary aerosol particles. Thirdly, it has been suggested that the expansion of built-up areas will lead to a decrease in wind speed over a certain area of the city center, which in turn increase the probability of stationary weather and increase haze pollution (Yang and Yan, 2021). Finally, the natural boundaries of cities usually have ecological green areas such as forests, wetlands, parks and grasslands that have a haze dissipating effect, and therefore, haze pollution is lower in suburban areas compared to urban areas. However, as urban land continues to expand it will destroy the green areas in the suburbs, leading to a decrease in the haze dissipating function of the city and exacerbating haze pollution (Zhang et al., 2020). Of course, in the process of urban land expansion, haze pollution will be mitigated by reducing the density of pollutant emissions per unit area, but in terms of the combined effect, the extensive land expansion in the current would increase haze pollutant emission, reduce the haze dissipation capacity of cities, and may exacerbate haze pollution. Then Hypothesis 2 put forward as: Hypothesis 2.The extensive urbanization which mainly in form of land expansion would increase haze pollution significantly.The inconsistency between population urbanization and land urbanization is essentially the unreasonable structure and allocation of factors (Lei et al., 2022). And in China, this inconsistency is particularly evident in the rapid urbanization of the land (Liu et al., 2021b), that land are over-allocated to cities and labour become relatively scarce. This distorted allocation of factors has led to extensive land use, which is useless for economic efficiency and environmental protection. In population’s effects on environmental pollution like haze, built-up area expansion would play a role. In here, the expansion of built-up areas would lead to the extension of urban space, leading to an increase in the spatial distance between the workforce and enterprises. This would strengthen the impact of population expansion in city on haze pollution. And more, emissions from pollutants brought about by population urbanization may be dissipated or weakened by nature’s air purification system. But urban land expansion would lead to the destruction of ecosystems such as forests around urban areas, that indirectly amplify the haze pollution from population urbanization (Li et al., 2022; Yang and Yan, 2021). Therefore, we put forward the Hypothesis 3: Hypothesis 3.The effect of urban population expansion on haze pollution is influenced by land expansion, and the incoordination of land urbanization and population urbanization would enlarge the negative effect of urbanization on haze pollution. 3 Model and Data 3.1 Model Considering the possible spatial correlation of haze pollution brought about by the regional correlation of stationary weather and moisture conditions, the mobility of haze pollution, and the correlation of economic activities between neighbouring cities and regions in the haze generation process, a spatial econometric model will be developed in this paper to test the correlation hypothesis. The general form of the spatial econometric model is the spatial Durbin model (SDM). The PM2.5 concentration of city i in period t is H a z e i,t H a z e i,t and X X is the correlation factor affecting urban haze pollution, so the SDM model for urban haze pollution is: H a z e i,t=ρ W Y i,t+β X i t+θ W X i,t+μ i,t μ=λ W μ+ε(1)H a z e i,t=ρ W Y i,t+β X i t+θ W X i,t+μ i,t μ=λ W μ+ε(1) In Eq. 1, H a z e i,t H a z e i,t is the explanatory variable and X X is the explanatory variable of interest, W W is the spatial weight matrix, μ μ is the random error term, W Y W Y is the spatial dependence of the dependent variable, and W X W X is the spatial dependence of the independent variable, ρ ρ is the spatial autoregressive coefficient and λ λ is the residual autoregressive coefficient. In addition, considering the possible existence of “time inertia” of haze pollution, this paper will also introduce the lag term of haze pollution in the spatial Durbin model and construct a dynamic spatial Durbin model as follows: H a z e i,t=α H a z e i,t−1+ρ W Y i,t+β X i t+θ W X i,t+μ i,t(2)H a z e i,t=α H a z e i,t−1+ρ W Y i,t+β X i t+θ W X i,t+μ i,t(2) In Eq. 2, H a z e i,t−1 H a z e i,t−1 is the haze pollution of city i i in period t−1 t−1, α α is the regression coefficient of variable H a z e i,t−1 H a z e i,t−1, and other variables and correlation coefficients have the same meaning as in Eq. 1. Equations 1, 2 are further expanded based on the environmental Kuznets curve to build the final SDM model for this paper: H a z e i.t=α H a z e i,t−1+β 1 g d p i t+β 2 g d p 2 i t+β 3 p o p u i t+β 4 b u a i t+β 5 f d i i t+β 6 e n e r g y i t+β 7 g r e e n l a n d i t+β 8 i n f o r+β 9 f i n i n c i t+β 10 f i n i t+β 11 b a n k i t+β 12 t b d i t+β 13 g o v i t+β 14 i s i t+ρ W Y i,t+θ W X i,t+μ i,t(3)H a z e i.t=α H a z e i,t−1+β 1 g d p i t+β 2 g d p i t 2+β 3 p o p u i t+β 4 b u a i t+β 5 f d i i t+β 6 e n e r g y i t+β 7 g r e e n l a n d i t+β 8 i n f o r+β 9 f i n i n c i t+β 10 f i n i t+β 11 b a n k i t+β 12 t b d i t+β 13 g o v i t+β 14 i s i t+ρ W Y i,t+θ W X i,t+μ i,t(3) In Eq. 3, H a z e i,t−1 H a z e i,t−1 is the lagged term of the explanatory variable, the core explanatory variables are population size (popu), built-up area (bua), primary (gdp) and secondary (gdp 2) indicators of economic development. The control variables added include: level of foreign direct investment (fdi), energy intensity (energy), greenland coverage (greenland), level of information technology development (infor), size of government (fiscal revenue fininc and fiscal expenditure share gov), level of financial development (savings fin, deposit balance bank), level of industrial structure (non-agricultural industry share is and service industry development level tbd) etc (Feng and Wang, 2020), When α=0 α=0, it is a static spatial panel model and when ρ=θ=0 ρ=θ=0, it is a dynamic spatial panel model. Newer literature has proposed models that can consider both the spatial error term and the spatial lag term, i.e. the spatial lag model with a spatial lag error term (SARAR), and to verify the robustness of the spatial Durbin model conclusions, the estimation results of the SARAR model are also presented in this paper. In addition, this paper will also develop a threshold model to test the threshold effect of urban sprawl on haze pollution. The threshold estimation method was proposed by Hansen B E (1999) and has been widely accepted by scholars to date. The basic form of the single threshold model is: Y=μ+β′1 X I(q≤γ)+β′2 X I(q>γ)+ε(4)Y=μ+β 1′X I(q≤γ)+β 2′X I(q>γ)+ε(4) In Eq. 4, Y Y is the dependent variable, which in this paper is urban haze pollution, X X is the independent variable, β′β′ is the value of the coefficient corresponding to the independent variable, q q represents the threshold variable, γ γ is the threshold value, ε ε is the random disturbance term, I(⋅)I(⋅) is the indicator functions. The determination of the number of threshold values and the test principles are not repeated. The threshold model developed for this study (a single threshold model with built-up area as the threshold variable) is: H a z e i t=μ i+β 1 p o p u i t I(b u a i t≤γ)+β 2 p o p u 2 i t I(b u a i t>γ)+φ 1 g d p i t+φ 2 g d p 2 i t+⋅⋅⋅⋅⋅⋅+ε i t(5)H a z e i t=μ i+β 1 p o p u i t I(b u a i t≤γ)+β 2 p o p u i t 2 I(b u a i t>γ)+φ 1 g d p i t+φ 2 g d p i t 2+⋅⋅⋅⋅⋅⋅+ε i t(5) The threshold model is consistent with the control variables of the spatial econometric model. 3.2 Data 3.2.1 The Data for This Paper are Derived From Two Sources The first is haze data. The haze data in this paper come from the Center for International Earth Science Information Network (CIESIN) at Columbia University, United States CIESIN at Columbia University relies on the Socio-Economic Data and Applications Center (SEDAC) to publish satellite-borne Moderate Resolution Imaging Spectrometer (MODIS) and Multi-angle Imaging Spectrometer (MISR) measurements to obtain aerosol optical thickness images, which can be transformed to obtain raster data of PM2.5 concentrations, which are widely used in haze pollution studies (Peter, 2008; Chen and Han, 2021) In this paper, ArcGIS 10.2 software was used to parse this data into PM2.5 concentration data for the corresponding year for sample cities above the prefecture level in mainland China. This is followed by urban expansion indicators and other related variables measuring economic development. The original data for urban expansion and other indicators were obtained from the China Urban Statistical Yearbook and the China Regional Economic Statistical Yearbook during the sample period, and the data for each price deflator were obtained from the publicly available provincial and municipal statistical yearbooks of each province in the corresponding year. In the process of data collection and collation, the balanced panel data of 283 mainland Chinese cities from 2010 to 2016 were finally collated in order to obtain balanced panel data, taking into account the availability of data. In the data quantification process. the explanatory variable haze pollution is selected as the annual average value of PM2.5 concentration; the core explanatory variables are selected as the population size, built-up area, economic development primary and secondary terms. The population size is taken from the total population of the municipal district at the end of the year, and the economic development index is taken as the logarithmic value of GDP per capita; other control variables include: foreign direct investment level, energy intensity, green space coverage, information development level, government size, financial development level and the industry structure level. Among them, the energy intensity indicators used the quantification of GDP output per unit of industrial electricity consumption (10,000RMB/10,000 kWh); the information development level was quantified by quantifying international internet users (household); the government size is quantified using the ratio of municipal revenues to fiscal expenditures; the consumption level indicators used the total retail sales of social consumer goods per capita in each city (10,000RMB/Per); the financial development level is quantified by the amount of savings and the balance of deposits; the industry structure level is quantified by the proportion of non-agricultural industries and the proportion of services. The capital stock indicator is obtained using the perpetual inventory method. The other variables are quantified by taking the original meaning of the indicator and are not repeated. All physical capital variables are taken as logarithmic values. The data are described in Table 1. TABLE 1 TABLE 1. Description of data. 4 Analysis of the Empirical Results 4.1 Impact of Land Expansion and Population Expansion on Urban Haze Pollution Before analyzing the impact of land expansion and population expansion on urban haze pollution, it is necessary to examine the spatial correlation between urban haze pollution. In this paper, As shown in Figure 1, this paper gives Moran scatter plots for 2010, 2012, 2014, and 2016. FIGURE 1 FIGURE 1. Moran scatter plot of haze pollution. From the Moran’s I index, haze pollution among Chinese cities shows a significant spatial correlation with a positive coefficient proving the existence of spatial spillover effects of haze pollution among cities, which is consistent with the findings of existing studies (Hao and Liu, 2016). The existence of strong spatial mobility of haze pollution, the existence of spatial clustering of economic activities in the haze generation process, and the existence of spatial similarity of climate in the haze generation process all contribute to the spatial correlation of haze pollution between cities. In this paper, the results of the spatial SARAR model and the spatial SDM model are presented, as well as the results of the models with and without control variables respectively, and the four sets of estimation results are given in Table 2. In the process of analysis, Stata 14.0 software was used. TABLE 2 TABLE 2. Impact of urban sprawl on haze pollution. From the model estimation results of the spatial SARAR and SDM, the spatial coefficients passed the significance test, further indicating that there is a spatial spillover effect of haze pollution, and the haze pollution in this region will have an impact on the haze pollution in neighboring regions, corroborating the results of the Moran scatter plot and confirming the spatial correlation of urban haze pollution. From the model estimation results, the population expansion and built-up area expansion of cities will significantly increase haze pollution (1% significance level), and this conclusion holds in all four models with strong robustness. The expansion of urban population will cause the concentration of economic activities in urban areas, which will increase haze pollution; at the same time, an increase in urban population will also lead to an increase in related consumption in terms of clothing, food, housing and transport, which will easily cause traffic congestion and generate more household waste, speed up energy consumption and thus increase haze pollution.The expansion of the built-up area will firstly cause an increase in the sources of haze pollution due to construction and building, etc. In addition, the expansion of the built-up area will increase the mileage of urban traffic and the probability of congestion, which will also increase the sources of haze pollution. The primary indicator of economic development has a significant positive effect on haze pollution (1% level of significance), indicating that economic development exacerbates haze pollution, but the secondary indicator of economic development has a significant negative effect on haze pollution (1% level of significance), indicating that there is an inverted “U" shaped relationship between economic development and haze pollution, confirming the existence of an environmental Kuznets curve in urban pollution in mainland China. This is consistent with the findings in the literature (Maddison, 2006; Müller-Fürstenberger and Wagner, 2007; Song et al., 2021). The reason for this phenomenon is that in the early stage of economic development, the economic activities are mainly industrial, and the government also relaxes the environmental control to attract more enterprises to invest, which causes a large amount of polluting gas and waste emissions, and thus aggravates the haze pollution. When the economic development enters the mature stage, the economy of scale has been formed, and it is no longer necessary to sacrifice the environment to promote development, instead, environmental protection should be the top priority to achieve sustainable development, so the economic development and haze pollution show an inverted “U" shape relationship. Considering the “temporal inertia” of haze pollution, this paper also presents the results of dynamic spatial Durbin model estimation. As shown in Table 3: Haze pollution in the lagged period has a significant effect on haze pollution in the current period (1% significant level), indicating that haze pollution has a significant time dependence, and haze pollution in the previous period will aggravate haze pollution in the current period. The other variables estimated are consistent with the baseline regression, the relationship between economic development and haze pollution still shows an inverted “U" shape feature, the expansion of population size and built-up area will increase haze pollution (1% significant level), and the spatial coefficient is still significant, which further confirms the robustness of the results of this paper. TABLE 3 TABLE 3. Impact of urban sprawl on haze pollution based on dynamic SDM. This paper focuses on the mechanisms involved, firstly, population expansion will increase resource consumption, change the consumption structure, create more pollution sources, and lead to haze pollution (Feng and Wang, 2020), secondly, population urbanization will inevitably increase housing demand, and construction will lead to an increase in solid pollutants such as dust and generate haze, and again, domestic waste disposal and winter heating will lead to pollutant gas emissions and intensify haze pollution (Wang et al., 2021).However, there may be a non-linear relationship between population size and haze pollution, as the government will continue to increase environmental management as urbanisation grows, while increased access to education will increase people’s environmental literacy and thus reduce haze pollution (Sun et al., 2020; Jiang et al., 2021), as will be further confirmed below. Land expansion will generate more solid particulate matter due to the continued construction of buildings and infrastructure, secondly, increased transport distances will lead to more vehicle emissions, and thirdly, the expansion of built-up areas will reduce wind speeds in city centres, which is detrimental to air movement, all of which will contribute to haze pollution (Yang and Yan, 2021). Some studies have pointed out that there may be inter-sample and inter-scale heterogeneity in the effect of expanding city size on haze pollution (Fan et al., 2019), suggesting that variables such as city size may have a non-linear effect on haze pollution, so this paper further adds the quadratic term of the city population size variable and the quadratic term of the built-up area of the city to the benchmark model for testing. The estimation results of the SARAR model and SDM model are also presented in this section. With the inclusion of the quadratic term, the spatial coefficients are still estimated to be significant (1% level of significance). From the results, the primary term of population size still has a significant positive effect on haze pollution, but its quadratic term is significantly negative, indicating that urban haze pollution will gradually increase as the population size increases, but after reaching a certain size, the increase in urban population will help to reduce urban haze pollution, i.e. there is also a significant inverse “U" shaped relationship between population size and urban haze pollution. The effect of population size on urban haze confirms the hypothesis of this paper. The positive effect of population size on haze pollution is easy to understand, and the explanation for the reduction in pollution brought about by further population size is that, firstly, the concentration of population in cities will improve the technical efficiency of cities through scale effects, etc., improving the quality of economic growth and reducing haze pollution; secondly, the expansion of city size will reduce the unit cost of the government to combat pollution such as haze, and will also improve the government’s ability to combat pollution; Thirdly, the concentration of population in the city will promote the development of the service industry on both the supply and demand sides, thus having a crowding-out effect on industries with high pollution emission intensity, which in turn will reduce haze pollution. The expansion of built-up area has a significant positive effect on urban haze pollution, but unlike population size, the secondary indicator of built-up area does not have a significant effect on haze pollution, indicating that the impact of urban area expansion on haze pollution is mainly monotonic and increasing. Possible reasons are that, on the one hand, the expansion of built-up areas will generate a large number of pollution sources, and also crowd the surrounding agricultural land, wetlands and woodlands, etc., breaking the ecological balance and reducing the ability of the environment to recover itself, on the other hand, the urbanization of people in China obviously lags behind the urbanization of land, and the continuous expansion of built-up areas further increases the problem of incompatible development between the two, indirectly aggravating haze pollution (Li et al., 2021; Zhang et al., 2022). Table 4 confirms that there is a non-linear effect of increasing population size on urban haze pollution, which suggests that there may be a threshold effect of population size indicators on urban haze pollution, and this paper will further construct a threshold model to test whether this threshold effect exists. In the process of analysis, this paper uses population size and built-up area as threshold variables respectively, and population size and its quadratic term as the core explanatory variables to test their effects on urban haze pollution. TABLE 4 TABLE 4. Non-linear effects of urban sprawl on haze pollution. 4.2 Threshold Effects of Land Expansion and Population Expansion on Urban Haze Pollution This paper uses Hansen’s (1999) method of minimizing the sum of squared residuals to test the number of thresholds. The test procedure used the Bootstrap Method to draw samples 300 times for significance testing of the threshold effect. From the test results in Table 5, the F-statistic of the single threshold with built-up area as the threshold variable was the largest and the most significant (5% significance level), indicating the existence of a significant threshold effect. In contrast, the test with population size as the threshold variable fails the significance test from the single threshold to all three thresholds, indicating that there is no threshold effect. This paper gives the estimation results of the two models, mainly based on the former analysis. TABLE 5 TABLE 5. Threshold effects of urban sprawl on haze pollution. Using the built-up area as the threshold variable, according to the test and estimation results of the threshold model, the primary and secondary terms of population size have significant effects on urban haze pollution in both intervals, with positive and negative effects respectively, which is consistent with the results of the spatial econometric model, indicating an inverted “U” shape relationship between population size and urban haze pollution. It is further observed that with the expansion of built-up area, the absolute values of the coefficients of the primary and secondary terms of population size on urban haze pollution are decreasing in different two threshold intervals, indicating that the impact of population expansion on haze pollution will be affected by the built-up area, and the joint effect of land urbanization and population urbanization will affect haze pollution. The impact of population size on haze pollution differs at different levels of land urbanization and built-up area size, and the impact of population size on haze pollution decreases with the expansion of built-up area, which is in line with our expectation. As the built-up area expands, urban space rises, population density decreases and therefore the concentration of haze pollution decreases. Of course, this conclusion does not mean that to reduce urban haze pollution and the impact of population expansion on haze pollution we need to increase the size of built-up areas. On the contrary, a prominent problem in the process of urban expansion in China is the incompatibility between population urbanization and land urbanization, which is manifested in the rapid urbanization of land (Liu et al., 2021), and it has been demonstrated in this paper and related literature that the expansion of urban built-up area can lead to increased haze pollution, the ideal scenario would be a coordinated expansion of population size and built-up area. Therefore, considering that the estimation results of the threshold model have demonstrated a causal relationship between population expansion, land expansion and haze pollution, and that the effect of population expansion on haze pollution varies with land expansion, this paper further incorporates the coordination of population urbanization and land urbanization into the analytical framework of this paper to verify its effect on urban haze pollution. 4.3 The Impact of Coordination Between Land Expansion and Population Expansion on Urban Haze Pollution This paper draws on the study of (Liao et al., 2020), where the coordination index of land urbanization and population urbanization is formulated as: C L T=∣∣L+T√2∣∣√L 2+T 2(6)C L T=\|L+T 2\|L 2+T 2(6) In Eq. 6, L L is the growth rate of urban population, T T is the growth rate of built-up area, C L T C L T representing the degree of coordination between the two. In terms of criteria, C L T<0.8 C L T<0.8 means that land urbanization and population urbanization are not coordinated; 0.8<C L T<0.9 0.8<C L T<0.9 means that they are basically coordinated; 0.9<C L T<1 0.9<C L T<1 means that they are coordinated. From the results of this paper, the average value of the coordination index between population urbanization and land urbanization in the sample cities is 0.7676, and since L<T, the urbanization of the sample cities has the incoherent problem of over-urbanization of land. This tendency of urbanization in favor of land rather than population leads to a series of problems, including inefficient land use, sloppy economic development and high urban transport costs (Liao et al., 2020). In this paper, both the SARAR model and the SDM model were constructed to test the effects of population urbanization and land urbanization coordination on urban haze pollution. The results are shown in Table 6, from the estimation results of the two models, the effect of coordination between population urbanization and land urbanization on urban haze pollution is not significant. Could it be that there is no effect of coordination between population urbanization and land urbanization on urban haze? It has been pointed out in the literature that the pattern of over-urbanization of land can lead to problems such as a sloppy economic development pattern, therefore it also leads to increased pollution such as haze, which is inconsistent with the present findings. The explanation given in this paper is that there may be a lagged effect of the coordinated degree of population urbanization and land urbanization on pollution such as haze. Therefore, this paper develops models with lag 1, lag 2 and lag 3 respectively to further test the effect of coordinated degree of population urbanization and land urbanization on urban haze pollution. TABLE 6 TABLE 6. Impact of land expansion and population expansion coordination on urban haze pollution. From the estimation results of the SARAR model and SDM model, the effect of coordination between population urbanization and land urbanization on urban haze pollution is significantly negative (1% significance level) in the lag 1 model, the estimation result of the lag 2 model is also significantly negative (1% significance level), and the estimation result of the lag 3 model is not significant, indicating that the lagged effect of population urbanization and land urbanization coordination on urban haze pollution is shown to be period 1 and period 2. In terms of specific impact mechanisms, the main reason for the uncoordinated development of population urbanization and land urbanization is that government-led land urbanization distorts the allocation of land factors, while unequal urban public services inhibit the development of population urbanization (Lei et al., 2022; Liao and; Wu et al., 2022), both of which manifest as inefficient factor allocation. The distortion of factor allocation will lead to a sloppy and inefficient way of economic development, with large-scale overbuilding such as industrial parks increasing haze pollution, while the low rate of allocation of factors such as labour to the urban sector will cause industrial parks to be idle. Empirical analyses have shown that the inconsistency between population urbanization and land urbanization due to rapid land urbanization increases haze pollution, which may be related to the sprawl of urban space due to the expansion of built-up areas and the consequent reduction in inter-city accessibility, or to the destruction of peri-urban ecosystems due to urban expansion and the consequent reduction in the ability of cities to dissipate haze (Yang and Yan, 2021), but these related effects do not manifest immediately in the current period and there is a significant lag period. 4.4 Sensitivity Analysis In order to better ensure the stability of the estimation results, a sensitivity analysis of the main explanatory variables was conducted in this paper based on the Delta Moment-Independent Analysis algorithm (E. Borgonovo, 2007). The variables selected are economic development, the secondary term of economic development, population size, built-up area and the urbanisation of population towns and land coordination. The final output parameters are shown in Table 7. TABLE 7 TABLE 7. Sensitivity analysis. The results report the sensitivity coefficients of the parameters, and the sobol first-order sensitivity coefficients based on variance, respectively, the sensitivity coefficients of each variable are positive and pass the 95% confidence test, which says that haze pollution has a certain sensitivity to each variable. As can be seen from Table 7, population is the most influential parameter on haze pollution, and its sensitivity coefficient is about 0.172, followed by built-up area with a sensitivity coefficient of 0.114, and again, are economic development and its secondary term, with the lowest sensitivity coefficient for the coordination of population urbanization and land urbanization. The sensitivity coefficients calculated based on the variance are basically consistent with the results of the total sensitivity coefficient. In general, all of the above variables have significant effects on haze pollution, which indicates that the results of this paper are robust. 5 Conclusion and Discussion China’s urbanization has achieved tremendous success with the expansion of urban scale, but it has also led to a series of economic, social and environmental problems (Fan and Zhao, 2012), among which there is a significant effect of population scale expansion and built-up area expansion and their coordination on urban haze pollution. Based on raster data of urban haze PM2.5 concentrations, this paper incorporates urban expansion and its quadratic term into the analytical framework of the environmental Kuznets curve and considers spatial effects to verify the influence of variables related to urban expansion on urban haze. The study finds that there is a significant spatial spillover effect on urban haze pollution, and the environmental Kuznets curve holds for urban haze pollution; there is an inverted “U" relationship between urban population size and urban haze pollution; the increase in built-up area significantly increases haze pollution; there is a threshold effect of population size on urban haze pollution, and as the built-up area increases, the effect of population size on haze pollution decreases; the coordination between population urbanization and land urbanization has a significant effect on haze pollution, and the incompatibility between population and land in China’s urbanization process exacerbates haze pollution, with a 1-2 period lag effect. While there is much debate in China about the optimal size of cities, this paper provides a new perspective on the consideration of haze pollution. Leaving aside the debate on city size, one point to note is that the coordinated development of land urbanization and population urbanization should be incorporated into the relevant discussion and analytical framework of urban environmental pollution. The policy implications of this paper include the following: Firstly, in addition to strengthening enforcement and supervision of haze pollution, the government should also work on strengthening spatial planning and implementation of urban development guidelines to curb “pie-spreading” urban development. First of all, the government should carry out scientific and reasonable urban development planning, and should develop the concept of spatial conservation, smart growth and compact city development to avoid unnecessary encroachment and over-exploitation of the space around the city, which will help reduce haze pollutant emissions and maintain haze dissipation capacity. In this process, the government should change the spatial structure of the land offered for sale, reduce the proportion of land on the urban fringe therein, and increase efforts to vacate and optimize the allocation of land in the city again to promote three-dimensional urban development. Secondly, the government should formulate corresponding implementation guidelines for development plans and strengthen the supervision of the implementation of the relevant guidelines to ensure that the corresponding plans are implemented. For example, the government can build a system of ecological isolation at the edge of the city, as a rigid bottom line for urban land expansion, to ensure that it is not breached. Secondly, the high quality and rapid development of population urbanization can help to cross the threshold value of its impact on urban environmental pollution and release the intensive effect of population concentration on environmental pollution. Therefore, the government should promote the high-quality development of population urbanization through initiatives such as promoting the equalization of public services in cities and towns and the orderly coniferization of the agricultural transfer population. Firstly, it should further deepen reforms to mend the shortcomings in the high-quality development of urbanization, break through the institutional and institutional barriers that restrict the development of population urbanization, achieve equalization in public services such as education, healthcare and pensions, and at the same time promote the work of settlement and the rational and orderly promotion of the citizenship of the agricultural professional population. Secondly, in response to the growing number of depopulated cities, reasonable and scientific analysis and planning should be carried out, and efforts should be made to promote the redevelopment of depopulated cities in terms of industrial cultivation and the introduction of talents. Thirdly, the government should base on the long term and strengthen the coordinated and balanced level of development of population urbanization and land urbanization to release their lagging effect on reducing pollution such as haze. The main characteristic of uncoordinated urbanization in cities is that land urbanization is ahead of its time while population urbanization is lagging behind. Therefore, promoting rapid population urbanization and curbing excessive land expansion is a feasible option to enhance the coordinated development of population urbanization and land urbanization. In addition, cities should determine their own urbanization development path based on their factor endowment advantages. Some cities should develop land-scarce and labour-intensive industries to absorb the employed population, correct the degree of coordinated development of population and land, optimise the level of factor allocation, improve efficiency and reduce pollution. Data Availability Statement The original contributions presented in the study are included in the article/supplementary material, further inquiries can be directed to the corresponding author. Author Contributions JJ and ZW conceptualized and designed the study, revising it critically for important intellectual content. JY coordinated and supervised data collection, carried out the initial analyses, and reviewed and revised the manuscript. ZW corrected the data analyses, drafted the manuscript, and critically reviewed the manuscript. All authors read and approved the final manuscript. Funding Scientific Research Funding Project of Liaoning Provincial Department of Education, Grant/Award Number: WSNQN202028 and WSNZK202003. Conflict of Interest The authors declare that the research was conducted in the absence of any commercial or financial relationships that could be construed as a potential conflict of interest. Publisher’s Note All claims expressed in this article are solely those of the authors and do not necessarily represent those of their affiliated organizations, or those of the publisher, the editors and the reviewers. Any product that may be evaluated in this article, or claim that may be made by its manufacturer, is not guaranteed or endorsed by the publisher. References Altıntaş, H., and Kassouri, Y. (2020). Is the Environmental Kuznets Curve in Europe Related to the Per-Capita Ecological Footprint or CO2 emissions?[J]. Ecol. Indic. 113, 106187. doi:10.1016/j.ecolind.2020.106187 CrossRef Full Text \| Google Scholar Borgonovo, E. (2007). A New Uncertainty Importance Measure. Reliab. Eng. Syst. Saf. 92 (6), 771–784. doi:10.1016/j.ress.2006.04.015 CrossRef Full Text \| Google Scholar Bradford, D. F., Fender, R. A., Shore, S. H., and Wagner, M. (2011). The Environmental Kuznets Curve: Exploring a Fresh Specification[J]. Contributions Econ. Analysis Policy 4 (1). doi:10.2202/1538-0645.1073 CrossRef Full Text \| Google Scholar Chen, Q-X., Han, X-L., Gu, Y., Yuan, Y., Jiang, J., Liou, K-N., et al. (2021). Evaluation of MODIS, MISR, and VIIRS Daily Level-3 Aerosol Optical Depth Products over Land[J]. Atmos. Res. 265, 105810. doi:10.1016/j.atmosres.2021.105810 CrossRef Full Text \| Google Scholar Cosimo, M., Marco, M., Giovanna, M., and Schneider, N. (2021). The Nexus between Information Technology and Environmental Pollution: Application of a New Machine Learning Algorithm to OECD Countries[J]. Util. Policy 72, 101256. doi:10.1016/j.jup.2021.101256 CrossRef Full Text \| Google Scholar Fan, Q., Yang, S., and Liu, S. (2019). Asymmetrically Spatial Effects of Urban Scale and Agglomeration on Haze Pollution in China. Int. J. Environ. Res. Public Health 16 (24), 4936. doi:10.3390/ijerph16244936 PubMed Abstract \| CrossRef Full Text \| Google Scholar Feng, Y., and Wang, X. (2020). Effects of Urban Sprawl on Haze Pollution in China Based on Dynamic Spatial Durbin Model during 2003–2016[J]. J. Clean. Prod. 242 (C), 118368. doi:10.1016/j.jclepro.2019.118368 CrossRef Full Text \| Google Scholar Hansen, B. E. (1999). Threshold Effects in Non-dynamic Panels: Estimation, Testing, and Inference. J. Econ. 93 (2), 345–368. doi:10.1016/s0304-4076(99)00025-1 CrossRef Full Text \| Google Scholar Hao, Y., and Liu, Y.-M. (2016). The Influential Factors of Urban PM2.5 Concentrations in China: a Spatial Econometric Analysis. J. Clean. Prod. 112, 1443–1453. doi:10.1016/j.jclepro.2015.05.005 CrossRef Full Text \| Google Scholar Jiang, X., Li, G., and Fu, W. (2021). Government Environmental Governance, Structural Adjustment and Air Quality: A Quasi-Natural Experiment Based on the Three-Year Action Plan to Win the Blue Sky Defense War. J. Environ. Manage 277, 111470. doi:10.1016/j.jenvman.2020.111470 PubMed Abstract \| CrossRef Full Text \| Google Scholar Kassouri, Y. (2021). Monitoring the Spatial Spillover Effects of Urbanization on Water, Built-Up Land and Ecological Footprints in Sub-saharan Africa. J. Environ. Manage 300, 113690. doi:10.1016/j.jenvman.2021.113690 PubMed Abstract \| CrossRef Full Text \| Google Scholar Katircioglu, S., Katircioglu, S., and Kilinc, C. C. (2018). Investigating the Role of Urban Development in the Conventional Environmental Kuznets Curve: Evidence from the Globe. Environ. Sci. Pollut. Res. Int. 25 (15), 15029–15035. doi:10.1007/s11356-018-1651-9 PubMed Abstract \| CrossRef Full Text \| Google Scholar Kim, J., and Kim, J. (2021). Regional Patterns of the Effect from Environmental Regulation: CO2 Emissions across Developed and Developing Countries[J]. APEC Stud. Assoc. Korea 13 (1), 85–98. doi:10.52595/jas.13.1.85 CrossRef Full Text \| Google Scholar Lan, H., and Pan, Y. (2019). Analysis and Research on Influencing Factors of Haze Weather[J]. J. Phys. Conf. Ser. 1267, 012031. doi:10.1088/1742-6596/1267/1/012031 CrossRef Full Text \| Google Scholar Lei, W., Jiao, L., and Xu, G. (2022). Understanding the Urban Scaling of Urban Land with an Internal Structure View to Characterize China’s Urbanization[J]. Land Use Policy 112, 105781. doi:10.1016/j.landusepol.2021.105781 CrossRef Full Text \| Google Scholar Li, W., Yang, G., and Li, X. (2021). Correlation between PM2.5 Pollution and its Public Concern in China: Evidence from Baidu Index[J]. J. Clean. Prod. 293, 126091. doi:10.1016/j.jclepro.2021.126091 CrossRef Full Text \| Google Scholar Li, X., Zhou, M., Zhang, W., Yu, K., and Meng, X. (2022). Study on the Mechanism of Haze Pollution Affected by Urban Population Agglomeration[J]. Atmosphere 13 (2), 278. doi:10.3390/atmos13020278 CrossRef Full Text \| Google Scholar Liao, S., Wu, Y., Wai Wong, S., and Shen, L. (2020). Provincial Perspective Analysis on the Coordination between Urbanization Growth and Resource Environment Carrying Capacity (RECC) in China[J]. Sci. Total Environ. 730, 138964. (prepublish). doi:10.1016/j.scitotenv.2020.138964 PubMed Abstract \| CrossRef Full Text \| Google Scholar Linberger, P. (2008). Center for International Earth Science Information Network (CIESIN)-Socioeconomic Data and Applications Center (SEDAC)[J]. J. Bus. Finance Librariansh. 5 (2). Google Scholar Liu, S., Liao, Q., Liang, Y., Li, Z., and Huang, C. (2021b). Spatio-Temporal Heterogeneity of Urban Expansion and Population Growth in China. Int. J. Environ. Res. Public Health 18 (24). doi:10.3390/ijerph182413031 CrossRef Full Text \| Google Scholar Liu, S., Shi, K., Wu, Y., and Chang, Z. (2021a). Remotely Sensed Nighttime Lights Reveal China's Urbanization Process Restricted by Haze Pollution[J]. Build. Environ. 206, 108350. doi:10.1016/j.buildenv.2021.108350 CrossRef Full Text \| Google Scholar Liu, X., and Ragusa, M. A. (2019). Effects of Urban Density and City Size on Haze Pollution in China: Spatial Regression Analysis Based on 253 Prefecture-Level Cities PM 2.5 Data[J]. Discrete Dyn. Nat. Soc. 2019, 1–8. doi:10.1155/2019/6754704 CrossRef Full Text \| Google Scholar Maddison, D. (2006). Environmental Kuznets Curves: A Spatial Econometric Approach. J. Environ. Econ. Manag. 51 (2), 218–230. doi:10.1016/j.jeem.2005.07.002 CrossRef Full Text \| Google Scholar Magazzino, C., Mele, M., Schneider, N., and Shahbaz, M. (2021). Can Biomass Energy Curtail Environmental Pollution? A Quantum Model Approach to Germany. J. Environ. Manage 287, 112293. doi:10.1016/j.jenvman.2021.112293 PubMed Abstract \| CrossRef Full Text \| Google Scholar Magazzino, C., Mele, M., and Schneider, N. (2022). Assessing a Fossil Fuels Externality with a New Neural Networks and Image Optimisation Algorithm: The Case of Atmospheric Pollutants as Confounders to COVID-19 Lethality. Epidemiol. Infect. 150, E1. doi:10.1017/s095026882100248x CrossRef Full Text \| Google Scholar Mele, M., Gurrieri, A. R., Morelli, G., and Magazzino, C. (2021). Nature and Climate Change Effects on Economic Growth: an LSTM Experiment on Renewable Energy Resources. Environ. Sci. Pollut. Res. Int. 28 (30), 41127–41134. doi:10.1007/s11356-021-13337-3 PubMed Abstract \| CrossRef Full Text \| Google Scholar Mele, M., Magazzino, C., Schneider, N., and Nicolai, F. (2021). Revisiting the Dynamic Interactions between Economic Growth and Environmental Pollution in Italy: Evidence from a Gradient Descent Algorithm. Environ. Sci. Pollut. Res. Int. 28 (37), 52188–52201. doi:10.1007/s11356-021-14264-z PubMed Abstract \| CrossRef Full Text \| Google Scholar Müller-Fürstenberger, G., and Wagner, M. (2007). Exploring the Environmental Kuznets Hypothesis: Theoretical and Econometric Problems[J]. Ecol. Econ. 62 (3-4), 648–660. doi:10.1016/j.ecolecon.2006.08.005 CrossRef Full Text \| Google Scholar Plischke, E., Borgonovo, E., and Smith, C. L. (2013). Global Sensitivity Measures from Given Data. Eur. J. Operational Res. 226 (3), 536–550. doi:10.1016/j.ejor.2012.11.047 CrossRef Full Text \| Google Scholar Song, W., Ye, C., Liu, Y., and Cheng, W. (2021). Do China's Urban-Environmental Quality and Economic Growth Conform to the Environmental Kuznets Curve? Int. J. Environ. Res. Public Health 18 (24). doi:10.3390/ijerph182413420 CrossRef Full Text \| Google Scholar Soo Hong, C., Huang, W., and Xun, L. (2021). Does Haze Cloud Decision Making? A Natural Laboratory Experiment[J]. J. Econ. Behav. Organ. 182, 132–161. doi:10.1016/j.jebo.2020.12.00 CrossRef Full Text \| Google Scholar Sun, L., Li, L., Li, Y., and Liu, B. (2020). Does Urbanization Promote Regional Industrial Environmental Efficiency? A Comparison of Economic Development-Oriented Regions and Environmental Governance-Oriented Regions [J]. Front. Energy Res. doi:10.3389/fenrg.2020.589733 CrossRef Full Text \| Google Scholar Wagner, M., Grabarczyk, P., and Hong, S. H. (2020). Fully Modified OLS Estimation and Inference for Seemingly Unrelated Cointegrating Polynomial Regressions and the Environmental Kuznets Curve for Carbon Dioxide Emissions[J]. J. Econ. 214 (1). doi:10.1016/j.jeconom.2019.05.012 CrossRef Full Text \| Google Scholar Wagner., M. (2015). The Environmental Kuznets Curve, Cointegration and Nonlinearity[J]. J. Appl. Econ. 30 (6). doi:10.1002/jae.2421 CrossRef Full Text \| Google Scholar Wang, C., Prasad, P., and Huang, C. (2021). The Impact of Renewable Energy, Urbanization, and Environmental Sustainability Ratings on the Environmental Kuznets Curve and the Pollution Haven Hypothesis[J]. Sustainability 13 (24). doi:10.3390/su132413747 CrossRef Full Text \| Google Scholar Wang, F., Wang, G., Liu, J., Ren, J., and Dong, M. (2021). Impact Paths of Land Urbanization on Haze Pollution: Spatial Nesting Structure Perspective[J]. Nat. Hazards 109 (1), 975–998. doi:10.1007/s11069-021-04864-w CrossRef Full Text \| Google Scholar Wu, R., Li, Y., and Wang, S. (2022). Will the Construction of High-Speed Rail Accelerate Urban Land Expansion? Evidences from Chinese Cities[J]. Land Use Policy 114, 105920. doi:10.1016/j.landusepol.2021.105920 CrossRef Full Text \| Google Scholar Wu, Y., Liao, M., Hu, M., Cao, J., Zhao, Z., Zhou, Y., et al. (2017). Atmospheric Levels and Cytotoxicity of Polycyclic Aromatic Hydrocarbons and Oxygenated-PAHs in PM2.5 in the Beijing-Tianjin-Hebei region)[J]. Ecol. Environ. Conservation 231, 1075–1084. doi:10.1016/j.envpol.2017.08.099 CrossRef Full Text \| Google Scholar%5BJ%5D&btnG=) Yacouba, K., and Oluyemi Adewole, O. (2022). Analysis of Spatio-Temporal Drivers and Convergence Characteristics of Urban Development in Africa[J]. Land Use Policy 112, 105868. doi:10.1016/j.landusepol.2021.105868 CrossRef Full Text \| Google Scholar Yang, Y., Tang, D., and Yang, X. (2020). Investigating the Spatio-Temporal Variations of the Impact of Urbanization on Haze Pollution Using Multiple Indicators[J]. Stoch. Environ. Res. Risk Assess. 35 (3), 703–717. doi:10.1007/s00477-020-01937-3 CrossRef Full Text \| Google Scholar Yang, Y., and Yan, D. (2021). Does Urban Sprawl Exacerbate Urban Haze Pollution? Environ. Sci. Pollut. Res. Int. 28 (40), 56522–56534. doi:10.1007/s11356-021-14559-1 PubMed Abstract \| CrossRef Full Text \| Google Scholar Yu, J., Wang, Y., and Liu, M. (2020). Mechanisms of an Extreme Fog and Haze Event in the Megacities of Central and Eastern China[J]. Meteorology Atmos. Phys. 133 (1), 123–139. doi:10.1007/s00703-020-00737-2 CrossRef Full Text \| Google Scholar Yu, X., Shen, M., Shen, W., and Zhang, X. (2020). Effects of Land Urbanization on Smog Pollution in China: Estimation of Spatial Autoregressive Panel Data Models[J]. Land 9 (9), 337. doi:10.3390/land9090337 CrossRef Full Text \| Google Scholar Zhang, C., Kuang, W., Wu, J., Liu, J., and Tian, H. (2020). Industrial Land Expansion in Rural China Threatens Environmental Securities[J]. Front. Environ. Sci. Eng. 15 (2). doi:10.1007/s11783-020-1321-2 CrossRef Full Text \| Google Scholar Zhang, M., Yue, L., Guo, R., and Yan, Y. (2022). Heterogeneous Effects of Urban Sprawl on Economic Development: Empirical Evidence from China[J]. Sustainability 14 (3), 1582. doi:10.3390/su14031582 CrossRef Full Text \| Google Scholar Zhang, S., Wang, Y., Liu, Z., and Yu, H. (2021). The Spatial Dynamic Relationship between Haze Pollution and Economic Growth: New Evidence from 285 Prefecture-Level Cities in China[J]. J. Environ. Plan. Manag. 64 (11). doi:10.1080/09640568.2020.1854694 CrossRef Full Text \| Google Scholar Keywords: urban sprawl, haze pollution, population urbanization, land urbanization, spatial Durbin model, coordination Citation: Wang Z, Yang J and Jiang J (2022) Urban Sprawl and Haze Pollution: Based on Raster Data of Haze PM2.5 Concentrations in 283 Cities in Mainland China. Front. Environ. Sci. 10:929558. doi: 10.3389/fenvs.2022.929558 Received: 27 April 2022; Accepted: 10 June 2022; Published: 30 June 2022. Edited by: Cosimo Magazzino, Roma Tre University, Italy Reviewed by: Yacouba Kassouri, Leipzig University, Germany Setareh Katircioglu, University of Kyrenia, Cyprus Robert Kunst, Institute for Advanced Studies (IHS), Austria Dimitrios Stathakis, University of Thessaly, Greece Copyright © 2022 Wang, Yang and Jiang. This is an open-access article distributed under the terms of the Creative Commons Attribution License (CC BY). The use, distribution or reproduction in other forums is permitted, provided the original author(s) and the copyright owner(s) are credited and that the original publication in this journal is cited, in accordance with accepted academic practice. No use, distribution or reproduction is permitted which does not comply with these terms. Correspondence: Jinqi Jiang, 2014500053@syau.edu.cn Disclaimer: All claims expressed in this article are solely those of the authors and do not necessarily represent those of their affiliated organizations, or those of the publisher, the editors and the reviewers. Any product that may be evaluated in this article or claim that may be made by its manufacturer is not guaranteed or endorsed by the publisher. 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15903	https://artofproblemsolving.com/wiki/index.php/Simon%27s_Favorite_Factoring_Trick?srsltid=AfmBOoqLVmUcnOAtgFLkPUTB-TtYzsG960AJoDJ927FcCm-rYOBVbhvU	Art of Problem Solving Simon's Favorite Factoring Trick - 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 Simon's Favorite Factoring Trick Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search Simon's Favorite Factoring Trick Simon's Favorite Factoring Trick (SFFT) is often used in Diophantine equations where factoring is needed, applicable for Diophantine equations of the form , for integer constants , , and , where there is a constant on one side of the equation and on the other side and a product of variables with each of those variables in linear terms. Simon's Favorite Factoring Trick is named after AoPS user ComplexZeta, or Dr. Simon Rubinstein-Salzedo. Contents 1 Statement 2 Applications 3 Problems 3.1 Introductory 3.2 Intermediate 3.3 Olympiad 4 See Also 4.1 External Links Statement Let's put it in general terms. We have an equation , where , , and are integer constants. Simon's Favorite Factoring Trick states that this equation can be factored into the equation For example, is the same as: Here is another way to look at it. Consider the equation .Let's start to factor the first group out: . How do we group the last term so we can factor by grouping? Notice that we can add to both sides. This yields . Now, we can factor as . This is important because this keeps showing up in number theory problems. Let's look at this problem below: Determine all possible ordered pairs of positive integers that are solutions to the equation . (2021 CEMC Galois #4b) Let's remove the denominators: . Then . Take out the : (notice how I artificially grouped up the terms by adding ). Now, (you can just do SFFT directly, but I am guiding you through the thinking behind SFFT). Now we use factor pairs to solve this problem. Look at all factor pairs of 20: . The first factor is for , the second is for . Solving for each of the equations, we have the solutions as . Applications This factorization frequently shows up on contest problems, especially those heavy on algebraic manipulation. Usually and are variables and are known constants. Sometimes, you have to notice that the variables are not in the form and . Additionally, you almost always have to subtract or add the and terms to one side so you can isolate the constant and make the equation factorable. It can be used to solve more than algebra problems, sometimes going into other topics such as number theory. When coefficient of is not , you can sometimes achieve an equation that can be factored by dividing the coefficient off of the equation. Problems Introductory Two different prime numbers between and are chosen. When their sum is subtracted from their product, which of the following numbers could be obtained? (Source) Intermediate AoPS Online If has a remainder of when divided by , and has a remainder of when divided by , find the value of the remainder when is divided by . are integers such that . Find . (Source) A rectangular floor measures by feet, where and are positive integers with . An artist paints a rectangle on the floor with the sides of the rectangle parallel to the sides of the floor. The unpainted part of the floor forms a border of width foot around the painted rectangle and occupies half of the area of the entire floor. How many possibilities are there for the ordered pair ? (Source) Olympiad The integer is positive. There are exactly ordered pairs of positive integers satisfying: Prove that is a perfect square. (Source) See Also Number Theory Diophantine equation Algebra Factoring External Links Video on AoPS Retrieved from " Categories: Number theory Theorems 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.
15904	https://askfilo.com/user-question-answers-smart-solutions/3-the-equation-has-roots-of-opposite-sign-if-lies-a-on-a-3137333734323033	Question asked by Filo student Views: 5,756 students Updated on: Dec 31, 2024 Text SolutionText solutionverified iconVerified Concepts: Quadratic equations, Roots of equations, Discriminant, Geometric loci Explanation: To determine the condition for the roots of the quadratic equation to have opposite signs, we need to analyze the given quadratic equation. The quadratic equation is given by: (1+a2)x2+2a2x+a2+b2−1=0. For the roots of this quadratic equation to have opposite signs, the product of the roots must be negative. Step by Step Solution: Step 1 The general form of a quadratic equation is Ax2+Bx+C=0. Here, A=1+a2, B=2a2, and C=a2+b2−1. Step 2 The product of the roots of a quadratic equation Ax2+Bx+C=0 is given by AC. Thus, the product of the roots for our equation is: 1+a2a2+b2−1. Step 3 For the roots to have opposite signs, the product of the roots must be negative: 1+a2a2+b2−1<0. Step 4 This inequality will hold if a2+b2−1<0. Therefore, a2+b2<1. This represents the interior of a circle with center at (0,0) and radius 1. Final Answer: The correct option is (b) Inside a circle of centre (0,0) and radius 1. Students who ask this question also asked Views: 5,709 Topic: Smart Solutions View solution Views: 5,625 Topic: Smart Solutions View solution Views: 5,625 Topic: Smart Solutions View solution Views: 5,271 Topic: Smart Solutions View solution Stuck on the question or explanation? Connect with our tutors online and get step by step solution of this question. \| \| \| --- \| \| Question Text \| 3. The equation (1+a2)x2+2a2x+a2+b2−1=0 has roots of opposite sign if (a,b) lies (a) On a straight line x+y=1 (b) Inside a circle of centre (0,0) and radius 1 (c) On a parabola of vertex (0,0) and focal length 1 (d) On a straight line x−y=1 \| \| Updated On \| Dec 31, 2024 \| \| Topic \| All topics \| \| Subject \| Smart Solutions \| \| Class \| Class 11 \| \| Answer Type \| Text solution:1 \| Are you ready to take control of your learning? Download Filo and start learning with your favorite tutors right away! Questions from top courses Explore Tutors by Cities Blog Knowledge © Copyright Filo EdTech INC. 2025
15905	https://www.fq.math.ca/Scanned/8-5/carlson-a.pdf	DETERMINATION OF HERONIAN TRIANGLES JOHN R.CARLSON San Diego State College, San Diego, California 1. A Pythagorean triangle is defined as any right-triangle having inte-gral sides. Using the well-known relationship a2 + b2 = c2 where a, b, and c are the two sides and the hypotenuse respectively, it is obvious that one of the sides must be even. Hence, the area of such a triangle is also an integer,, In his book , Ore introduces the generalization of this situation: a triangle is called Heronian if it has integral sides and area. He further com-ments that, "although we know a considerable number of Heronian triangles, we have no general formula giving them all.?? In this paper, we propose to find all such triangles and prove a few basic properties concerning them. 2. Since every Pythagorean triangle is Heronian, and since Pytha-gorean triangles are completely described by the well-known formulas for the sides u2 + v2, u2 - v2, and 2uv, the real problem is to characterize all non-right-angled Heronian triangles. We first give an obvious property. Lemma 1. Let a, b, c, and n all be integers. Then the triangle with sides of na, nb, and nc is Heronian if and only if the "reduced" t r i -angle with sides a , b , c is Heronian. Proof. We shall use Heron1 s formula for the area of a triangle (1) A = -JsTs - a)(s - b)(s - c), where s = - (a + b + c) . Let A be the area of triangle a , b , c and let Af be the area of na, nb, nce Then Eq. (1) shows us immediately that Oystein Ore, Invitation to Number Theory, pp. 59-60, Random House, New York, 1967. 499 500 DETERMINATION OF HERONIAN TRIANGLES [Dec. (2) Af = n2A . Hence, if A is an integer, so is AT. For the converse, suppose that Af is integral. Then Eq. (2) insures that A is at least rationaL On the other hand, Eq. (1) implies that A is the square root of an integer, which is well known to be either integral, or irrational. Thus we conclude that A must be an integer and the lemma is proven. Before we proceed to our first theorem, we illustrate with two examples. Suppose we juxtapose (or "adjoin") the two Pythagorean triangles 5, 12, 13 and 9, 12, 15 so that their common-length sides coincide. Clearly a (non-right-angled) Heronian triangle results with sides of 13, 14, 15 and area equalling 84. As a second example, we adjoin the triangles 12, 16, 20 and 16, 63, 65 (one of which is primative) to obtain the Heronian triangle 20, 65, 75 which may be reduced to 4, 13, 15, a primitive Heronian triangle with area equal-ling 24. That these two examples illustrate all possible events is the content of our first theorem. Theorem 1. A triangle is Heronian if and only if it is the adjunction of two Pythagorean triangles along a common side, or a reduction of such an adjunction. Proof. One direction is clear: since every Pythagorean triangle is Heronian, so is every adjunction of two. The previous lemma guarantees that every reduction is also Heronian, Thus, let us suppose that the triangle a,b,c is Heronian and try to prove that it is either the adjunction or a reduction of an adjunction of two Pythagorean triangles. First we assume the obvious: that from some vertex we may draw a perpendicular to the opposite side, thus dividing the given tri-angle into two "adjoined" right triangles. That such is possible is easily shown. Let the length of the altitude so constructed be x and let the base c be thus divided into segments c1 and c2, so that c = cA + c2. Since the triangle is Heronian, it is clear that is rational. Let 1970] DETERMINATION OF HERONIAN TRIANGLES 5 0 1 m x = — n be reduced to lowest terms. By the law of cosines b2 = a2'+ c2 - 2ac cos wi and thus a2 + c2 - b2 C 0 S wi = Wc is rational. Hence both Cj = b cos wj and c2 = c - cA are also rational If the numbers x, c1? and c2 are in fact all integers, we are done. Other-wise we look at the triangle having sides na, nb, and nc. From elementary geometry, this triangle is similar to the original one and thus the new altitude is equal to nx = m, an integer,, But nc1? still rational, is given by v^ which is9 as before, either integral or irrational, and thus must be integraL Likewise nc2 is integrals and the new enlarged triangle is the adjunction of two Pythagorean triangles. Thus the original is a reduction of an adjunction and we are through. Since the sides of any Pythagorean triangle are given by u2 + v2, u2 -v2, and 2uvs we now have a method for finding all Heronian triangles. Corollary 1. A triangle is Heronian if and only if its sides are given by either (3) u2 + v2, r2 + s2, and u2 - v2 + r2 - s2; where rs = uv; or (4) u2 + v2, r2 + s2, and 2(uv + rs); where r2 - s2 = u2 - v2; or (5) a re-duction by any constant factor in either case (3) or (4). 3. Although the preceding theorem and its corollary give formulations for finding all Heronian triangles, there are many properties of Heronian tri-angles that are not obvious from examination of the special subset of right-angled triangles. Some of these properties will be given here. Lemma 2. A primitive Heronian triangle is isosceles if and only if it has sides given by (3), (4), or (5) with r = u and s = v. Proof. Since a triangle is primitive only when one side is even, the equal sides of the isosceles triangle must be odd? say 2m + 1. Let the even 502 DETERMINATION OF HERONIAN TRIANGLES [Dec. side be 2n. Then the semiperimeter is given by 2m + n + 1 and the area, given by Eqe (1) becomes ^f(2m"+ n + l ) ^ T T ^ T l J ( n j f o ) , so that A = n)/(2m + I) 2 - n2 . If this is to be an integer, there must be an integer Q such that (2m + I)2 - n2 = Q2 . Thus the number 2m + 1 is the hypotenuse of a Pythagorean triangle, which meansj of course, that 2m + 1 is as given in Corollary 1. Conversely, every triangle described by those formulae will be isosceles if r = u and s = v. We note in particular that any number of the form 4n + 2 may be used as the even side of a primitive isosceles Heronian triangle, by using sides 2n2 + 2n + 1, 2n2 + 2n + 1, and 4n + 2, We will show in fact, that any inte-ger greater than two may be used as the side of a primitive non-right-angled Heronian triangle, Before we do, we shall establish the following Lemma 39 No Heronian triangle has a side of either 1 or 20 Proof, Since the sides of the triangle must be integers, the difference between two sides is either 0 or an integer ^>1, This latter case would p r e -clude the use of 1 as a side. But an isosceles triangle with side one is also impossiblej since in that event, we must have two sides equal to 1, and hence the third side either 0 or 2 Thus we have only to show that 2 cannot be used as the side of an Heronian triangle. Suppose, to the contrary, that we do have a triangLe with sides 2, a, and b, Then the area as given by (1) is A - \ s ( s - 2)(s - " a M s ^ T ) , where a + b + 2 = 2s, The only values of a and b which satisfy this last equation are a = b = s - 1. Thus the area becomes 1970] DETERMINATION OF HERONIAN TRIANGLES 503 A = Vs(s - 2)(1)(1) and we must have s(s - 2) = Q2 for some Q But this Is impossible so we are donee Using the formulae for the sides of a Pythagorean triangle, It Is easy to show that every integer greater than two can be used as a side In a finite number of Pythagorean triangles. This observation has the following r e -markable generalization,, Theorem 2a Let a be an integer greater than twoe Then there exists an infinitude of primitive Heronian triangles with one side of length ae Proof, If a is odd we may use sides given by (6) a9 i(at - 1), and \| ( a t + 1) where t is a solution of the Pellian equation (7) t2 - (a2 - l)y2 = 1 . Since a2 - 1 is even and never a perfect square, Eq (7) has an infinitude of solutions for t, each of them odd9 so that (6) lists only Integers,, Since \| ( a t - 1) + 1 = \| ( a t + 1), the triangle is obviously primitive We compute the area of the triangle by (1) and find A2 = \| ( t + l ) a \| ( t - l ) a \| ( a + l ) \| ( a - 1) = (f)2a2(a2 - l)(t2 - 1) . But, by Eq (7), we have t2 - 1 = (a2 - l)y2 so that A2 = (\|)2a2(a2 - l)2y2 and thus A = -g-(a - D ^ ( a + Day , 504 DETERMINATION OF HERONIAN TRIANGLES [Dec. an integer, and hence the triangle is Heronian. Next, suppose that a is even, say a = 2n, where n is odd. Then we may use (8) a, tn - 2, and tn + 2 , where t is any odd solution of (9) t2 - (n2 - 4)y2 = 1 . Since n is odd, n2 - 4 is also odd and thus an infinitude of odd values of t is available. That (8) forms a primitive triangle is clear; we prove it is Heronian by computing A2 = (t + l)n(t - l)n(n + 2)(n - 2) = n2(n2 - 4)(t2 - 1) = n2(n2 - 4)2(y)2 so that A is an integer, Lastly, suppose that a = 2n, where n is even. Then we may use (10) a, tn - 1, and tn + 1, where t is any solution of (11) t2 - (n2 - l)y2 = 1 . The proof follows the same lines as that just given. Thus our theorem is proven for all cases. It should be obvious that the formulations given in the above proof are simply chosen from an infinitude of possibilities. Thus we could also have shown, for example, that the triangle having sides a, ^-a(x - 1) + 1, and 1970] DETERMINATION OF HERONIAN TRIANGLES 505 ~a(x + 1) - 1, where x is found from x2 - (a - l)y 2 - 1, which will have an infinitude of solutions as long as a - 1 is not a perfect square, 4. We conclude with a few observations and a short list of examples., It is easy to show that every primitive Heronian triangle has exactly one even side. A simple check of all the possibilities obtained from the adjunction of two Pythagorean triangles (which are known to have either one or three even sides) will suffice to prove this. From this, we conclude that the area of any Heronian. triangle is divisible by 2, since, if s is even, a factor of 2 divides A2 (and thus also A) while, if s is odd, then s - a will be even where a is an odd side of the triangle, and so again 2 divides the area, Since the area of any Pythagorean triangle is given by A = uv(u - v)(u + v) it is easy to show that such a triangle has area divisible by three, A simple analysis of adjunctions and possible reductions will then show that every Heronian triangle has area divisible by three, Following is alist of the "first" few primitive (non-right-angled) Heron-ian triangles: a = 3 b = 25 c = 26 4 13 15 5 5 6 5 5 8 5 29 30 6 25 29 7. 15 20 8 29 35 506 DETERMINATION OF HERONIAN TRIANGLES [Dec. a = 9 b = _10 c = 17 9 65 70 10 13 13 10 17 21 11 13 20 12 17 25 13 13 24 13 14 15 13 20 21 13 37 40 14 25 25 15 28 41 15 37 44 15 41 52 16 17 17 17 17 30 17 17 25 28 18 41 41 19 20 37 20 37 51 21 85 104 22 61 61 23 212 225 24 37 37 25 25 48 25 29 36 25 39 40 25 51 74 26 85 85 27 676 701 28 85 111 29 29 40 29 29 42 [Continued on page 5510 ]
15906	https://open.lib.umn.edu/humanbiology/chapter/1-7-the-evolution-of-primates/	1.7 The Evolution of Primates – Human Biology Skip to content Menu Primary Navigation Home Read Sign in Search in book: Search Want to create or adapt books like this? Learn more about how Pressbooks supports open publishing practices. Book Contents Navigation Contents Theme 1: What Makes Us Unique? 1.1 Structural Organization of the Human Body 1.2 DNA Overview 1.3 The Genetic Basis of Evolution 1.4 Mechanisms of Evolution 1.5 Introduction to Phylogenies 1.6 How Phylogenies are Made 1.7 The Evolution of Primates 1.8 Cells of the Nervous System 1.9 The Brain and Spinal Cord 1.10 How Memory Functions 1.11 Parts of the Brain Involved with Memory 1.12 Problems with Memory: Eyewitness Testimony Theme 2: How Does Blood and Organ Donation Work? 2.1 Human Genetics 2.2 Components of the Blood 2.3 Transcription 2.4 Translation 2.5 How Genes are Regulated 2.6 Innate Immunity 2.7 Adaptive Immunity 2.8 Ethics of Research 2.9 The Process of Science Theme 3: What Causes Stress? 3.1 What is Stress? 3.2 Parts of the Nervous System 3.3 Circulatory and Respiratory Systems 3.4 Atherosclerosis, blood lipids, and stress Theme 4: How Do Diet, Exercise and Weight Affect Health? 4.1 Biological Molecules 4.2 Nutrition and Diet 4.3 The Digestive System 4.4 ATP: Adenosine Triphosphate 4.5 Energy and Metabolism 4.6 Enzymes 4.7 Musculoskeletal System 4.8 Hunger, Eating, and Weight Theme 5: How Do We Control Our Fertility? 5.1 Human Reproductive Anatomy Human Reproductive Anatomy 5.2 Meiosis and Gametogenesis 5.3 Hormones and the Endocrine System 5.4 Hormonal Control of Human Reproduction 5.5 Fertilization and Early Embryonic Development 5.6 Human Pregnancy and Birth Theme 6: What Causes Cancer? 6.1 The Cell Cycle 6.2 Cancer and the Cell Cycle 6.3 DNA Replication and Repair Mechanisms 6.4 Cancer Treatment Theme 7: Resources 7.1: Geological Time 7.2 The Periodic Table 7.3 Measurements and the Metric System 7.4 Essential Mathematics Human Biology Theme 1: What Makes Us Unique? 1.7 The Evolution of Primates Order Primatesof class Mammalia includes lemurs, tarsiers, monkeys, apes, and humans. Non-human primates live primarily in the tropical or subtropical regions of South America, Africa, and Asia. They range in size from the mouse lemur at 30 grams (1 ounce) to the mountain gorilla at 200 kilograms (441 pounds). The characteristics and evolution of primates is of particular interest to us as it allows us to understand the evolution of our own species. Characteristics of Primates All primate species possess adaptations for climbing trees, as they all descended from tree-dwellers. This arboreal heritage of primates has resulted in hands and feet that are adapted for brachiation, or climbing and swinging through trees. These adaptations include, but are not limited to: 1) a rotating shoulder joint, 2) a big toe that is widely separated from the other toes and thumbs, which are widely separated from fingers (except humans), which allow for gripping branches, 3) stereoscopic vision, two overlapping fields of vision from the eyes, which allows for the perception of depth and gauging distance. Other characteristics of primates are brains that are larger than those of most other mammals, claws that have been modified into flattened nails, typically only one offspring per pregnancy, and a trend toward holding the body upright. Order Primates is divided into two groups: prosimians and anthropoids. Prosimians include the bush babies of Africa, the lemurs of Madagascar, and the lorises, pottos, and tarsiers of Southeast Asia.Anthropoidsinclude monkeys, apes, and humans. In general, prosimians tend to be nocturnal (in contrast to diurnal anthropoids) and exhibit a smaller size and smaller brain than anthropoids. Evolution of Primates The first primate-like mammals are referred to as proto-primates. They were roughly similar to squirrels and tree shrews in size and appearance. The existing fossil evidence (mostly from North Africa) is very fragmented. These proto-primates remain largely mysterious creatures until more fossil evidence becomes available. The oldest known primate-like mammals with a relatively robust fossil record is Plesiadapis(although some researchers do not agree that Plesiadapis was a proto-primate). Fossils of this primate have been dated to approximately 55 million years ago. Plesiadapiforms were proto-primates that had some features of the teeth and skeleton in common with true primates. They were found in North America and Europe in the Cenozoic and went extinct by the end of the Eocene. The first true primates were found in North America, Europe, Asia, and Africa in the Eocene Epoch. These early primates resembled present-day prosimians such as lemurs. Evolutionary changes continued in these early primates, with larger brains and eyes, and smaller muzzles being the trend. By the end of the Eocene Epoch, many of the early prosimian species went extinct due either to cooler temperatures or competition from the first monkeys. Anthropoid monkeys evolved from prosimians during the Oligocene Epoch. By 40 million years ago, evidence indicates that monkeys were present in the New World (South America) and the Old World (Africa and Asia). New World monkeys are also called Platyrrhini—a reference to their broad noses (Figure 1). Old World monkeys are called Catarrhini—a reference to their narrow noses. There is still quite a bit of uncertainty about the origins of the New World monkeys. At the time the platyrrhines arose, the continents of South American and Africa had drifted apart. Therefore, it is thought that monkeys arose in the Old World and reached the New World either by drifting on log rafts or by crossing land bridges. Due to this reproductive isolation, New World monkeys and Old World monkeys underwent separate adaptive radiations over millions of years. The New World monkeys are all arboreal, whereas Old World monkeys include arboreal and ground-dwelling species. Figure 1. The howler monkey is native to Central and South America. It makes a call that sounds like a lion roaring. (credit: Xavi Talleda) Apes evolved from the catarrhines in Africa midway through the Cenozoic, approximately 25 million years ago. Apes are generally larger than monkeys and they do not possess a tail. All apes are capable of moving through trees, although many species spend most their time on the ground. Apes are more intelligent than monkeys, and they have relatively larger brains proportionate to body size. The apes are divided into two groups. The lesser apes comprise the family Hylobatidae, including gibbons and siamangs. The great apes include the genera Pan(chimpanzees and bonobos) (Figure 2a), Gorilla(gorillas),Pongo (orangutans), andHomo(humans) (Figure 2b). The very arboreal gibbons are smaller than the great apes; they have low sexual dimorphism (that is, the sexes are not markedly different in size); and they have relatively longer arms used for swinging through trees. Figure 2.The (a) chimpanzee is one of the great apes. It possesses a relatively large brain and has no tail. (b) All great apes have a similar skeletal structure. (credit a: modification of work by Aaron Logan; credit b: modification of work by Tim Vickers) Human Evolution The family Hominidae of order Primates includes the hominoids: the great apes (Figure 3). Evidence from the fossil record and from a comparison of human and chimpanzee DNA suggests that humans and chimpanzees diverged from a common hominoid ancestor approximately 6 million years ago. Several species evolved from the evolutionary branch that includes humans, although our species is the only surviving member. The term homininis used to refer to those species that evolved after this split of the primate line, thereby designating species that are more closely related to humans than to chimpanzees. Hominins were predominantly bipedal and include those groups that likely gave rise to our species—including Australopithecus,Homo habilis, and Homo erectus—and those non-ancestral groups that can be considered “cousins” of modern humans, such as Neanderthals. Determining the true lines of descent in hominins is difficult. In years past, when relatively few hominin fossils had been recovered, some scientists believed that considering them in order, from oldest to youngest, would demonstrate the course of evolution from early hominins to modern humans. In the past several years, however, many new fossils have been found, and it is clear that there was often more than one species alive at any one time and that many of the fossils found (and species named) represent hominin species that died out and are not ancestral to modern humans. Figure 3. This chart shows the evolution of modern humans. Very Early Hominins Three species of very early hominids have made news in the past few years. The oldest of these,Sahelanthropus tchadensis, has been dated to nearly 7 million years ago. There is a single specimen of this genus, a skull that was a surface find in Chad. The fossil, informally called “Toumai,” is a mosaic of primitive and evolved characteristics, and it is unclear how this fossil fits with the picture given by molecular data, namely that the line leading to modern humans and modern chimpanzees apparently bifurcated about 6 million years ago. It is not thought at this time that this species was an ancestor of modern humans. A second, younger species,Orrorin tugenensis, is also a relatively recent discovery, found in 2000. There are several specimens of Orrorin. It is not known whether Orrorin was a human ancestor, but this possibility has not been ruled out. Some features of Orrorin are more similar to those of modern humans than are the australopiths, although Orrorin is much older. A third genus,Ardipithecus, was discovered in the 1990s, and the scientists who discovered the first fossil found that some other scientists did not believe the organism to be a biped (thus, it would not be considered a hominid). In the intervening years, several more specimens of Ardipithecus, classified as two different species, demonstrated that the organism was bipedal. Again, the status of this genus as a human ancestor is uncertain. Early Hominins: Genus Australopithecus Australopithecus(“southern ape”) is a genus of hominin that evolved in eastern Africa approximately 4 million years ago and went extinct about 2 million years ago. This genus is of particular interest to us as it is thought that our genus, genus Homo, evolved from a common ancestor shared with Australopithecus about 2 million years ago (after likely passing through some transitional states).Australopithecus had a number of characteristics that were more similar to the great apes than to modern humans. For example, sexual dimorphism was more exaggerated than in modern humans. Males were up to 50 percent larger than females, a ratio that is similar to that seen in modern gorillas and orangutans. In contrast, modern human males are approximately 15 to 20 percent larger than females. The brain size of Australopithecus relative to its body mass was also smaller than modern humans and more similar to that seen in the great apes. A key feature that Australopithecus had in common with modern humans was bipedalism, although it is likely that Australopithecus also spent time in trees. Hominin footprints, similar to those of modern humans, were found in Laetoli, Tanzania and dated to 3.6 million years ago. They showed that hominins at the time of Australopithecus were walking upright. There were a number of Australopithecus species, which are often referred to as australopiths.Australopithecus anamensis lived about 4.2 million years ago. More is known about another early species,Australopithecus afarensis, which lived between 3.9 and 2.9 million years ago. This species demonstrates a trend in human evolution: the reduction of the dentition and jaw in size.A.afarensis(Figure 4) had smaller canines and molars compared to apes, but these were larger than those of modern humans. Its brain size was 380–450 cubic centimeters, approximately the size of a modern chimpanzee brain. It also had prognathic jaws, which is a relatively longer jaw than that of modern humans. In the mid-1970s, the fossil of an adult female A.afarensis was found in the Afar region of Ethiopia and dated to 3.24 million years ago (Figure 5). The fossil, which is informally called “Lucy,” is significant because it was the most complete australopith fossil found, with 40 percent of the skeleton recovered. Figure 4. The skull of (a) Australopithecus afarensis, an early hominid that lived between two and three million years ago, resembled that of (b) modern humans but was smaller with a sloped forehead and prominent jaw. Figure 5. This adult female Australopithecus afarensis skeleton, nicknamed Lucy, was discovered in the mid 1970s. (credit: “120”/Wikimedia Commons) Australopithecus africanus lived between 2 and 3 million years ago. It had a slender build and was bipedal, but had robust arm bones and, like other early hominids, may have spent significant time in trees. Its brain was larger than that of A.afarensis at 500 cubic centimeters, which is slightly less than one-third the size of modern human brains. Two other species,Australopithecus bahrelghazali and Australopithecus garhi, have been added to the roster of australopiths in recent years. A Dead End: Genus Paranthropus The australopiths had a relatively slender build and teeth that were suited for soft food. In the past several years, fossils of hominids of a different body type have been found and dated to approximately 2.5 million years ago. These hominids, of the genus Paranthropus, were muscular, stood 1.3-1.4 meters tall, and had large grinding teeth. Their molars showed heavy wear, suggesting that they had a coarse and fibrous vegetarian diet as opposed to the partially carnivorous diet of the australopiths.Paranthropus includes Paranthropus robustus of South Africa, and Paranthropus aethiopicus and Paranthropus boisei of East Africa. The hominids in this genus went extinct more than 1 million years ago and are not thought to be ancestral to modern humans, but rather members of an evolutionary branch on the hominin tree that left no descendants. Early Hominins: Genus Homo The human genus,Homo, first appeared between 2.5 and 3 million years ago. For many years, fossils of a species called H.habilis were the oldest examples in the genus Homo, but in 2010, a new species called Homo gautengensis was discovered and may be older. Compared to A.africanus,H.habilis had a number of features more similar to modern humans.H.habilis had a jaw that was less prognathic than the australopiths and a larger brain, at 600–750 cubic centimeters. However,H.habilis retained some features of older hominin species, such as long arms. The name H.habilis means “handy man,” which is a reference to the stone tools that have been found with its remains. H.erectus appeared approximately 1.8 million years ago (Figure 6). It is believed to have originated in East Africa and was the first hominin species to migrate out of Africa. Fossils of H.erectus have been found in India, China, Java, and Europe, and were known in the past as “Java Man” or “Peking Man.”H.erectus had a number of features that were more similar to modern humans than those of H.habilis.H.erectus was larger in size than earlier hominins, reaching heights up to 1.85 meters and weighing up to 65 kilograms, which are sizes similar to those of modern humans. Its degree of sexual dimorphism was less than earlier species, with males being 20 to 30 percent larger than females, which is close to the size difference seen in our species.H.erectus had a larger brain than earlier species at 775–1,100 cubic centimeters, which compares to the 1,130–1,260 cubic centimeters seen in modern human brains.H.erectus also had a nose with downward-facing nostrils similar to modern humans, rather than the forward facing nostrils found in other primates. Longer, downward-facing nostrils allow for the warming of cold air before it enters the lungs and may have been an adaptation to colder climates. Artifacts found with fossils of H.erectus suggest that it was the first hominin to use fire, hunt, and have a home base.H.erectus is generally thought to have lived until about 50,000 years ago. Figure 6. Homo erectus had a prominent brow and a nose that pointed downward rather than forward. Humans:Homo sapiens A number of species, sometimes called archaic Homo sapiens, apparently evolved from H.erectus starting about 500,000 years ago. These species include Homo heidelbergensis,Homo rhodesiensis, and Homo neanderthalensis. These archaic H.sapiens had a brain size similar to that of modern humans, averaging 1,200–1,400 cubic centimeters. They differed from modern humans by having a thick skull, a prominent brow ridge, and a receding chin. Some of these species survived until 30,000–10,000 years ago, overlapping with modern humans (Figure 7). Figure 7. The Homo neanderthalensis used tools and may have worn clothing. There is considerable debate about the origins of anatomically modern humans or Homo sapiens sapiens. As discussed earlier,H.erectus migrated out of Africa and into Asia and Europe in the first major wave of migration about 1.5 million years ago. It is thought that modern humans arose in Africa from H.erectus and migrated out of Africa about 100,000 years ago in a second major migration wave. Then, modern humans replaced H.erectus species that had migrated into Asia and Europe in the first wave. This evolutionary timeline is supported by molecular evidence. One approach to studying the origins of modern humans is to examine mitochondrial DNA (mtDNA) from populations around the world. Because a fetus develops from an egg containing its mother’s mitochondria (which have their own, non-nuclear DNA), mtDNA is passed entirely through the maternal line. Mutations in mtDNA can now be used to estimate the timeline of genetic divergence. The resulting evidence suggests that all modern humans have mtDNA inherited from a common ancestor that lived in Africa about 160,000 years ago. Another approach to the molecular understanding of human evolution is to examine the Y chromosome, which is passed from father to son. This evidence suggests that all men today inherited a Y chromosome from a male that lived in Africa about 140,000 years ago. Section Summary All primate species possess adaptations for climbing trees, as they all probably descended from tree-dwellers, although not all species are arboreal. Other characteristics of primates are brains that are larger than those of other mammals, claws that have been modified into flattened nails, typically only one young per pregnancy, stereoscopic vision, and a trend toward holding the body upright. Primates are divided into two groups: prosimians and anthropoids. Monkeys evolved from prosimians during the Oligocene Epoch. Apes evolved from catarrhines in Africa during the Miocene Epoch. Apes are divided into the lesser apes and the greater apes. Hominins include those groups that gave rise to our species, such as Australopithecus and H.erectus, and those groups that can be considered “cousins” of humans, such as Neanderthals. Fossil evidence shows that hominins at the time of Australopithecus were walking upright, the first evidence of bipedal hominins. A number of species, sometimes called archaic H.sapiens, evolved from H.erectus approximately 500,000 years ago. There is considerable debate about the origins of anatomically modern humans or H.sapiens sapiens. Glossary anthropoid monkeys, apes, and humans Australopithecus genus of hominins that evolved in eastern Africa approximately 4 million years ago brachiation movement through trees branches via suspension from the arms Catarrhini clade of Old World monkeys Gorilla genus of gorillas hominin species that are more closely related to humans than chimpanzees hominoid pertaining to great apes and humans Homo genus of humans Homo sapiens sapiens anatomically modern humans Hylobatidae family of gibbons Pan genus of chimpanzees and bonobos Platyrrhini clade of New World monkeys Plesiadapis oldest known primate-like mammal Pongo genus of orangutans Primates order of lemurs, tarsiers, monkeys, apes, and humans prognathic jaw long jaw prosimian division of primates that includes bush babies of Africa, lemurs of Madagascar, and lorises, pottos, and tarsiers of Southeast Asia stereoscopic vision two overlapping fields of vision from the eyes that produces depth perception Previous/next navigation Previous: 1.6 How Phylogenies are Made Next: 1.8 Cells of the Nervous System Back to top License Human Biology Copyright © by Sarah Malmquist and Kristina Prescott is licensed under a Creative Commons Attribution-NonCommercial 4.0 International License, except where otherwise noted. 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15907	https://readingfeynman.org/tag/relativistic-mass/	relativistic mass – Reading Feynman Skip to content Reading Feynman Menu Home About Matter Atoms Light Interactions Math Metaphysics Tag: relativistic mass The nature of time: an easy explanation of relativity My manuscript offers a somewhat sacrilegious but intuitive explanation of (special) relativity theory (The Emperor Has No Clothes: the force law and relativity, p. 24-27). It is one of my lighter and more easily accessible pieces of writing. The argument is based on the idea that we may define infinity or infinite velocities as some kind of limit (or some kind of limiting idea), but that we cannot really imagine it: it leads to all kinds of logical inconsistencies. Let me give you a very simple example here to illustrate these inconsistencies: if something is traveling at an infinite velocity, then it is everywhere and nowhere at the same time, and no theory of physics can deal with that. Now, if I would have to rewrite that brief introduction to relativity theory, I would probably add another logical argument. One that is based on our definition or notion of time itself. What is the definition of time, indeed? When you think long and hard about this, you will have to agree we can only measure time with reference to some fundamental cycle in Nature, right? It used to be the seasons, or the days or nights. Later, we subdivided a day into hours, and now we have atomic clocks. Whatever you can count and meaningfully communicate to some other intelligent being who happens to observe the same cyclical phenomenon works just fine, right? Hence, if we would be able to communicate to some other intelligent being in outer space, whose position we may or may not know but both he/she/it (let us think of a male Martian for ease of reference) and we/me/us are broadcasting our frequency- or amplitude-modulated signals wide enough so as to ensure ongoing communication, then we would probably be able to converge on a definition of time in terms of the fundamental frequency of an elementary particle – let us say an electron to keep things simple. We could, therefore, agree on an experiment where he – after receiving a pre-agreed start signal from us – would starting counting and send us a stop signal back after, say, three billion electron cycles (not approximately, of course, but three billion exactly). In the meanwhile, we would be capable, of course, to verify that, inbetween sending and receiving the start and stop signal respectively (and taking into account the time that start and stop signal needs to travel between him and us), his clock seems to run somewhat differently than ours. So that is the amazing thing, really. Our Martian uses the same electron clock, but our/his motion relative to his/ours leads us to the conclusion his clock works somewhat differently, and Einstein’s (special) relativity theory tells us how, exactly: time dilation, as given by the Lorentz factor. Does this explanation make it any easier to truly understand relativity theory? Maybe. Maybe not. For me, it does, because what I am describing here is nothing but the results of the Michelson-Morley experiment in a slightly more amusing context which, for some reason I do not quite understand, seems to make them more comprehensible. At the very least, it shows Galilean relativity is as incomprehensible – or as illogical or non-intuitive, I should say – as the modern-day concept of relativity as pioneered by Albert Einstein. You may now think (or not): OK, but what about relativistic mass? That concept is, and will probably forever remain, non-intuitive. Right? Time dilation and length contraction are fine, because we can now somehow imagine the what and why of this, but how do you explain relativistic mass, really? The only answer I can give you here it to think some more about Newton’s law: mass is a measure of inertia, so that is a resistance to a change in the state of motion of an object. Motion and, therefore, your measurement of any acceleration or deceleration (i.e. a change in the state of motion) will depend on how you measure time and distance too. Therefore, mass has to be relativistic too. QED: quod erat demonstrandum. In fact, it is not a proof, so I should not say it’s QED. It’s SE: a s atisfactory e xplanation. Why is an explanation and not a proof? Because I take the constant speed of light for granted, and so I kinda derive the relativity of time, distance and mass from my point of departure (both figuratively and literally speaking, I’d say). Post scriptum: For the mentioned calculation, we do need to know the (relative) position of the Martian, of course. Any event in physics is defined by both its position as well as its timing. That is what (also) makes it all very consistent, in fact. I should also note this short story here (I mean my post) is very well aligned with Einstein’s original 1905 article, so you can (also) go there to check the math. The main difference between his article and my explanation here is that I take the constant speed of light for granted, and then all that’s relative derives its relativity from that. Einstein looked at it the other way around, because things were not so obvious then. Jean Louis Van BellePhilosophy of science, PhysicsLeave a commentSeptember 15, 2020 September 15, 20203 Minutes Field energy and field momentum This post goes to the heart of the E = m c 2, equation. It’s kinda funny, because Feynman just compresses all of it in a sub-section of his Lectures. However, as far as I am concerned, I feel it’s a very crucial section. Pivotal, I’d say, which would fit with its place in all of the 115 Lectures that make up the three volumes, which is sort of mid-way, which is where we are here.So let’s get go for it. Let’s first recall what we wrote about the Poynting vector S, which we calculate from the magnetic and electric field vectors E and B by taking their cross-product: This vector represents the energy flow, per unit area and per unit time, in electrodynamical situations. If E and/or Bare zero (which is the case in electrostatics, for example, because we don’t have magnetic fields in electrostatics), then S is zero too, so there is no energy flow then. That makes sense, because we have no moving charges, so where would the energy go to? I also made it clear we should think of S as something physical, by comparing it to the heat flow vectorh, which we presented when discussing vector analysis and vector operators. The heat flow out of a surface element da is the area times the component of h perpendicular to da, so that’s (h•n)·da = h n·da. Likewise, we can write(S•n)·da = S n·da. The units of S and h are also the same:joule per second and per square meter or, using the definition of the watt(1 W = 1 J/s), in watt per square meter.In fact, if you google a bit, you’ll find that both h and S are referred to as a flux density: The heat flow vector h is the heat flux density vector, from which we get the heat flux through an area through the (h•n)·da = h n·da product. The energy flow Sis the energy flux density vector, from which we get the energy flux through the (S•n)·da = S n·da product. So that should be enough as an introduction to what I want to talk about here. Let’s first look at the energy conservation principle once again. Local energy conservation In a way, you can look atmy previous postas being all about the equation below, which we referred to as the ‘local’ energy conservation law: Of course, it is not the complete energy conservation law. The local energy is not only in the field. We’ve got matter as well, and so that’s what I want to discuss here: we want to look at the energy in the field as well as the energy that’s in the matter. Indeed, field energy is conserved, and then it isn’t: if the field is doing work on matter, or matter is doing work on the field, then… Well… Energy goes from one to the other, i.e. from the field to the matter or from the matter to the field. So we need to include matter in our analysis, which we didn’t do in our last post. Feynman gives the following simple example: we’re in a dark room, and suddenly someone turns on the light switch. So now the room is full of field energy—and, yes, I just mean it’s not dark anymore. :-). So that means some matter out there must have radiated its energy out and, in the process, it must have lost the equivalent mass of that energy. So, yes, we had matter losing energy and, hence, losing mass. Now, we know that energy and momentum are related. Respecting and incorporating relativity theory, we’ve got two equivalent formulas for it: E 2− p 2 c 2= m 0 2 c 4 p c= E·(v/c)⇔ p = v·E/c 2= m·v The E= m c 2 and m = ·m 0·(1−v 2/c 2)−1/2 formulas connect both expressions. So we can look at it in either of two ways. We could use the energy conservation law, but Feynman prefers the conservation of momentum approach, so let’s see where he takes us. If the field has some energy(and, hence, some equivalent mass) per unit volume, and if there’s some flow, so if there’s some velocity(which there is: that’s what our previous post was all about), then it will have a certain momentum per unit volume. [Remember: momentum is mass times velocity.] That momentum will have a direction, so it’s a vector, just like p = mv. We’ll write it as g, so we defineg as: g is the momentum of the field per unit volume. What units would we express it in? We’ve got a bit of choice here. For example, because we’re relating everything to energy here, we may want to convert our kilogram into eV/c 2 or J/c 2 units, using the mass-energy equivalence relation E= m c 2. Hmm… Let’s first keep the kg as a measure of inertia though. So we write: [g] = [m]·[v]/m 3= (kg·m/s)/m 3. Hmm… That doesn’t show it’s energy, so let’s replace the kg with a unit that’s got newton and meter in it, cf. the F = ma law. So we write:[g] = (kg·m/s)/m 3=(kg/s)/m 2= [(N·s 2/m)/s]/m 2=N·s/m 3. Well… OK. The newton·second is the unit of momentum indeed, and we can re-write it including the joule (1 J = 1 N·m), so then we get [g] = (J·s/m 4), so what’s that? Well… Nothing much. However, I do note it happens to be the dimension of S/c 2, so that’s [S/c 2] = [J/(s·m 2)]·(s 2/m 2) =(J·s/m 4). Let’s continue the discussion. Now, momentum is conserved, and each component of it is conserved. So let’s look at the x-direction. We should have something like: If you look at this carefully, you’ll probably say: “OK. I understood the thing with the dark room and light switch. Mass got converted into field energy, but what’s that second term of the left?” Good. Smart. Right remark. Perfect. […] Let me try to answer the question. While all of the quantities above are expressed per unit volume, we’re actually looking at the same infinitesimal volume element here, so the example of the light switch is actually an example of a ‘momentum outflow’, so it’s actually an example of that second term of the left-hand side of the equation above kicking in! Indeed, the first term just sort of reiterates the mass-energy equivalence: the energy that’s in the matter can become field energy, so to speak, in our infinitesimal volume element itself, and vice versa. But if it doesn’t, then it should get out and, hence, become ‘momentum outflow’. Does that make sense? No? Hmm… What to say? You’ll need to look at that equation a couple of times more, I guess. But I need to move on, unfortunately. [Don’t get put off when I say things like this: I am basically talking to myself, so it means I’ll need to re-visit this myself. :-/] Let’s look at all of the three terms: The left-hand side (i.e. the time rate-of-change of the momentum of matter) is easy. It’s just the force on it, which we know is equal to F =q(E+v×B). Do we know that? OK… I’ll admit it.Sometimes it’s easy to forget where we are in an analysis like this, but so we’re looking at the electromagnetic force here. As we’re talking infinitesimals here and, therefore, charge density rather than discrete charges, we should re-write this as the force per unit volume which is ρE+j×B. [This is an interesting formula which I didn’t use before, so you should double-check it. :-)] The first term on the right-hand side should be equally obvious, or… Well… Perhaps somewhat less so. But with all my rambling on the Uncertainty Principleand/or the wave-particle duality, it should make sense. If we scrap the second term on the right-hand side, we basically have an equation that is equivalent to the E = m c 2 equation. No? Sorry. Just look at it, again and again. You’ll end up understanding it. So it’s that second term on the right-hand side. What the hell does that say? Well… I could say: it’s the local energy or momentum conservation law. If the energy or momentum doesn’t stay in, it has to go out. But that’s not very satisfactory as an answer, of course. However, please just go along with this‘temporary’ answer for a while. So what is that second term on the right-hand side? As we wrote it, it’s an x-component – or, let’s put it differently, it is or was part of the x-component of the momentum density –but, frankly, we should probably allow it to go out in any direction really, as the only constraint on the left-hand side is a per second rate of change of something. Hence, Feynman suggest to equate it to something like this: What a, b and c? The components of some vector? Not sure. We’re stuck. This piece really requires very advanced math. In fact, as far as I know, this is the only time where Feynman says: “Sorry. This is too advanced. I’ll just give you the equation.Sorry.” So that’s what he does. He explains the philosophy of the argument, which is the following: On the left-hand side, we’ve got the time rate-of-change of momentum, so that obeys the F = dp/dt = d(mv)/dt law, with the force F,per unit volume, being equal to F(unit volume) =ρE+j×B. On the right-hand side, we’ve got something that can be written as: So we’d need to find a way to ρE+j×Bin terms of EandBonly– eliminating ρ andjby using Maxwell’s equations or whatever other trick – and then juggle terms and make substitutions to get it into a form that looks like the formula above, i.e. the right-hand side of that equation. But so Feynman doesn’t show us how it’s being done. He just mentions some theorem in physics, which says that the energy that’s flowing through a unit area per unit time divided by c 2– so that’s E/c 2 per unit area and per unit time– mustbe equal to the momentum per unit volume in the space, so we write: g = S/c 2 He illustrates the general theorem that’s used to get the equation above by giving two examples: OK. Two good examples. However, it’s still frustrating to not see how we get the g = S/c 2 in the specific context of the electromagnetic force, so let’s do a dimensional analysis at least. In my previous post, I showed that the dimension ofSmust be J/(m 2·s), so [S/c 2] = [J/(m 2·s)]/(m 2/s 2) = [N·m/(m 2·s)]·(s 2/m 2) = [N·s/m 3]. Now, we know that the unit of mass is 1 kg = N/(m/s 2). That’s just the force law: a force of 1 newton will give a mass of 1 kg an acceleration of 1 m/s per second, so 1 N = 1 kg·(m/s 2). So the[N·s/m 3] dimension is equal to [kg·(m/s 2)·s/m 3] = [(kg·(m/s)/m 3] =[(kg·(m/s)]/m 3, which is the dimension of momentum (p = mv) per unit volume, indeed.So, yes, the dimensional analysis works out, and it’s also in line with thep = v·E/c 2= m·v equation, but… Oh… We did a dimensional analysis already, where we also showed that [g]= [S/c 2] = (J·s/m 4). Well… In any case… It’s a bit frustrating to not see the detail here, but let us note the the Grand Resultonce again: The Poynting vectorSgives us the energy flow as well as the momentum density g= S/c 2. But what does it all mean, really? Let’s go through Einstein’s illustration of the principle. That will help us a lot. Before we do, however, I’d like to note something. I’ve always wondered a bit about that dichotomy between energy and momentum. Energy is force times distance: 1 joule is 1 newton× 1 meter indeed (1 J = 1 N·m). Momentum is force times time, as we can express it in N·s. Planck’s constant h combines all three in the dimension of action, which is force times distance times time: h≈6.6×10−34 N·m·s, indeed. I like that unity. In this regard, you should, perhaps, quickly review that post in which I explain that h is the energy per cycle, i.e. per wavelength or per period, of a photon, regardless of its wavelength. So it’s really something very fundamental. We’ve got something similar here: energy and momentum coming together, and being shown as one aspect of the same thing: some oscillation. Indeed, just see what happens with the dimensions when we ‘distribute’ the 1/c 2 factor on the right-hand side over the two sides, so we write:c·g= S/c and work out the dimensions: [c·g]= (m/s)·(N·s)/m 3= N/m 2= J/m 3. [S/c] =(s/m)·(N·m)/(s·m 2) = N/m 2= J/m 3. Isn’t that nice? Both sides of the equation now have a dimension like ‘the force per unit area’, or ‘the energy per unit volume’. To get that, we just re-scaledg and S, by c and 1/c respectively. As far as I am concerned, this shows an underlying unity we probably tend to mask with our ‘related but different’ energy and momentum concepts. It’s like E and B: I just love it we can write them together in our Poynting formula S= ε 0 c 2E×B. In fact, let me show something else here, which you should think about. You know that c 2= 1/(ε 0 μ 0), so we can write Salso as S =E×B/μ 0. That’s nice, but what’s nice too is the following: S/c = c·g= ε 0 cE×B= E×B/μ0c S/g= c 2= 1/(ε 0 μ 0) So, once again, Feynman may feel the Poynting vector is sort of counter-intuitive when analyzing specific situations but, as far as I am concerned, I feel the Poyning vector makes things actually easier to understand. Instead of two E and B vectors, and two concepts to deal with ‘energy’ (i.e. energy and momentum), we’re sort of unifying things here. In that regard – i.e in regard of feeling we’re talking the same thing really – I’d really highlight the S/g= c 2 = 1/(ε 0 μ 0) equation. Indeed, the universal constant c acts just like the fine-structure constant here: it links everything to everything. And, yes, it’s also about time we introduce the so-called principle of least action to explain things, because action, as a concept, combines force, distance and time indeed, so it’s a bit more promising than just energy, of just momentum. Having said that, you’ll see in the next section that it’s sometimes quite useful to have the choice between one formula or the other. But… Well…Enough talk. Let’s look at Einstein’s car. Einstein’s car Einstein’s car is a wonderful device: it rolls without any friction and it moves with a little flashlight. That’s all it needs. It’s pictured below. So the situation is the following: the flashlight shoots some light out from one side, which is then stopped at the opposite end of the car. When the light is emitted, there must be some recoil. In fact, we know it’s going to be equal to 1/c times the energy because all we need to do is apply the p c= E·(v/c) formula for v = c, so we know that p = E/c. Of course, this momentum now needs to move Einstein’s car. It’s frictionless, so it should work, but still… The car has some mass M, and so that will determine its recoil velocity: v = p/M. We just apply the general p = mv formula here, and v is not equal to c here, of course! Of course, then the light hits the opposite end of the car and delivers the same momentum, so that stops the car again. However, it did move over some distance x = vt. So we could flash our light again and get to wherever we want to get. [Never mind the infinite accelerations involved!] So… Well… Great! Yes, but Einstein didn’t like this car when he first saw it. In fact, he still doesn’t like it, because he knows it won’t take you very far. The problem is that we seem to be moving the center of gravity of this car by fooling around on the inside only. Einstein doesn’t like that. He thinks it’s impossible. And he’s right of course. The thing is: the center of gravity did not change. What happened here is that we’ve got some blob of energy, and so that blob has some equivalent mass (which we’ll denote by U/c 2), and so that equivalent mass moved all the way from one side to the other, i.e. over the length of the car, which we denote by L. In fact, it’s stuff like this that inspired the whole theory of the field energy and field momentum, and how it interacts with matter. What happens here is like switching the light on in the dark room: we’ve got matter doing work on the field, and so matter loses mass, and the field gains it, through its momentum and/or energy. To calculate how much, we could integrate S/c or c·gover the volume of our blob, and we’d get something in joule indeed, but there’s a simpler way here. The momentum conservation says that the momentum of our car and the momentum of our blob must be equal, so if T is the time that was needed for our blob to go to the other side – and so that’s, of course, also the time during which our car was rolling–then M·v = M·x/T must be equal to (U/c 2)·c=(U/c 2)·L/T. The 1/T factor on both sides cancel, so we write:M·x = (U/c 2)·L.Now, what is x? Yes. In case you were wondering, that’s what we’re looking for here. Here it is: x = vT = vL/c = (p/M)·(L/c) = [U/c)/M]·(L/c) = (U/c 2)·(L/M) So what’s next? Well… Now we need to show that the center-of-mass actually did not move with this ‘transfer’ of the blob. I’ll leave the math to you here: it should all work out. And you can also think through the obvious questions: Where is the energy and, hence, the mass of our blob after it stops the car? Hint: think about excited atoms and imagine they might radiate some light back. As the car did move a little bit, we should be able to move it further and further away from its center of gravity, until the center of gravity is no longer in the car. Hint: think about batteries and energy levels going down while shooting light out. It just won’t happen. Now, what about a blob of light going from the top to the bottom of the car? Well… That involves the conservation of angular momentum: we’ll have more mass on the bottom, but on a shorter lever-arm, so angular momentum is being conserved. It’s a very good question though, and it led Einstein to combine the center-of-gravity theorem with the angular momentum conservation theorem to explain stuff like this. It’s all fascinating, and one can think of a great many paradoxes that, at first, seem to contradict the Grand Principles we used here, which means that they would contradict all that we have learned so far. However, a careful analysis of those paradox reveals that they are paradoxes indeed:propositions which sound true but are, in the end,self-contradictory. In fact, when explaining electromagnetism over his various Lectures, Feynman tasks his readers with a rather formidable paradox when discussing the laws of induction, he solves it here, ten chapters later, after describing what we described above. You can busy yourself with it but… Well… I guess you’ve got something better to do. If so, just take away the key lesson: there’s momentum in the field, and it’s also possible to build up angular momentum in a magnetic field and, if you switch it off, the angular momentum will be given back, somehow, as it’s stored energy. That’s also why the seemingly irrelevant circulation of Swe discussed in my previous post, where we had a charge next to an ordinary magnet, and where we found that there was energy circulating around, is not so queer. The energy is there, in the circulating field, and it’s real. As real as can be. Some content on this page was disabled on June 16, 2020 as a result of a DMCA takedown notice from The California Institute of Technology. You can learn more about the DMCA here: Some content on this page was disabled on June 16, 2020 as a result of a DMCA takedown notice from The California Institute of Technology. You can learn more about the DMCA here: Advertisement Jean Louis Van BelleMathematics, Philosophy of science, Physics3 CommentsSeptember 29, 2015 September 30, 201514 Minutes On (special) relativity: what’s relative? Pre-scriptum (dated 26 June 2020): These posts on elementary math and physics have not suffered much the attack by the dark force—which is good because I still like them. While my views on the true nature of light, matter and the force or forces that act on them have evolved significantly as part of my explorations of a more realist (classical) explanation of quantum mechanics, I think most (if not all) of the analysis in this post remains valid and fun to read. In fact, I find the simplest stuff is often the best. Original post: This is my third and final post about special relativity. In the previous posts, I introduced the general idea and the Lorentz transformations. I present these Lorentz transformations once again below, next to their Galilean counterparts. [Note that I continue to assume, for simplicity, that the two reference frames move with respect to each other along the x- axis only, so the y- and z-component of u is zero. It is not all that difficult to generalize to three dimensions (especially not when using vectors) but it makes an intuitive understanding of what’s relativity all about more difficult.] As you can see, under a Lorentz transformation, the new ‘primed’ space and time coordinates are a mixture of the ‘unprimed’ ones. Indeed, the new x’ is a mixture of x and t, and the new t’ is a mixture as well. You don’t have that under a Galilean transformation: in the Newtonian world, space and time are neatly separated, and time is absolute, i.e. it is the same regardless of the reference frame. In Einstein’s world – our world – that’s not the case: time is relative, or local as Hendrik Lorentz termed it,and so it’s space-time– i.e. ‘some kind of union of space and time’ as Minkowski termed it – that transforms. In practice, physicists will use so-called four-vectors, i.e. vectors with four coordinates, to keep track of things. These four-vectors incorporate both the three-dimensional space vector as well as the time dimension. However, we won’t go into the mathematical details of that here. What else is relative? Everything, except the speed of light. Of course, velocity is relative, just like in the Newtonian world, but the equation to go from a velocity as measured in one reference frame to a velocity as measured in the other, is different: it’s not a matter of just adding or subtracting speeds. In addition, besides time, mass becomes a relative concept as well in Einstein’s world, and that was definitely not the case in the Newtonian world. What about energy? Well… We mentioned that velocities are relative in the Newtonian world as well, so momentum and kinetic energy were relative in that world as well: what you would measure for those two quantities would depend on your reference frame as well. However, here also, we get a different formula now. In addition, we have this weird equivalence between mass and energy in Einstein’s world, about which I should also say something more. But let’s tackle these topics one by one. We’ll start with velocities. Relativistic velocity In the Newtonian world, it was easy. From the Galilean transformation equations above, it’s easy to see that v’ = dx’/dt’ = d(x– ut)/dt = dx/dt– d(ut)/dt = v– u So, in the Newtonian world, it’s just a matter of adding/subtracting speeds indeed: if my car goes 100 km/h (v), and yours goes 120 km/h, then you will see my car falling behind at a speed of (minus) 20 km/h. That’s it. In Einstein’s world, it is not so simply. Let’s take the spaceship example once again. So we have a man on the ground (the inertial or ‘unprimed’ reference frame) and a man in the spaceship (the primed reference frame), which is moving away from us with velocity u. Now, suppose an object is moving inside the spaceship (along the x-axis as well) with a (uniform) velocity v x’, as measured from the point of view of the man inside the spaceship. Then the displacement x’ will be equal to x’ =v x’ t’. To know how that looks from the man on the ground, we just need to use the opposite Lorentz transformations: just replace u by –u everywhere (to the man in the spaceship, it’s like the man on the ground moves away with velocity –u), and note that the Lorentz factor does not change because we’re squaring and (–u)2 =u 2. So we get: Hence,x’=v x’ t’ can be written as x = γ(v x’ t’ + ut’). Now we should also substitute t’, because we want to measure everything from the point of view of the man on the ground. Now, t = γ(t’ + u _v_ _x’_ _t’_/c 2). Because we’re talking uniform velocities, v x(i.e. the velocity of the object as measured by the man on the ground) will be equal to x divided by t(so we don’t need to take the time derivative of x), and then, after some simplifying and re-arranging (note, for instance, how the t’ factor miraculously disappears), we get: What does this rather complicated formula say? Just put in some numbers: Suppose the object is moving at half the speed of light, so 0.5 c, and that the spaceship is moving itself also at 0.5 c, then we get the rather remarkable result that, from the point of view of the observer on the ground, that object is not going as fast as light, but only at v x= (0.5 c + 0.5 c)/(1 + 0.5·0.5) = 0.8 c. Or suppose we’re looking at a light beam inside the spaceship, so something that’s traveling at speed c itself in the spaceship. How does that look to the man on the ground? Just put in the numbers: v x= (0.5 c+ c)/(1 + 0.5·1) = c! So the speed of light is not dependent on the reference frame: it looks the same– both to the man in the ship as well as to the man on the ground. As Feynman puts it: “This is good, for it is, in fact, what the Einstein theory of relativity was designed to do in the first place–so it had better work!” It’s interesting to note that, even if u has no y– or z-component, velocity in the y direction will be affected too. Indeed, if an object is moving upward in the spaceship, then the distance of travel of that object to the man on the ground will appear to be larger. See the triangle below: if that object travels a distance Δs’ = Δy’ = Δy = v’Δt’ with respect to the man in the spaceship, then it will have traveled a distance Δs = vΔt to the man on the ground, and that distance is longer. I won’t go through the process of substituting and combining the Lorentz equations (you can do that yourself) but the grand result is the following: v y=(1/γ)v y’ 1/γ is the reciprocal of the Lorentz factor, and I’ll leave it to you to work out a few numeric examples. When you do that, you’ll find the rather remarkable result that v y is actually less than v y’. For example, for u = 0.6 c, 1/γ will be equal to 0.8, so v y will be 20% less than v y’. How is that possible? The vertical distance is what it is (Δy’ =Δy), and that distance is not affected by the ‘length contraction’ effect (y’ = y). So how can the vertical velocity be smaller?The answer is easy to state, but not so easy to understand: it’s the time dilation effect: time in the spaceship goes slower. Hence, the object will cover the same vertical distance indeed– for both observers –but, from the point of view of the observer on the ground, the object will apparently need more time to cover that distance than the time measured by the man in the spaceship: Δt > Δt’. Hence, the logical conclusion is that the vertical velocity of that object will appear to be less to the observer on the ground. How much less? The time dilation factor is the Lorentz factor. Hence,Δt = γΔt’. Now, if u= 0.6 c, then γ will be equal to 1.25 and Δt = 1.25Δt’. Hence, if that object would need, say, one second to cover that vertical distance, then, from the point of view of the observer on the ground, it would need 1.25 seconds to cover the same distance. Hence, its speed as observed from the ground is indeed only 1/(5/4) = 4/5 = 0.8 of its speed as observed by the man in the spaceship. Is that hard to understand? Maybe. You have to think through it. One common mistake is that people think that length contraction and/or time dilation are, somehow, related to the fact that we are looking at things from a distance and that light needs time to reach us. Indeed, on the Web, you can find complicated calculations using the angle of view and/or the line of sight (and tons of trigonometric formulas) as, for example, shown in the drawing below. These have nothing to do with relativity theory and you’ll never get the Lorentz transformation out of them. They are plain nonsense: they are rooted in an inability of these youthful authors to go beyond Galilean relativity. Length contraction and/or time dilation are not some kind of visual trick or illusion. If you want to see how one can derive the Lorentz factor geometrically, you should look for a good description of the Michelson-Morley experiment in a good physics handbook such as, yes :-), Feynman’s Lectures. So, I repeat: illustrations that try to explain length contraction and time dilation in terms of line of sight and/or angle of view are useless and will not help you to understand relativity. On the contrary, they will only confuse you. I will let you think through this and move on to the next topic. Relativistic mass and relativistic momentum Einstein actually stated two principles in his (special) relativity theory: The first is the Principle of Relativity itself, which is basically just the same as Newton’s principle of relativity. So that was nothing new actually: “If a system of coordinates K is chosen such that, in relation to it, physical laws hold good in their simplest form, then the same laws must hold good in relation to any other system of coordinates K’ moving in uniform translation relatively to K.” Hence, Einstein did not change the principle of relativity– quite on the contrary: he re-confirmed it–but he did change Newton’s Laws, as well as the Galilean transformation equations that came with them. He also introduced a new ‘law’, which is stated in the second ‘principle’, and that the more revolutionary one really: The Principle of Invariant Light Speed: “Light is always propagated in empty space with a definite velocity [speed]c which is independent of the state of motion of the emitting body.” As mentioned above, the most notable change in Newton’s Laws – the only change, in fact –is Einstein’s relativistic formula for mass: m v= γm 0 This formula implies that the inertia of an object, i.e. its mass, also depends on the reference frame of the observer. If the object moves (but velocity is relative as we know: an object will not be moving if we move with it), then its mass increases. This affects its momentum. As you may or may not remember, the momentum of an object is the product of its mass and its velocity. It’s a vector quantity and, hence, momentum has not only a magnitude but also a direction: pv= m vv= γm 0v As evidenced from the formula above, the momentum formula is a relativistic formula as well, as it’s dependent on the Lorentz factor too. So where do I want to go from here? Well… In this section (relativistic mass and momentum), I just want to show that Einstein’s mass formula is not some separate law or postulate: it just comes with the Lorentz transformation equations (and the above-mentioned consequences in terms of measuring horizontal and vertical velocities). Indeed, Einstein’s relativistic mass formula can be derived from the momentum conservation principle, which is one of the ‘physical laws’ that Einstein refers to. Look at the elastic collision between two billiard balls below. These balls are equal – same mass and same speed from the point of view of an inertial observer – but not identical: one is red and one is blue. The two diagrams show the collision from two different points of view: left, we have the inertial reference frame, and, right, we have a reference frame that is moving with a velocity equal to the horizontal component of the velocity of the blue ball. The points to note are the following: The total momentum of such elastic collision before and after the collision must be the same. Because the two balls have equal mass (in the inertial reference frame at least), the collision will be perfectly symmetrical. Indeed, we may just turn the diagram ‘upside down’ and change the colors of the balls, as we do below, and the values w, u and v (as well as the angle α) are the same. As mentioned above, the velocity of the blue and red ball and, hence, their momentum, will depend on the frame of reference. In the diagram on the left, we’re moving with a velocity equal to the horizontal component of the velocity of the blue ball and, therefore, in this particular frame of reference,the velocity (and the momentum) of the blue ball consists of a vertical component only, which we refer to as w. From this point of view (i.e. the reference frame moving with, the velocity (and, hence, the momentum) of the red ball will have both a horizontal as well as a vertical component. If we denote the horizontal component by u, then it’s easy to show that the vertical velocity of the red ball must be equal to sin(α)v. Now, because u = cos(α)v, this vertical component will be equal to tan(α)u. But so what is tan(α)u? Now, you’ll say, that is quite evident: tan(α)u must be equal to w, right? No. That’s Newtonian physics. The red ball is moving horizontally with speed u with respect to the blue ball and, hence, its vertical velocity will not be quite equal to w. Its vertical velocity will be given by the formula which we derived above:v y=(1/γ)v y’, so it will be a little bit slower than the w we see in the diagram on the right which is, of course, the same w as in the diagram on the left. [If you look carefully at my drawing above, then you’ll notice that the w vector is a bit longer indeed.] Huh? Yes. Just think about it: tan(α)u=(1/γ)w. But then… How can momentum be conserved if these speeds are not the same? Isn’t the momentum conservation principle supposed to conserve both horizontal as well as vertical momentum? It is, and momentum is being conserved. Why? Because of the relativistic mass factor. Indeed, the change in vertical momentum (Δp) of the blue ball in the diagram on the left or – which amounts to the same – the red ball in the diagram on the right (i.e. the vertically moving ball) is equal to Δp blue = 2m w w. [The factor 2 is there because the ball goes down and then up (or vice versa) and, hence, the total change in momentum must be twice the m w w amount.] Now, that amount must be equal to Δp red, which is equal to Δp blue= 2m v(1/γ)w . Equating both yields the following grand result: m v/m _w_=γ⇔m v=γm w What does this mean? It means that mass of the red ball in the diagram on the left is larger than the mass of the blue ball. So here we have actually derived Einstein’s relativistic mass formula from the momentum conservation principle ! Of course you’ll say: not quite. This formula is not the m u=γm 0 formula that we’re used to ! Indeed, it’s not. The blue ball has some velocity w itself, and so the formula links two velocities v and w. However, we can derive m v=γm 0 formula as a limit of m v=γm w for w going to zer0. How can w become infinitesimally small? If the angle α becomes infinitesimally small. It’s obvious, then, that v and u will be practically equal. In fact, if w goes to zero, then m w will be equal to m 0 in the limiting case, and m v will be equal to m u. So, then, indeed, we get the familiar formula as a limiting case: m u=γm 0 Hmm… You’ll probably find all of this quite fishy. I’d suggest you just think about it. What I presented above, is actually Feynman’s presentation of the subject, but with a bit more verbosity. Let’s move on to the final. Relativistic energy From what I wrote above (and from what I wrote in my two previous posts on this topic), it should be obvious, by now, that energy also depends on the reference frame. Indeed, mass and velocity depend on the reference frame (moving or not), and both appear in the formula for kinetic energy which, as you’ll remember, is K.E. = m c 2– m 0 c 2= (m–m 0)c 2= γm 0 c 2– m 0 c 2= m 0 c 2(γ– 1). Now, if you go back to the post where I presented that formula, you’ll see that we’re actually talking the change in kinetic energy here: if the mass is at rest, it’s kinetic energy is zero (because m = m 0), and it’s only when the mass is moving, that we can observe the increase in mass. [If you wonder how, think about the example of the fast-moving electrons in an electron beam: we see it as an increase in the inertia: applying the same force does no longer yield the same acceleration.] Now, in that same post, I also noted that Einstein added an equivalent rest mass energy(E 0= m 0 c 2) to the kinetic energy above, to arrive at the total energy of an object: E =E 0+ K.E. =m c 2 Now, what does this equivalence actually mean? Is mass energy? Can we equate them really? The short answer to that is: yes. Indeed, in one of my older posts (Loose Ends), I explained that protons and neutrons are made of quarks and, hence, that quarks are the actual matter particles, not protons and neutrons. However, the mass of a proton – which consists of two up quarks and one down quark – is 938 MeV/c 2(don’t worry about the units I am using here: because protons are so tiny, we don’t measure their mass in grams), but the mass figure you get when you add the rest mass of two u‘s and one d, is 9.6 MeV/c 2 only: about one percent of 938 ! So where’s the difference? The difference is the equivalent mass (or inertia) of the binding energy between the quarks. Indeed, the so-called ‘mass’ that gets converted into energy when a nuclear bomb explodes is not the mass of quarks. Quarks survive: nuclear power is binding energy between quarks that gets converted into heat and radiation and kinetic energy and whatever else a nuclear explosion unleashes. In short, 99% of the ‘mass’ of a proton or an electron is due to the strong force.So that’s ‘potential’ energy that gets unleashed in a nuclear chain reaction. In other words, the rest mass of the proton is actually the inertia of the system of moving quarks and gluons that make up the particle. In such atomic system, even the energy of massless particles (e.g. the virtual photons that are being exchanged between the nucleus and its electron shells) is measured as part of the rest mass of the system. So, yes, mass is energy. As Feynman put it, long before the quark model was confirmed and generally accepted: “We do not have to know what things are made of inside; we cannot and need not justify, inside a particle, which of the energy is rest energy of the parts into which it is going to disintegrate. It is not convenient and often not possible to separate the total mc 2 energy of an object into (1) rest energy of the inside pieces, (2) kinetic energy of the pieces, and (3) potential energy of the pieces; instead we simply speak of the total energy of the particle. We ‘shift the origin’ of energy by adding a constant m 0 c 2 to everything, and say that the total energy of a particle is the mass in motion times c 2, and when the object is standing still, the energy is the mass at rest times c 2.” (Richard Feynman’s Lectures on Physics, Vol. I, p. 16-9) So that says it all, I guess, and, hence, that concludes my little ‘series’ on (special) relativity. I hope you enjoyed it. Post scriptum: Feynman describes the concept of space-time with a nice analogy: “When we move to a new position, our brain immediately recalculates the true width and depth of an object from the ‘apparent’ width and depth. But our brain does not immediately recalculate coordinates and time when we move at high speed, because we have had no effective experience of going nearly as fast as light to appreciate the fact that time and space are also of the same nature. It is as though we were always stuck in the position of having to look at just the width of something, not being able to move our heads appreciably one way or the other; if we could, we understand now, we would see some of the other man’s time—we would see “behind”, so to speak, a little bit. Thus, we shall try to think of objects in a new kind of world, of space and time mixed together, in the same sense that the objects in our ordinary space-world are real, and can be looked at from different directions. We shall then consider that objects occupying space and lasting for a certain length of time occupy a kind of a “blob” in a new kind of world, and that when we look at this “blob” from different points of view when we are moving at different velocities. This new world, this geometrical entity in which the “blobs” exist by occupying position and taking up a certain amount of time, is called space-time.” If none of what I wrote could convey the general idea, then I hope the above quote will. Apart from that, I should also note that physicists will prefer to re-write the Lorentz transformation equations by measuring time and distance in so-called equivalent units: velocities will be expressed not in km/h but as a ratio of c and, hence, c= 1 (a pure number) and so u will also be a pure number between 0 and 1. That can be done by expressing distance in light-seconds ( a light-second is the distance traveled by light in one second or, alternatively, by expressing time in ‘meter’. Both are equivalent but, in most textbooks, it will be time that will be measured in the ‘new’ units. So how do we express time in meter? It’s quite simple: we multiply the old seconds with c and then we get:time expressed in meters= time expressed in seconds multiplied by 3×10 8 meters per second. Hence, as the ‘second’ the first factor and the ‘per second’ in the second factor cancel out, the dimension of the new time unit will effectively be the meter. Now, if both time and distance are expressed in meter, then velocity becomes a pure number without any dimension, because we are dividing distance expressed in meter by time expressed in meter, and it should be noted that it will be a pure number between 0 and 1 (0 ≤ u ≤ 1), because 1 ‘time second’ = 1/(3×10 8) ‘time meters’. Also, c itself becomes the pure number 1. The Lorentz transformation equations then become: They are easy to remember in this form (cf. the symmetry between x–ut and t–ux) and, if needed, we can always convert back to the old units to recover the original formulas. I personally think there is no better way to illustrate how space and time are ‘mere shadows’ of the same thing indeed: if we express both time and space in the same dimension (meter), we can see how, as result of that, velocity becomes a dimensionless number between zero and one and, more importantly, how the equations for x’ and t’ then mirror each other nicely. I am not sure what ‘kind of union’ between space and time Minkowski had in mind, but this must come pretty close, no? Final note: I noted the equivalence of mass and energy above. In fact, mass and energy can also be expressed in the same units, and we actually do that above already. If we say that an electron has a rest mass of 0.511 MeV/c 2(a bit less than a quarter of the mass of the u quark), then we express the mass in terms of energy. Indeed, the eV is an energy unit and so we’re actually using the m = E/c 2 formula when we express mass in such units. Expressing mass and energy in equivalent units allows us to derive similar ‘Lorentz transformation equations’ for the energy and the momentum of an object as measured under an inertial versus a moving reference frame. Hence, energy and momentum also transform like our space-time four-vectors and – likewise – the energy and the momentum itself, i.e. the components of the (four-)vector, are less ‘real’ than the vector itself. However, I think this post has become way too long and, hence, I’ll just jot these four equations down – please note, once again, the nice symmetry between (1) and (2)– but then leave it at that and finish this post. Jean Louis Van BellePhysics3 CommentsMay 30, 2014 June 26, 202017 Minutes Another post for my kids: introducing (special)relativity Pre-scriptum (dated 26 June 2020): These posts on elementary math and physics have not suffered much the attack by the dark force—which is good because I still like them. While my views on the true nature of light, matter and the force or forces that act on them have evolved significantly as part of my explorations of a more realist (classical) explanation of quantum mechanics, I think most (if not all) of the analysis in this post remains valid and fun to read. In fact, I find the simplest stuff is often the best. Original post: In my previous post, I talked about energy, and I tried to keep it simple– but also accurate. However, to be completely accurate, one must, of course, introduce relativity at some point. So how does that work? What’s ‘relativistic’ energy? Well… Let me try to convey a few ideas here. The first thing to note is that the energy conservation law still holds: special theory or not, the sum of the kinetic and potential energies in a (closed) system is always equal to some constant C.What constant? That doesn’t matter: Nature does not care about our zero point and, hence, we can add or subtract any(other) constant to the equation K.E. + P.E. = T + U = C. That being said, in my previous post, I pointed out that the constant depends on the reference point for the potential energy term U: we will usually take infinity as the reference point (for a force that attracts) and associate it with zero potential (U = 0). We then get a function U(x) like the one below: for gravitational energy we have U(x) =–GMm/x, and for electrical charges, we have U(x) = q 1 q 2/4πε 0 x. The mathematical shape is exactly the same but, in the case of the electromagnetic forces, you have to remember that likes repel, and opposites attract,so we don’t need the minus sign: the sign of the charges takes care of it. Minus sign? In case you wonder why we need that minus sign for the potential energy function, well… I explained that in my previous post and so I’ll be brief on that here: potential energy is measured by doing work against the force. That’s why. So we have an infinite sum (i.e. an integral) over some trajectory or path looking like this: U =– ∫F·ds. For kinetic energy, we don’t need any minus sign: as an object picks up speed, it’s the force itself that is doing the work as its potential energy is converted into kinetic energy, so the change in kinetic energy will equal the change in potential energy, but with opposite sign: as the object loses potential energy, it gains kinetic energy. Hence, we write ΔT =–ΔU = ∫F·ds.. That’s all kids stuff obviously. Let’s go beyond this and ask some questions. First, why can we add or subtract any constant to the potential energy but not to the kinetic energy? The answer is… Well… We actually can add or subtract a ‘constant’ to the kinetic energy as well. Now you will shake your head: Huh? Didn’t we have that T = m v 2/2 formula for kinetic energy? So how and why could one add or subtract some number to that? Well… That’s where relativity comes into play. The velocity v depends on your reference frame. If another observer would move with and/or alongside the object, at the same speed, that observer would observe a velocity equal to zero and, hence, its kinetic energy – as that observer would measure it – would also be zero. You will object to that, saying that a change of reference frame does not change the force, and you’re right: the force will cause the object to accelerate or decelerate indeed, and if the observer is not subject to the same force, then he’ll see the object accelerate or decelerate indeed, regardless of his reference frame is a moving or inertial frame. Hence, both the inertial as well as the moving observer will see an increase(or decrease) in its kinetic energy and, therefore, both will conclude that its potential energy decreases(or increases)accordingly. In short, it’s the change in energy that matters, both for the potential as well as for the kinetic energy. The reference point itself, i.e. the point from where we start counting so to say, does not: that’s relative. [This also shows in the derivation for kinetic energy which I’ll do below.] That brings us to the second question. We all learned in high school that mass and energy are related through Einstein’s mass-energy relation, E = m c 2, which establishes an equivalence between the two: the mass of an object that’s picking up speed increases, and so we need to look at both speed and mass as a function of time. Indeed, remember Newton’s Law: force is the time rate of change of momentum: F = d(mv)/dt. When the speed is low (i.e. non-relativistic), then we can just treat m as a constant and write thatF= mdv/dt = ma(the mass times the acceleration). Treating m as a constant also allows us to derive the classical (Newtonian) formula for kinetic energy: So if we assume that the velocity of the object at point O is equal to zero (so v o= 0), then ΔT will be equal to T and we get what we were looking for: the kinetic energy at point P will be equal to T = m v 2/2. Now, y ou may wonder why we can’t do that same derivation for a non-constant mass? The answer to that question is simple: taking the m factor out of the integral can only be done if we assume it is a constant. If not, then we should leave it inside. It’s similar to taking a derivative. If m would not be constant, then we would have to apply the product rule to calculate d(mv)/dt, so we’d write d(m v)/dt = (dm/dt)v + m(dv/dt). So we have two terms here and it’s only when m is constant that we can reduce it to d(m v)/dt = m(dv/dt). So we have our classical kinetic energy function. However, when the velocity gets really high – i.e. if it’s like the same order of magnitude as the velocity of light – then we cannot assume that mass is constant. Indeed, the same high-school course in physics that taught you that E = m c 2 equation will probably also have taught you that an object can never go faster than light, regardless of the reference frame. Hence, as the object goes faster and faster, it will pick up more momentum, but its rate of acceleration should (and will) go down in such way that the object can never actually reach the speed of light. Indeed, if Newton’s Law is to remain valid, we need to correct it such a way that m is no longer constant: m itself will increase as a function of its velocity and, hence, as a function of time. You’ll remember the formula for that: This is often written as m =γm 0, with m 0 denoting the mass of the object at rest (in your reference frame that is) and γ = (1 –v 2/c 2)–1/2 the so-called Lorentz factor. The Lorentz factor is named after a Dutch physicist who introduced it near the end of the 19th century in order to explain why the speed of light is always c, regardless of the frame of reference (moving or not), or – in other words – why the speed of light is not relative. Indeed, while you’ll remember that there is no such thing as an absolute velocity according to the (special) theory of relativity, the velocity of light actually is absolute ! That means you will always see light traveling at speed c regardless of your reference frame. To put it simply, you can never catch up with light and, if you would be traveling away from some star in a spaceship with a velocity of 200,000 km per second, and a light beam from that star would pass you, you’d measure the speed of that light beam to be equal to 300,000 km/s, not 100,000 km/s. So c is an absolute speed that acts as an absolute speed limit regardless of your reference frame. [Note that we’re talking only about reference frames moving at a uniform speed: when acceleration comes into play, then we need to refer to the general theory of relativity and that’s a somewhat different ball game.] The graph below shows how γ varies as a function of v. As you can see, the mass increase only becomes significant at speeds of like 100,000 km per second indeed.Indeed, for v = 0.3 c, the Lorentz factor is 1.048, so the increase is about 5% only. For v= 0.5 c, it’s still limited to an increase of some 15%. But then it goes up rapidly: for v= 0.9 c, the mass is more than twice the rest mass: m≈ 2.3m 0; for v= 0.99 c, the mass increase is 600%: m ≈ 7m 0; and so on. For v= 0.999 c – so when the speed of the object differs from c only by 1 part in 1,000 – the mass of the object will be more than twenty-two times the rest mass (m ≈ 22.4m 0). You probably know that we can actually reach such speeds and, hence, verify Einstein’s correction of Newton’s Law in particle accelerators: the electrons in an electron beam in a particle accelerator get usually pretty close to c and have a mass that’s like 2000 times their rest mass. How do we know that? Because the magnetic field needed to deflect them is like 2000 times as great as their (theoretical) rest mass. So how fast do they go? For their mass to be 2000 times m 0, 1 –v 2/c 2 must be equal to 1/4,000,000. Hence, their velocity v differs from c only by one part in 8,000,000. You’ll have to admit that’s very close. Other effects of relativistic speeds So we mentioned the thing that’s best known about Einstein’s (special) theory of relativity: the mass of an object, as measured by the inertial observer, increases with its speed. Now, you may or may not be familiar with two other things that come out of relativity theory as well: The first is length contraction: objects are measured to be shortened in the direction of motion with respect to the (inertial) observer. The formula to be used incorporates the reciprocal of the Lorentz factor: L = (1/γ)L 0. For example, a meter stick in a space ship moving at a velocity v = 0.6 c will appear to be only 80 cm to the external/inertial observer seeing it whizz past… That is if he can see anything at all of course: he’d have to take like a photo-finish picture as it zooms past ! The second is time dilation, which is also rather well known – just like the mass increase effect – because of the so-called twin paradox: time will appear to be slower in that space ship and, hence, if you send one of two twins away on a space journey, traveling at such relativistic speed, he will come back younger than his brother. The formula here is a bit more complicated, but that’s only because we’re used to measure time in seconds. If we would take a more natural unit, i.e. the time it takes light to travel a distance of 1 m, then the formula will look the same as our mass formula: t= γt 0 and, hence, one ‘second’ in the space ship will be measured as 1.25 ‘seconds’ by the external observer. Hence, the moving clock will appear to run slower – to the external (inertial) observer that is. Again, the reality of this can be demonstrated. You’ll remember that we introduced the muon in previous posts: muons resemble electrons in the sense that they have the same charge, but their mass is more than 200 times the mass of an electron. As compared to other unstable particles, their average lifetime is quite long: 2.2 micro seconds. Still, that would not be enough to travel more than 600 meters or so– even at the speed of light (2.2 μs × 300,000 km/s = 660 m). But so we do detect muons in detectors down here that come all the way down from the stratosphere, where they are created when cosmic rays hit the Earth’s atmosphere some 10 kilometers up. So how do they get here if they decay so fast? Well, those that actually end up in those detectors, do indeed travel very close to the speed of light and, hence, while from their own point of view they live only like two millionths of a second, they live considerably longer from our point of view. Relativistic energy: E = m c 2 Let’s go back to our main story line: relativistic energy. We wrote above that it’s the change of energy that matters really. So let’s look at that. You may or may not remember that the concept of work in physics is closely related to the concept of power. In fact, you may actually remember that power, in physics at least, is defined as the work done per second.Indeed, we defined work as the (dot) product of the force and the distance. Now, when we’re talking a differential distance only (i.e. an infinitesimally small change only), then we can write dT = F·ds, but when we’re talking something larger, then we have to do that integral: ΔT = ∫F·ds. However, we’re interested in the time rate of change of T here, and so that’s the time derivative dT/dt which, as you easily verify, will be equal to dT/dt = (F·ds)/dt = F·(ds/dt) = F·vand so we can use that differential formula and we don’t need the integral. Now, that (dot) product of the force and the velocity vectors is what’s referred to as the power. [Note that only the component of the force in the direction of motion contributes to the work done and, hence, to the power.] OK. What am I getting at? Well… I just want to show an interesting derivation: if we assume, with Einstein, that mass and energy are equivalent and, hence, that the total energy of a body always equals E = m c 2, then we can actually derive Einstein’s mass formula from that. How? Well… If the time rate of change of the energy of an object is equal to the power expended by the forces acting on it, then we can write: dE/dt = d(m c 2)/dt = F·v Now, we can not take the mass out of those brackets after the differential operator (d) because the mass is not a constant in this case (relativistic speeds) and, hence, dm/dt≠ 0. However, we can take out c 2(that’s an absolute constant, remember?) and we can also substitute F using Newton’s Law (F = d(m v)/dt), again taking care to leave m between the brackets, not outside. So then we get: d(m c 2)/dt = c 2 dm/dt = [d(m v)/dt]·v = v· d(m v)/dt In case you wonder why we can replace the vectors (bold face) v and d(mv) by their magnitudes (or lengths) v and d(m v):v and mvhave the same direction and, hence, the angle θ between them is zero, and so v·v =│v││v│cosθ =v 2. Likewise, d(mv) and v also have the same direction and so we can just replace the dot product by the product of the magnitudes of those two vectors. Now, let’s not forget the objective: we need to solve this equation for m and, hopefully, we’ll find Einstein’s mass formula, which we need to correct Newton’s Law. How do we do that? We’ll first multiply both sides by 2m. Why? Because we can then apply another mathematical trick, as shown below: c 2(2m)·dm/dt = 2m v· d(m v)/dt⇔d(m 2 c 2)/dt = d(m 2 v 2)/dt However, if the derivatives of two quantities are equal, then the quantities themselves can only differ by a constant, say C. So we integrate both sides and get: m 2 c 2=m 2 v 2+ C Be patient: we’re almost there. The above equation must be true for all velocities v and, hence, we can choose the special case where v = 0 and call this mass m 0, and then substitute, so we get m 0 c 2=m 0 0 2+ C = C. Now we put this particular value for C back in the more general equation above and we get: m c 2=m v 2+ m 0 c 2⇔m=m v 2/c 2+m 0⇔m(1–v 2/c 2) = m 0⇔m = m 0/(1–v 2/c 2)–1/2 So there we are: we have just shown that we get the relativistic mass formula (it’s on the right-hand side above) if we assume that Einstein’s mass-energy equivalence relation holds. Now, you may wonder why that’s significant. Well… If you’re disappointed, then, at the very least, you’ll have to admit that it’s nice to show how everything is related to everything in this theory: from E =m c 2, we get m 0/(1–v 2/c 2)–1/2. I think that’s kinda neat! In addition, let us analyze that mass-energy relation in another way. It actually allows us to re-definekinetic energy as the excess of a particle over its rest mass energy, or – it’s the same expression really – or the difference between its total energy and its rest energy. How does that work? Well… When we’re looking at high-speed or high-energy particles, we will write the kinetic energy as: K.E. = m c 2– m 0 c 2= (m–m 0)c 2= γm 0 c 2– m 0 c 2= m 0 c 2(γ– 1). Now, we can expand that Lorentz factor γ = (1 –v 2/c 2)–1/2 into a binomial series (the binomial series is an infinite Taylor series, so it’s not to be confused with the (finite) binomial expansion: just check it online if you’re in doubt). If we do that, we we can write γ as an infinite sum of the following terms: γ = 1 + (1/2)v 2/c 2+ (3/8)v 4/c 4+ (5/16)v 6/c 6+ … Now, when we plug this back into our (relativistic) kinetic energy equation, we can scrap a few things (just do it) to get where I wanted to get: K.E. = (1/2)m 0 v 2+ (3/8)m 0 v 4/c 2+ (5/16)m 0 v 6/c 4+ … Again, you’ll wonder: so what? Well… See how the non-relativistic formula for kinetic energy (K.E. = m 0 v 2/2) appears here as the first term of this series and, hence, how the formula above shows that our ‘Newtonian’ formula is just an approximation. Of course,at low speeds, the second, third etcetera terms represent close to nothing and, hence, then our Newtonian ‘approximation is obviously pretty good of course ! OK… But… Now you’ll say: that’s fine, but how did Einstein get inspired to write E = m c 2 in the first place? Well, truth be told, the relativistic mass formula was derived first (i.e. before Einstein wrote his E = m c 2 equation),out of a derivation involving the momentum conservation law and the formulas we must use to convert the space-time coordinates from one reference frame to another when looking at phenomena (i.e. the so-called Lorentz transformations). And it was only afterwards that Einstein noted that, when expanding the relativistic mass formula, that the increase in mass of a body appeared to be equal to the increase in kinetic energy divided by c 2(Δm = Δ(K.E.)/c 2). Now, that, in turn, inspired him to also assign an equivalent energy to the rest mass of that body: E 0= m 0 c 2. […] At least that’s how Feynman tells the story in his 1965 Lectures… But so we’ve actually been doing it the other way around here! Hmm… You will probably find all of this rather strange, and you may also wonder what happened to our potential energy. Indeed, that concept sort of ‘disappeared’ in this story: from the story above, it’s clear that kinetic energy has an equivalent mass, but what about potential energy? That’s a very interesting question but, unfortunately, I can only give a rather rudimentary answer to that. Let’s suppose that we have two masses M and m. According to the potential energy formula above, the potential energy U between these two masses will then be equal to U =–GMm/r. Now, that energy is not interpreted as energy of either M or m, but as energy that is part of the (M, m)system, which includes the system’s gravitational field. So that energy is considered to be stored in that gravitational field. If the two masses would sit right on top of each other, then there would be no potential energy in the (M, m) system and, hence, the system as a whole would have less energy. In contrast, when we separate them further apart, then we increase the energy of the system as a whole, and so the system’s gravitational field then increases. So, yes, the potential energy does impact the (equivalent) mass of the system, but not the individual masses M and m. Does that make sense? For me , it does, but I guess you’re a bit tired by now and, hence, I think I should wrap up here. In my next (and probably last) post on relativity, I’ll present those Lorentz transformations that allow us to ‘translate’ the space and time coordinates from one reference frame to another, and in that post I’ll also present the other derivation of Einstein’s relativistic mass formula, which is actually based on those transformations.In fact, I realize I should have probably started with that (as mentioned above, that’s how Feynman does it in his Lectures) but, then, for some reason, I find the presentation above more interesting, and so that’s why I am telling the story starting from another angle. I hope you don’t mind. In any case, it should be the same, because everything is related to everything in physics – just like in math. That’s why it’s important to have a good teacher. Jean Louis Van BellePhysicsLeave a commentMay 24, 2014 June 26, 202015 Minutes Blog Stats 519,347 hits Search for: Top Posts Maxwell-Boltzmann, Bose-Einstein and Fermi-Dirac statistics Cargo cult science The Breit-Wheeler process: can matter be created out of light? Feynman as the Great Teacher? The dark force An easy piece: introducing quantum mechanics and the wave function The Hamiltonian for a two-state system: the ammonia example Maxwell, Lorentz, gauges and gauge transformations Re-visiting the matter wave (I) The de Broglie relations, the wave equation, and relativistic length contraction Create a free website or blog at WordPress.com. SubscribeSubscribed Reading Feynman Join 184 other subscribers Sign me up Already have a WordPress.com account? Log in now. Reading Feynman SubscribeSubscribed Sign up Log in Report this content View site in Reader Manage subscriptions Collapse this bar Loading Comments... Write a Comment... 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15908	https://www.wyzant.com/resources/answers/891195/a-theater-has-15-rows-of-seats-with-30-in-the-first-row-33-in-the-second-ro	WYZANT TUTORING Abby C. A theater has 15 rows of seats with 30 in the first row, 33 in the second row, 36 in the third row, and so on in the same pattern. How many TOTAL seats are in the theater? 3 Answers By Expert Tutors Touba M. answered • 04/05/22 High School and College Tutor Hi, In fact it is arithmetic series, 15 rows it means n= 15 with 30 in the first row it means a = 30 same pattern it means d = 3 How many TOTAL seats it means you need to use sum formula s = n/2 [ 2a + (n-1)d] s = 15/2 [ 230 + ( 15-1) 3 ] = 765 seats I hope it is useful, Minoo Raymond B. answered • 04/05/22 Math, microeconomics or criminal justice 15 rows 1st row has 30 2nd has 33 3rd has 36 nth row has 30+3(n-1) 15th has 30 + 3(14) = 30+42 = 72 an = a1 +d(n-1) an = 30 +3(n-1) sum of n terms s1 = 30 s2 = 30+33 = 63 = n/2(30+33) = 63 s3 = 63+36 = 99 s4 = 99+ 39 = 138 = 4/2(30+39)= 138 s5 = 138+42 = 180 = 5/2(30+42) = 5(36) = 180 s6 = 180+45 = 225 = 6/2(30+45)=225 s7 = 225+48 = 273 s8 = 273+51 = 324 s9= 324+54= 378 = s10 = 378+57 = 435 s11 = 435+60 = 495 = 11/2(30+60) = 11/2(90)= 11(45) = 495 s12 = 495+63= 558 = 12/2(30+63) = 6(93) = 558 s13 = 558+66 = 624 s14= 624+69= 693 s15 = 693+72= 765 sn = (n/2)(a1 +an) a1=30, an=30+d(n-1) = 30+3(14) = 30+42 = 72 s15 =(15/2)(30+72) = (15/2)(102) = 15(51) = 765 Brenda D. 04/05/22 Touba M. 04/06/22 Touba M. 04/09/22 Samantha U. answered • 04/05/22 Mathematics Tutor First, write the pattern as a sequence row 1 row 2 row 3 30 seats 33 seats 36 seats Second, find the equation that makes the sequence y=mx+b m will be the difference in the sequence numbers b will be what you + or - to make the equation work if needed 1) difference in numbers 36-33 = 3 33-30 = 3 so the difference is 3 2) y=3x Check y=3(1) the first number in sequence is supposed to be 30 but 3(1) is just 3 so the equation is not right yet 3) + or - to make it work y=3x+27 check y=3(2)+27 the second number is supposed to be 33 3(2)+27 = 33 so equation works Now, you want the 15th row so substitute 15 into the equation to find the amount of seats. Still looking for help? Get the right answer, fast. Get a free answer to a quick problem. Most questions answered within 4 hours. OR Choose an expert and meet online. No packages or subscriptions, pay only for the time you need. RELATED TOPICS RELATED QUESTIONS graph, find x and y intercepts and test for symmetry x=y^3 Answers · 3 -2x/x+6+5=-x/x+6 Answers · 7 Find the mean and standard deviation for the random variable x given the following distribution Answers · 4 using interval notation to show intervals of increasing and decreasing and postive and negative Answers · 5 trouble spots for the domain may occur where the denominator is ? or where the expression under a square root symbol is negative Answers · 4 RECOMMENDED TUTORS Abigail C. Jennifer M. Drew L. find an online tutor Download our free app A link to the app was sent to your phone. Get to know us Learn with us Work with us Download our free app Let’s keep in touch Need more help? Learn more about how it works Tutors by Subject Tutors by Location IXL Comprehensive K-12 personalized learning Rosetta Stone Immersive learning for 25 languages Education.com 35,000 worksheets, games, and lesson plans TPT Marketplace for millions of educator-created resources Vocabulary.com Adaptive learning for English vocabulary ABCya Fun educational games for kids SpanishDictionary.com Spanish-English dictionary, translator, and learning Inglés.com Diccionario inglés-español, traductor y sitio de aprendizaje Emmersion Fast and accurate language certification
15909	https://stats.libretexts.org/Bookshelves/Probability_Theory/Probability_Mathematical_Statistics_and_Stochastic_Processes_(Siegrist)/08%3A_Set_Estimation/8.03%3A_Estimation_in_the_Bernoulli_Model	8.3: Estimation in the Bernoulli Model - Statistics 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 8: Set Estimation Probability, Mathematical Statistics, and Stochastic Processes (Siegrist) { } { "8.01:_Introduction_to_Set_Estimation" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "8.02:_Estimation_the_Normal_Model" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "8.03:_Estimation_in_the_Bernoulli_Model" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "8.04:_Estimation_in_the_Two-Sample_Normal_Model" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "8.05:_Bayesian_Set_Estimation" : "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:_Foundations" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "02:_Probability_Spaces" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "03:_Distributions" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "04:_Expected_Value" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "05:_Special_Distributions" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "06:_Random_Samples" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "07:_Point_Estimation" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "08:_Set_Estimation" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "09:_Hypothesis_Testing" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "10:_Geometric_Models" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "11:_Bernoulli_Trials" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "12:_Finite_Sampling_Models" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "13:_Games_of_Chance" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "14:_The_Poisson_Process" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "15:_Renewal_Processes" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "16:_Markov_Processes" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "17:_Martingales" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "18:_Brownian_Motion" : "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" } Sun, 24 Apr 2022 06:26:00 GMT 8.3: Estimation in the Bernoulli Model 10202 10202 admin { } Anonymous Anonymous 2 false false [ "article:topic", "license:ccby", "authorname:ksiegrist", "licenseversion:20", "source@ ] [ "article:topic", "license:ccby", "authorname:ksiegrist", "licenseversion:20", "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. Probability Theory 4. Probability, Mathematical Statistics, and Stochastic Processes (Siegrist) 5. 8: Set Estimation 6. 8.3: Estimation in the Bernoulli Model Expand/collapse global location 8.3: Estimation in the Bernoulli Model Last updated Apr 24, 2022 Save as PDF 8.2: Estimation the Normal Model 8.4: Estimation in the Two-Sample Normal Model Page ID 10202 Kyle Siegrist University of Alabama in Huntsville via Random Services ( \newcommand{\kernel}{\mathrm{null}\,}) Table of contents 1. Introduction 2. The One-Sample Model 1. Preliminaries 2. Basic Confidence Intervals 3. Simplified Confidence Intervals 4. Conservative Confidence Intervals The Two-Sample Model Preliminaries Simplified Confidence Intervals Conservative Confidence Intervals Computational Exercises Introduction Recall that an indicator variable is a random variable that just takes the values 0 and 1. In applications, an indicator variable indicates which of two complementary events in a random experiment has occurred. Typical examples include A manufactured item subject to unavoidable random factors is either defective or acceptable. A voter selected from a population either supports a particular candidate or does not. A person selected from a population either does or does not have a particular medical condition. A student in a class either passes or fails a standardized test. A sample of radioactive material either does or does not emit an alpha particle in a specified ten-second period. Recall also that the distribution of an indicator variable is known as the Bernoulli distribution, named for Jacob Bernoulli, and has probability density function given by P⁡(X=1)=p, P⁡(X=0)=1−p, where p∈(0,1) is the basic parameter. In the context of the examples above, p is the probability that the manufactured item is defective. p is the proportion of voters in the population who favor the candidate. p is the poportion of persons in the population that have the medical condition. p is the probability that a student in the class will pass the exam. p is the probability that the material will emit an alpha particle in the specified period. Recall that the mean and variance of the Bernoulli distribution are E⁡(X)=p and var⁡(X)=p⁢(1−p). Often in statistical applications, p is unknown and must be estimated from sample data. In this section, we will see how to construct interval estimates for the parameter from sample data. A parallel section on Tests in the Bernoulli Model is in the chapter on Hypothesis Testing. The One-Sample Model Preliminaries Suppose that X=(X 1,X 2,…,X n) is a random sample from the Bernoulli distribution with unknown parameter p∈[0,1]. That is, X is a squence of Bernoulli trials. From the examples in the introduction above, note that often the underlying experiment is to sample at random from a dichotomous population. When the sampling is with replacement, X really is a sequence of Bernoulli trials. When the sampling is without replacement, the variables are dependent, but the Bernoulli model is still approximately valid if the population size is large compared to the sample size n. For more on these points, see the discussion of sampling with and without replacement in the chapter on Finite Sampling Models. Note that the sample mean of our data vector X, namely (8.3.1)M=1 n⁢∑i=1 n X i is the sample proportion of objects of the type of interest. By the central limit theorem, the standard score (8.3.2)Z=M−p p⁢(1−p)/n has approximately a standard normal distribution and hence is (approximately) a pivot variable for p. For a given sample size n, the distribution of Z is closest to normal when p is near 1 2 and farthest from normal when p is near 0 or 1 (extreme). Because the pivot variable is (approximately) normally distributed, the construction of confidence intervals for p in this model is similar to the construction of confidence intervals for the distribution mean μ in the normal model. But of course all of the confidence intervals so constructed are approximate. As usual, for r∈(0,1), let z⁡(r) denote the quantile of order r for the standard normal distribution. Values of z⁡(r) can be obtained from the special distribution calculator, or from most statistical software packages. Basic Confidence Intervals For α∈(0,1), the following are approximate 1−α confidence sets for p: {p∈[0,1]:M−z⁢(1−α/2)⁢p⁢(1−p)/n≤p≤M+z⁢(1−α/2)⁢p⁢(1−p)/n} {p∈[0,1]:p≤M+z⁢(1−α)⁢p⁢(1−p)/n} {p∈[0,1]:M−z⁢(1−α)⁢p⁢(1−p)/n≤p} Proof From our discussion above, (M−p)/p⁢(1−p)/n has approximately a standard normal distribution. Hence by definition of the quantiles, P⁡[−z⁢(1−α/2)≤(M−p)/p⁢(1−p)/n≤z⁢(1−α/2)]≈1−α P⁡[−z⁢(1−α)≤(M−p)/p⁢(1−p)/n]≈1−α P⁡[(M−p)/p⁢(1−p)/n≤z⁢(1−α)]≈1−α Solving the inequalities for p in the numerator of (M−p)/p⁢(1−p)/n for each event gives the corresponding confidence set. These confidence sets are actually intervals, known as the Wilson intervals, in honor of Edwin Wilson. The confidence sets for p in (1) are intervals. Let (8.3.3)U⁡(z)=n n+z 2⁢(M+z 2 2⁢n+z⁢M⁢(1−M)n+z 2 4⁢n 2) Then the following have approximate confidecne level 1−α for p. The two-sided interval [U⁡[−z⁢(1−α/2)],U⁡[z⁢(1−α/2)]]. The upper bound U⁡[z⁢(1−α)]. The lower bound U⁡[−z⁢(1−α)]. Proof This follows by solving the inequalities in (1) for p. For each inequality, we can isolate the square root term, and then square both sides. This gives quadratic inequalities, which can be solved using the quadratic formula. As usual, the equal-tailed confidence interval in (a) is not the only two-sided 1−α confidence interval for p. We can divide the α probability between the left and right tails of the standard normal distribution in any way that we please. For α,r∈(0,1), an approximate two-sided 1−α confidence interval for p is [U⁡[z⁢(α−r⁢α)],U⁡[z⁢(1−r⁢α)]] where U is the function in (2). Proof As in the proof of (1), (8.3.4)P⁡[z⁢(α−r⁢α)≤M−p p⁢(1−p)/n≤z⁢(1−r⁢α)]≈1−α Solving for p with the help of the quadratic formula gives the result. In practice, the equal-tailed 1−α confidence interval in part (a) of (2), obtained by setting r=1 2, is the one that is always used. As r↑1, the right enpoint converges to the 1−α confidence upper bound in part (b), and as r↓0 the left endpoint converges to the 1−α confidence lower bound in part (c). Simplified Confidence Intervals Simplified approximate 1−α confidence intervals for p can be obtained by replacing the distribution mean p by the sample mean M in the extreme parts of the inequalities in (1). For α∈(0,1), the following have approximate confidence level 1−α for p: The two-sided interval with endpoints M±z⁢(1−α/2)⁢M⁢(1−M)/n. The upper bound M+z⁢(1−α)⁢M⁢(1−M)/n. The lower bound M−z⁢(1−α)⁢M⁢(1−M)/n. Proof As noted, these results follows from the confidence set in (1) by replacing p with M in the expression p⁢(1−p)/n. These confidence intervals are known as Wald intervals, in honor of Abraham Wald.. Note that the Wald interval can also be obtained from the Wilson intervals in (2) by assuming that n is large compared to z, so that n/(n+z 2)≈1, z 2/2⁢n≈0, and z 2/4⁢n 2≈0. Note that this interval in (c) is symmetric about the sample proportion M but that the length of the interval, as well as the center is random. This is the two-sided interval that is normally used. Use the simulation of the proportion estimation experiment to explore the procedure. Use various values of p and various confidence levels, sample sizes, and interval types. For each configuration, run the experiment 1000 times and compare the proportion of successful intervals to the theoretical confidence level. As always, the equal-tailed interval in (4) is not the only two-sided, 1−α confidence interval. For α,r∈(0,1), an approximate two-sided 1−α confidence interval for p is (8.3.5)[M−z⁢(1−r⁢α)⁢M⁢(1−M)n,M−z⁢(α−r⁢α)⁢M⁢(1−M)n] The interval with smallest length is the equal-tail interval with r=1 2. Conservative Confidence Intervals Note that the function p↦p⁢(1−p) on the interval [0,1] is maximized when p=1 2 and thus the maximum value is 1 4. We can obtain conservative confidence intervals for p from the basic confidence intervals by using this fact. For α∈(0,1), the following have approximate confidence level at least 1−α for p: The two-sided interval with endpoints M±z⁢(1−α/2)⁢1 2⁢n. The upper bound M+z⁢(1−α)⁢1 2⁢n. The lower bound M−z⁢(1−α)⁢1 2⁢n. Proof As noted, these results follows from the confidence sets in (1) by replacing p with 1 2 in the expression p⁢(1−p)/n. Note that the confidence interval in (a) is symmetric about the sample proportion M and that the length of the interval is deterministic. Of course, the conservative confidence intervals will be larger than the approximate simplified confidence intervals in (4). The conservative estimate can be used to design the experiment. Recall that the margin of error is the distance between the sample proportion M and an endpoint of the confidence interval. A conservative estimate of the sample size n needed to estimate p with confidence 1−α and margin of error d is (8.3.6)n=⌈z α 2 4⁢d 2⌉ where z α=z⁢(1−α/2) for the two-sided interval and z α=z⁢(1−α) for the confidence upper or lower bound. Proof With confidence level 1−α, the margin of error is z α⁢1 2⁢n. Setting this equal to the prescribed value d and solving gives the result. As always, the equal-tailed interval in (7) is not the only two-sided, conservative, 1−α confidence interval. For α,r∈(0,1), an approximate two-sided, conservative 1−α confidence interval for p is (8.3.7)[M−z⁢(1−r⁢α)⁢1 2⁢n,M−z⁢(α−r⁢α)⁢1 2⁢n] The interval with smallest length is the equal-tail interval with r=1 2. The Two-Sample Model Preliminaries Often we have two underlying Bernoulli distributions, with parameters p 1,p 2∈[0,1] and we would like to estimate the difference p 1−p 2. This problem could arise in the following typical examples: In a quality control setting, suppose that p 1 is the proportion of defective items produced under one set of manufacturing conditions while p 2 is the proportion of defectives under a different set of conditions. In an election, suppose that p 1 is the proportion of voters who favor a particular candidate at one point in the campaign, while p 2 is the proportion of voters who favor the candidate at a later point (perhaps after a scandal has erupted). Suppose that p 1 is the proportion of students who pass a certain standardized test with the usual test preparation methods while p 2 is the proportion of students who pass the test with a new set of preparation methods. Suppose that p 1 is the proportion of unvaccinated persons in a certain population who contract a certain disease, while p 2 is the proportion of vaccinated person who contract the disease. Note that several of these examples can be thought of as treatment-control problems. Of course, we could construct interval estimates I 1 for p 1 and I 2 for p 2 separately, as in the subsections above. But as we noted in the Introduction, if these two intervals have confidence level 1−α, then the product set I 1×I 2 has confidence level (1−α)2 for (p 1,p 2). So if p 1−p 2 is our parameter of interest, we will use a different approach. Simplified Confidence Intervals Suppose now that X=(X 1,X 2,…,X n 1) is a random sample of size n 1 from the Bernoulli distribution with parameter p 1, and Y=(Y 1,Y 2,…,Y n 2) is a random sample of size n 2 from the Bernoulli distribution with parameter p 2. We assume that the samples X and Y are independent. Let (8.3.8)M 1=1 n 1⁢∑i=1 n 1 X i,M 2=1 n 2⁢∑i=1 n 2 Y i denote the sample means (sample proportions) for the samples X and Y. A natural point estimate for p 1−p 2, and the building block for our interval estimate, is M 1−M 2. As noted in the one-sample model, if n i is large, M i has an approximate normal distribution with mean p i and variance p i⁢(1−p i)/n i for i∈{1,2}. Since the samples are independent, so are the sample means. Hence M 1−M 2 has an approximate normal distribution with mean p 1−p 2 and variance p 1⁢(1−p 1)/n 1+p 2⁢(1−p 2)/n 2. We now have all the tools we need for a simplified, approximate confidence interval for p 1−p 2. For α∈(0,1), the following have approximate confidence level 1−α for p 1−p 2: The two-sided interval with endpoints (M 1−M 2)±z⁢(1−α/2)⁢M 1⁢(1−M 1)/n 1+M 2⁢(1−M 2)/n 2. The lower bound (M 1−M 2)−z⁢(1−α)⁢M 1⁢(1−M 1)/n 1+M 2⁢(1−M 2)/n 2. The upper bound (M 1−M 2)+z⁢(1−α)⁢M 1⁢(1−M 1)/n 1+M 2⁢(1−M 2)/n 2. Proof As noted above, if n 1 and n 2 are large, (8.3.9)(M 1−M 2)−(p 1−p 2)p 1⁢(1−p 1)/n 1+p 2⁢(1−p 2)/n 2 has approximatle a standard normal distribution, and hence so does (8.3.10)Z=(M 1−M 2)−(p 1−p 2)M 1⁢(1−M 1)/n 1+M 2⁢(1−M 2)/n 2 P⁡[−z⁢(1−α/2)≤Z≤z⁢(1−α/2)]≈1−α. Solving for p 1−p 2 gives the two-sided confidence interval. P⁡[−z⁢(1−α)≤Z]≈1−α. Solving for p 1−p 2 gives the confidence upper bound. P⁡[Z≤z⁢(1−α/2)]≈1−α. Solving for p 1−p 2 gives the confidence lower bound. As always, the equal-tailed interval in (a) is not the only approximate two-sided 1−α confidence interval. For α,r∈(0,1), an approximate 1−α confidence set for p 1−p 2 is (8.3.11)[(M 1−M 2)−z⁢(1−r⁢α)⁢M 1⁢(1−M 1)/n 1+M 2⁢(1−M 2)/n 2,(M 1−M 2)−z⁢(α−r⁢α)⁢M 1⁢(1−M 1)/n 1+M 2⁢(1−M 2)/n 2] Proof As noted in the proof of the previous theorem, (8.3.12)Z=(M 1−M 2)−(p 1−p 2)M 1⁢(1−M 1)/n 1+M 2⁢(1−M 2)/n 2 has approximately a standard normal distribution if n 1 and n 2 are large. Hence P⁡[−z⁢(α−r⁢α)≤Z≤z⁢(1−r⁢α)]≈1−α. Solving for p 1−p 2 gives the two-sided confidence interval. Conservative Confidence Intervals Once again, p↦p⁢(1−p) is maximized when p=1 2 with maximum value 1 4. We can use this to construct approximate conservative confidence intervals for p 1−p 2. For α∈(0,1), the following have approximate confidence level at least 1−α for p 1−p 2: The two-sided interval with endpoints (M 1−M 2)±1 2⁢z⁢(1−α/2)⁢1/n 1+1/n 2. The lower bound (M 1−M 2)−1 2⁢z⁢(1−α)⁢1/n 1+1/n 2. The upper bound (M 1−M 2)+1 2⁢z⁢(1−α)⁢1/n 1+1/n 2. Proof These results follow from the previous theorem by replacing M 1⁢(1−M 1) and M 2⁢(1−M 2) each with 1 4. Computational Exercises In a poll of 1000 registered voters in a certain district, 427 prefer candidate X. Construct the 95% two-sided confidence interval for the proportion of all registered voters in the district that prefer X. Answer (0.396,0.458) A coin is tossed 500 times and results in 302 heads. Construct the 95% confidence lower bound for the probability of heads. Do you believe that the coin is fair? Answer 0.579. No, the coin is almost certainly not fair. A sample of 400 memory chips from a production line are tested, and 30 are defective. Construct the conservative 90% two-sided confidence interval for the proportion of defective chips. Answer (0.034,0.116) A drug company wants to estimate the proportion of persons who will experience an adverse reaction to a certain new drug. The company wants a two-sided interval with margin of error 0.03 with 95% confidence. How large should the sample be? Answer 1068 An advertising agency wants to construct a 99% confidence lower bound for the proportion of dentists who recommend a certain brand of toothpaste. The margin of error is to be 0.02. How large should the sample be? Answer 3382 The Buffon trial data set gives the results of 104 repetitions of Buffon's needle experiment. Theoretically, the data should correspond to Bernoulli trials with p=2/π, but because real students dropped the needle, the true value of p is unknown. Construct a 95% confidence interval for p. Do you believe that p is the theoretical value? Answer (0.433,0.634). The theoretical value is approximately 0.637, which is not in the confidence interval. A manufacturing facility has two production lines for a certain item. In a sample of 150 items from line 1, 12 are defective. From a sample of 130 items from line 2, 10 are defective. Construct the two-sided 95% confidence interval for p 1−p 2, where p i is the proportion of defective items from line i, for i∈{1,2} Answer [−0.050,0.056] The vaccine for influenza is tailored each year to match the predicted dominant strain of influenza. Suppose that of 500 unvaccinated persons, 45 contracted the flu in a certain time period. Of 300 vaccinated persons, 20 contracted the flu in the same time period. Construct the two-sided 99% confidence interval for p 1−p 2, where p 1 is the incidence of flu in the unvaccinated population and p 2 the incidence of flu in the vaccinated population. This page titled 8.3: Estimation in the Bernoulli Model is shared under a CC BY 2.0 license and was authored, remixed, and/or curated by Kyle Siegrist (Random Services) via source content that was edited to the style and standards of the LibreTexts platform. Back to top 8.2: Estimation the Normal Model 8.4: Estimation in the Two-Sample Normal Model Was this article helpful? Yes No Recommended articles 8.1: Introduction to Set Estimation 8.2: Estimation the Normal Model 8.4: Estimation in the Two-Sample Normal Model 8.5: Bayesian Set Estimation 1: FoundationsIn this chapter we review several mathematical topics that form the foundation of probability and mathematical statistics. These include the algebra o... Article typeSection or PageAuthorKyle SiegristLicenseCC BYLicense Version2.0 Tags source@ © Copyright 2025 Statistics LibreTexts Powered by CXone Expert ® ? The LibreTexts libraries arePowered by NICE CXone Expertand are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. Privacy Policy. Terms & Conditions. 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15910	https://brilliant.org/wiki/coloring-definition/	Coloring (Parity Arguments) Sign up with Facebook or Sign up manually Already have an account? Log in here. Soumava Pal, Sameer Kailasa, and Jimin Khim contributed Coloring is a technique in combinatorics that can be used to solve board-tiling problems, specifically to prove certain tilings are impossible. Generally, one assigns a specific color or label to each square on a board and shows that the tiles cannot satisfy constraints set by the coloring. Examples and Problems Suppose one has an 8-by-8 chessboard, and one wishes to tile it with 1-by-2 dominoes. This is certainly possible, by simply placing 4 dominoes horizontally in each row. Now, suppose the top right hand corner square of the chessboard has been removed. Is it still possible to tile the board with dominoes? No: each domino covers two squares of the board, so the total number of squares covered by any tiling must be even, but there are 63 squares in the "mutilated chessboard." Additionally, suppose the bottom left hand corner square of the chessboard has been removed. Is it still possible to tile this chessboard (which now has two antipodal corner squares removed)? Unfortunately, a parity argument will not work as before, since there are now 62 squares. This will take a clever insight. Consider the mutilated chessboard in the image below. Is it possible to tile this board entirely with 31 1-by-2 dominoes? Two antipodal corner squares have been removed from this board. Is it still possible to tile it with dominoes? We color the board as shown in the picture, alternating white and black squares. Any domino placed on the board will cover exactly one white square and exactly one black square. Thus, for any tiling of the board by dominoes, the number of white squares covered equals the number of black squares colored. The mutilated chessboard has 32 white squares and 30 black squares, since the two antipodal corner squares have been removed. If there were a covering of this board with dominoes, then the implication would be that 32 equals 30, which is absurd. Thus, it is impossible to tile the board with 1-by-2 dominoes. □ Yes; sometimes Yes; always No; never Suppose that one black square and one white square of an 8-by-8 chessboard are removed. Is it possible to tile what remains of the board with 1-by-2 dominoes? The correct answer is: Yes; always The chessboard coloring from above permits a generalization that solves the following problem: Given an m-by-n rectangular board, can it be tiled by 1-by-p dominoes (where m, n, and p are all positive integers, and p is prime)? Choose p colors, labeled 1 through p. Color the first row of the board {1,2,⋯,p,1,2,⋯,}, the second row {2,3,⋯,p,1,2,⋯}, the third row {3,4,⋯,p,1,2,⋯}, etc. By this construction, any 1-by-p domino must cover precisely one tile of each color. Hence, if the board can be covered by the dominoes, there are an equal number of squares of each color. This implies p divides mn. Since p is prime, it follows p divides m or p divides n. Conversely, if p divides m or n, then one may fill out each (without loss of generality) row with dominoes, and hence it is possible to tile the entire board. Thus, the criterion is that the tiling is possible if and only if p divides m or p divides n. □ However, in some situations, a more complicated coloring will be necessary. Is it possible to tile a 10-by-10 board with 1-by-4 dominoes? Choose 4 colors, labeled 1 through 4. Color each odd-numbered row of the board with {1,2,1,2,⋯,1,2}, and each even-numbered row with {3,4,3,4,⋯,3,4}. By this construction, any 1-by-4 domino must cover an even number of squares of each color. Hence, if the board can be covered by the dominoes, there are an even number of squares of each color. But there are precisely 25 squares of each color; contradiction! □ Here is another problem that uses the idea of a coloring solution for easy visualization: Coloring problems are, I think, one of the most elegant ways of visualising a problem, that might otherwise be impossible or increasingly difficult to handle. Here I present a problem, and my coloring solution that really lifted my spirits. The problem was something like this: On a 9×9 grid, two squares are said to be neighboring if they share a common side. There are 65 bugs, each occupying a single square. At t=0, each bug moves to any of the neighboring squares of the square it was previously in. After t=0, each of the bug moves into a square that is to the right of it, or to the left of the square it is presently in. The orientation of the bug, is decided in the following way: Suppose a bug is at the square A at first, and at t=0, it decides to move into square C (It could have moved to B or I or F as well at t=0). The head of the bug is shown by the pointer at its tip. Now it is facing east and it has two options, either to move to D or to E, which are to the left and right of it at this moment. (Note that the initial position of the pointer of the bug is pointed towards B but then it decides to go towards the right; this is because at A its choices are totally random, so it can actually go to any of the neighboring squares.) We need to prove that at some point of time t, there will exist a square containing more than one bug. Solution First we begin by noting that by the given move scheme, the bug will be on H or D or E or G square after 2 moves. So after 2 moves, the bug will have moved one place along a diagonal in any direction. According to that scheme here is my coloring: We set the notation by defining that the set of all squares colored red is R, that consisting of orange colored squares is O, and correspondingly Y and G for those containing yellow and green colored squares. We note that all the bugs from R will migrate to O after 2 moves and vice versa, and all the bugs from Y will migrate to G after 2 moves and vice versa. So we have by our colouring ∣R∣=25,∣O∣=16,∣Y∣=20,∣G∣=20. Now by the Pigeonhole Principle we have that there are 65 bugs and 4 kinds of coloured squares they can be put in. So there are at least 17 bugs all of which occupy squares of the same color. In the next two cases we are going to talk about these 17 bugs exclusively. Case 1 If they are all initially in O, then we are done, because 17 bugs in 16 squares ensures more than one bug in at least one square by the pigeonhole principle. Case 2 If all of them are initially in R, then after 2 moves all 17 will be in O, whence we are done again by the pigeonhole principle as in Case 1. Another case can happen for which we make the following couple of observations. After one move, a bug from R or O will always land in Y or G. Case 3 So, in the third case, we apply the pigeonhole principle in the following way: We have 65 bugs and two classes of colored squares: A (consisting of the squares from R and O) and B (consisting of the squares from Y and G). So there must be at least 33 bugs in one of the two classes A and B. If the 33 bugs are in class A then there are a total of 33 bugs in R and O, so by the pigeonhole principle again, there is at least 17 bugs in one of them, and the first 2 cases ensure that more than one bug can be located in the same square in such cases. If the 33 bugs are in B then after one move they will have moved to A, by the observation given above. So after one move we have 33 bugs in A, whence the argument immediately before this ensures more than one bug in one square. Hence we are done! □ Note: An interesting thing to note from this proof is that we can achieve the goal of having more than one bug in one square after a maximum of three moves. So at t=3 we can actually ensure that there is more than one bug in some square. That's the beauty of coloring proofs, a way to visualize. Cite as: Coloring (Parity Arguments). Brilliant.org. Retrieved 03:00, September 28, 2025, from
15911	https://geoinfo.nmt.edu/publications/about/authors/style/capitalization-punctuation.html	home page New Mexico Bureau of Geology & Mineral Resources Terms & Standards Capitalization & Punctuation Citations & References New Mexico Bureau of Geology Editorial Style Guide Capitalization & Punctuation We are a research and service division of: Jump to: capitalization punctuation [back to top] Last updated July 26, 2023 This page is a test of concept. Internal links may not work properly. Rules governing capitalization are numerous and subject to many exceptions. The goal is to avoid random acts of capitalization! Be consistent throughout a publication. Biological classification Capitalize the name of a biological phylum, class, order, family, or genus, but not species, subspecies, or variety (for example, Foraminifera, Arecaceae, Tyrannosaurus rex, Homo sapiens, Prunus virginiana var. demissa). Callouts to figures, tables, appendices, and chapters Capitalize the words figure, table, appendix, and chapter when they’re given in the text to refer readers to a subsequent item (for example, the features in Figure 1 illustrate, refer to Chapter 2). When calling out to another chapter in the same publication, use the same numbering convention as the chapter itself, that is, if the chapters are numbered in Roman numerals, use Roman numerals in the callouts. If another author’s figure, table, appendix, or chapter is being called out, use lowercase (for example, At its reference section in the Rio Salado Valley (Hook et al., 1983, table 3), the formation is divided into three members.). See figure callouts. Chemical elements and molecules The names of elements and molecules are not capitalized in text. When using chemical symbols (see abbreviations and acronyms), use the capitalization style on the periodic table. Compass directions Use lowercase for compass directions when written out (for example, north, south, northwest, north-northwest) and for descriptive terms that denote direction or position (for example, southeastern New Mexico). Abbreviations for compass directions should be capitalized (for example, N, S, E, W, NE, SW, NNE). Capitalize directions only when they designate regions: Indianapolis is northwest of Lexington. Earthquakes are more common on the West Coast. Southerners eat more fried foods than Northerners. The climate is wet in the Northwest. Geography Proper geographic names are capitalized (for example, Monument Valley, Wasatch Range, Great Salt Lake). Names that are not recognized as formal may be capitalized in some cases, for example, if they are the subject of a publication. This guide contains individual entries for some geographic names whose capitalizations differs from these rules due to common usage. For terms not listed here, authors should refer to the U.S. Geological Survey Geographic Names Information System (GNIS) to determine if a geographic name is formal and should be capitalized. More specifics: Areas of indefinite extent are lowercase (for example, Salt Lake City area, Bonneville basin). Mining districts, oil fields, gas fields, fields, mines, smelters, and other terms related to mineral or hydrocarbon extraction and processing are not capitalized when included as part of a proper name (for example, City Creek mining district, Grassy Trail Creek oil field, Wasatch Plateau coalfield, Bingham Canyon mine, Black Rock smelter). Structural geologic terms are not capitalized when included as part of a geologic name (for example, Sulphur Creek anticline, Wasatch fault zone). Even when used with a proper name, lowercase is preferred for anticline, arch, area, basalt flow, batholith, claim, coal bed, coalfield, coal seam, cone, cyclothem, deposit, dome, embayment, escarpment, facies, fault, fault zone, homocline, mill, mine, mining district, monocline, oil field, pluton, principal meridian, prospect, quadrangle, region, rift, sag, syncline, uplift, well, and zone. See geographic names. Geologic time/stratigraphy Capitalize the names of eons, eras, periods, epochs, ages, eonothems, erathems, systems, series, and stages. The geochronologic terms early, middle, and late and the chronostratigraphic terms lower, middle, and upper are capitalized only as part of a formal geologic time or rock series name (for example, Early Jurassic, Upper Devonian). They are lowercase for informal designations (for example, early Holocene, middle Eocene). For geochronologic use, authors are encouraged to follow the Geological Society of America’s Geologic Time Scale. For chronostratigraphic use, authors are encouraged to follow the International Commission on Stratigraphy’s International Chronostratigraphic Chart. Geochronologic (time) Capitalize the names of geologic eons, eras, periods, epochs, and ages. All subdivisions of ages are informal (for example, early Maastrichtian, late Albian). Capitalize early, middle, and late when used with Jurassic, Triassic, Pennsylvanian, Mississippian, Devonian, and Ordovician. Capitalize early and late with Cretaceous (no middle). Lowercase early, middle, and late when used with Quaternary, Neogene, Paleogene, Permian, and Silurian. Chronostratigraphic (position) Capitalize the names of geologic eonothems, erathems, systems, series, and stages. All subdivisions of stages are informal (for example, lower Maastrichtian, upper Albian). Capitalize lower, middle, and upper when used with Jurassic, Triassic, Pennsylvanian, Mississippian, Devonian, and Ordovician. Capitalize lower and upper with Cretaceous (no middle). Lowercase lower, middle, and upper when used with Quaternary, Neogene, Paleogene, Permian, and Silurian. Geologic units The names of formal geologic units are proper nouns and are capitalized (for example, Chinle Formation, Navajo Draw Member of Arroyo Ojito Formation of Santa Fe Group). Authors should consult the U.S. Geological Society National Geologic Map Database (aka Geolex) to determine if a unit is formally recognized. In the names of informal geologic units, don’t capitalize unit terms for lithology or formation (for example, Huckleberry Ridge ash bed, Left Creek quartzite, formation of Aurora). For more, see geologic names/geologic units. Headings Use title case capitalization for headings, that is, capitalize all words except for minor words (articles [for example, a, an, the, etc.], prepositions [for example, on, for, after, etc.], and short conjunctions [for example, and, but, or, etc.]) that are not the first or last word of the title. Legislative, administrative, and judicial bodies The full names of legislative, administrative, and judicial bodies, departments, bureaus, and offices are capitalized (for example, U.S. Department of Energy). Abbreviated names are not (for example, the energy department [in relation to the U.S. Department of Energy]). Mineral names Mineral names are not capitalized unless they begin a sentence; see minerals. Professional titles Capitalize professional titles only when they come before a person’s name (for example, Pat Jones is the state geologist, State Geologist Pat Jones is president of the Association of American State Geologists). Proper nouns Always capitalize proper nouns, including hyphenated words in titles and headings (for example, Pat Jones, Open-File Report 88-546). Don’t capitalize conjunctions, short prepositions (for example, of, for, from), or articles (for example, the, a, an) in long proper names (for example, New Mexico Bureau of Geology and Mineral Resources). In the report text, capitalize all important words in the titles of books, journal articles, and reports (for example, Geologic History of Utah by Lehi F. Hintze; see title case); for capitalization of reference entries, see references. Singular nouns used to replace proper nouns are not capitalized (for example, Grand Staircase–Escalante National Monument, but the monument; Zion National Park, but the park). Seasons Don’t capitalize spring, summer, fall, autumn, or winter unless part of a formal name (for example, sampling was carried out in spring and winter, the program was held in the fall, the New Mexico Geological Society Fall Field Conference was in October). Titles Use title case capitalization for titles in the main text of a publication, that is, capitalize all words except for minor words (articles [for example, a, an, the, etc.], prepositions [for example, on, for, after, etc.], and short conjunctions [for example, and, but, or, etc.]) that are not the first or last word of the title. punctuation apostrophe ' Apostrophes have two primary uses: to indicate possession and to make contractions. To indicate possession: Add ’s to a singular noun to make it possessive (for example, the layer’s thickness). Add only an apostrophe to plural nouns ending in s (for example, the layers’ thicknesses). Add ’s to singular nouns ending in s (for example, Dr. Jones’s research). This is an exception to some style guides, but it indicates the correct pronunciation. Don’t use an apostrophe with possessive pronouns: ours, yours, his, hers, its, theirs, whose. Don’t use an apostrophe with a word ending in s when its use is descriptive rather than possessive (for example, Masters swimming, Beatles music, Titans game). Don’t use an apostrophe in place names unless it is part of the formal name (for example, Pikes Peak, Tates Creek, Martha’s Vineyard). To make contractions: Use an apostrophe to form contractions and to indicate omitted letters or numerals (for example, don’t, won’t, can’t, rock ’n’ roll, class of ’74). Avoid contractions in technical writing, but use them sensibly for a general audience to prevent the prose from being overly formal. Don’t use ’s to pluralize numerals or multiple-letter combinations (for example, 1890s, not 1890’s; ABCs, not ABC’s). Don’t use an apostrophe as a substitute for the prime symbol (′ – Alt + 8242). Don’t use an apostrophe as an abbreviation for feet (for example, 6 ft, not 6’). brackets [ ] Brackets serve the same function as parentheses—to enclose incidental information or explanatory material—but they’re restricted to situations where it’s necessary to enclose information within an existing set of parentheses: ...or thin dolomitic beds (note, however, that Bauch [1982, p. 42] included salt-cast-bearing gray siltstones in the upper Abo Formation). Brackets are also used in equations and formulas, and to insert brief editorial comments or corrections into quoted material (see quotations). colon : Use a colon (not a semicolon) to introduce lists, long quotations, and for emphasis: These are the three classes of rocks: igneous, sedimentary, and metamorphic. Vanderbilt coach Bryce Drew said: “I’m really proud of our effort tonight. We could have ducked our heads when we got 10 points down, but we didn’t. We came back and became the aggressor late in the second half. That turned the game around for us.” Malik Monk: All he does is win. Don’t capitalize the first word after a colon unless it’s a proper noun or the beginning of a complete sentence: The best songwriting team ever: Lennon and McCartney. The worst thing that can happen: defeat. You can count on one thing: The replay booth will rule against Vanderbilt. Colons are also used after salutations in a formal letter (for example, Dear Dr. Jones:), to separate hours from minutes (for example, 3:45 p.m.), to separate the two halves of a ratio (for example, 4:1), and in reference entries to separate the title of a source from its publisher (see references). For help with structuring lists with colons, see lists. comma , Commas have many uses. To delineate items in a series: John, Paul, George, and Ringo are the Beatles. In lists of three items or more in technical publications, use a serial comma (also known as the Oxford comma) before the last item and its conjunction. In the above example, this is the comma after George. Before and after a state name when a city is given: Nashville, Tennessee, is my hometown. To separate adjectives that are equal in rank (adjectives that can switch locations in the sentence with one another such that the sentence will still make sense): tan, brown to reddish-brown, well-sorted silt and sand No comma is needed before the last adjective if it’s critical to the meaning: a stretched-out, transparent team swimsuit (not just any suit, but the suit designated for team members) To set off introductory phrases: When we arrived at the pool, we were overcome by chlorine fumes. To set off two phrases joined by a conjunction when each phrase could stand alone as a sentence: She came to Savannah to score points, but the highest she placed was the dreaded 11th place. To introduce short quotations: Phelps said, “I’m pleased with my time.” In numbers greater than three digits (but not after a decimal): The attendance was 12,060 The first 10 digits of pi are 3.141592653 (not 3.141,592,653) To identify month, date, and year: On September 5, 1893, the New Mexico School of Mines opened. Don’t use a comma to separate months and years: October 1956 (not October, 1956) Don’t use a comma with Inc.: Pendant Publishing Inc. (not Pendant Publishing, Inc.) Don’t use a comma with Jr. and Sr.: Garland Dever Jr. (not Garland Dever, Jr.) dash – — There are two types of dashes: en dashes (–) and em dashes (—). An en dash (so called because it’s the width of the letter n) is used to connect inclusive words and numbers such as dates, page ranges, and ranges of values when given as equivalent units of measure in parentheses: I usually swim 2,000–3,000 yards in a workout. 1945–1978 p. 312–327 20 to 30 ft (6–9 m) cross section A–A’ snowpack–runoff relationship north–south trajectory As a test for when an en dash is appropriate over a hyphen, substitute the words to, through, or up to and including between the items. An em dash (so called because it’s the width of the letter m) is used to set off extra information, such as examples, explanatory or descriptive phrases, or supplemental facts: One can piece together from different locales—particularly the Caballo Mountains and Sierra de las Uvas—a seemingly conformable sequence of these rock units. In central New Mexico, most now regard the Glorieta as a distinct formation—a practice followed in this study. Don’t put spaces before or after either dash. Don’t use a hyphen where a dash would be the correct choice and vice versa. In reference list entries, use an em dash to replace a colon that appears within the document title to avoid possible confusion with the colon that is used to mark the end of the document title (see references): Basins of the Rio Grande Rift—Structure, Stratigraphy, and Tectonic Setting: Geological Society of America Special Paper 291 Don’t use en dashes to substitute for the words to or and when used with from or between, respectively, to indicate a range of values. Don’t use an en dash when the two items in the range are hyphenated. Nitrate concentration ranges from 2.6 to 9.3 mg/L. (not from 2.6–9.3 mg/L) Fine-grained sediments accumulated in a shallow lake between 45 and 40 million years ago. (not between 45–40 million years ago) We measured several 31- to 66-m-long profiles. (not 31–66-m-long) PC/Mac controls for en dash: PC: Alt + 0150 (Note: This method works only for keyboards that include a 10-key numeric pad; use a double hyphen [--] if the alt code doesn’t work.) Mac: Option + Dash (-) PC/Mac controls for em dash: PC: Alt + 0151 (Note: This method works only for keyboards that include a 10-key numeric pad; use a triple hyphen [---] if the alt code doesn’t work.) Mac: Shift + Option + Dash (-) ellipsis … An ellipsis indicates that one or more words have been omitted from quoted material: A previous author wrote, “We will examine the development of Brough’s Tunnel … within Clifty Falls State Park.” Ellipses are formed by typing three periods in a row. Include a space before the first period and after the last period. Ellipses are not used before the first word of a quotation, even if the beginning of the original sentence is omitted, or after the last word of a quotation, even if the end of the original sentence is omitted. Ellipses are not meant to indicate a break in thought; don’t use them instead of an em dash. hyphen - A hyphen joins two or more words that work together to modify a subsequent noun: a well-defined aquifer pale-green shale reddish-brown sandstone light-gray limestone 60-meter-wide graben fine-grained sandstone fine- to coarse-grained sandstone 1/2-inch crystals northwest-trending fault east-central New Mexico Don’t use a hyphen when the noun comes first: an aquifer that is well defined shale of pale green color a graben that is 60 m wide the sandstone is fine grained sandstone that is fine to coarse grained crystals of 1/2 inch a fault trending northwest Don’t use a hyphen with very or adverbs ending in ly: a very porous layer a finely crystalline rock Don’t use a hyphen when the modifying words do not function as one unit and could be separated by a comma: a small, blue mineral a large, dry riverbed Use a hyphen to prevent doubled vowels and tripled consonants: anti-intellectual well-liked When describing rock units in stratigraphic sections, well logs, and other lists, compound modifiers following a rock name are traditionally hyphenated in the geologic literature: Sandstone: blue-gray, thinly bedded, coarse-grained Welded tuff: reddish-brown, flow-banded A number of compound nouns exist in geological vocabulary. Some are hyphenated (acre-foot, cross-bed, cross-stratification, meta-arkose), and others are not (cross section, dip angle, dike swarm, solution banding). Consult the Glossary of Geology for the proper forms of compound nouns. Compound numbers between twenty-one and ninety-nine and fractions are hyphenated when written out: thirty-three four-fifths one-third Use a hyphen to separate noninclusive numbers, such as telephone numbers and serial numbers: 801-537-3300 no. 14558-789-D NMBGMR Miscellaneous Publication 03-7 Use a hyphen where numbers and units of measure form a compound modifier: A 4-foot-thick fossiliferous limestone bed is at the top of the interval. See prefixes and suffixes for more on the use of hyphens. parentheses ( ) Parentheses enclose incidental information or explanatory material and are the most common type of punctuation found in scientific reports after commas and periods. Parentheses are used in the following ways. To enclose abbreviations the first time they appear: Differences in the timing of the most recent event (MRE) along a fault zone are used to define fault segments. For in-text citations (see brackets): Wilpolt and Wanek (1951) were the first to map the geology of the Quebradas region. Terranes accreted to the southern margin of Laurentia during the Proterozoic (Karlstrom et al., 2004). To provide alternative units of measurement: Deposition was rapid during the Pennsylvanian and Permian, resulting in a total accumulation of 25,000 to 30,000 ft (7,600–9,100 m) of marine sediments. To call out figures, tables, chapters, and appendices: In the Blackington Hills, the tuff is 90–150 m thick (Fig. 63). To provide additional information: The Emery high (or Piute platform as it is now called) and the Kaibab uplift (a poorly defined band of uplifts) have been identified by the thinning or absence of Pennsylvanian formations. To enumerate points in a list: Much of the unit is (1) red, pink, or gray, (2) medium to coarse grained, and (3) equigranular or slightly porphyritic. Don’t use back-to-back parentheses; use a semicolon to separate items, for example, (Fig. 2; Hook, 1983), or rewrite to avoid back-to-back parentheses or ambiguity. quotation marks “ ” Quotation marks set off direct speech and material quoted verbatim from other sources (see quotations): Hook and Cobban (2015, p. 27) defined it as “that portion of the Mancos Shale lying between the undifferentiated or main body of the Dakota Sandstone and the Tres Hermanos Formation.” In text (but not reference entries), quotation marks enclose the titles of journal articles, book chapters, and newspaper articles and editorials. Italics are used for the names of the larger works these items are published in (see italics): “The House Range, Western Utah—Cambrian Mecca” by Hintze and Robison (1987) in Rocky Mountain Section of the Geological Society of America Centennial Field Guide, Volume 2 provides a succinct description of an important fossil locality in Utah. Quotation marks also enclose words used in a special way: “Caliche” usually refers to an indurated layer of calcium carbonate accumulation in a soil; “hardpan” is a more general term that refers to any indurated soil layer resulting from the precipitation of soluble materials in the soil profile. Commas and periods are always placed inside closing quotation marks; colons and semicolons are always placed outside closing quotation marks. All other punctuation marks are placed inside quotation marks only if they are part of the quoted material. Don’t use a quotation mark as a substitute for the double-prime symbol (″) (i.e., seconds of angle; see prime symbol). Don’t use a quotation mark as an abbreviation for inches (for example, 6 in., not 6”). semicolon ; A semicolon indicates a pronounced separation between grammatical units. It’s stronger than a comma and almost as full as a period, but it implies a closer relationship between the thoughts than a period would. A semicolon separates two or more complete thoughts (independent clauses) within a single sentence: These bentonites range in thickness from 0.6 to 35.6 cm; most are white but weather orange. When a semicolon is used to link complete thoughts, it doesn’t require a conjunction (and, but, or, if, so, etc.); however, a semicolon is necessary when an adverbial conjunction (accordingly, besides, consequently, furthermore, hence, however, instead, moreover, nevertheless, otherwise, still, therefore, thus) connects the thoughts: The full extent of the Escalante silver vein was known only after extensive exploration; consequently, a number of smaller ore deposits in the district were overlooked for many years. Don’t use a semicolon with the simple coordinating conjunctions and, or, for, nor, yet, so, and but. Instead use a comma. Semicolons are also used to separate items in a series when one or more of the items requires a comma (see lists). slash / Slashes are used to signify alternatives (for example, and/or, his/hers), in abbreviations (for example, m/s, acre-ft/yr), or in fractions (for example, 1/3, 1/4).
15912	https://stats.libretexts.org/Courses/City_University_of_New_York/Introductory_Statistics_with_Probability_(CUNY)/02%3A_Descriptive_Statistics/2.07%3A_Measures_of_Spread-_Variance_and_Standard_Deviation	2.7: Measures of Spread- Variance and Standard Deviation - Statistics 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 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"license:ccby", "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. City University of New York 4. Introductory Statistics with Probability (CUNY) 5. 2: Descriptive Statistics 6. 2.7: Measures of Spread- Variance and Standard Deviation Expand/collapse global location 2.7: Measures of Spread- Variance and Standard Deviation Last updated Sep 10, 2021 Save as PDF 2.6: Box Plots 2.8: Skewness and the Mean, Median, and Mode Page ID 26030 OpenStax OpenStax ( \newcommand{\kernel}{\mathrm{null}\,}) Table of contents 1. The Range 2. The Variance and the Standard Deviation 1. Calculating the Standard Deviation WeBWorK Problems The number of standard deviations a data value is from the mean can be used as a measure of the closeness of a data value to the mean. Because the standard deviation measures the spread of the data this gives a uniform measure for any data set. Explanation of the standard deviation calculation shown in the table Standard deviation of Grouped Frequency Tables Comparing Values from Different Data Sets References Review Formula Review Glossary Contributors and Attributions [NOTE from VS: The following is pulled from Shafer and Zhang] Look at the two data sets in Table 2.7.1 and the graphical representation of each, called a _dot plot_, in Figure 2.7.1. Table 2.7.1 _: Two Data Sets_\| Data Set I: \| 40 \| 38 \| 42 \| 40 \| 39 \| 39 \| 43 \| 40 \| 39 \| 40 \| \| Data Set II: \| 46 \| 37 \| 40 \| 33 \| 42 \| 36 \| 40 \| 47 \| 34 \| 45 \| The two sets of ten measurements each center at the same value: they both have mean, median, and mode equal to 40. Nevertheless a glance at the figure shows that they are markedly different. In Data Set I the measurements vary only slightly from the center, while for Data Set II the measurements vary greatly. Just as we have attached numbers to a data set to locate its center, we now wish to associate to each data set numbers that measure quantitatively how the data either scatter away from the center or cluster close to it. These new quantities are called measures of variability, and we will discuss three of them. Figure 2.7.1:Dot Plots of Data Sets The Range First we discuss the simplest measure of variability. Definition: range The range R of a data set is difference between its largest and smallest values (2.7.1)R=x max−x min where x max is the largest measurement in the data set and x min is the smallest. Example 2.7.1: Identifyig the Range of a dataset Find the range of each data set in Table 2.7.1. Solution: For Data Set I the maximum is 43 and the minimum is 38, so the range is R=43−38=5. For Data Set II the maximum is 47 and the minimum is 33, so the range is R=47−33=14. The range is a measure of variability because it indicates the size of the interval over which the data points are distributed. A smaller range indicates less variability (less dispersion) among the data, whereas a larger range indicates the opposite. The range is very limited in the information it gives us, as it is only based on the largest and smallest values. Anything can happen in between and the range tells us nothing about these. In order to get information about how all of the data points are spread out we can compare each one to the mean. We do this with the "Variance" and "Standard Deviation". The Variance and the Standard Deviation The other two measures of variability that we will consider are the Variance and the Standard Deviation. They are intimately connected, as the standard deviation is just the square root of the variance. The word "deviation" gives us the clue of what we are trying to do. In order to measure how much variation there is in the data, we use the mean as the central value and then calculate all of the differences ("deviations") of each data value from the mean. The Variance is easier to calculate because it does not involve the square root. It has a drawback in that the quantities used are squared, so it will not represent the correct units for the data. The Standard Deviation takes the square root of the Variance and so the squared units are returned to regular units (such as inches, pounds and so forth based on the sampled data0. Calculating the Standard Deviation If x is a number, then the difference "x – mean" is called its deviation. In a data set, there are as many deviations as there are items in the data set. The deviations are used to calculate the standard deviation. If the numbers belong to a population, in symbols a deviation is x−μ. For sample data, in symbols a deviation is x−x¯. To calculate the standard deviation, we need to calculate the variance first, and then take the square root. The variance is the averageof the squares of the deviations (the x−x¯ values for a sample, or the x−μ values for a population). The symbol σ 2 represents the population variance; the population standard deviation σ is the square root of the population variance. The symbol s 2 represents the sample variance; the sample standard deviation s is the square root of the sample variance. You can think of the standard deviation as a special average of the deviations. If the numbers come from a census of the entire population and not a sample, when we calculate the average of the squared deviations to find the variance, we divide by N, the number of items in the population. If the data are from a sample rather than a population, when we calculate the average of the squared deviations, we divide by n – 1, one less than the number of items in the sample. In summary, the procedure to calculate the standard deviation depends on whether the numbers are the entire population or are data from a sample. The calculations are the same except that for the sample we divide by "sample size - 1: n-1" and for the population we divide by "Population size N". Therefore the symbol used to represent the standard deviation depends on whether it is calculated from a population or a sample. The lower case letter s represents the sample standard deviation and the Greek letter σ (sigma, lower case) represents the population standard deviation. If the sample has the same characteristics as the population, then s should be a good estimate of σ. Formulas for the Sample Standard Deviation (2.7.2)s=∑(x−x¯)2 n−1 For the sample standard deviation, the denominator is n−1, that is the sample size MINUS 1. Formulas for the Population Standard Deviation (2.7.3)σ=∑(x−μ)2 N For the population standard deviation, the denominator is N, the number of items in the population. The Sample Variance is the calculation before taking the square root. Definition: sample variance and sample Standard Deviation The sample variance of a set of n sample data is the number s 2 defined by the formula (2.7.4)s 2=∑(x−x¯)2 n−1 An algebraically equivalent formula is sometimes used, because the calculations are easier to perform: (2.7.5)s 2=∑x 2−1 n⁢(∑x)2 n−1 The square root s of the sample variance is called the sample standard deviation of a set of n sample data. It is given by the formulas (2.7.6)s=s 2=∑(x−x¯)2 n−1=∑x 2−1 n⁢(∑x)2 n−1. Although the first formula in each case looks less complicated than the second, the latter is easier to use in hand computations, and is called a shortcut formula. Example 2.7.2: Identifying the Variance and Standard Deviation of a Dataset Find the sample variance and the sample standard deviation of Data Set II in Table 2.7.1. Solution To use the defining formula (the first formula) in the definition we first compute for each observation x its deviation x−x¯ from the sample mean. Since the mean of the data is x¯=40, we obtain the ten numbers displayed in the second line of the supplied table x 46 37 40 33 42 36 40 47 34 45 x−x¯−6−3 0−7 2−4 0 7−6 5 Thus ∑(x−x¯)2=6 2+(−3)2+0 2+(−7)2+2 2+(−4)2+0 2+7 2+(−6)2+5 2=224 so the variance is s 2=∑(x−x¯)2 n−1=224 9=24.⁢8¯ and the standard deviation is s=24.⁢8¯≈4.99 The student is encouraged to compute the ten deviations for Data Set I and verify that their squares add up to 20, so that the sample variance and standard deviation of Data Set I are the much smaller numbers (2.7.7)s 2=20/9=2.⁢2¯ and (2.7.8)s=20/9≈1.49 The standard deviation provides a numerical measure of the overall amount of variation in a data set, and can be used to determine whether a particular data value is close to or far from the mean. WeBWorK Problems 2.7.1 2.7.2 2.7.3 2.7.4 2.7.5 The number of standard deviations a data value is from the mean can be used as a measure of the closeness of a data value to the mean. Because the standard deviation measures the spread of the data this gives a uniform measure for any data set. For example:suppose that Rosa and Binh both shop at supermarket A. Rosa waits at the checkout counter for seven minutes and Binh waits for one minute. At supermarket A, the mean waiting time is five minutes and the standard deviation is two minutes. Rosa waits for seven minutes: This is two minutes longer than the average wait time. Two minutes is the same as one standard deviation. Rosa's wait time of seven minutes is one standard deviation above the average of five minutes. Binh waits for one minute. This is four minutes less than the average of five; four minutes is equal to two standard deviations. Binh's wait time of one minute is two standard deviations below the average of five minutes. A "rule of thumb" is that more than two standard deviations away from the average is considered "far from the average". In general, the shape of the distribution of the data affects how much of the data is further away than two standard deviations. (You will learn more about this in later chapters.) The number line may help you understand standard deviation. If we were to put five and seven on a number line, seven is to the right of five. We say, then, that seven is one standard deviation to the right of five because 5+(1)⁢(2)=7. If one were also part of the data set, then one is two standard deviations to the left of five because 5+(−2)⁢(2)=1. Figure 2.7.1 In general, a value = mean + (#ofSTDEV)(standard deviation) where #ofSTDEVs = the number of standard deviations ofSTDEV does not need to be an integer One is two standard deviations less than the mean of five because: 1=5+(−2)⁢(2). The equation value = mean + (#ofSTDEVs)(standard deviation) can be expressed for a sample and for a population. sample: x = \bar{x} + \text{(#ofSTDEV)(s)} Population: x = \mu + \text{(#ofSTDEV)(s)} The lower case letter s represents the sample standard deviation and the Greek letter σ (sigma, lower case) represents the population standard deviation. The symbol x¯ is the sample mean and the Greek symbol μ is the population mean. In practice, USE A CALCULATOR OR COMPUTER SOFTWARE TO CALCULATE THE STANDARD DEVIATION. If you are using a TI-83, 83+, 84+ calculator, you need to select the appropriate standard deviation σ x or s x from the summary statistics. We will concentrate on using and interpreting the information that the standard deviation gives us. However you should study the following step-by-step example to help you understand how the standard deviation measures variation from the mean. (The calculator instructions appear at the end of this example.) Example 2.7.1 In a fifth grade class, the teacher was interested in the average age and the sample standard deviation of the ages of her students. The following data are the ages for a SAMPLE of n = 20 fifth grade students. The ages are rounded to the nearest half year: 9; 9.5; 9.5; 10; 10; 10; 10; 10.5; 10.5; 10.5; 10.5; 11; 11; 11; 11; 11; 11; 11.5; 11.5; 11.5; x¯=9+9.5⁢(2)+10⁢(4)+10.5⁢(4)+11⁢(6)+11.5⁢(3)20=10.525 The average age is 10.53 years, rounded to two places. The variance may be calculated by using a table. Then the standard deviation is calculated by taking the square root of the variance. We will explain the parts of the table after calculating s. \| Data \| Freq. \| Deviations \| Deviations 2 \| (Freq.)(Deviations 2) \| --- --- \| x \| f \| (x – x¯) \| (x – x¯)2 \| (f)(x – x¯)2 \| \| 9 \| 1 \| 9 – 10.525 = –1.525 \| (–1.525)2 = 2.325625 \| 1 × 2.325625 = 2.325625 \| \| 9.5 \| 2 \| 9.5 – 10.525 = –1.025 \| (–1.025)2 = 1.050625 \| 2 × 1.050625 = 2.101250 \| \| 10 \| 4 \| 10 – 10.525 = –0.525 \| (–0.525)2 = 0.275625 \| 4 × 0.275625 = 1.1025 \| \| 10.5 \| 4 \| 10.5 – 10.525 = –0.025 \| (–0.025)2 = 0.000625 \| 4 × 0.000625 = 0.0025 \| \| 11 \| 6 \| 11 – 10.525 = 0.475 \| (0.475)2 = 0.225625 \| 6 × 0.225625 = 1.35375 \| \| 11.5 \| 3 \| 11.5 – 10.525 = 0.975 \| (0.975)2 = 0.950625 \| 3 × 0.950625 = 2.851875 \| \| \| \| \| \| The total is 9.7375 \| The sample variance, s 2, is equal to the sum of the last column (9.7375) divided by the total number of data values minus one (20 – 1): s 2=9.7375 20−1=0.5125 The sample standard deviations is equal to the square root of the sample variance: s=0.5125=0.715891 and this is rounded to two decimal places, s=0.72. Typically, you do the calculation for the standard deviation on your calculator or computer. The intermediate results are not rounded. This is done for accuracy. For the following problems, recall that value = mean + (#ofSTDEVs)(standard deviation). Verify the mean and standard deviation or a calculator or computer. For a sample: x = x¯ + (#ofSTDEVs)(s) For a population: x = μ + (#ofSTDEVs)σ For this example, use x = x¯ + (#ofSTDEVs)(s) because the data is from a sample Verify the mean and standard deviation on your calculator or computer. Find the value that is one standard deviation above the mean. Find (x¯ + 1s). Find the value that is two standard deviations below the mean. Find (x¯ – 2s). Find the values that are 1.5 standard deviations from (below and above) the mean. Solution Clear lists L1 and L2. Press STAT 4:ClrList. Enter 2nd 1 for L1, the comma (,), and 2nd 2 for L2. Enter data into the list editor. Press STAT 1:EDIT. If necessary, clear the lists by arrowing up into the name. Press CLEAR and arrow down. Put the data values (9, 9.5, 10, 10.5, 11, 11.5) into list L1 and the frequencies (1, 2, 4, 4, 6, 3) into list L2. Use the arrow keys to move around. Press STAT and arrow to CALC. Press 1:1-VarStats and enter L1 (2nd 1), L2 (2nd 2). Do not forget the comma. Press ENTER. x¯ = 10.525 Use Sx because this is sample data (not a population): Sx=0.715891 (x¯+1⁢s)=10.53+(1)⁢(0.72)=11.25 (x¯−2⁢s)=10.53–(2)⁢(0.72)=9.09 (x¯−1.5⁢s)=10.53–(1.5)⁢(0.72)=9.45 (x¯+1.5⁢s)=10.53+(1.5)⁢(0.72)=11.61 Explanation of the standard deviation calculation shown in the table The deviations show how spread out the data are about the mean. The data value 11.5 is farther from the mean than is the data value 11 which is indicated by the deviations 0.97 and 0.47. A positive deviation occurs when the data value is greater than the mean, whereas a negative deviation occurs when the data value is less than the mean. The deviation is –1.525 for the data value nine. If you add the deviations, the sum is always zero. (For Example 2.7.1, there are n=20 deviations.) So you cannot simply add the deviations to get the spread of the data. By squaring the deviations, you make them positive numbers, and the sum will also be positive. The variance, then, is the average squared deviation. The variance is a squared measure and does not have the same units as the data. Taking the square root solves the problem. The standard deviation measures the spread in the same units as the data. Notice that instead of dividing by n=20, the calculation divided by n−1=20−1=19 because the data is a sample. For the sample variance, we divide by the sample size minus one (n−1). Why not divide by n? The answer has to do with the population variance. The sample variance is an estimate of the population variance. Based on the theoretical mathematics that lies behind these calculations, dividing by (n−1) gives a better estimate of the population variance. Your concentration should be on what the standard deviation tells us about the data. The standard deviation is a number which measures how far the data are spread from the mean. Let a calculator or computer do the arithmetic. The standard deviation, s or σ, is either zero or larger than zero. When the standard deviation is zero, there is no spread; that is, all the data values are equal to each other. The standard deviation is small when the data are all concentrated close to the mean, and is larger when the data values show more variation from the mean. When the standard deviation is a lot larger than zero, the data values are very spread out about the mean; outliers can make s or σ very large. The standard deviation, when first presented, can seem unclear. By graphing your data, you can get a better "feel" for the deviations and the standard deviation. You will find that in symmetrical distributions, the standard deviation can be very helpful but in skewed distributions, the standard deviation may not be much help. The reason is that the two sides of a skewed distribution have different spreads. In a skewed distribution, it is better to look at the first quartile, the median, the third quartile, the smallest value, and the largest value. Because numbers can be confusing, always graph your data. Display your data in a histogram or a box plot. Example 2.7.2 Use the following data (first exam scores) from Susan Dean's spring pre-calculus class: 33; 42; 49; 49; 53; 55; 55; 61; 63; 67; 68; 68; 69; 69; 72; 73; 74; 78; 80; 83; 88; 88; 88; 90; 92; 94; 94; 94; 94; 96; 100 Create a chart containing the data, frequencies, relative frequencies, and cumulative relative frequencies to three decimal places. Calculate the following to one decimal place using a TI-83+ or TI-84 calculator: The sample mean The sample standard deviation The median The first quartile The third quartile IQR Construct a box plot and a histogram on the same set of axes. Make comments about the box plot, the histogram, and the chart. Answer See Table The sample mean = 73.5 The sample standard deviation = 17.9 The median = 73 The first quartile = 61 The third quartile = 90 IQR = 90 – 61 = 29 The x-axis goes from 32.5 to 100.5; y-axis goes from -2.4 to 15 for the histogram. The number of intervals is five, so the width of an interval is (100.5−32.5) divided by five, is equal to 13.6. Endpoints of the intervals are as follows: the starting point is 32.5, 32.5+13.6=46.1, 46.1+13.6=59.7, 59.7+13.6=73.3, 73.3+13.6=86.9, 86.9+13.6=100.5= the ending value; No data values fall on an interval boundary. Figure 2.7.2. The long left whisker in the box plot is reflected in the left side of the histogram. The spread of the exam scores in the lower 50% is greater (73−33=40) than the spread in the upper 50% (100−73=27). The histogram, box plot, and chart all reflect this. There are a substantial number of A and B grades (80s, 90s, and 100). The histogram clearly shows this. The box plot shows us that the middle 50% of the exam scores (IQR = 29) are Ds, Cs, and Bs. The box plot also shows us that the lower 25% of the exam scores are Ds and Fs. \| Data \| Frequency \| Relative Frequency \| Cumulative Relative Frequency \| --- --- \| \| 33 \| 1 \| 0.032 \| 0.032 \| \| 42 \| 1 \| 0.032 \| 0.064 \| \| 49 \| 2 \| 0.065 \| 0.129 \| \| 53 \| 1 \| 0.032 \| 0.161 \| \| 55 \| 2 \| 0.065 \| 0.226 \| \| 61 \| 1 \| 0.032 \| 0.258 \| \| 63 \| 1 \| 0.032 \| 0.29 \| \| 67 \| 1 \| 0.032 \| 0.322 \| \| 68 \| 2 \| 0.065 \| 0.387 \| \| 69 \| 2 \| 0.065 \| 0.452 \| \| 72 \| 1 \| 0.032 \| 0.484 \| \| 73 \| 1 \| 0.032 \| 0.516 \| \| 74 \| 1 \| 0.032 \| 0.548 \| \| 78 \| 1 \| 0.032 \| 0.580 \| \| 80 \| 1 \| 0.032 \| 0.612 \| \| 83 \| 1 \| 0.032 \| 0.644 \| \| 88 \| 3 \| 0.097 \| 0.741 \| \| 90 \| 1 \| 0.032 \| 0.773 \| \| 92 \| 1 \| 0.032 \| 0.805 \| \| 94 \| 4 \| 0.129 \| 0.934 \| \| 96 \| 1 \| 0.032 \| 0.966 \| \| 100 \| 1 \| 0.032 \| 0.998 (Why isn't this value 1?) \| Standard deviation of Grouped Frequency Tables Recall that for grouped data we do not know individual data values, so we cannot describe the typical value of the data with precision. In other words, we cannot find the exact mean, median, or mode. We can, however, determine the best estimate of the measures of center by finding the mean of the grouped data with the formula: (2.7.9)Mean of Frequency Table=∑f⁡m∑f where f interval frequencies and m= interval midpoints. Just as we could not find the exact mean, neither can we find the exact standard deviation. Remember that standard deviation describes numerically the expected deviation a data value has from the mean. In simple English, the standard deviation allows us to compare how “unusual” individual data is compared to the mean. Example 2.7.3 Find the standard deviation for the data in Table 2.7.3. Table 2.7.3\| Class \| Frequency, f \| Midpoint, m \| m 2 \| x¯ \| fm 2 \| Standard Deviation \| --- --- --- \| 0–2 \| 1 \| 1 \| 1 \| 7.58 \| 1 \| 3.5 \| \| 3–5 \| 6 \| 4 \| 16 \| 7.58 \| 96 \| 3.5 \| \| 6–8 \| 10 \| 7 \| 49 \| 7.58 \| 490 \| 3.5 \| \| 9–11 \| 7 \| 10 \| 100 \| 7.58 \| 700 \| 3.5 \| \| 12–14 \| 0 \| 13 \| 169 \| 7.58 \| 0 \| 3.5 \| \| 15–17 \| 2 \| 16 \| 256 \| 7.58 \| 512 \| 3.5 \| For this data set, we have the mean, x¯ = 7.58 and the standard deviation, s x = 3.5. This means that a randomly selected data value would be expected to be 3.5 units from the mean. If we look at the first class, we see that the class midpoint is equal to one. This is almost two full standard deviations from the mean since 7.58 – 3.5 – 3.5 = 0.58. While the formula for calculating the standard deviation is not complicated, s x=f⁡(m−x¯)2 n−1 where s x = sample standard deviation, x¯ = sample mean, the calculations are tedious. It is usually best to use technology when performing the calculations. Comparing Values from Different Data Sets The standard deviation is useful when comparing data values that come from different data sets. If the data sets have different means and standard deviations, then comparing the data values directly can be misleading. For each data value, calculate how many standard deviations away from its mean the value is. Use the formula: value = mean + (#ofSTDEVs)(standard deviation); solve for #ofSTDEVs. \text{#ofSTDEVs} = \dfrac{\text{value-mean}}{\text{standard deviation}} Compare the results of this calculation. ofSTDEVs is often called a "z-score"; we can use the symbol z. In symbols, the formulas become: Sample x=x¯+z⁢s Population x=μ+z⁢σ Example 2.7.4 Two students, John and Ali, from different high schools, wanted to find out who had the highest GPA when compared to his school. Which student had the highest GPA when compared to his school? \| Student \| GPA \| School Mean GPA \| School Standard Deviation \| --- --- \| \| John \| 2.85 \| 3.0 \| 0.7 \| \| Ali \| 77 \| 80 \| 10 \| Answer For each student, determine how many standard deviations (#ofSTDEVs) his GPA is away from the average, for his school. Pay careful attention to signs when comparing and interpreting the answer. z = \text{#ofSTDEVs} = \left(\dfrac{\text{value-mean}}{\text{standard deviation}}\right) = \left(\dfrac{x + \mu}{\sigma}\right) \nonumber For John, z = \text{#ofSTDEVs} = \left(\dfrac{2.85-3.0}{0.7}\right) = -0.21 \nonumber For Ali, z = \text{#ofSTDEVs} = (\dfrac{77-80}{10}) = -0.3 \nonumber John has the better GPA when compared to his school because his GPA is 0.21 standard deviations below his school's mean while Ali's GPA is 0.3 standard deviations below his school's mean. John's z-score of –0.21 is higher than Ali's z-score of –0.3. For GPA, higher values are better, so we conclude that John has the better GPA when compared to his school. The following lists give a few facts that provide a little more insight into what the standard deviation tells us about the distribution of the data. For ANY data set, no matter what the distribution of the data is: At least 75% of the data is within two standard deviations of the mean. At least 89% of the data is within three standard deviations of the mean. At least 95% of the data is within 4.5 standard deviations of the mean. This is known as Chebyshev's Rule. For data having a distribution that is BELL-SHAPED and SYMMETRIC: Approximately 68% of the data is within one standard deviation of the mean. Approximately 95% of the data is within two standard deviations of the mean. More than 99% of the data is within three standard deviations of the mean. This is known as the Empirical Rule. It is important to note that this rule only applies when the shape of the distribution of the data is bell-shaped and symmetric. We will learn more about this when studying the "Normal" or "Gaussian" probability distribution in later chapters. References Data from Microsoft Bookshelf. King, Bill.“Graphically Speaking.” Institutional Research, Lake Tahoe Community College. Available online at www.ltcc.edu/web/about/institutional-research (accessed April 3, 2013). Review The standard deviation can help you calculate the spread of data. There are different equations to use if are calculating the standard deviation of a sample or of a population. The Standard Deviation allows us to compare individual data or classes to the data set mean numerically. s=∑(x−x¯)2 n−1 or s=∑f⁡(x−x¯)2 n−1 is the formula for calculating the standard deviation of a sample. To calculate the standard deviation of a population, we would use the population mean, μ, and the formula σ=∑(x−μ)2 N or σ=∑f⁡(x−μ)2 N.∑f(x−μ)2 N−−−−−−−−−√. Formula Review (2.7.10)s x=∑f⁡m 2 n−x¯2 where s x sample standard deviation and x¯=sample mean Use the following information to answer the next two exercises: The following data are the distances between 20 retail stores and a large distribution center. The distances are in miles. 29; 37; 38; 40; 58; 67; 68; 69; 76; 86; 87; 95; 96; 96; 99; 106; 112; 127; 145; 150 Glossary Standard Deviation a number that is equal to the square root of the variance and measures how far data values are from their mean; notation: s for sample standard deviation and σ for population standard deviation.Variance mean of the squared deviations from the mean, or the square of the standard deviation; for a set of data, a deviation can be represented as x – x¯ where x is a value of the data and x¯ is the sample mean. The sample variance is equal to the sum of the squares of the deviations divided by the difference of the sample size and one. Contributors and Attributions Barbara Illowsky and Susan Dean (De Anza College) with many other contributing authors. Content produced by OpenStax College is licensed under a Creative Commons Attribution License 4.0 license. Download for free at This page titled 2.7: Measures of Spread- Variance and Standard Deviation 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 2.6: Box Plots 2.8: Skewness and the Mean, Median, and Mode Was this article helpful? Yes No Recommended articles 3.2: Measures of VariationAn important characteristic of any set of data is the variation in the data. In some data sets, the data values are concentrated closely near the mean... 3.6: Measures of the Spread of the DataAn important characteristic of any set of data is the variation in the data. In some data sets, the data values are concentrated closely near the mean... 2.8: Measures of the Spread of the DataAn important characteristic of any set of data is the variation in the data. In some data sets, the data values are concentrated closely near the mean... 2.7: Measures of the Spread of the DataAn important characteristic of any set of data is the variation in the data. In some data sets, the data values are concentrated closely near the mean... 9.8: Measures of the Spread of the DataAn important characteristic of any set of data is the variation in the data. In some data sets, the data values are concentrated closely near the mean... Article typeSection or PageAuthorOpenStaxLicenseCC BYLicense Version4.0OER program or PublisherOpenStaxShow TOCno Tags Population Standard Deviation sample Standard Deviation source@ standard deviation © Copyright 2025 Statistics LibreTexts Powered by CXone Expert ® ? The LibreTexts libraries arePowered by NICE CXone Expertand are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. Privacy Policy. Terms & Conditions. Accessibility Statement.For more information contact us atinfo@libretexts.org. Support Center How can we help? Contact Support Search the Insight Knowledge Base Check System Status×
15913	https://www.youtube.com/watch?v=MrXzWiSDqiU	Counting Possible SURJECTIVE (onto) Functions Given Constraints Gresty Academy 57300 subscribers 23 likes Description 492 views Posted: 5 Mar 2024 Occasionally Maths Olympiads and College Entrance Tests ask a question along the lines of 'how many possible SURJECTIVE (onto) functions are there from set A to set B which satisfy certain constraints?'. This type of question is more complex than those asking to count injective or bijective functions (see below playlists for example videos of those) in that we need to use the 'Stirling Number of the Second Kind' when the number of elements in the codomain is fewer than the number of elements in the domain (for surjective/onto functions with same sized domain and codomain it is far easier to consider it as a 'bijective' function question - see below playlist for a video explaining why). In this video we tackle a fairly tricky JEE Main question with a constraint in order to explain how to tackle surjective/onto questions when the number of elements in the codomain is fewer than the number of elements in the domain - note that this question is identical to counting distinct objects shared between distinct people where each gets at least one - see the video for more on that. Counting is not easy - it is quite difficult not to double count or omit possibilities, so care really needs to be taken. There are huge overlaps between surjective function counting and other types of 'counting' - indeed some of the questions are actually identical even though they don't appear so. As such, it may be beneficial to try out some of our other 'Counting' videos in our playlist 'Counting' For more videos on surjective (onto) functions, see our playlist 'Injective (one-to-one), Surjective (onto) and Bijective Functions' For more videos on JEE Main see our playlist 'JEE Aspirant' For more videos on functions generally see our playlist 'Functions' For more videos on Maths Olympiads, see our playlist 'Maths Olympiads' For more videos on College Entrance Tests see our playlist 'UPCAT and Other CETs' 8 comments Transcript: today we're going to be dealing with counting surjective or onto functions okay so um just a quick um uh revisit we have that a function is where every element in the input domain has to match with exactly one element in the output Co domain a surjective or onto function every element in the output is mapped to by at least one element uh in the input so for example let's uh say that there are M elements here in the domain and there are n elements here in the co- domain well if m is less than n let's say for example that uh there are only two elements here well it is impossible uh that a function can be subjective here because there are fewer elements in the domain than there are in the co- domain so it is not possible that these two can map to at least one of those three if m equals n so for example if we have three elements in the domain and three in the co-domain well then actually yes that could be onto we could have a uh like that um but it's actually bjective um as well because um it's subjective and injective and we'll deal with that one in the next video so what we're really interested in um in this particular video is when there are where m is greater than n I.E there are Moree elements in the domain than there are in the co-domain and basically the question always becomes with these extra elements where do they go and how do we count them okay so um anyway let's let's just have a quick look um because in actual fact there are many overlaps between counting surjective functions and Counting the number of ways uh of Distributing M distinct objects among n people so let's just have a a quick look so let's imagine that we have uh here uh uh M or well in this case five distinct objects whatever the objects are um and here we have for example three boxes or three people or whatever well the number uh where each of the person or box must receive at least one object this little puzzle is identical to the surjective problem that we're about to talk about where for example example we have the domain here is a b CDE e and the co-domain is 1 2 3 and we ask um how many functions are there which are surjective I.E where at least uh one of these maps to each one of these so it's they're identical puzzles and to you to solve both of them we use the Sterling number of the second kind and this is the formula for the Sterling number of the second kind which does look a little bit horrific but it's not where this n is the number of elements in the co-domain n and M is the number of elements in the domain and basically if we want to find out how many surjective functions there are between the M elements in the domain and the N elements in the co-domain we have to use the Sterling number of the second kind remembering that M must be greater than n um because otherwise it would be bjective or there cannot be any subjective so um let's just um basically Give an example um so let's imagine um that the are we have a b c and d in our uh domain and we have FG and H in our co- domain and we could say um you know identically we could say these are four distinct objects and we could say these are three people or three boxes it's exactly the same puzzle okay so how would we use the Sterling formula well basically m in this case is going to be four and N in this case is going to be three so what we do is we plug that into the Sterling formula and then we get basically 3 ^ 4 minus 3 C1 2 ^ 4 add 3 C2 1^ 4 and minus I mean we don't need to put this in but I'll put it in anyway just in case anybody's uh wants to to follow for completion and that is equal to the number of subjective functions well 3 4 is 81 - 3 C1 is 3 2 4 and 3 C2 is also 3 1 4 which is 1 and obviously that is 0 which equals 81- 48 add 3 which equals 36 so in this particular example here there are 36 surjective functions from this domain here which has four elements to this co-domain here um which has three elements now in case you're thinking well this this doesn't really make any sense in actual fact it does because we can look at this um and and at least get some vague idea as to why this Sterling formula uh Sterling number works so let's just uh put this back in here so we have a b c and d uh and here we had FG and H okay and the answer was 3 to ^ 4us 3 2 ^ 4 add 3 1 ^ 4 that's the number of surjective functions okay well let's imagine that there was We we forgot about uh subjective functions and there's no restriction how many mappings would there be if they could map to anything well clearly this one can map to any of the three and this one can map to any of the three and this can map to any of the three and this can map to any of the three so that's where we get the 3 to the^ of four from that's if there were no restrictions what happens if one of them is empty let's say f is empty well basically um the probability that one of these goes to here is 2/3 so the probability that they all go um to there is 2/3 to the power of four and so therefore given that we know there are three to the four mappings well that basically means that there are two to the four cases where f is empty and we have to multiply that by three because there could be a case where G is empty and a case where H is empty and that is where we get this bit from and then finally well what we've done is we've actually double counted because in calculating the number of um functions where f is empty we've also calculated the number of functions where f and g are empty so we have to add that back in which is this bit here so it does sort of vaguely make um some sort of uh sense okay so now before we move on to the The The J main question uh there is a fairly common question uh that we can answer by inspection so let's just uh draw a little thing here and let's say that there are M elements in here and let's just say there are two elements here in the codomain um and the question often asks how many surjective functions are there between the domain here and uh the codomain here well we can use the Sterling formula so we would say okay well basically there are 2^ of M minus 2 C1 1 the^ of M which is 2 the M - 2 um which is the correct answer that is how many surjective functions there are between uh this domain and this co- domain but in actual fact if we look at this this is actually obvious because if you think about it doesn't matter how many elements there are in here there are only two posses whereby it's not subjective and that's if all of them map to one or if all of them map to two all of the other ones which which is basically 2 the^ of M functions will be subjective so we could do by inspection that any domain of M elements mapping to two there will always be 2 the N minus 2 that's just a little bit of a Sidetrack it is quite a common uh question in uh in math Olympiad so now let's get on to the J main question anyway here it is um anybody wishes to do it please pause the video now okay so we have basically a b c d and e is our domain and that clearly has m equal to 5 uh which is the number of elements in the domain and that maps to 1 2 3 4 which has n equal four there are four elements uh in the domain and basically the Restriction they're making this even harder they're not just asking us how many surjective functions there are between uh R and S they're actually adding a restriction which does make it more complicated that F cannot map a to one okay so the way we're going to do this this is our plan of attack firstly let's how see how many uh subjectives functions there are ignoring this then let's see how many of the functions have a mapping to one and then we take one from the other and that leaves us with how many subjective functions there are such that F of a is not equal to one okay so first of all let's use the uh the Sterling number and um work out how many functions there are with Nal 5 and Nal 4 well that would be 4 to ^ 5 take away 4 C1 3 to ^ 5 add 4 C2 2 ^ 5 - 4 C3 1 ^ 5 which equals uh 1024 - 4 3 ^ 5 is 2 43 add 4 4 C2 is 4 3 / 2 which is 6 2 5 is 32 take away 4 C 3 is is 4 1 5 is 1 uh and all of that lot 1024 take away um 972 add uh 1 92 take away 4 equals 240 so we know that there are 240 surjective functions from R to S if we don't have this uh con constraint that F of a cannot map uh to one okay now what about when F of a maps to one now we actually have to do this in two steps okay step one um is that let's let's just redraw this again because this is really important so we got a b c d and e and we got here 1 2 3 and four so basically we're trying to find out how many subjective functions there are where a does map to one so either a uh this one is the only element that maps to one which means that we are are left with these four elements must map to these three and that is again a different subjective function that we're going to have to calculate or B one of these B C D or E also maps to one and then we are left with the other three whatever the other three are having a bjective function between whatever BC and uh BC and D for example or 2 three and four now basically as we will look at in the video tomorrow the number of bced functions that there are so let's say uh between a three element and a three element the number of bced functions there are basically is three choices two choices and one choice which is three factorial and we have that either B C D or E could map to one as well as a so that's times four so the number of uh subjective functions where BC D or E also mapped to one as well as a is basically equal to 4 3 factorial which is um uh sorry 4 3 torial which is 24 the second case is where a maps to one but BC D and don't map to one so let's just redraw that bit and we basically now have BC DN mapping to 2 3 and four as a subjective function well again we have to use the Sterling formula but this time we M = 4 and N = 3 uh and basically there we have 3 ^ 4 - 3 2 ^ 4 add 3 1 ^ 4 uh which equals 3 4 is uh well we've already done that one in our example before wasn't it that was 81 take away 48 add three which equals 36 okay so basically we know that just to confirm so we have 36 cases where a maps to one and the others do not map to one and we have 24 cases where a maps to one and one of the other one maps to one as well uh and we have 240 cases in total so therefore the answer to our question which is the total number of on two functions where a does not map to one is equal to 240 takeway 24 takeway 36 which equals 180 um it's not an easy one that uh does need a little bit of uh thinking about now in the next video we are going to look at um bjective functions uh which are quite easy and also we're going to include in that surjective or onto functions where the domain and co-domain are the same size because they are far easier doing it that way than doing it this way okay well I hope you found this useful if you did please like the video and subscribe to the gesia Academy YouTube channel thank you
15914	https://www.youtube.com/watch?v=1uh-7m-SzYk	Graphing Equations by Plotting Points ThinkwellVids 109000 subscribers 113 likes Description 10890 views Posted: 11 Feb 2014 From Thinkwell's College Algebra Chapter 3 Coordinates and Graphs, Subchapter 3.1 Functions and Their Graphs 14 comments Transcript: okay now throughout this course together we've been thinking about a lot of equations and we've seen uh equations with sometimes more than one variable like x's and y's and so forth where in fact one unknown actually leads to the value of another unknown and so two things are actually related somehow maybe they're proportional maybe you have some equation where they're both in it or something or whatever well a lot of times in fact more times than not having a visual image a visual idea about that relationship between two unknowns could be extremely powerful in understanding sort of what's going on and so this really gets into a whole realm of of mathematics which we could think of just as graphing or or visual mathematics and so I want want I now want to begin by just talking about graphing equations so what does it mean to graph an equation in the plane so of course we have the the x- axis and the Y look I'm doing X it's the x-axis here and the Y AIS here and we want to do is go over some number of X values and then up or down some number of yv values and we can graph things and we can start to get a picture image for what something is looking like okay and the relationship so now I want to illustrate this by looking at a variety of really important types of equations and the way we'll start to graph them just for for now anyway is just to sort of see some points and see how the relationship goes for example suppose I have that y = 2x + 1 so what that means is that for this thing to be satisfied I have to find x's and y's that make this true so one way to do that is to make a little table of possibilities so for example what I could do here is make a little chart and put like X values here and then figure out the corresponding y values so I could pick for example some negative values maybe like -2 -1 zero one two and so forth just to get some values and now for each of these X values I'll plug that value into here for x and find out what the associated y value would be for example if I plug in -2 here I see -2 2 is -4 + 1 would be -3 here when I plug in -1 for X I would see 2 1 which would be -2 + 1 would be1 if I plug in a zero here I would just see that plus one if I plug in a 1 here I'd see 2 + 1 which is three if I plug in a two in here I would see 2 2 is 4 + 1 is 5 so now I have a chart of points you can think of these as ordered pairs -2 comma -3 -1 comma -1 0 comma 1 1 comma 3 2 comma 5 and I can actually plot those points now on a graph and we can see what this thing looks like so let's try to do that right now so here's the y- axis here's the x-axis and so what what do we see here -2 comma minus 3 so let's say this is min-2 1 2 that's -2 in the x minus 3 down 1 2 3 so I put a dot right there that's -2 comma -3 then I have -1 comma min-1 -1 comma minus one put a dot there then I have zero so that's just on the y- axis if x is zero we're on the y- AIS one then I have 1 three so I go one over and three up 1 2 3 oops one two three and then I have two one two and I go up five one 2 3 four five so one thing I can do is just connect those points and see what this thing visually looks like and if I connect those points what you see is what looks like a straight line a straight line seems to pass very well right through all those points so in fact this kind of object where you see an X and A Y all to the first Power like that will represent something that graphically looks like a straight line and I want us to all start getting in the habit of looking at this and just thinking Some Kind of straight line I don't know maybe know what kind there are a lot of different straight lines this straight line turns out to be the one associated with this so there's our first really basic kind of equation it's a straight line okay let's take a look at another one together okay the next one I want to take a look at is the following how about y = 3 x^2 and again I'll make a table just to see what these values look like as I change x what happens to Y so I'll pick let's say -2 -1 0o 1 and two you can pick other points too I'm just picking these just to illustrate what's going on so for each of these X values I have to square it multiply by 3 so - 2^ 2 is -4 new - 2^ 2 is actually 4 3 is 12 - 1 2 is 1 3 is 3 0 2ar is 0 3 is 0 1 2ar is 1 3 is 3 2^ 2 is 4 3 is 12 so now I have all these values and let's see what happens if we plot them so if we plot them I have minus two so one 2 in the negative Direction 12 uhoh 1 2 3 4 5 6 7 8 9 10 11 12 so this point right here is the point Min - 2 comma 12 and then -1 comma 3 so minus one comma 3 1 2 3 look how steep that is it really drops really drops a lot pretty dramatic 0 0 so then I go right to the origin 1 three so then I go one over and three up and then 22 is way up here so it sort of comes down and then goes up look how symetric by the way those points are in fact they line up perfectly like a little mirror right across these things sort of interesting okay so what would the graph what would it look like if you sort of connect those points well if we connect those points I'd have to sort of Bend this thing up to fit them and it sort of looks like this doesn't it sort of look like this and actually this is an example of a parabola and in fact whenever you see an equation like this where one of the variables just is to the first power but the other variable is a square then you're going to have a parabola and and it turns out that if this coefficient here is very large the parabola will be very very tight it'll be very very tight like this is very very tight if that number were smaller the the the effect would be to make this thing a little bow out more and if it's really small it would even go like this and the parabola would become very wide and so forth if that that number increases then the parabola tightens up like this and wants to close up like that so in fact this is a great example of a parabola and you can see I found that just by plotting points okay let's try another one okay uh I wish I had that back let me bring this back up here because the next one is so close to this one I wanted you to see the difference so the next one is actually this Y = 3x^ 2 + 1 so in fact it looks like the exact same equation as this but I just added one to the to the end here and in fact what I'll do is I'll just make my chart in Black so you know the black corresponds to the black equation and this purple corresponds to the purple stuff here and again I'll put down - 2 -1 0 1 2 if I put in minus two in here and square it and multiply by 3 I get 12 but now I have to add one so I have 13 minus one in here produces a 1 3 + 1 is 4 zero in here makes a 0 + 1 is 1 one in here makes 3 + 1 is four and then two in here makes a 12 + 1 is 13 you'll notice that in fact these values are exactly the same as the Y values but they've all been increased by 1 12 + 1 is 13 3 + 1 is 4 0 + 1 is and that makes sense because in each case all I did was take the old answer but increase the Y by one so each of these values are one more than their counterparts so what would happen if I graph that if I graph that what I would see is every point would just go up by one you see so this would be up by one unit this would be up by one unit this would be up by one unit up by one unit and up by one unit so if I looked at the graph of that it would would be the exact same graph that we had before this was the graph we had before let me remind you but now I have to move it up by one Watch What [Music] Happens exact same graph but just shift it up by one so notice that it's still a parabola it's just that now it moved up a little bit because I added one to the very end without adding one I get that I add one I get that what do you think would happen if I put a two in there that's right it would just go up by two see that is what if I had a minus one at the very end then a minus one would actually instead of being here I'd push down by one so you can see this by adding how the effect it has on a graph let's try another example x equal y^ 2 let's make a table here X and Y now notice something here if I start to plug in a value like I put in two here it's going to be sort of hard to figure out what the Y's are if I put in a three here okay what's that it would be easier to actually put the Y values in and then fill in the X values so let me actually do that I'll put in -2 minus1 0 1 and two here these are y values and I now want to see the corresponding x value because it's easier if I put in a minus two for y if I Square it I see four - 1^ 2ar is 1 0 SAR is z 1 squ is 1 2 squ is four so now I have these points let's see what they look like so here's the axis and let's graph graph away here we go so I have four in the X 1 2 3 4 and I go -2 in the Y one two so I put a point right there then I have one minus one so I put a point right here then I have 0 0 so I put a point right there then I have 1 comma 1 so I put a point right there and then I have four 4 comma 2 1 2 so I put a point right there so if we connect those things what do we see well we actually see a curve that looks sort of like this you see and again we see a shape that looks like a parabola this Parabola curve and notice something interesting in the previous example where the parabola went like this notice that was a case where we had the X squar now we have the Y squ and notice the parabola is going in this direction and that's not a fluke in fact if you have y squs the parabola will either be like this or like this whereas if we have X squs and the Y is just appearing to the first Power the parabas would be either up or down so parabas with y^2 will look sort of like this or like this and with x squs the parabis will look like this or like this so it's sort of neat in this case this Parabola looks just like that sort of a sideways Parabola okay I thought I'd do one last one with you and this last one is y equals the absolute value of x so let's make a table of points here X and Y and see what happens Min - 2 - 1 0 1 2 so we're just taking absolute value of x so absolute value of -2 that's just two absolute value of negative 1 is one absolute value of zero is zero absolute value of one is one two is two that was pretty fast what does this look like minus 2 1 two and I go up two 1 2 minus one 1 and 0 0 looks like a nice straight line doesn't it but let's see what happens now 1 one uhoh so much for the straight line two two so it looks like almost sort of a parabola like thing but in fact it's not it goes straight down like this and then comes straight out so in fact the look is something like this it comes straight down and then goes straight out so it makes this sort of sharp V here and this is what the absolute value equation looks like it usually is made up of two wings and they're both very sharp like this one goes straight down one goes straight up like that these are some very very basic equations where we're getting a sense of what the picture of these relationships look like by making a little table and graphing up next we'll start learning how to graph things in a slightly more sophisticated way but for now visual with each picture with each with each equation sorry see you soon
15915	https://files.ele-math.com/articles/mia-12-21.pdf	Mathematical Inequalities & Applications Volume 12, Number 2 (2009), 255–264 A NEW REFINED JORDAN’S INEQUALITY AND ITS APPLICATION RAVI P. AGARWAL, YOUNG-HO KIM AND S. K. SEN (Communicated by R. Verma) Abstract. New refined lower and upper bound forms of Jordan’s inequality are proved. As an application, the lower bound form is shown to improve L. Yang’s inequality that plays a pivotal role in the theory of distribution of values of functions. Some numerical results are included. 1. Introduction The celebrated Jordan’s inequality states that if x ∈(0, π/2], then 2 π ⩽sin x x < 1, (1.1) where the left-hand side inequality becomes equality if and only if x = π/2 . Jordan’s inequality and its subsequent refinements are useful in several mathemat-ical areas such as calculus and trigonometry, where specifically the applications of the theory of limits are involved. These are important tools in approximating Riemann zeta function ζ(x) , in improving Yang Le’s inequality and its generalization which play an important role in the theory of distribution of values of functions [6, 9, 10]. During the past few years many authors [1, 4, 5, 7, 8] established several refined forms of Jordan’s inequality. One of these forms was due to ¨ Ozban in 2006 . He obtained a new lower bound for the function sin x x . He showed that 2 π + 1 π3 (π2 −4x2) + 4(π −3) π3 x −π 2 2 ⩽sin x x , x ∈ 0, π 2 (1.2) where the equality holds if and only if x = π/2 . Almost at the same time, a new interesting refined form of Jordan’s inequality was established by Zhu . He proved the following theorem (Theorem 1.1). THEOREM 1.1. If 0 < x ⩽π/2 , then 2 π +π2−4x2 π3 +4(π−3) π3 x−π 2 2 ⩽sin x x ⩽2 π +π2−4x2 π3 +12−π2 π3 x−π 2 2 , (1.3) where both the inequalities become equalities if and only if x = π/2 Furthermore, 4(π−3) π3 and 12−π2 π3 are the best constants in (1.3). Mathematics subject classiﬁcation (2000): 26D15. Keywords and phrases: distance function, Jordan’s inequality, Taylor’s formula, Yang’s inequality. c ⃝ , Zagreb Paper MIA-12-21 255 256 RAVI P. AGARWAL, YOUNG-HO KIM AND S. K. SEN The main aim of the present paper is to establish one more new refined lower bound form as well as one more new upper bound form of Jordan’s inequality. These forms are illustrated to demonstrate their usefulness in the background of most recent contributions. 2. Main results Our main result, viz., the new refined inequality, follows from Theorem 2.1 below. THEOREM 2.1. If 0 < x ⩽π/2 , then 1 −B1x −B2x2 −B3x3 ⩽sin x x , (2.1) where the equality holds if and only if x = π/2 , where B1 = 4 π2 (−66 + 43π −7π2), B2 = 4 π3 (124 −83π + 14π2), B3 = 4 π4 (12 −4π). Proof. Define a function f : (0, π/2] →R by f (x) = x −sin x x2 −B1 −B2x −B3x2. (2.2) Then, we have f (4)(x) = 1 x6 (24x + 96x cos x + 36x2 sin x −8x3 cos x −x4 sin x −120 sin x). (2.3) Now consider the function g : [0, π/2] →R defined by g(x) = x6f (4)(x). Then, we get g′′(x) = x4 sin x. Clearly, g′′(x) > 0 for all x ∈(0, π/2], this implies that g′(x) is strictly increasing on (0, π/2]. Using the equality g′(0) = 0 , we find g′(x) > 0 for all x ∈(0, π/2]. Thus, g(x) is strictly increasing with g(0) = 0 , it follows that g(x) > 0 for all x ∈(0, π/2]. Now combining the function g and the equality (2.3), we obtain f 4(x) > 0 for all x ∈(0, π/2]. Hence f ′′′(x) is strictly increasing and f ′′′(x) < 0 on x ∈(0, π/2]. On the other hand, using the Taylor’s formula, we have for ξ ∈(x, π/4), f (x) = f π 4 + f ′π 4 x −π 4 + f ′′(π/4) 2 x −π 4 2 + f ′′′(ξ) 3! x −π 4 3 ⩽f π 4 + f ′π 4 x −π 4 + f ′′(π/4) 2 x −π 4 2 + lim ξ→0 f ′′′(ξ) 3! x −π 4 3 = f π 4 + f ′π 4 x −π 4 + f ′′(π/4) 2 x −π 4 2 −(x −π/4)3 120 , (2.4) A NEW REFINED JORDAN’S INEQUALITY AND ITS APPLICATION 257 where x ∈(0, π/4). Now define an auxiliary function p1 : [0, π/4] →R by p1(x) = f π 4 + f ′π 4 x −π 4 + f ′′(π/4) 2 x −π 4 2 −(x −π/4)3 120 . It is obvious that f (x) ⩽p1(x) and p1(x) is continuous with p′′′ 1 (x) = −1/20 < 0, x ∈ [0, π/4]. This implies that p′′ 1 (x) is strictly decreasing on [0, π/4]. Hence p′′ 1 (x) > 0 for all x ∈[0, π/4] with p′′ 1(π/4) > 0 . Therefore, p′ 1(x) is strictly increasing with p′ 1(π/4) > 0 and p′ 1(0) < 0 . Now consider the fixed point x1 ∈[0, π/4] such that p′ 1(x1) = 0 . Notice that p′ 1(x) > 0 for all x ∈(x1, π/4], this implies that p1(x) is strictly increasing on (x1, π/4]. Since p1(π/4) = f (π/4) < 0 , we have p1(x) < 0 for all x ∈(x1, π/4]. Notice that p′ 1(x) < 0 for all x ∈[0, x1). This implies that p1(x) is strictly decreasing on x ∈[0, x1). Since p1(0) < 0 , we have p1(x) < 0 for all x ∈[0, x1). Therefore, p1(x) < 0 for all x ∈[0, π/4]. Thus from the inequality (2.4) and the function p1(x), we get f (x) ⩽0 for all x ∈[0, π/4]. Now we consider the case x ∈[π/4, π/2] below. Using once again the Taylor’s formula for ξ ∈(x, π/2), we obtain f (x) = f π 2 + f ′π 2 x−π 2 + f ′′(π/2) 2 x−π 2 2 + f ′′′(ξ) 3! x−π 2 3 ⩽f π 2 + f ′π 2 x−π 2 + f ′′(π/2) 2 x−π 2 2 + f ′′′(π/4) 3! x−π 2 3 , (2.5) where x ∈[π/4, π/2]. Define an auxiliary function p2 : [π/4, π/2] →R by p2(x) = f π 2 + f ′π 2 x −π 2 + f ′′(π/2) 2 x −π 2 2 + f ′′′(π/4) 3! x −π 2 3 . It is obvious that f (x) ⩽p2(x) and p2(x) is continuous with p′′′ 2 (x) = f ′′′(π/4) < 0 . This implies that p′′ 2 (x) is strictly decreasing on [π/4, π/2] with p′′ 2 (π/4) > 0 and p′′ 2 (π/2) < 0 . Let the fixed point x2 ∈[π/4, π/2] be such that p′′ 2 (x2) = 0 . Notice that p′′ 2 (x) > 0 for all x ∈[π/4, x2). This implies that p′ 2(x) is strictly increasing and p′ 2(x) > p′ 2(π/4) > 0 for all x ∈[π/4, x2). Therefore p2(x) is strictly increasing on the semi-open interval -[π/4, x2). From the value p2(x2) = f ′π 2 x2 −π 2 + 1 2 x2 −π 2 2 k′′(x2) −1 3 x2 −π 2 3 f ′′′π 4 = f ′π 2 x2 −π 2 −1 3 x2 −π 2 3 f ′′′π 4 < 0, where x2 = π 2 −f ′′( π 2 ) f ′′′( π 2 ), f ′π 2 = −8(66 −43π + 7π2) π3 , 258 RAVI P. AGARWAL, YOUNG-HO KIM AND S. K. SEN f ′′′π 4 = 8(1536 √ 2 −96(2 + 3 √ 2)π −24 √ 2π2 + √ 2π3 π5 , we have p2(x) < 0 for all x ∈[π/4, x2). Observe that p′′ 2 (x) < 0 for all x ∈(x2, π/2]. This implies that p′ 2(x) is strictly decreasing and p′ 2(x) > p′ 2(π/2) > 0 for all x ∈ (x2, π/2). Thus p2(x) is strictly increasing with p2(x) < p2(π/2) = 0 . Consequently p2(x) ⩽0 for all x ∈[π/4, π/2]. Now combining the inequality (2.5) and the function p2(x), we get f (x) ⩽0 for all x ∈[π/4, π/2]. Hence, we obtain f (x) ⩽0 for all x ∈(0, π/2]. Now, multiplying f (x) by x we get the desired result. □ REMARK 2.1. A graph of the distance function y(x) = h1(x) −h(x), where h(x) = (sin x)/x , h1(x) = [1 −B1x −B2x2 −B3x3], (2.6) is given in Fig. 1. 0 0.25 0.5 0.75 1 1.25 1.5 -0.0025 -0.002 -0.0015 -0.001 -0.0005 0 y y=h1- h Fig. 1 A graph of the error represented by the function e(x) = h1(x) −h2(x), where h2(x) = 2/π + (π2 −4x2)/π3 + 4(π −3)(x −π/2)2/π3 (2.7) is given in Fig .2. These imply that h2(x) ⩽h1(x) ⩽(sin x)/x , where the equalities hold if and only if x = π/2 . Thus, the inequality (2.1) is a new refined form of Jordan’s inequality for all x ∈(0, π/2]. 0 0.25 0.5 0.75 1 1.25 1.5 0 0.002 0.004 0.006 0.008 0.01 0.012 e e=h1- h2 Fig. 2 A NEW REFINED JORDAN’S INEQUALITY AND ITS APPLICATION 259 THEOREM 2.2. If x ∈(0, π/2], then sin x x ⩽1 −C1x + C2x2 −B3x3 (2.8) where the equality holds if and only if x = π/2 , where C1 = 4(−75 + 49π −8π2) π2 , C2 = 4(−142 + 95π −16π2) π3 , and B3 is as deﬁned in Theorem 2.1. Proof. Define a function f : (0, π/2] →R by f (x) = x −sin x x3 −C1 1 x + C2 −B3(π)x. (2.9) Then, we have f (4)(x) = 24x[300x −196πx + π2(5 + 32x)] −12π2x(−20 + x2) cos x π2x7 −π2(360 −72x2 + x4) sin x π2x7 . (2.10) Now consider a function g : [0, π/2] →R defined by g(x) = x7f (4)(x). Then, we get g′′′(x) = x4 cos x. Clearly, g′′′(x) > 0 for all x ∈(0, π/2], this implies that g′′(x) is strictly increasing on (0, π/2]. Using the equality g′′(0) = 192(75−49π+8π2) π2 , we find g′′(0) > 0 . This implies that g′′(x) > 0 for all x ∈(0, π/2]. Hence g′(x) is strictly increasing with g′(0) = 0 , implying g′(x) > 0 for all x ∈(0, π/2]. Therefore, g(x) is strictly increasing with g(0) = 0 . Thus g(x) > 0 for all x ∈(0, π/2]. Now combining the function g and equality (2.10), we obtain f 4(x) > 0 for all x ∈(0, π/2]. Hence f ′′′(x) is strictly increasing and f ′′′(x) < 0 on x ∈(0, π/2]. On the other hand, using the Taylor’s formula for ξ ∈(x, π/4), we have f (x) = f π 4 + f ′π 4 x−π 4 + f ′′(π/4) 2 x−π 4 2 + f ′′′(ξ) 3! x−π 4 3 ⩾f π 4 + f ′π 4 x−π 4 + f ′′(π/4) 2 x−π 4 2 + f ′′′(π/4) 3! x−π 4 3 , (2.11) where x ∈(0, π/4). Now define an auxiliary function k1 : [0, π/4] →R by k1(x) = f π 4 + f ′π 4 x −π 4 + f ′′(π/4) 2 x −π 4 2 + f ′′′(π/4) 3! x −π 4 3 . It is obvious that f (x) ⩽k1(x) and the function k1(x) is continuous with k′′′ 1 (x) = f ′′′(π/4) < 0 , x ∈(0, π/4], this implies that k′′ 1 (x) is strictly decreasing on (0, π/4]. 260 RAVI P. AGARWAL, YOUNG-HO KIM AND S. K. SEN Clearly k′′ 1 (π/4) = f ′′(π/4) > 0 for all x ∈(0, π/4]. Hence k′′ 1 (x) > 0 for all x ∈(0, π/4]. That is, k′ 1(x) is strictly increasing with k′ 1(π/4) < 0 . So, we have k′ 1(x) < 0 , x ∈(0, π/4]. This implies that p(x) is strictly decreasing on (0, π/4]. Since k1(π/4) = f (π/4) > 0 , k1(x) > 0 for all x ∈(0, π/4]. Now combining the inequality (2.11) and the function k1(x), we get f (x) ⩾0 for all x ∈(0, π/4]. Consider the case x ∈[π/4, π/2] below. Using once again the Taylor’s formula for ξ ∈(x, π/2), we obtain f (x) = f π 2 + f ′π 2 x−π 2 + f ′′(π/2) 2 x−π 2 2 + f ′′′(ξ) 3! x−π 2 3 ⩾f π 2 + f ′π 2 x−π 2 + f ′′(π/2) 2 x−π 2 2 + f ′′′(π/2) 3! x−π 2 3 , (2.12) where x ∈[π/4, π/2]. Define an auxiliary function k2 : [π/4, π/2] →R by k2(x) = f π 2 + f ′π 2 x −π 2 + f ′′(π/2) 2 x −π 2 2 + f ′′′(π/2) 3! x −π 2 3 . Evidently, f (x) ⩽k2(x) and k2(x) is continuous with k′′′ 2 (x) = f ′′′(π/4) < 0 . This impliesthat k′′ 2 (x) isstrictly decreasing on [π/4, π/2] with k′′ 2 ( π 4 ) = 4(2664−1888π+331π2) π5 < 0 . Thus we have k′′ 2 (x) < 0, x ∈[π/4, π/2]. Consequently, k′ 2(x) is strictly decreasing and we get k′ 2 π 4 = −6168 −4216π + 717π2 2π4 > 0, k′ 2 π 2 = −16(75 −49π + 8π2) π4 < 0. Now let the fixed point x3 ∈[π/4, π/2] be such that k′ 2(x3) = 0 . Notice that k′ 2(x) > 0 for all x ∈[π/4, x3), which implies that k2(x) is strictly increasing and we obtain k2(x) > k2(π/4), x ∈[π/4, x3) where k2(π/4) = 4024−2704π+453π2 8π3 > 0 . Therefore k2(x) > 0 on [π/4, x2). Observe that k′ 2(x) < 0 for all x ∈(x3, π/2]. This implies that k2(x) is strictly decreasing and we obtain k2(x) > k2(π/2), x ∈(x3, π/2], where k2(π/2) = f (π/2) = 0 . Therefore k2(x) ⩾0 on x ∈(x3, π/2]. Consequently, k2(x) ⩾0, x ∈[π/4, π/2]. Combining the inequality (2.12) and the function k2(x), we get f (x) ⩾0 for all x ∈[π/4, π/2]. Hence, we obtain f (x) ⩾0 for all x ∈(0, π/2]. Now, multiplying f (x) by x2 we get the desired result. □ A NEW REFINED JORDAN’S INEQUALITY AND ITS APPLICATION 261 REMARK 2.2. A graph of the distance function y(x) = h3(x) −h(x), where h(x) = (sin x)/x , h3(x) = 1 −C1(π)x + C2(π)x2 −B3(π)x3, (2.13) is given in Fig. 3. 0 0.25 0.5 0.75 1 1.25 1.5 0 0.0002 0.0004 0.0006 0.0008 0.001 y y=h3- h Fig. 3 A graph of the error represented by the function e(x) = h3(x) −h4(x), where h4(x) = 2 π + 1 π3 (π2 −4x2) + 12 −π2 π3 x −π 2 2 , (2.14) is given in Fig. 4. This implies (sin x)/x < h3(x) < h4(x), x ∈(0, 1.2739) and (sin x)/x ⩽h4(x) ⩽h3(x), x ∈(1.2739, π/2) where the equalities hold if and only if x = π/2 . Thus, the inequality (2.8) is one refined form of Jordan’s inequality for all x ∈(0, π/2]. 0 0.25 0.5 0.75 1 1.25 1.5 -0.08 -0.06 -0.04 -0.02 0 e e=h3- h4 Fig. 4 3. Application Yang’s inequality and its generalization play an important role in the theory of distribution of values of functions [6,9]. In this section, using the inequality (2.1), we show that our result can be used to improve Yang’s inequality. 262 RAVI P. AGARWAL, YOUNG-HO KIM AND S. K. SEN THEOREM 3.1. Let n ⩾2 be a natural number, Ai > 0(i = 1, 2, ..., n) with n i=1 Ai ⩽π and 0 ⩽λ ⩽1 . Then U(λ) ⩽ 1⩽i<j⩽n sin2 λπ ⩽ 1⩽i<j⩽n Hij ⩽ 1⩽i<j⩽n λ 2π2, (3.1) where Hij = cos2 λAi + cos2 λAj −2 cos λAi cos λAj cos λπ , and U(λ) = n(n −1) 2 λ 2[B(λ; π)]2 cos2 λ 2 π with B(λ; π) = π + 2(66 −43π + 7π2)λ −(124 −83π + 14π2)λ 2 + 2(π −3)λ 3 . Proof. Substituting x = λπ/2 in (2.1), the inequality can be written as sin λ 2 π ⩾ λ 2 [B(λ; π)]. Hence λ 2 [B(λ; π)] ⩽sin λ 2 π ⩽λ 2 π or λ 2 4 [B(λ; π)]2 ⩽sin2 λ 2 π ⩽λ 2 4 π2. (3.2) On the other hand, using the inequality sin2 λπ ⩽Hij ⩽4 sin2 λ 2 π, (3.3) and noting sin2 λπ = 4 sin2 λ 2 π cos2 λ 2 π and the inequality (3.2), we get λ 2[B(λ; π)]2 cos2 λ 2 π ⩽sin2 λπ ⩽Hij ⩽λ 2π2. (3.4) Let 1 ⩽i < j ⩽n . Introducing the summation in the inequality (3.4), we obtain U(λ) ⩽ 1⩽i<j⩽n sin2 λπ ⩽ 1⩽i<j⩽n Hij ⩽ 1⩽i<j⩽n λ 2π2, (3.5) where U(λ) = 1⩽i<j⩽n λ 2[B(λ; π)]2 cos2 λ 2 π . Hence the theorem. □ REMARK 3.1. It has been shown that S(λ) ⩽ 1⩽i<j⩽n sin2 λπ , where S(λ) = n(n −1) 2 λ 2[π + (6 −2π)λ + (π −4)λ 2]2 cos2 λ 2 π . On the other hand, for 0 ⩽λ ⩽1 we have (6 −2π)λ + (π −4)λ 2 ⩽B(λ; π). (3.6) A NEW REFINED JORDAN’S INEQUALITY AND ITS APPLICATION 263 From (3.6), it can be readily seen that S(λ) ⩽U(λ) ⩽ 1⩽i<j⩽n sin2 λπ ⩽ 1⩽i<j⩽n Hij (3.7) which shows that the inequality (2.1) is a strengthened version of the inequality given by ¨ Ozban in for λ ∈[0, 1]. In conclusion, we present Table 1, which enables us to compare the numerical values of lower bounds S(λ) and U(λ) and those of 1⩽i<j⩽n sin2 λπ for some values of n and λ . λ S(λ) U(λ) 1⩽i<j⩽n sin2 λπ 0.25 7.28441 7.47568 7.5 0.5 14.54708 14.9125 15.00000 n = 6 0.624 12.5343 12.7678 12.8365 0.8 5.13029 5.16792 5.182372 0.95 0.3667546 0.366948 0.3670761 0.25 92.2692 94.6919 95.000 0.5 184.2630 188.892 190.0000 n = 20 0.624 158.768 161.726 162.596 0.8 64.98368 65.4604 65.64338 0.95 4.645558 4.64801 4.649630 Table 1. Comparison S(λ) , U(λ) and 1⩽i<j⩽n sin2 λπ R E F E R E N C E S L. DEBNATH, C.-J. ZHAO, New strengthened Jordan’s inequality and its applications, Appl. Math. Lett. 16 (4), (2003) 557–560. Q. M. LUO, Z. L. WEI, F. QI, Lower and upper bounds of ζ(3) , RGMIA Research Report Collection 4, (2001) 565–569. D. S. MITRINOVIC, Integral Analytic Inequalities, Springer-Verlag, 1970. A. Y. ¨ OZBAN, A new reﬁned form of Jordan’s inequality and its applications, Appl. Math. Lett. 19, (2006) 155–160. S. H. WU, On generalizations and reﬁnements of Jordan type inequality, RGMIA Research Report Collection 7, (2004) Supplement, Article 2. L. YANG, Distribution of Values and New Research, Science Press, Beijing, 1982 (in Chinese). L. ZHU, Sharpening Jordan’s inequality and the Yang Le inequality, II, Appl. Math. Lett. 19, (2006) 990–994. L. ZHU, Sharpening of Jordan’s inequality and its applications, Math. Ineq. Appl. Vol. 9, No. 1, (2006) 103–106. 264 RAVI P. AGARWAL, YOUNG-HO KIM AND S. K. SEN C.-J. ZHAO, Generalization and strengthen of Yang Le inequality, Mathematics in Practice and Theory 30 (4), (2000) 493–497. C.-J. ZHAO, On several new inequalities, Chinese Quarterly Journal of Mathematics No. 16 (2), (2001) 42–46. (Received November 29, 2007) Ravi P. Agarwal Department of Mathematical Sciences Florida Institute of Technology 150 West University Boulevard Melbourne, FL 32901–6975 USA e-mail: agarwal@fit.edu Young-Ho Kim Department of Applied Mathematics Changwon National University Changwon, Kyeong–Nam 641–773 Republic of Korea e-mail: yhkim@changwon.ac.kr S. K. Sen Department of Mathematical Sciences Florida Institute of Technology 150 West University Boulevard Melbourne, FL 32901–6975 USA e-mail: sksen@fit.edu Mathematical Inequalities & Applications www.ele-math.com mia@ele-math.com
15916	https://www.tsc.fl.edu/media/divisions/learning-commons/resources-by-subject/math/foundational-math/inequalities-/Solving-Inequalities-in-the-Form-x--a-less-than-b-and-ax-less-than-b.pdf	Solving Inequalities in the Form x + a < b and ax < b Inequalities are different from equations (or equalities). The difference is in the solutions or solution sets. If we restrict ourselves to linear equations and inequalities in one variable, we can compare the solutions in the following way. x = 2 -- this is an equality -- the left side of the equation has the same value as the right side. -- x has only one value -- x can only be equal to 2 x > 2 -- this is an inequality -- the left side is greater than the right side -- x has an infinite number of values -- x can be any number greater than 2 x ≥ 2 -- this is an inequality -- the left side may be equal to the right side but it may also be greater than the right side -- x has an infinite number of values -- x can be equal to 2 or any number greater than 2 x < 2 -- this is an inequality -- the left side is less than the right side -- x has an infinite number of values -- x can be any number less than 2 x ≤ 2 -- this is an inequality -- the left side may be equal to the right side or it may be less than the right side -- x has an infinite number of values -- x can be equal to 2 or any number less than 2 The solution sets are as follows: x = 2 equality 0 1 2 Solution set is 2, which is shown as one point on the number line. x > 2 • ( 2 0 1 2 because 2 is not a solution. x ≥ 2 0 1 2 because 2 is a solution. x < 2 inequalities 0 1 2 ) because 2 is not a solution. x ≤ 2 0 1 2 Solution set is 2 and all numbers less than 2. Use ] because 2 is a solution. To graph the solution set of an inequality, we find the point on the number line where the solution set begins. If the number itself is included in the solution set ( ≤ or ≥ ) we put a bracket, with an arrow going in either the positive or negative direction. The direction of the arrows will depend upon the inequality. If it is “≤” it will go in the negative direction and if it is “≥” it will go in the positive direction. If the number itself is not in the solution set we put a parenthesis at the number. The direction of the arrow and the direction of the bracket or the parenthesis will depend upon whether the inequality is “<” or “>”. We solve an inequality in much the same way we do an equation. The Addition Property of Inequalities allows us to add the same number to both sides of an inequality without changing the solution set. 5 > 2 the left side is greater than the right side. 5 + 4 > 2 + 4 9 > 6 the left side is still greater than the right side. -8 < 4 the left side is less than the right side. -8 + (-2) < 4 + (-2) -10 < 2 the left side is still less than the right side. EXAMPLE: Solve: x - Solution set is all numbers greater than 2. Use ( [ Solution set is 2 and all numbers greater than 2. Use [ ) Solution set is all numbers less than 2. Use ] 3 x - Add to both sides. x + 0 ≤ 10 8 x ] EXAMPLE: x + 7 > -4 x + 7 + (-7) > -4 + (-7) Add (-7) to both sides. x + 0 > -11 x > -11 ( -11 The Multiplication Property of Inequalities allows us to multiply an inequality by the same positive number without changing the solution set. 6 > 2 <----the left side is greater than or equal to the right side 3(6) > 2(3) <--the left side is still greater than or equal to the right side The Multiplication Property of Inequalities also says that if we multiply both sides of an inequality by the same negative number we must reverse the inequality symbol so that the solution set stays the same. 5 < 8 <--- the left side is less than the right side ↓ 5(-2) > 8(-2) Notice the inequality symbol was reversed at the time each side was multiplied by the same negative number. -10 > -16 <--the left side is no longer less than the right side We must reverse the symbol to make the statement true. SOLVE: 3x > -9 The coefficient of x is positive. · 3x > -9 · Multiply both sides by positive. x > -3 Keep the same symbol. ( 4 -3 EXAMPLE: Solve: (-3/4)x> 12 (The coefficient of x is negative) (-4/3)(-3/4)x < 12 (-4/3) Multiply both sides by -¾ and reverse the symbol x < -16 ) -16 NOTE: We only reverse the inequality symbol if the coefficient of the variable term is negative. EXERCISES: Solve and graph the solution set. 1. -6x ≤ 18 2. y + 3x ≤ 6 3. - 7 4. x + 5 ≥ 3 2y 5. > 4 4 6. x - 3 > -2 3x 7. - < 8 5x 8. 6 < -20 (Answers on next page) KEYS: 1. x ≥ -3 [ 5 -3 2. y < ) 3. 4. 5. 6. x ≥ -14 x ≥ -2 y > 8 x > 1 [ -14 [ -2 ( 8 ( 1 7. x > - ( - 8. x < -24 ) -24
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Richiedi un preventivo su misura per le tue esigenze +39 0514593138ContattaciRichiedi un preventivo I servizi del Laboratorio Metrologico Kiwa, accreditato LAT n°052 da ACCREDIA, ente di accreditamento Italiano, si rivolgono a tutte le aziende dell'industria manifatturiera, con particolare attenzione ai settori dell'Automotive&Racing, dell'Aerospace e della Difesa, del Medicale e del Packaging, che necessitano di conoscere e controllare la qualità - riconosciuta e riferibile - delle misure dei loro strumenti. I blocchetti di riscontro pianparalleli sono definiti, per quanto riguarda le caratteristiche dimensionali e qualitative, dalla norma ISO 3650, che fornisce anche i principali termini e definizioni. La norma ISO 3650 definisce il blocchetto di riscontro come un parallelepipedo retto, costruito con materiale resistente all’usura, con due facce di misura piane e tra loro parallele. La caratteristica dei blocchetti pianparalleli è che le loro facce di misura hanno superfici con un grado di finitura tale che possono aderire alle facce di misura di altri blocchetti pianparalleli oppure a superfici della stessa qualità (questa proprietà è chiamata “adesione”). La taratura dei blocchetti di riscontro pianparalleli si esegue per confronto con altri campioni che riportano un’accuratezza migliore. Kiwa Italia è in grado di offrire la taratura con emissione del Certificato di taratura ACCREDIA per blocchetti di riscontro in qualunque materiale,con lunghezza nominale fino a 1000 mm. 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15918	https://artofproblemsolving.com/wiki/index.php/P-adic_valuation?srsltid=AfmBOoo1bWkZA3XEBWdxsGOIixAZC6rpC0NyjlOXpWpsDYBkCrOwYzb4	Art of Problem Solving p-adic valuation - 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 Wikip-adic valuation Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search p-adic valuation The title of this article has been capitalized due to technical restrictions. The correct title should be -adic valuation. For some integer and prime, the -adic valuation of n, denoted , represents the largest power of which divides. In other words, it is the value of the exponent of in the prime factorization of . Contents [hide] 1 Basic Examples 2 Properties 3 Extension to Rational Numbers 4 See Also Basic Examples . . . Properties For positive integers and , This property follows from the fact that . Furthermore, This follows because we can factor out copies of from the sum . Note that equality holds if , because, in this case, after factoring out copies of from the sum , the remaining factor cannot be congruent to modulo, because one of the terms will be congruent to , while the other will not (because all common factors of have already been factored out). If is a positive integer, because , we deduce that because logarithms are monotone increasing for all bases greater than , which includes all primes. Lifting the Exponent: A series of identities, among which the most prominent is: for odd primes if . Legendre's Formula: . Extension to Rational Numbers is defined to be infinite. Furthermore, as seen in the properties above, From this inspiration, we can define fractional inputs as follows: Note that it does not matter if is simplified or not, because ν p(k x k y)=ν p(k x)−ν p(k y)=(ν p(k)+ν p(x))−(ν p(k)+ν p(y))=ν p(x)−ν p(y)=ν p(x y). See Also Lifting the Exponent Legendre's Formula p-adic number This article is a stub. Help us out by expanding it. Retrieved from " Categories: Titles incorrect due to Limitations Number theory Definition Stubs 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.
15919	https://www.khanacademy.org/math/algebra2/x2ec2f6f830c9fb89:exp-model/x2ec2f6f830c9fb89:construct-exp/v/constructing-exponential-models-half-life	Constructing exponential models: half life (video) \| Khan Academy Skip to main content If you're seeing this message, it means we're having trouble loading external resources on our website. If you're behind a web filter, please make sure that the domains .kastatic.org and .kasandbox.org are unblocked. Explore Browse By Standards Explore Khanmigo Math: Pre-K - 8th grade Math: High school & college Math: Multiple grades Math: Illustrative Math-aligned Math: Eureka Math-aligned Math: Get ready courses Test prep Science Economics Reading & language arts Computing Life skills Social studies Partner courses Khan for educators Select a category to view its courses Search AI for Teachers FreeDonateLog inSign up Search for courses, skills, and videos Help us do more We'll get right to the point: we're asking you to help support Khan Academy. 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Skip to lesson content Algebra 2 Course: Algebra 2>Unit 7 Lesson 2: Constructing exponential models according to rate of change Constructing exponential models Constructing exponential models: half life Constructing exponential models: percent change Construct exponential models Math> Algebra 2> Exponential models> Constructing exponential models according to rate of change © 2025 Khan Academy Terms of usePrivacy PolicyCookie NoticeAccessibility Statement Constructing exponential models: half life CCSS.Math: HSA.CED.A.2, HSF.BF.A.1, HSF.BF.A, HSF.LE.A.2 Google Classroom Microsoft Teams About About this video Transcript Sal models the decay of a Carbon-14 sample using an exponential function. Skip to end of discussions Questions Tips & Thanks Want to join the conversation? Log in Sort by: Top Voted frank 9 years ago Posted 9 years ago. Direct link to frank's post “If carbon-14's mass gets ...” more If carbon-14's mass gets halved every 5730 years, then wouldn't carbon-14 never disappear? If the current mass gets halved, there would always be something to half for the next 5730 years. Answer Button navigates to signup page •Comment Button navigates to signup page (12 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Jonan 9 years ago Posted 9 years ago. Direct link to Jonan's post “When we say that a mass i...” more When we say that a mass is "halved", we don't mean that, after 5730 years, half suddenly disappears. We're saying that atoms of carbon-14 (let's say there are 1000) are disappearing at a certain rate (we don't know the exact rate) which is decelerating (so we lose less every 5730 years). So, after 5730 years, we will have in total lost 500 atoms of carbon-14, leaving us at 500. But this happened gradually. Maybe we lost a couple atoms every 10 years or so. Eventually, we get down to the point where we have very few atoms, say 3. If atoms are gradually disappearing, can you really say that after 5730 years we'll have 1.5 atoms (which doesn't make sense)? No, you'd say that maybe we lost 2 atoms over all that time. 1/2 is not a perfect rate. It's just really, really close. it's also worth noting that, eventually, when you reach only 1 atom, you can't really divide that anymore, so you're stuck at 1 atom. It would be silly to keep halving that, because then we wouldn't have carbon-14 anymore. Comment Button navigates to signup page (36 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Show more... David Liu 3 years ago Posted 3 years ago. Direct link to David Liu's post “I was so curious how can ...” more I was so curious how can we use Exponential function in our real life? Answer Button navigates to signup page •Comment Button navigates to signup page (7 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer KLaudano 3 years ago Posted 3 years ago. Direct link to KLaudano's post “One common use for expone...” more One common use for exponential models is calculating compounding interest. When you leave money in a bank, the bank may give you interest, which is additional money based on the amount you already have. When the amount of interest you get increases as you gain interest, then we call it compounding interest. CI = P (1 - r/n) ^ (n t) - P where CI is the compound interest, P is the original amount of money (principal amount), r is the interest rate, n is the number of times the interest is calculated (compounded), and t is the length of time the money is left in the bank. Comment Button navigates to signup page (15 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more pieter 6 years ago Posted 6 years ago. Direct link to pieter's post “On 2:18 Sal wrote 1⁄2^2 b...” more On 2:18 Sal wrote 1⁄2^2 but shouldn't he do it with parentheses? Answer Button navigates to signup page •Comment Button navigates to signup page (6 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Kim Seidel 6 years ago Posted 6 years ago. Direct link to Kim Seidel's post “Yes, Sal should be using ...” more Yes, Sal should be using parentheses. Technically, his version only applies the exponent to the 1, not the whole fraction. Comment Button navigates to signup page (6 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Mohammad a year ago Posted a year ago. Direct link to Mohammad's post “How come you have half of...” more How come you have half of a element after certain years did it vanished with the air? Answer Button navigates to signup page •Comment Button navigates to signup page (4 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer TheReal3A a year ago Posted a year ago. Direct link to TheReal3A's post “It's not that the element...” more It's not that the element vanished, that would violate the Law of Conservation of Mass. Rather, radioactive atoms of an element have a chance over time to undergo some sort of decay. This changes the atom so that it's not that element anymore; not a vanishing! For math, it's not important to know what it is, but just know that they decay. If you're curious, see Here, a sample of atoms of a particular element will decay, so over time there will be less and less of that element as it gets converted into other elements. As atoms are so small, there's so many in even a small sample; the number of remaining atoms over time is predictable and can be accurately modelled. The time it takes for half a sample of an element to decay is known as the half-life. Happy learning! Comment Button navigates to signup page (6 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Show more... Tzviofen ✡ 4 years ago Posted 4 years ago. Direct link to Tzviofen ✡'s post “Just saying that at 2:16 ...” more Just saying that at 2:16 Sal should have put parentheses around the 1/2. Answer Button navigates to signup page •Comment Button navigates to signup page (4 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Jayanthika a year ago Posted a year ago. Direct link to Jayanthika's post “Yeah, as the way he wrote...” more Yeah, as the way he wrote it could imply that it only affects the 1, which is wrong. Comment Button navigates to signup page (2 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Ghassan O. Najjar 9 years ago Posted 9 years ago. Direct link to Ghassan O. Najjar's post “At 1:29 why didn't you mu...” more At 1:29 why didn't you multiply 741 by 1.5 instead of 0.5? Answer Button navigates to signup page •Comment Button navigates to signup page (1 vote) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Kim Seidel 9 years ago Posted 9 years ago. Direct link to Kim Seidel's post “Read the problem carefull...” more Read the problem carefully. It tells you that the carbon-14 loses half its mass. Your method of multiplying by 1.5 would mean that its mass is increasing by 1/2. Hope this helps. Comment Button navigates to signup page (8 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more coolhatz 2 months ago Posted 2 months ago. Direct link to coolhatz's post “wait guys why is the numb...” more wait guys why is the number e used to model exponential radioactive decay , rate of chemical reactions Answer Button navigates to signup page •Comment Button navigates to signup page (1 vote) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Leo Polanscak 6 years ago Posted 6 years ago. Direct link to Leo Polanscak's post “I was trying to expand th...” more I was trying to expand this question little bit and think further and got stuck ... need help ... I was asking my self when will carbon -14 sample loose 1/8 of its mass or when it will loose 12,5% which is same as multiplying by 0,875 right ? - after how many years? so if M(t) = 741(1/2)^t/5730 what I was asking my self is actually for what t will this equation be true (1/2)^t/5730 = 0,875 ( because for that ''t'' it would be same as multiplying by 0,875) right ? did I setup everything correct for my problem ? - If yes - how can I solve for ''t'' in the equation? Answer Button navigates to signup page •1 comment Comment on Leo Polanscak's post “I was trying to expand th...” (0 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Video transcript [Voiceover] We're told carbon-14 is an element which loses exactly half of its mass every 5730 years. The mass of a sample of carbon-14 can be modeled by a function, M, which depends on its age, t, in years. We measure that the initial mass of a sample of carbon-14 is 741 grams. Write a function that models the mass of the carbon-14 sample remaining t years since the initial measurement. Alright, so, like always, pause the video and see if you can come up with this function, M, that is going to be a function of t, the years since the initial measurement. Alright, let's work through it together. What I like to do is, I always like to start off with a little bit of a table to get a sense of things. So let's think about t, how much time, how many years have passed since the initial measurement, and what the amount of mass we're going to have. Well, we know that the initial, we know that the initial mass of a sample of carbon-14 is 741 grams, so at t equals zero, our mass is 741. Now, what's another interesting t that we could think about? Well, we know at every 5730 years, we lose exactly half of our mass of carbon-14. Every 5730 years. So let's think about what happens when t is 5730. Well, we're going to lose half of our mass, so we're going to multiply this times 1/2. So this is going to be 741 times 1/2. I'm not even gonna calculate what that is right now. And then let's say we have another 5730 years take place, so that's going to be, and I'm just gonna write two times 5730. I could calculate what it's going to be. 10,000, 11,460 or something like that. Alright, but let's just go with two times 5730. Is it 10? Yeah. 10,000 plus 1400 so 11,400 plus 60. Yeah. So 11,460. But let's just leave it like this. Well, then, it's gonna be this times 1/2. So it's gonna be 741 times 1/2 times 1/2. So we're gonna multiply by 1/2 again. And so this is the same thing as 741 times 1/2 squared. And then, let's just think about if we wait another 5730 years, so three times 5730. Well, then it's going to be 1/2 times this. So it's going to be 741, this times 1/2 is gonna be 1/2 to the third power. So you might notice a little bit of a pattern here. However many half-lifes we have, we're gonna multiply, we're gonna raise 1/2 to that power and then we multiply it times our initial mass. This is one half-life has gone by, two half-lifes, we have an exponent of two, three half-lifes, we multiply by three. Sorry, we multiply by 1/2 three times. So what's going to be a general way to express M of t? Well, M of t is going to be our initial value, 741, times, and you might already be identifying this as an exponential function, we're going to multiply times this number, which we could call our common ratio, as many half-lifes has passed by. So how do we know how many half-lives have passed by? Well, we could take t, and we could divide it by the half-life. And try to test this out. When t equals zero, it's gonna be 1/2 to the zeroth power, which is just one, and we're just gonna have 741. When t is equal to 5730, this exponent is going to be one, which we want it to be. We're just gonna multiply our initial value by 1/2 once. When this exponent is two times 5730, when t is two times 5730, well then the exponent is going to be two, and we're gonna multiply by 1/2 twice. It's going to be 1/2 to the second power. And it's going to work for everything in between. When we are a fraction of a half-life along, we're gonna get a non-integer exponent, and that, too, will work out. And so this is our function. We are, we are done. We have written our function, M, that models the mass of carbon-14 remaining t years since the initial measurement. Creative Commons Attribution/Non-Commercial/Share-AlikeVideo on YouTube Up next: video Use of cookies Cookies are small files placed on your device that collect information when you use Khan Academy. Strictly necessary cookies are used to make our site work and are required. Other types of cookies are used to improve your experience, to analyze how Khan Academy is used, and to market our service. 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15920	https://www.nextgenscience.org/topic-arrangement/hsweather-and-climate	HS.Weather and Climate \| Next Generation Science Standards Skip to main content Read the latest on NGSS NOW Newsletterhere. Utility menu Glossary Contact Staff News FAQ Subscribe Toggle navigation Search Main navigation The Standards Read the Standards Appendices Understanding the Standards Developing the Standards Instruction and Assessment Evaluating NGSS Design Lesson Screener EQuIP Rubric for Science Peer Review Panel NGSS Design Badge Task Screening Tools Understanding NGSS Design Resource Library Video Hub NGSS Lessons and Units Assessment Resources Planning and Communication State and District Implementation Communicating About the Standards Utility menu Glossary Contact Staff News FAQ Subscribe HS.Weather and Climate Breadcrumb Home Body 1 Body 2 Body 3 HS.Weather and Climate Students who demonstrate understanding can: \| HS-ESS2-4. \| Use a model to describe how variations in the flow of energy into and out of Earth’s systems result in changes in climate. [Clarification Statement: Examples of the causes of climate change differ by timescale, over 1-10 years: large volcanic eruption, ocean circulation; 10-100s of years: changes in human activity, ocean circulation, solar output; 10-100s of thousands of years: changes to Earth's orbit and the orientation of its axis; and 10-100s of millions of years: long-term changes in atmospheric composition.] [Assessment Boundary: Assessment of the results of changes in climate is limited to changes in surface temperatures, precipitation patterns, glacial ice volumes, sea levels, and biosphere distribution.] \| \| HS-ESS3-5. \| Analyze geoscience data and the results from global climate models to make an evidence-based forecast of the current rate of global or regional climate change and associated future impacts to Earth's systems.[Clarification Statement: Examples of evidence, for both data and climate model outputs, are for climate changes (such as precipitation and temperature) and their associated impacts (such as on sea level, glacial ice volumes, or atmosphere and ocean composition).] [Assessment Boundary: Assessment is limited to one example of a climate change and its associated impacts.] \| The performance expectations above were developed using the following elements from the NRC document A Framework for K-12 Science Education: Science and Engineering Practices Developing and Using Models Modeling in 9–12 builds on K–8 experiences and progresses to using, synthesizing, and developing models to predict and show relationships among variables between systems and their components in the natural and designed world(s). Use a model to provide mechanistic accounts of phenomena. (HS-ESS2-4) Analyzing and Interpreting Data Analyzing data in 9–12 builds on K–8 experiences and progresses to introducing more detailed statistical analysis, the comparison of data sets for consistency, and the use of models to generate and analyze data. Analyze data using computational models in order to make valid and reliable scientific claims. (HS-ESS3-5) Connections to Nature of Science Scientific Investigations Use a Variety of Methods Science investigations use diverse methods and do not always use the same set of procedures to obtain data. (HS-ESS3-5) New technologies advance scientific knowledge. (HS-ESS3-5) Scientific Knowledge is Based on Empirical Evidence Science knowledge is based on empirical evidence. (HS-ESS3-5) Science arguments are strengthened by multiple lines of evidence supporting a single explanation. (HS-ESS2-4),(HS-ESS3-5)Disciplinary Core Ideas ESS1.B: Earth and the Solar System Cyclical changes in the shape of Earth’s orbit around the sun, together with changes in the tilt of the planet’s axis of rotation, both occurring over hundreds of thousands of years, have altered the intensity and distribution of sunlight falling on the earth. These phenomena cause a cycle of ice ages and other gradual climate changes. (secondary to HS-ESS2-4) ESS2.A: Earth Materials and Systems The geological record shows that changes to global and regional climate can be caused by interactions among changes in the sun’s energy output or Earth’s orbit, tectonic events, ocean circulation, volcanic activity, glaciers, vegetation, and human activities. These changes can occur on a variety of time scales from sudden (e.g., volcanic ash clouds) to intermediate (ice ages) to very long-term tectonic cycles. (HS-ESS2-4) ESS2.D: Weather and Climate The foundation for Earth’s global climate systems is the electromagnetic radiation from the sun, as well as its reflection, absorption, storage, and redistribution among the atmosphere, ocean, and land systems, and this energy’s re-radiation into space. (HS-ESS2-4) Changes in the atmosphere due to human activity have increased carbon dioxide concentrations and thus affect climate. (HS-ESS2-6),(HS-ESS2-4) ESS3.D: Global Climate Change Though the magnitudes of human impacts are greater than they have ever been, so too are human abilities to model, predict, and manage current and future impacts. (HS-ESS3-5)Crosscutting Concepts Cause and Effect Empirical evidence is required to differentiate between cause and correlation and make claims about specific causes and effects.(HS-ESS2-4) Stability and Change Change and rates of change can be quantified and modeled over very short or very long periods of time. Some system changes are irreversible. (HS-ESS3-5) Connections to other DCIs in this grade-band: HS.PS3.A (HS-ESS2-4); HS.PS3.B (HS-ESS2-4),(HS-ESS3-5); HS.PS3.D (HS-ESS3-5); HS.LS1.C (HS-ESS3-5); HS.LS2.C (HS-ESS2-4); HS.ESS1.C (HS-ESS2-4); HS.ESS2.D (HS-ESS3-5); HS.ESS3.C (HS-ESS2-4); HS.ESS3.D (HS-ESS2-4) Articulation of DCIs across grade-bands: MS.PS3.A (HS-ESS2-4); MS.PS3.B (HS-ESS2-4),(HS-ESS3-5); MS.PS3.D (HS-ESS2-4),(HS-ESS3-5); MS.PS4.B (HS-ESS2-4); MS.LS1.C (HS-ESS2-4); MS.LS2.B (HS-ESS2-4); MS.LS2.C (HS-ESS2-4); MS.ESS2.A (HS-ESS2-4),(MS-ESS3-5); MS.ESS2.B (HS-ESS2-4); MS.ESS2.C (HS-ESS2-4); MS.ESS2.D (HS-ESS2-4),(HS-ESS3-5); MS.ESS3.B (HS-ESS3-5); MS.ESS3.C (HS-ESS2-4),(HS-ESS3-5); MS.ESS3.D (HS-ESS2-4),(HS-ESS3-5) Common Core State Standards Connections: ELA/Literacy - RST.11-12.1Cite specific textual evidence to support analysis of science and technical texts, attending to important distinctions the author makes and to any gaps or inconsistencies in the account.(HS-ESS3-5) RST.11-12.2Determine the central ideas or conclusions of a text; summarize complex concepts, processes, or information presented in a text by paraphrasing them in simpler but still accurate terms.(HS-ESS3-5) RST.11-12.7Integrate and evaluate multiple sources of information presented in diverse formats and media (e.g., quantitative data, video, multimedia) in order to address a question or solve a problem.(HS-ESS3-5) SL.11-12.5Make strategic use of digital media (e.g., textual, graphical, audio, visual, and interactive elements) in presentations to enhance understanding of findings, reasoning, and evidence and to add interest.(HS-ESS2-4) Mathematics - MP.2Reason abstractly and quantitatively.(HS-ESS2-4),(HS-ESS3-5) MP.4Model with mathematics.(HS-ESS2-4) HSN.Q.A.1Use units as a way to understand problems and to guide the solution of multi-step problems; choose and interpret units consistently in formulas; choose and interpret the scale and the origin in graphs and data displays.(HS-ESS2-4),(HS-ESS3-5) HSN.Q.A.2Define appropriate quantities for the purpose of descriptive modeling.(HS-ESS2-4),(HS-ESS3-5) HSN.Q.A.3Choose a level of accuracy appropriate to limitations on measurement when reporting quantities.(HS-ESS2-4),(HS-ESS3-5) HS.Weather and Climate Students who demonstrate understanding can: \| HS-ESS2-4. \| Use a model to describe how variations in the flow of energy into and out of Earth’s systems result in changes in climate.[Clarification Statement: Examples of the causes of climate change differ by timescale, over 1-10 years: large volcanic eruption, ocean circulation; 10-100s of years: changes in human activity, ocean circulation, solar output; 10-100s of thousands of years: changes to Earth's orbit and the orientation of its axis; and 10-100s of millions of years: long-term changes in atmospheric composition.] [Assessment Boundary: Assessment of the results of changes in climate is limited to changes in surface temperatures, precipitation patterns, glacial ice volumes, sea levels, and biosphere distribution.] \| \| HS-ESS3-5. \| Analyze geoscience data and the results from global climate models to make an evidence-based forecast of the current rate of global or regional climate change and associated future impacts to Earth's systems.[Clarification Statement: Examples of evidence, for both data and climate model outputs, are for climate changes (such as precipitation and temperature) and their associated impacts (such as on sea level, glacial ice volumes, or atmosphere and ocean composition).] [Assessment Boundary: Assessment is limited to one example of a climate change and its associated impacts.] \| The performance expectations above were developed using the following elements from the NRC document A Framework for K-12 Science Education: Science and Engineering Practices Developing and Using Models Modeling in 9–12 builds on K–8 experiences and progresses to using, synthesizing, and developing models to predict and show relationships among variables between systems and their components in the natural and designed world(s). Use a model to provide mechanistic accounts of phenomena. (HS-ESS2-4) Analyzing and Interpreting Data Analyzing data in 9–12 builds on K–8 experiences and progresses to introducing more detailed statistical analysis, the comparison of data sets for consistency, and the use of models to generate and analyze data. Analyze data using computational models in order to make valid and reliable scientific claims. (HS-ESS3-5) Connections to Nature of Science Scientific Investigations Use a Variety of Methods Science investigations use diverse methods and do not always use the same set of procedures to obtain data. (HS-ESS3-5) New technologies advance scientific knowledge. (HS-ESS3-5) Scientific Knowledge is Based on Empirical Evidence Science knowledge is based on empirical evidence. (HS-ESS3-5) Science arguments are strengthened by multiple lines of evidence supporting a single explanation. (HS-ESS2-4),(HS-ESS3-5)Disciplinary Core Ideas ESS1.B: Earth and the Solar System Cyclical changes in the shape of Earth’s orbit around the sun, together with changes in the tilt of the planet’s axis of rotation, both occurring over hundreds of thousands of years, have altered the intensity and distribution of sunlight falling on the earth. These phenomena cause a cycle of ice ages and other gradual climate changes. (secondary to HS-ESS2-4) ESS2.A: Earth Materials and Systems The geological record shows that changes to global and regional climate can be caused by interactions among changes in the sun’s energy output or Earth’s orbit, tectonic events, ocean circulation, volcanic activity, glaciers, vegetation, and human activities. These changes can occur on a variety of time scales from sudden (e.g., volcanic ash clouds) to intermediate (ice ages) to very long-term tectonic cycles. (HS-ESS2-4) ESS2.D: Weather and Climate The foundation for Earth’s global climate systems is the electromagnetic radiation from the sun, as well as its reflection, absorption, storage, and redistribution among the atmosphere, ocean, and land systems, and this energy’s re-radiation into space. (HS-ESS2-4) Changes in the atmosphere due to human activity have increased carbon dioxide concentrations and thus affect climate. (HS-ESS2-6),(HS-ESS2-4) ESS3.D: Global Climate Change Though the magnitudes of human impacts are greater than they have ever been, so too are human abilities to model, predict, and manage current and future impacts. (HS-ESS3-5)Crosscutting Concepts Cause and Effect Empirical evidence is required to differentiate between cause and correlation and make claims about specific causes and effects.(HS-ESS2-4) Stability and Change Change and rates of change can be quantified and modeled over very short or very long periods of time. Some system changes are irreversible. (HS-ESS3-5) Connections to other DCIs in this grade-band: HS.PS3.A (HS-ESS2-4); HS.PS3.B (HS-ESS2-4),(HS-ESS3-5); HS.PS3.D (HS-ESS3-5); HS.LS1.C (HS-ESS3-5); HS.LS2.C (HS-ESS2-4); HS.ESS1.C (HS-ESS2-4); HS.ESS2.D (HS-ESS3-5); HS.ESS3.C (HS-ESS2-4); HS.ESS3.D (HS-ESS2-4) Articulation of DCIs across grade-bands: MS.PS3.A (HS-ESS2-4); MS.PS3.B (HS-ESS2-4),(HS-ESS3-5); MS.PS3.D (HS-ESS2-4),(HS-ESS3-5); MS.PS4.B (HS-ESS2-4); MS.LS1.C (HS-ESS2-4); MS.LS2.B (HS-ESS2-4); MS.LS2.C (HS-ESS2-4); MS.ESS2.A (HS-ESS2-4),(MS-ESS3-5); MS.ESS2.B (HS-ESS2-4); MS.ESS2.C (HS-ESS2-4); MS.ESS2.D (HS-ESS2-4),(HS-ESS3-5); MS.ESS3.B (HS-ESS3-5); MS.ESS3.C (HS-ESS2-4),(HS-ESS3-5); MS.ESS3.D (HS-ESS2-4),(HS-ESS3-5) Common Core State Standards Connections: ELA/Literacy - RST.11-12.1Cite specific textual evidence to support analysis of science and technical texts, attending to important distinctions the author makes and to any gaps or inconsistencies in the account.(HS-ESS3-5) RST.11-12.2Determine the central ideas or conclusions of a text; summarize complex concepts, processes, or information presented in a text by paraphrasing them in simpler but still accurate terms.(HS-ESS3-5) RST.11-12.7Integrate and evaluate multiple sources of information presented in diverse formats and media (e.g., quantitative data, video, multimedia) in order to address a question or solve a problem.(HS-ESS3-5) SL.11-12.5Make strategic use of digital media (e.g., textual, graphical, audio, visual, and interactive elements) in presentations to enhance understanding of findings, reasoning, and evidence and to add interest.(HS-ESS2-4) Mathematics - MP.2Reason abstractly and quantitatively.(HS-ESS2-4),(HS-ESS3-5) MP.4Model with mathematics.(HS-ESS2-4) HSN.Q.A.1Use units as a way to understand problems and to guide the solution of multi-step problems; choose and interpret units consistently in formulas; choose and interpret the scale and the origin in graphs and data displays.(HS-ESS2-4),(HS-ESS3-5) HSN.Q.A.2Define appropriate quantities for the purpose of descriptive modeling.(HS-ESS2-4),(HS-ESS3-5) HSN.Q.A.3Choose a level of accuracy appropriate to limitations on measurement when reporting quantities.(HS-ESS2-4),(HS-ESS3-5) HS.Weather and Climate Students who demonstrate understanding can: \| HS-ESS2-4. \| Use a model to describe how variations in the flow of energy into and out of Earth’s systems result in changes in climate. [Clarification Statement: Examples of the causes of climate change differ by timescale, over 1-10 years: large volcanic eruption, ocean circulation; 10-100s of years: changes in human activity, ocean circulation, solar output; 10-100s of thousands of years: changes to Earth's orbit and the orientation of its axis; and 10-100s of millions of years: long-term changes in atmospheric composition.] [Assessment Boundary: Assessment of the results of changes in climate is limited to changes in surface temperatures, precipitation patterns, glacial ice volumes, sea levels, and biosphere distribution.] \| \| HS-ESS3-5. \| Analyze geoscience data and the results from global climate models to make an evidence-based forecast of the current rate of global or regional climate change and associated future impacts to Earth's systems.[Clarification Statement: Examples of evidence, for both data and climate model outputs, are for climate changes (such as precipitation and temperature) and their associated impacts (such as on sea level, glacial ice volumes, or atmosphere and ocean composition).] [Assessment Boundary: Assessment is limited to one example of a climate change and its associated impacts.] \| The performance expectations above were developed using the following elements from the NRC document A Framework for K-12 Science Education: Science and Engineering Practices Developing and Using Models Modeling in 9–12 builds on K–8 experiences and progresses to using, synthesizing, and developing models to predict and show relationships among variables between systems and their components in the natural and designed world(s). Use a model to provide mechanistic accounts of phenomena. (HS-ESS2-4) Analyzing and Interpreting Data Analyzing data in 9–12 builds on K–8 experiences and progresses to introducing more detailed statistical analysis, the comparison of data sets for consistency, and the use of models to generate and analyze data. Analyze data using computational models in order to make valid and reliable scientific claims. (HS-ESS3-5) Connections to Nature of Science Scientific Investigations Use a Variety of Methods Science investigations use diverse methods and do not always use the same set of procedures to obtain data. (HS-ESS3-5) New technologies advance scientific knowledge. (HS-ESS3-5) Scientific Knowledge is Based on Empirical Evidence Science knowledge is based on empirical evidence. (HS-ESS3-5) Science arguments are strengthened by multiple lines of evidence supporting a single explanation. (HS-ESS2-4),(HS-ESS3-5)Disciplinary Core Ideas ESS1.B: Earth and the Solar System Cyclical changes in the shape of Earth’s orbit around the sun, together with changes in the tilt of the planet’s axis of rotation, both occurring over hundreds of thousands of years, have altered the intensity and distribution of sunlight falling on the earth. These phenomena cause a cycle of ice ages and other gradual climate changes. (secondary to HS-ESS2-4) ESS2.A: Earth Materials and Systems The geological record shows that changes to global and regional climate can be caused by interactions among changes in the sun’s energy output or Earth’s orbit, tectonic events, ocean circulation, volcanic activity, glaciers, vegetation, and human activities. These changes can occur on a variety of time scales from sudden (e.g., volcanic ash clouds) to intermediate (ice ages) to very long-term tectonic cycles. (HS-ESS2-4) ESS2.D: Weather and Climate The foundation for Earth’s global climate systems is the electromagnetic radiation from the sun, as well as its reflection, absorption, storage, and redistribution among the atmosphere, ocean, and land systems, and this energy’s re-radiation into space. (HS-ESS2-4) Changes in the atmosphere due to human activity have increased carbon dioxide concentrations and thus affect climate. (HS-ESS2-6),(HS-ESS2-4) ESS3.D: Global Climate Change Though the magnitudes of human impacts are greater than they have ever been, so too are human abilities to model, predict, and manage current and future impacts. (HS-ESS3-5)Crosscutting Concepts Cause and Effect Empirical evidence is required to differentiate between cause and correlation and make claims about specific causes and effects.(HS-ESS2-4) Stability and Change Change and rates of change can be quantified and modeled over very short or very long periods of time. Some system changes are irreversible. (HS-ESS3-5) Connections to other DCIs in this grade-band: HS.PS3.A (HS-ESS2-4); HS.PS3.B (HS-ESS2-4),(HS-ESS3-5); HS.PS3.D (HS-ESS3-5); HS.LS1.C (HS-ESS3-5); HS.LS2.C (HS-ESS2-4); HS.ESS1.C (HS-ESS2-4); HS.ESS2.D (HS-ESS3-5); HS.ESS3.C (HS-ESS2-4); HS.ESS3.D (HS-ESS2-4) Articulation of DCIs across grade-bands: MS.PS3.A (HS-ESS2-4); MS.PS3.B (HS-ESS2-4),(HS-ESS3-5); MS.PS3.D (HS-ESS2-4),(HS-ESS3-5); MS.PS4.B (HS-ESS2-4); MS.LS1.C (HS-ESS2-4); MS.LS2.B (HS-ESS2-4); MS.LS2.C (HS-ESS2-4); MS.ESS2.A (HS-ESS2-4),(MS-ESS3-5); MS.ESS2.B (HS-ESS2-4); MS.ESS2.C (HS-ESS2-4); MS.ESS2.D (HS-ESS2-4),(HS-ESS3-5); MS.ESS3.B (HS-ESS3-5); MS.ESS3.C (HS-ESS2-4),(HS-ESS3-5); MS.ESS3.D (HS-ESS2-4),(HS-ESS3-5) Common Core State Standards Connections: ELA/Literacy - RST.11-12.1Cite specific textual evidence to support analysis of science and technical texts, attending to important distinctions the author makes and to any gaps or inconsistencies in the account.(HS-ESS3-5) RST.11-12.2Determine the central ideas or conclusions of a text; summarize complex concepts, processes, or information presented in a text by paraphrasing them in simpler but still accurate terms.(HS-ESS3-5) RST.11-12.7Integrate and evaluate multiple sources of information presented in diverse formats and media (e.g., quantitative data, video, multimedia) in order to address a question or solve a problem.(HS-ESS3-5) SL.11-12.5Make strategic use of digital media (e.g., textual, graphical, audio, visual, and interactive elements) in presentations to enhance understanding of findings, reasoning, and evidence and to add interest.(HS-ESS2-4) Mathematics - MP.2Reason abstractly and quantitatively.(HS-ESS2-4),(HS-ESS3-5) MP.4Model with mathematics.(HS-ESS2-4) HSN.Q.A.1Use units as a way to understand problems and to guide the solution of multi-step problems; choose and interpret units consistently in formulas; choose and interpret the scale and the origin in graphs and data displays.(HS-ESS2-4),(HS-ESS3-5) HSN.Q.A.2Define appropriate quantities for the purpose of descriptive modeling.(HS-ESS2-4),(HS-ESS3-5) HSN.Q.A.3Choose a level of accuracy appropriate to limitations on measurement when reporting quantities.(HS-ESS2-4),(HS-ESS3-5) The performance expectations marked with an asterisk integrate traditional science content with engineering through a Practice or Disciplinary Core Idea. The section entitled “Disciplinary Core Ideas” is reproduced verbatim fromA Framework for K-12 Science Education: Practices, Cross-Cutting Concepts, and Core Ideas. Integrated and reprinted with permission from the National Academy of Sciences. Download PDF Storyline Viewing Options Hide Popup Black and White Practices and Core Ideas Practices and Crosscutting Concepts Use browser zoom to increase text size (ctrl + on PC, command + on Mac) Related Evidence Statements HS.Weather-1 Evidence Statements HS.Weather-2 Evidence Statements Related Quality Units High School: OpenSciEd Unit C.1: Thermodynamics in Earth’s Systems High School: OpenSciEd Unit P.5 Electromagnetic Radiation Related Example Bundles High School Conceptual Progression Model Course 2: Bundle 2 High School Conceptual Progression Model Course 3: Bundle 4 High School Domains Model Course 1: Chemistry: Bundle 2 High School Domains Model Course 2: Physics: Bundle 3 How to Read the Standards The standards integrate three dimensions within each standard and have intentional connections across standards. More... 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15921	https://www.thesaurus.com/browse/relinquishes	Daily Crossword Word Puzzle Word Finder All games Word of the Day Word of the Year New words Language stories All featured Slang Emoji Memes Acronyms Gender and sexuality All culture Writing tips Writing hub Grammar essentials Commonly confused All writing tips Games Featured Culture Writing tips Advertisement Advertisement relinquishes verb as in give up, let go Strongest matches abandon abdicate cede drop out forgo hand over quit renounce surrender vacate waive withdraw yield Strong matches abnegate cast desert discard ditch drop dump forbear forsake forswear kick leave release repudiate resign sacrifice shed Weak matches back down cast off cut loose drop like hot potato kiss good-bye lay aside opt out quit cold turkey retire from stand down swear off take the oath take the pledge Advertisement Quiz Q: Instead of the term COTTON CANDY, British English speakers are more likely to use the word … candycotton. candyfloss. candyfluff. Take the full quiz.Go to all quizzes From Roget's 21st Century Thesaurus, Third Edition Copyright © 2013 by the Philip Lief Group. Advertisement Advertisement Advertisement Browse
15922	https://people.mpim-bonn.mpg.de/zagier/files-restricted/doi/10.1007/978-3-540-74119-0/fulltext.pdf	Jan Hendrik Bruinier Gerard van der Geer Günter Harder Don Zagier The 1-2-3 of Modular Forms Lectures at a Summer School in Nordfjordeid, Norway 123 Jan Hendrik Bruinier Fachbereich Mathematik Technische Universität Darmstadt Schloßgartenstraße 7 64289 Darmstadt, Germany bruinier@mathematik.tu-darmstadt.de Gerard van der Geer Korteweg-de Vries Instituut Universiteit van Amsterdam Plantage Muidergracht 24 1018 TV Amsterdam, The Netherlands geer@science.uva.nl Editor: Kristian Ranestad Department of Mathematics University of Oslo P.O. Box 1053 Blindern 0316 Oslo, Norway ranestad@math.uio.no Günter Harder Mathematisches Institut Universität Bonn Beringstraße 1 53115 Bonn, Germany harder@math.uni-bonn.de and Max-Planck-Institut für Mathematik Vivatsgasse 7 53111 Bonn, Germany harder@mpim-bonn.mpg.de Don Zagier Max-Planck-Institut für Mathematik Vivatsgasse 7 53111 Bonn, Germany zagier@mpim-bonn.mpg.de and Collège de France 3, rue d’Ulm 75231 Paris Cedex 05, France don.zagier@college-de-france.fr ISBN 978-3-540-74117-6 e-ISBN 978-3-540-74119-0 DOI 10.1007/978-3-540-74119-0 Library of Congress Control Number: 2007939406 Mathematics Subject Classiﬁcation (2000): 14-01, 11Gxx, 14Gxx c ⃝2008 Springer-Verlag Berlin Heidelberg This work is subject to copyright. All rights are reserved, whether the whole or part of the material is concerned, speciﬁcally the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microﬁlm or in any other way, and storage in data banks. Duplication of this publication or parts thereof is permitted only under the provisions of the German Copyright Law of September 9, 1965, in its current version, and permission for use must always be obtained from Springer. Violations are liable to prosecution under the German Copyright Law. The use of general descriptive names, registered names, trademarks, etc. in this publication does not imply, even in the absence of a speciﬁc statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. Typesetting and Production: LE-T EX Jelonek, Schmidt & Vöckler GbR, Leipzig, Germany Cover design: WMX Design GmbH, Heidelberg, Germany Printed on acid-free paper 9 8 7 6 5 4 3 2 1 springer.com Preface This book grew out of lectures given at the summer school on “Modular Forms and their Applications” at the Sophus Lie Conference center in Nordfjordeid in June 2004. This center, set beautifully in the fjords of the west coast of Norway, has been the site of annual summer schools in algebra and algebraic geometry since 1996. The schools are a joint eﬀort between the universities in Bergen, Oslo, Tromsø and Trondheim. They are primarily aimed at graduate students in Norway, but also attract a large number of students from other parts of the world. The theme varies among central topics in contemporary mathematics, but the format is the same: three leading experts give indepen-dent but connected series of lectures, and give exercises that the students work on in evening sessions. In 2004 the organizing committee consisted of Stein Arild Strømme (Bergen), Geir Ellingsrud and Kristian Ranestad (Oslo) and Alexei Rudakov (Trondheim). We wanted to have a summer school that introduced the stu-dents both to the beauty of modular forms and to their varied applications in other areas of mathematics, and were very fortunate to have Don Zagier, Jan Bruinier and Gerard van der Geer give the lectures. The lectures were organized in three series that are reﬂected in the title of this book both by their numbering and their content. The ﬁrst series treats the classical one-variable theory and some of its many applications in number theory, algebraic geometry and mathematical physics. The second series, which has a more geometric ﬂavor, gives an introduction to the theory of Hilbert modular forms in two variables and to Hilbert modular surfaces. In particular, it discusses Borcherds products and some geometric and arithmetic applications. The third gives an introduction to Siegel modular forms, both scalar- and vector-valued, especially Siegel modular forms of degree 2, which are functions of three complex variables. It presents a beautiful application of the theory of curves over ﬁnite ﬁelds to Siegel modular forms by providing evidence for VI Preface a conjecture of Harder on congruences between elliptic and Siegel modular forms. Günter Harder came forward with this conjecture in a colloquium lecture in Bonn in 2003. He kindly allowed us to include his notes for this colloquium talk in Bonn on the subject. Even though the three lecture series are strongly connected, each of them is self contained and can be read independently of the others. There is quote ascribed (perhaps apocryphally) to Martin Eichler, saying that there are ﬁve fundamental operations in mathematics: addition, sub-traction, multiplication, division and modular forms. We hope this book will help convince newcomers and oldtimers alike that this is only partially an exaggeration. Oslo, July 2007 Kristian Ranestad Contents Elliptic Modular Forms and Their Applications Don Zagier . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 Foreword . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 1 Basic Deﬁnitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 1.1 Modular Groups, Modular Functions and Modular Forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 1.2 The Fundamental Domain of the Full Modular Group . . . . . . . . 5 ♠ Finiteness of Class Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . 7 1.3 The Finite Dimensionality of Mk(Γ) . . . . . . . . . . . . . . . . . . . . . . . 8 2 First Examples: Eisenstein Series and the Discriminant Function . . . 12 2.1 Eisenstein Series and the Ring Structure of M∗(Γ1) . . . . . . . . . . 12 2.2 Fourier Expansions of Eisenstein Series . . . . . . . . . . . . . . . . . . . . . 15 ♠ Identities Involving Sums of Powers of Divisors . . . . . . . . . . 18 2.3 The Eisenstein Series of Weight 2 . . . . . . . . . . . . . . . . . . . . . . . . . . 18 2.4 The Discriminant Function and Cusp Forms. . . . . . . . . . . . . . . . . 20 ♠ Congruences for τ(n) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23 3 Theta Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24 3.1 Jacobi’s Theta Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25 ♠ Sums of Two and Four Squares . . . . . . . . . . . . . . . . . . . . . . . . 26 ♠ The Kac–Wakimoto Conjecture . . . . . . . . . . . . . . . . . . . . . . . 31 3.2 Theta Series in Many Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31 ♠ Invariants of Even Unimodular Lattices . . . . . . . . . . . . . . . . 33 ♠ Drums Whose Shape One Cannot Hear. . . . . . . . . . . . . . . . . 36 4 Hecke Eigenforms and L-series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37 4.1 Hecke Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37 4.2 L-series of Eigenforms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39 4.3 Modular Forms and Algebraic Number Theory . . . . . . . . . . . . . . 41 ♠ Binary Quadratic Forms of Discriminant −23 . . . . . . . . . . . 42 4.4 Modular Forms Associated to Elliptic Curves and Other Varieties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44 VIII Contents ♠ Fermat’s Last Theorem. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46 5 Modular Forms and Diﬀerential Operators . . . . . . . . . . . . . . . . . . . . . . . 48 5.1 Derivatives of Modular Forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48 ♠ Modular Forms Satisfy Non-Linear Diﬀerential Equations . 49 ♠ Moments of Periodic Functions . . . . . . . . . . . . . . . . . . . . . . . . 50 5.2 Rankin–Cohen Brackets and Cohen–Kuznetsov Series . . . . . . . . 53 ♠ Further Identities for Sums of Powers of Divisors . . . . . . . . 56 ♠ Exotic Multiplications of Modular Forms . . . . . . . . . . . . . . . 56 5.3 Quasimodular Forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58 ♠ Counting Ramiﬁed Coverings of the Torns . . . . . . . . . . . . . . 60 5.4 Linear Diﬀerential Equations and Modular Forms . . . . . . . . . . . . 61 ♠ The Irrationality of ζ(3) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64 ♠ An Example Coming from Percolation Theory . . . . . . . . . . 66 6 Singular Moduli and Complex Multiplication . . . . . . . . . . . . . . . . . . . . 66 6.1 Algebraicity of Singular Moduli . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67 ♠ Strange Approximations to π . . . . . . . . . . . . . . . . . . . . . . . . . 73 ♠ Computing Class Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . 74 ♠ Explicit Class Field Theory for Imaginary Quadratic Fields 75 ♠ Solutions of Diophantine Equations . . . . . . . . . . . . . . . . . . . . 76 6.2 Norms and Traces of Singular Moduli . . . . . . . . . . . . . . . . . . . . . . 77 ♠ Heights of Heegner Points . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79 ♠ The Borcherds Product Formula . . . . . . . . . . . . . . . . . . . . . . . 83 6.3 Periods and Taylor Expansions of Modular Forms . . . . . . . . . . . . 83 ♠ Two Transcendence Results . . . . . . . . . . . . . . . . . . . . . . . . . . . 85 ♠ Hurwitz Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85 ♠ Generalized Hurwitz Numbers. . . . . . . . . . . . . . . . . . . . . . . . . 89 6.4 CM Elliptic Curves and CM Modular Forms . . . . . . . . . . . . . . . . 90 ♠ Factorization, Primality Testing, and Cryptography . . . . . . 92 ♠ Central Values of Hecke L-Series . . . . . . . . . . . . . . . . . . . . . . 95 ♠ Which Primes are Sums of Two Cubes? . . . . . . . . . . . . . . . . 97 References and Further Reading . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99 Hilbert Modular Forms and Their Applications Jan Hendrik Bruinier. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105 Introduction. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105 1 Hilbert Modular Surfaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 106 1.1 The Hilbert Modular Group . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 106 1.2 The Baily–Borel Compactiﬁcation . . . . . . . . . . . . . . . . . . . . . . . . . 109 Siegel Domains . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111 1.3 Hilbert Modular Forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113 1.4 Mk(Γ) is Finite Dimensional . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 118 1.5 Eisenstein Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 119 Restriction to the Diagonal . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122 The Example Q( √ 5). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123 Contents IX 1.6 The L-function of a Hilbert Modular Form . . . . . . . . . . . . . . . . . . 125 2 The Orthogonal Group O(2, n) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 127 2.1 Quadratic Forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 128 2.2 The Cliﬀord Algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 129 2.3 The Spin Group . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133 Quadratic Spaces in Dimension Four. . . . . . . . . . . . . . . . . . . . 135 2.4 Rational Quadratic Spaces of Type (2, n) . . . . . . . . . . . . . . . . . . . 136 The Grassmannian Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 136 The Projective Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 137 The Tube Domain Model. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 137 Lattices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 138 Heegner Divisors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 140 2.5 Modular Forms for O(2, n) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 140 2.6 The Siegel Theta Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141 2.7 The Hilbert Modular Group as an Orthogonal Group. . . . . . . . . 143 Hirzebruch–Zagier Divisors . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145 3 Additive and Multiplicative Liftings . . . . . . . . . . . . . . . . . . . . . . . . . . . . 146 3.1 The Doi–Naganuma Lift . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 146 3.2 Borcherds Products . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 150 Local Borcherds Products . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 150 The Borcherds Lift . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 154 Obstructions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 158 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 160 3.3 Automorphic Green Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 162 A Second Approach . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 167 3.4 CM Values of Hilbert Modular Functions . . . . . . . . . . . . . . . . . . . 168 Singular Moduli. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 168 CM Extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171 CM Cycles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172 CM Values of Borcherds Products . . . . . . . . . . . . . . . . . . . . . . 173 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 175 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 176 Siegel Modular Forms and Their Applications Gerard van der Geer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 181 1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 181 2 The Siegel Modular Group . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183 3 Modular Forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 187 4 The Fourier Expansion of a Modular Form . . . . . . . . . . . . . . . . . . . . . . 189 5 The Siegel Operator and Eisenstein Series . . . . . . . . . . . . . . . . . . . . . . . 192 6 Singular Forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 194 7 Theta Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 195 8 The Fourier–Jacobi Development of a Siegel Modular Form . . . . . . . . 196 9 The Ring of Classical Siegel Modular Forms for Genus Two . . . . . . . . 198 X Contents 10 Moduli of Principally Polarized Complex Abelian Varieties . . . . . . . . 201 11 Compactiﬁcations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 204 12 Intermezzo: Roots and Representations. . . . . . . . . . . . . . . . . . . . . . . . . . 207 13 Vector Bundles Deﬁned by Representations . . . . . . . . . . . . . . . . . . . . . . 209 14 Holomorphic Diﬀerential Forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 210 15 Cusp Forms and Geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 212 16 The Classical Hecke Algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 213 17 The Satake Isomorphism . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 215 18 Relations in the Hecke Algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 218 19 Satake Parameters. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 219 20 L-functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 220 21 Liftings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 221 22 The Moduli Space of Principally Polarized Abelian Varieties . . . . . . . 226 23 Elliptic Curves over Finite Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 226 24 Counting Points on Curves of Genus 2 . . . . . . . . . . . . . . . . . . . . . . . . . . 230 25 The Ring of Vector-Valued Siegel Modular Forms for Genus 2 . . . . . . 232 26 Harder’s Conjecture . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 235 27 Evidence for Harder’s Conjecture . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 237 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 241 A Congruence Between a Siegel and an Elliptic Modular Form Günter Harder. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 247 1 Elliptic and Siegel Modular Forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 247 2 The Hecke Algebra and a Congruence . . . . . . . . . . . . . . . . . . . . . . . . . . . 250 3 The Special Values of the L-function . . . . . . . . . . . . . . . . . . . . . . . . . . . 252 4 Cohomology with Coeﬃcients . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 253 5 Why the Denominator? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 257 6 Arithmetic Implications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 258 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 259 Appendix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 260 Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 263 Elliptic Modular Forms and Their Applications Don Zagier Max-Planck-Institut für Mathematik, Vivatsgasse 7, 53111 Bonn, Germany E-mail: zagier@mpim-bonn.mpg.de Foreword These notes give a brief introduction to a number of topics in the classical theory of modular forms. Some of theses topics are (planned) to be treated in much more detail in a book, currently in preparation, based on various courses held at the Collège de France in the years 2000–2004. Here each topic is treated with the minimum of detail needed to convey the main idea, and longer proofs are omitted. Classical (or “elliptic”) modular forms are functions in the complex upper half-plane which transform in a certain way under the action of a discrete subgroup Γ of SL(2, R) such as SL(2, Z). From the point of view taken here, there are two cardinal points about them which explain why we are interested. First of all, the space of modular forms of a given weight on Γ is ﬁnite dimen-sional and algorithmically computable, so that it is a mechanical procedure to prove any given identity among modular forms. Secondly, modular forms occur naturally in connection with problems arising in many other areas of mathematics. Together, these two facts imply that modular forms have a huge number of applications in other ﬁelds. The principal aim of these notes – as also of the notes on Hilbert modular forms by Bruinier and on Siegel modular forms by van der Geer – is to give a feel for some of these applications, rather than emphasizing only the theory. For this reason, we have tried to give as many and as varied examples of interesting applications as possible. These applications are placed in separate mini-subsections following the relevant sections of the main text, and identiﬁed both in the text and in the table of contents by the symbol ♠. (The end of such a mini-subsection is correspond-ingly indicated by the symbol ♥: these are major applications.) The subjects they cover range from questions of pure number theory and combinatorics to diﬀerential equations, geometry, and mathematical physics. The notes are organized as follows. Section 1 gives a basic introduction to the theory of modular forms, concentrating on the full modular group 2 D. Zagier Γ1 = SL(2, Z). Much of what is presented there can be found in standard textbooks and will be familiar to most readers, but we wanted to make the exposition self-contained. The next two sections describe two of the most im-portant constructions of modular forms, Eisenstein series and theta series. Here too most of the material is quite standard, but we also include a number of concrete examples and applications which may be less well known. Sec-tion 4 gives a brief account of Hecke theory and of the modular forms arising from algebraic number theory or algebraic geometry whose L-series have Eu-ler products. In the last two sections we turn to topics which, although also classical, are somewhat more specialized; here there is less emphasis on proofs and more on applications. Section 5 treats the aspects of the theory con-nected with diﬀerentiation of modular forms, and in particular the diﬀerential equations which these functions satisfy. This is perhaps the most important single source of applications of the theory of modular forms, ranging from irrationality and transcendence proofs to the power series arising in mirror symmetry. Section 6 treats the theory of complex multiplication. This too is a classical theory, going back to the turn of the (previous) century, but we try to emphasize aspects that are more recent and less familiar: formulas for the norms and traces of the values of modular functions at CM points, Borcherds products, and explicit Taylor expansions of modular forms. (The last topic is particularly pretty and has applications to quite varied problems of num-ber theory.) A planned seventh section would have treated the integrals, or “periods,” of modular forms, which have a rich combinatorial structure and many applications, but had to be abandoned for reasons of space and time. Apart from the ﬁrst two, the sections are largely independent of one another and can be read in any order. The text contains 29 numbered “Propositions” whose proofs are given or sketched and 20 unnumbered “Theorems” which are results quoted from the literature whose proofs are too diﬃcult (in many cases, much too diﬃcult) to be given here, though in many cases we have tried to indicate what the main ingredients are. To avoid breaking the ﬂow of the exposition, references and suggestions for further reading have not been given within the main text but collected into a single section at the end. Notations are standard (e.g., Z, Q, R and C for the integers, rationals, reals and complex numbers, respectively, and N for the strictly positive integers). Multiplication precedes division hierarchically, so that, for instance, 1/4π means 1/(4π) and not (1/4)π. The presentation in Sections 1–5 is based partly on notes taken by Chris-tian Grundh, Magnus Dehli Vigeland and my wife, Silke Wimmer-Zagier, of the lectures which I gave at Nordfjordeid, while that of Section 6 is partly based on the notes taken by John Voight of an earlier course on complex multiplication which I gave in Berekeley in 1992. I would like to thank all of them here, but especially Silke, who read each section of the notes as it was written and made innumerable useful suggestions concerning the exposition. And of course special thanks to Kristian Ranestad for the wonderful week in Nordfjordeid which he organized. Elliptic Modular Forms and Their Applications 3 1 Basic Deﬁnitions In this section we introduce the basic objects of study – the group SL(2, R) and its action on the upper half plane, the modular group, and holomorphic modular forms – and show that the space of modular forms of any weight and level is ﬁnite-dimensional. This is the key to the later applications. 1.1 Modular Groups, Modular Functions and Modular Forms The upper half plane, denoted H, is the set of all complex numbers with positive imaginary part: H = z ∈C \| I(z) > 0 . The special linear group SL(2, R) acts on H in the standard way by Möbius transformations (or fractional linear transformations): γ = a b c d : H →H , z →γz = γ(z) = az + b cz + d . To see that this action is well-deﬁned, we note that the denominator is non-zero and that H is mapped to H because, as a sinple calculation shows, I(γz) = I(z) \|cz + d\|2 . (1) The transitivity of the action also follows by direct calculations, or alterna-tively we can view H as the set of classes of ω1 ω2 ∈C2 \| ω2 ̸= 0, I(ω1/ω2)>0 under the equivalence relation of multiplication by a non-zero scalar, in which case the action is given by ordinary matrix multiplication from the left. Notice that the matrices ±γ act in the same way on H, so we can, and often will, work instead with the group PSL(2, R) = SL(2, R)/{±1}. Elliptic modular functions and modular forms are functions in H which are either invariant or transform in a speciﬁc way under the action of a dis-crete subgroup Γ of SL(2, R). In these introductory notes we will consider only the group Γ1 = SL(2, Z) (the “full modular group”) and its congruence subgroups (subgroups of ﬁnite index of Γ1 which are described by congruence conditions on the entries of the matrices). We should mention, however, that there are other interesting discrete subgroups of SL(2, R), most notably the non-congruence subgroups of SL(2, Z), whose corresponding modular forms have rather diﬀerent arithmetic properties from those on the congruence sub-groups, and subgroups related to quaternion algebras over Q, which have a compact fundamental domain. The latter are important in the study of both Hilbert and Siegel modular forms, treated in the other contributions in this volume. 4 D. Zagier The modular group takes its name from the fact that the points of the quotient space Γ1\H are moduli (= parameters) for the isomorphism classes of elliptic curves over C. To each point z ∈H one can associate the lattice Λz = Z.z + Z.1 ⊂C and the quotient space Ez = C/Λz, which is an elliptic curve, i.e., it is at the same time a complex curve and an abelian group. Conversely, every elliptic curve over C can be obtained in this way, but not uniquely: if E is such a curve, then E can be written as the quotient C/Λ for some lattice (discrete rank 2 subgroup) Λ ⊂C which is unique up to “homotheties” Λ →λΛ with λ ∈C∗, and if we choose an oriented basis (ω1, ω2) of Λ (one with I(ω1/ω2) > 0) and use λ = ω−1 2 for the homothety, then we see that E ∼ = Ez for some z ∈H, but choosing a diﬀerent oriented basis replaces z by γz for some γ ∈Γ1. The quotient space Γ1\H is the simplest example of what is called a moduli space, i.e., an algebraic variety whose points classify isomorphism classes of other algebraic varieties of some ﬁxed type. A complex-valued function on this space is called a modular function and, by virtue of the above discussion, can be seen as any one of four equivalent objects: a function from Γ1\H to C, a function f : H →C satisfying the transformation equation f(γz) = f(z) for every z ∈H and every γ ∈Γ1, a function assigning to every elliptic curve E over C a complex number depending only on the isomorphism type of E, or a function on lattices in C satisfying F(λΛ) = F(Λ) for all lattices Λ and all λ ∈C×, the equivalence between f and F being given in one direction by f(z) = F(Λz) and in the other by F(Λ) = f(ω1/ω2) where (ω1, ω2) is any oriented basis of Λ. Generally the term “modular function”, on Γ1 or some other discrete subgroup Γ ⊂SL(2, R), is used only for meromorphic modular functions, i.e., Γ-invariant meromorphic functions in H which are of exponential growth at inﬁnity (i.e., f(x + iy) = O(eCy) as y →∞and f(x + iy) = O(eC/y) as y →0 for some C > 0), this latter condition being equivalent to the requirement that f extends to a meromorphic function on the compactiﬁed space Γ\H obtained by adding ﬁnitely many “cusps” to Γ\H (see below). It turns out, however, that for the purposes of doing interesting arithmetic the modular functions are not enough and that one needs a more general class of functions called modular forms. The reason is that modular functions have to be allowed to be meromorphic, because there are no global holomorphic functions on a compact Riemann surface, whereas modular forms, which have a more ﬂexible transformation behavior, are holomorphic functions (on H and, in a suitable sense, also at the cusps). Every modular function can be repre-sented as a quotient of two modular forms, and one can think of the modular functions and modular forms as in some sense the analogues of rational num-bers and integers, respectively. From the point of view of functions on lattices, modular forms are simply functions Λ →F(Λ) which transform under homo-theties by F(λΛ) = λ−kF(Λ) rather than simply by F(λΛ) = F(Λ) as before, where k is a ﬁxed integer called the weight of the modular form. If we translate this back into the language of functions on H via f(z) = F(Λz) as before, then we see that f is now required to satisfy the modular transformation property Elliptic Modular Forms and Their Applications 5 f az + b cz + d = (cz + d)k f(z) (2) for all z ∈H and all a b c d ∈Γ1; conversely, given a function f : H →C satis-fying (2), we can deﬁne a funcion on lattices, homogeneous of degree −k with respect to homotheties, by F(Z.ω1 + Z.ω2) = ω−k 2 f(ω1/ω2). As with modular functions, there is a standard convention: when the word “modular form” (on some discrete subgroup Γ of SL(2, R)) is used with no further adjectives, one generally means “holomorphic modular form”, i.e., a function f on H satisfy-ing (2) for all a b c d ∈Γ which is holomorphic in H and of subexponential growth at inﬁnity (i.e., f satisﬁes the same estimate as above, but now for all rather than some C > 0). This growth condition, which corresponds to holomorphy at the cusps in a sense which we do not explain now, implies that the growth at inﬁnity is in fact polynomial; more precisely, f automatically satisﬁes f(z) = O(1) as y →∞and f(x + iy) = O(y−k) as y →0. We denote by Mk(Γ) the space of holomorphic modular forms of weight k on Γ. As we will see in detail for Γ = Γ1, this space is ﬁnite-dimensional, eﬀectively com-putable for all k, and zero for k < 0, and the algebra M∗(Γ) := k Mk(Γ) of all modular forms on Γ is ﬁnitely generated over C. If we specialize (2) to the matrix 1 1 0 1 , which belongs to Γ1, then we see that any modular form on Γ1 satisﬁes f(z + 1) = f(z) for all z ∈H, i.e., it is a periodic function of period 1. It is therefore a function of the quantity e2πiz, traditionally denoted q ; more precisely, we have the Fourier development f(z) = ∞ n=0 an e2πinz = ∞ n=0 an qn z ∈H, q = e2πiz , (3) where the fact that only terms qn with n ≥0 occur is a consequence of (and in the case of Γ1, in fact equivalent to) the growth conditions on f just given. It is this Fourier development which is responsible for the great importance of modular forms, because it turns out that there are many examples of modular forms f for which the Fourier coeﬃcients an in (3) are numbers that are of interest in other domains of mathematics. 1.2 The Fundamental Domain of the Full Modular Group In the rest of §1 we look in more detail at the modular group. Because Γ1 contains the element −1 = −1 0 0 −1 which ﬁxes every point of H, we can also consider the action of the quotient group Γ 1 = Γ1/{±1} = PSL(2, Z) ⊂ PSL(2, R) on H. It is clear from (2) that a modular form of odd weight on Γ1 (or on any subgroup of SL(2, R) containing −1) must vanish, so we can restrict our attention to even k. But then the “automorphy factor” (cz + d)k in (2) is unchanged when we replace γ ∈Γ1 by −γ, so that we can consider equation (2) for k even and a b c d ∈Γ 1. By a slight abuse of notation, we will use the same notation for an element γ of Γ1 and its image ±γ in Γ 1, and, for k even, will not distinguish between the isomorphic spaces Mk(Γ1) and Mk(Γ 1). 6 D. Zagier The group Γ 1 is generated by the two elements T = 1 1 0 1 and S = 0 1 −1 0 , with the relations S2 = (ST )3 = 1. The actions of S and T on H are given by S : z →−1/z , T : z →z + 1 . Therefore f is a modular form of weight k on Γ1 precisely when f is periodic with period 1 and satisﬁes the single further functional equation f −1/z = zkf(z) (z ∈H) . (4) If we know the value of a modular form f on some group Γ at one point z ∈H, then equation (2) tells us the value at all points in the same Γ1-orbit as z. So to be able to completely determine f it is enough to know the value at one point from each orbit. This leads to the concept of a fundamental domain for Γ, namely an open subset F ⊂H such that no two distinct points of F are equivalent under the action of Γ and every point z ∈H is Γ-equivalent to some point in the closure F of F. Proposition 1. The set F1 = z ∈H \| \|z\| > 1, \|ℜ(z)\| < 1 2 is a fundamental domain for the full modular group Γ1. (See Fig. 1A.) Proof. Take a point z ∈H. Then {mz + n \| m, n ∈Z} is a lattice in C. Every lattice has a point diﬀerent from the origin of minimal modulus. Let cz + d be such a point. The integers c, d must be relatively prime (otherwise we could divide cz + d by an integer to get a new point in the lattice of even smaller modulus). So there are integers a and b such that γ1 = a b c d ∈Γ1. By the transformation property (1) for the imaginary part y = I(z) we get that I(γ1z) is a maximal member of {I(γz) \| γ ∈Γ1}. Set z∗= T nγ1z = γ1z + n, Fig. 1. The standard fundamental domain for Γ 1 and its neighbors Elliptic Modular Forms and Their Applications 7 where n is such that \|ℜ(z∗)\| ≤1 2. We cannot have \|z∗\| < 1, because then we would have I(−1/z∗) = I(z∗)/\|z∗\|2 > I(z∗) by (1), contradicting the maximality of I(z∗). So z∗∈F1, and z is equivalent under Γ1 to z∗. Now suppose that we had two Γ1-equivalent points z1 and z2 = γz1 in F1, with γ ̸= ±1. This γ cannot be of the form T n since this would contradict the condition \|ℜ(z1)\|, \|ℜ(z2)\| < 1 2, so γ = a b c d with c ̸= 0. Note that I(z) > √ 3/2 for all z ∈F1. Hence from (1) we get √ 3 2 < I(z2) = I(z1) \|cz1 + d\|2 ≤ I(z1) c2 I(z1)2 < 2 c2√ 3 , which can only be satisﬁed if c = ±1. Without loss of generality we may assume that Iz1 ≤Iz2. But \|±z1+d\| ≥\|z1\| > 1, and this gives a contradiction with the transformation property (1). Remarks. 1. The points on the borders of the fundamental region are Γ1-equivalent as follows: First, the points on the two lines ℜ(z) = ± 1 2 are equiva-lent by the action of T : z →z + 1. Secondly, the points on the left and right halves of the arc \|z\| = 1 are equivalent under the action of S : z →−1/z. In fact, these are the only equivalences for the points on the boundary. For this reason we deﬁne F1 to be the semi-closure of F1 where we have added only the boundary points with non-positive real part (see Fig. 1B). Then every point of H is Γ1-equivalent to a unique point of F1, i.e., F1 is a strict fundamental domain for the action of Γ1. (But terminology varies, and many people use the words “fundamental domain” for the strict fundamental domain or for its closure, rather than for the interior.) 2. The description of the fundamental domain F1 also implies the above-mentioned fact that Γ1 (or Γ 1) is generated by S and T . Indeed, by the very deﬁnition of a fundamental domain we know that F1 and its translates γF1 by elements γ of Γ1 cover H, disjointly except for their overlapping boundaries (a so-called “tesselation” of the upper half-plane). The neighbors of F1 are T −1F1, SF1 and T F1 (see Fig. 1C), so one passes from any translate γF1 of F1 to one of its three neighbors by applying γSγ−1 or γT ±1γ−1. In parti-cular, if the element γ describing the passage from F1 to a given translated fundamental domain F′ 1 = γF1 can be written as a word in S and T , then so can the element of Γ1 which describes the motion from F1 to any of the neighbors of F′ 1. Therefore by moving from neighbor to neighbor across the whole upper half-plane we see inductively that this property holds for every γ ∈Γ1, as asserted. More generally, one sees that if one has given a fundamen-tal domain F for any discrete group Γ, then the elements of Γ which identify in pairs the sides of F always generate Γ. ♠ Finiteness of Class Numbers Let D be a negative discriminant, i.e., a negative integer which is congruent to 0 or 1 modulo 4. We consider binary quadratic forms of the form Q(x, y) = 8 D. Zagier Ax2 + Bxy + Cy2 with A, B, C ∈Z and B2 −4AC = D. Such a form is deﬁnite (i.e., Q(x, y) ̸= 0 for non-zero (x, y) ∈R2) and hence has a ﬁxed sign, which we take to be positive. (This is equivalent to A > 0.) We also assume that Q is primitive, i.e., that gcd(A, B, C) = 1. Denote by QD the set of these forms. The group Γ1 (or indeed Γ 1) acts on QD by Q →Q ◦γ, where (Q ◦γ)(x, y) = Q(ax + by, cx + dy) for γ = ± a b c d ∈Γ1. We claim that the number of equivalence classes under this action is ﬁnite. This number, called the class number of D and denoted h(D), also has an interpretation as the number of ideal classes (either for the ring of integers or, if D is a non-trivial square multiple of some other discriminant, for a non-maximal order) in the imaginary quadratic ﬁeld Q( √ D), so this claim is a special case – historically the ﬁrst one, treated in detail in Gauss’s Disquisitiones Arithmeticae – of the general theorem that the number of ideal classes in any number ﬁeld is ﬁnite. To prove it, we observe that we can associate to any Q ∈QD the unique root zQ = (−B + √ D)/2A of Q(z, 1) = 0 in the upper half-plane (here √ D = +i \|D\| by deﬁnition and A > 0 by assumption). One checks easily that zQ◦γ = γ−1(zQ) for any γ ∈Γ 1, so each Γ 1-equivalence class of forms Q ∈QD has a unique representative belonging to the set Qred D = [A, B, C] ∈QD \| −A < B ≤A < C or 0 ≤B ≤A = C (5) of Q ∈QD for which zQ ∈ F1 (the so-called reduced quadratic forms of discriminant D), and this set is ﬁnite because C ≥A ≥\|B\| implies \|D\| = 4AC −B2 ≥3A2, so that both A and B are bounded in absolute value by \|D\|/3, after which C is ﬁxed by C = (B2 −D)/4A. This even gives us a way to compute h(D) eﬀectively, e.g., Qred −47 = {[1, 1, 12], [2, ±1, 6], [3, ±1, 4]} and hence h(−47) = 5. We remark that the class numbers h(D), or a small modiﬁcation of them, are themselves the coeﬃcients of a modular form (of weight 3/2), but this will not be discussed further in these notes. ♥ 1.3 The Finite Dimensionality of Mk(Γ ) We end this section by applying the description of the fundamental domain to show that Mk(Γ1) is ﬁnite-dimensional for every k and to get an upper bound for its dimension. In §2 we will see that this upper bound is in fact the correct value. If f is a modular form of weight k on Γ1 or any other discrete group Γ, then f is not a well-deﬁned function on the quotient space Γ\H, but the transformation formula (2) implies that the order of vanishing ordz(f) at a point z ∈H depends only on the orbit Γz. We can therefore deﬁne a local order of vanishing, ordP (f), for each P ∈Γ\H. The key assertion is that the total number of zeros of f, i.e., the sum of all of these local orders, depends only on Γ and k. But to make this true, we have to look more carefully at the geometry of the quotient space Γ\H, taking into account the fact that some points (the so-called elliptic ﬁxed points, corresponding to the points Elliptic Modular Forms and Their Applications 9 z ∈H which have a non-trivial stabilizer for the image of Γ in PSL(2, R)) are singular and also that Γ\H is not compact, but has to be compactiﬁed by the addition of one or more further points called cusps. We explain this for the case Γ = Γ1. In §1.2 we identiﬁed the quotient space Γ1\H as a set with the semi-closure F1 of F1 and as a topological space with the quotient of F1 obtained by identifying the opposite sides (lines ℜ(z) = ± 1 2 or halves of the arc \|z\| = 1) of the boundary ∂F1. For a generic point of F1 the stabilizer subgroup of Γ 1 is trivial. But the two points ω = 1 2(−1 + i √ 3) = e2πi/3 and i are stabilized by the cyclic subgroups of order 3 and 2 generated by ST and S respectively. This means that in the quotient manifold Γ1\H, ω and i are singular. (From a metric point of view, they have neighborhoods which are not discs, but quotients of a disc by these cyclic subgroups, with total angle 120◦or 180◦ instead of 360◦.) If we deﬁne an integer nP for every P ∈Γ1\H as the order of the stabilizer in Γ 1 of any point in H representing P, then nP equals 2 or 3 if P is Γ1-equivalent to i or ω and nP = 1 otherwise. We also have to consider the compactiﬁed quotient Γ1\H obtained by adding a point at inﬁnity (“cusp”) to Γ1\H. More precisely, for Y > 1 the image in Γ1\H of the part of H above the line I(z) = Y can be identiﬁed via q = e2πiz with the punctured disc 0 < q < e−2πY . Equation (3) tells us that a holomorphic modular form of any weight k on Γ1 is not only a well-deﬁned function on this punctured disc, but extends holomorphically to the point q = 0. We therefore deﬁne Γ1\H = Γ1\H ∪{∞}, where the point “∞” corresponds to q = 0, with q as a local parameter. One can also think of Γ1\H as the quotient of H by Γ1, where H = H∪Q∪{∞} is the space obtained by adding the full Γ1-orbit Q∪{∞} of ∞to H. We deﬁne the order of vanishing at inﬁnity of f, denoted ord∞(f), as the smallest integer n such that an ̸= 0 in the Fourier expansion (3). Proposition 2. Let f be a non-zero modular form of weight k on Γ1. Then P ∈Γ1\H 1 nP ordP (f) + ord∞(f) = k 12 . (6) Proof. Let D be the closed set obtained from F1 by deleting ε-neighborhoods of all zeros of f and also the “neighborhood of inﬁnity” I(z) > Y = ε−1, where ε is chosen suﬃciently small that all of these neighborhoods are disjoint (see Fig. 2.) Since f has no zeros in D, Cauchy’s theorem implies that the integral of d log f(z) = f ′(z) f(z) dz over the boundary of D is 0. This boundary consists of several parts: the horizontal line from −1 2 + iY to 1 2 + iY , the two vertical lines from ω to −1 2 +iY and from ω +1 to 1 2 +iY (with some ε-neighborhoods removed), the arc of the circle \|z\| = 1 from ω to ω + 1 (again with some ε-neighborhoods deleted), and the boundaries of the ε-neighborhoods of the zeros P of f. These latter have total angle 2π if P is not an elliptic ﬁxed point 10 D. Zagier Fig. 2. The zeros of a modular form (they consist of a full circle if P is an interior point of F1 and of two half-circles if P corresponds to a boundary point of F1 diﬀerent from ω, ω + 1 or i), and total angle π or 2π/3 if P ∼i or ω. The corresponding contributions to the integral are as follows. The two vertical lines together give 0, because f takes on the same value on both and the orientations are opposite. The horizontal line from −1 2 +iY to 1 2 +iY gives a contribution 2πi ord∞(f), because d(log f) is the sum of ord∞(f) dq/q and a function of q which is holomorphic at 0, and this integral corresponds to an integral around a small circle \|q\| = e−2πY around q = 0. The integral on the boundary of the deleted ε-neighborhood of a zero P of f contributes 2πi ordP (f) if nP = 1 by Cauchy’s theorem, because ordP (f) is the residue of d(log f(z)) at z = P, while for nP > 1 we must divide by nP because we are only integrating over one-half or one-third of the full circle around P. Finally, the integral along the bottom arc contributes πik/6, as we see by breaking up this arc into its left and right halves and applying the formula d log f(Sz) = d log f(z)+kdz/z, which is a consequence of the transformation equation (4). Combining all of these terms with the appropriate signs dictated by the orientation, we obtain (6). The details are left to the reader. Corollary 1. The dimension of Mk(Γ1) is 0 for k < 0 or k odd, while for even k ≥0 we have dim Mk(Γ1) ≤ [k/12] + 1 if k ̸≡2 (mod 12) [k/12] if k ≡2 (mod 12) . (7) Proof. Let m = [k/12] + 1 and choose m distinct non-elliptic points Pi ∈ Γ1\H. Given any modular forms f1, . . . , fm+1 ∈Mk(Γ1), we can ﬁnd a linear Elliptic Modular Forms and Their Applications 11 combination f of them which vanishes in all Pi, by linear algebra. But then f ≡0 by the proposition, since m > k/12, so the fi are linearly dependent. Hence dim Mk(Γ1) ≤m. If k ≡2 (mod 12) we can improve the estimate by 1 by noticing that the only way to satisfy (6) is to have (at least) a simple zero at i and a double zero at ω (contributing a total of 1/2 + 2/3 = 7/6 to ordP (f)/nP ) together with k/12 −7/6 = m −1 further zeros, so that the same argument now gives dim Mk(Γ1) ≤m −1. Corollary 2. The space M12(Γ1) has dimension ≤2, and if f, g ∈M12(Γ1) are linearly independent, then the map z →f(z)/g(z) gives an isomorphism from Γ1\H ∪{∞} to P1(C). Proof. The ﬁrst statement is a special case of Corollary 1. Suppose that f and g are linearly independent elements of M12(Γ1). For any (0, 0) ̸= (λ, μ) ∈ C2 the modular form λf −μg of weight 12 has exactly one zero in Γ1\H∪{∞} by Proposition 2, so the modular function ψ = f/g takes on every value (μ : λ) ∈P1(C) exactly once, as claimed. We will make an explicit choice of f, g and ψ in §2.4, after we have introduced the “discriminant function” Δ(z) ∈M12(Γ1). The true interpretation of the factor 1/12 multiplying k in equation (6) is as 1/4π times the volume of Γ1\H, taken with respect to the hyperbolic metric. We say only a few words about this, since these ideas will not be used again. To give a metric on a manifold is to specify the distance be-tween any two suﬃciently near points. The hyperbolic metric in H is deﬁned by saying that the hyperbolic distance between two points in a small neigh-borhood of a point z = x + iy ∈H is very nearly 1/y times the Euclidean distance between them, so the volume element, which in Euclidean geometry is given by the 2-form dx dy, is given in hyperbolic geometry by dμ = y−2dx dy. Thus Vol Γ1\H) = F1 dμ = 1/2 −1/2 ∞ √ 1−x2 dy y2 dx = 1/2 −1/2 dx √ 1 −x2 = arcsin(x) 1/2 −1/2 = π 3 . Now we can consider other discrete subgroups of SL(2, R) which have a fun-damental domain of ﬁnite volume. (Such groups are usually called Fuchsian groups of the ﬁrst kind, and sometimes “lattices”, but we will reserve this lat-ter term for discrete cocompact subgroups of Euclidean spaces.) Examples are the subgroups Γ ⊂Γ1 of ﬁnite index, for which the volume of Γ\H is π/3 times the index of Γ in Γ1 (or more precisely, of the image of Γ in PSL(2, R) in Γ 1). If Γ is any such group, then essentially the same proof as for Pro-position 2 shows that the number of Γ-inequivalent zeros of any non-zero 12 D. Zagier modular form f ∈Mk(Γ) equals k Vol(Γ\H)/4π, where just as in the case of Γ1 we must count the zeros at elliptic ﬁxed points or cusps of Γ with appropriate multiplicities. The same argument as for Corollary 1 of Propos-ition 2 then tells us Mk(Γ) is ﬁnite dimensional and gives an explicit upper bound: Proposition 3. Let Γ be a discrete subgroup of SL(2, R) for which Γ\H has ﬁnite volume V . Then dim Mk(Γ) ≤kV 4π + 1 for all k ∈Z. In particular, we have Mk(Γ) = {0} for k < 0 and M0(Γ) = C, i.e., there are no holomorphic modular forms of negative weight on any group Γ, and the only modular forms of weight 0 are the constants. A further consequence is that any three modular forms on Γ are algebraically dependent. (If f, g, h were algebraically independent modular forms of positive weights, then for large k the dimension of Mk(Γ) would be at least the number of monomials in f, g, h of total weight k, which is bigger than some positive multiple of k2, contradicting the dimension estimate given in the proposition.) Equivalent-ly, any two modular functions on Γ are algebraically dependent, since every modular function is a quotient of two modular forms. This is a special case of the general fact that there cannot be more than n algebraically indepen-dent algebraic functions on an algebraic variety of dimension n. But the most important consequence of Proposition 3 from our point of view is that it is the origin of the (unreasonable?) eﬀectiveness of modular forms in number theory: if we have two interesting arithmetic sequences {an}n≥0 and {bn}n≥0 and conjecture that they are identical (and clearly many results of number theory can be formulated in this way), then if we can show that both anqn and bnqn are modular forms of the same weight and group, we need only verify the equality an = bn for a ﬁnite number of n in order to know that it is true in general. There will be many applications of this principle in these notes. 2 First Examples: Eisenstein Series and the Discriminant Function In this section we construct our ﬁrst examples of modular forms: the Eisenstein series Ek(z) of weight k > 2 and the discriminant function Δ(z) of weight 12, whose deﬁnition is closely connected to the non-modular Eisenstein series E2(z). 2.1 Eisenstein Series and the Ring Structure of M∗(Γ1) There are two natural ways to introduce the Eisenstein series. For the ﬁrst, we observe that the characteristic transformation equation (2) of a modular Elliptic Modular Forms and Their Applications 13 form can be written in the form f\|kγ = f for γ ∈Γ, where f\|kγ : H →C is deﬁned by f kg (z) = (cz + d)−k f az + b cz + d z ∈C, g = a b c d ∈SL(2, R) . (8) One checks easily that for ﬁxed k ∈Z, the map f →f\|kg deﬁnes an operation of the group SL(2, R) (i.e., f\|k(g1g2) = (f\|kg1)\|kg2 for all g1, g2 ∈SL(2, R)) on the vector space of holomorphic functions in H having subexponential or polynomial growth. The space Mk(Γ) of holomorphic modular forms of weight k on a group Γ ⊂SL(2, R) is then simply the subspace of this vector space ﬁxed by Γ. If we have a linear action v →v\|g of a ﬁnite group G on a vector space V , then an obvious way to construct a G-invariant vector in V is to start with an arbitrary vector v0 ∈V and form the sum v = g∈G v0\|g (and to hope that the result is non-zero). If the vector v0 is invariant under some sub-group G0 ⊂G, then the vector v0\|g depends only on the coset G0g ∈G0\G and we can form instead the smaller sum v = g∈G0\G v0\|g, which again is G-invariant. If G is inﬁnite, the same method sometimes applies, but we now have to be careful about convergence. If the vector v0 is ﬁxed by an inﬁnite subgroup G0 of G, then this improves our chances because the sum over G0\G is much smaller than a sum over all of G (and in any case g∈G v\|g has no chance of converging since every term occurs inﬁnitely often). In the context when G = Γ ⊂SL(2, R) is a Fuchsian group (acting by \|k) and v0 a ra-tional function, the modular forms obtained in this way are called Poincaré series. An especially easy case is that when v0 is the constant function “1” and Γ0 = Γ∞, the stabilizer of the cusp at inﬁnity. In this case the series Γ∞\Γ 1\|kγ is called an Eisenstein series. Let us look at this series more carefully when Γ = Γ1. A matrix a b c d ∈ SL(2, R) sends ∞to a/c, and hence belongs to the stabilizer of ∞if and only if c = 0. In Γ1 these are the matrices ± 1 n 0 1 with n ∈Z, i.e., up to sign the matrices T n. We can assume that k is even (since there are no modular forms of odd weight on Γ1) and hence work with Γ1 = PSL(2, Z), in which case the stabilizer Γ ∞is the inﬁnite cyclic group generated by T . If we multiply an arbitrary matrix γ = a b c d on the left by 1 n 0 1 , then the resulting matrix γ′ = a+nc b+nd c d has the same bottom row as γ. Conversely, if γ′ = a′ b′ c d ∈Γ1 has the same bottom row as γ, then from (a′−a)d−(b′−b)c = det(γ)−det(γ′) = 0 and (c, d) = 1 (the elements of any row or column of a matrix in SL(2, Z) are coprime!) we see that a′ −a = nc, b′ −b = nd for some n ∈Z, i.e., γ′ = T nγ. Since every coprime pair of integers occurs as the bottom row of a matrix in SL(2, Z), these considerations give the formula Ek(z) = γ∈Γ∞\Γ1 1 kγ = γ∈Γ ∞\Γ 1 1 kγ = 1 2 c, d∈Z (c,d) = 1 1 (cz + d)k (9) 14 D. Zagier for the Eisenstein series (the factor 1 2 arises because (c d) and (−c −d) give the same element of Γ1\Γ 1). It is easy to see that this sum is absolutely convergent for k > 2 (the number of pairs (c, d) with N ≤\|cz + d\| < N + 1 is the number of lattice points in an annulus of area π(N + 1)2 −πN 2 and hence is O(N), so the series is majorized by ∞ N=1 N 1−k), and this absolute convergence guarantees the modularity (and, since it is locally uniform in z, also the holomorphy) of the sum. The function Ek(z) is therefore a modular form of weight k for all even k ≥4. It is also clear that it is non-zero, since for I(z) →∞all the terms in (9) except (c d) = (±1 0) tend to 0, the convergence of the series being suﬃciently uniform that their sum also goes to 0 (left to the reader), so Ek(z) = 1 + o(1) ̸= 0. The second natural way of introducing the Eisenstein series comes from the interpretation of modular forms given in the beginning of §1.1, where we identiﬁed solutions of the transformation equation (2) with functions on lat-tices Λ ⊂C satisfying the homogeneity condition F(λΛ) = λ−kF(Λ) under homotheties Λ →λΛ. An obvious way to produce such a homogeneous func-tion – if the series converges – is to form the sum Gk(Λ) = 1 2 λ∈Λ∖0 λ−k of the (−k)th powers of the non-zero elements of Λ. (The factor “ 1 2” has again been introduce to avoid counting the vectors λ and −λ doubly when k is even; if k is odd then the series vanishes anyway.) In terms of z ∈H and its associated lattice Λz = Z.z + Z.1, this becomes Gk(z) = 1 2 m, n∈Z (m,n)̸=(0,0) 1 (mz + n)k (k > 2, z ∈H) , (10) where the sum is again absolutely and locally uniformly convergent for k > 2, guaranteeing that Gk ∈Mk(Γ1). The modularity can also be seen directly by noting that (Gk\|kγ)(z) = m,n(m′z + n′)−k where (m′, n′) = (m, n)γ runs over the non-zero vectors of Z2 ∖{(0, 0)} as (m, n) does. In fact, the two functions (9) and (10) are proportional, as is easily seen: any non-zero vector (m, n) ∈Z2 can be written uniquely as r(c, d) with r (the greatest common divisor of m and n) a positive integer and c and d coprime integers, so Gk(z) = ζ(k) Ek(z) , (11) where ζ(k) = r≥1 1/rk is the value at k of the Riemann zeta function. It may therefore seem pointless to have introduced both deﬁnitions. But in fact, this is not the case. First of all, each deﬁnition gives a distinct point of view and has advantages in certain settings which are encountered at later points in the theory: the Ek deﬁnition is better in contexts like the famous Rankin-Selberg method where one integrates the product of the Eisenstein series with another modular form over a fundamental domain, while the Gk deﬁnition is better for analytic calculations and for the Fourier development given in §2.2. Elliptic Modular Forms and Their Applications 15 Moreover, if one passes to other groups, then there are σ Eisenstein series of each type, where σ is the number of cusps, and, although they span the same vector space, they are not individually proportional. In fact, we will actually want to introduce a third normalization Gk(z) = (k −1)! (2πi)k Gk(z) (12) because, as we will see below, it has Fourier coeﬃcients which are rational numbers (and even, with one exception, integers) and because it is a normal-ized eigenfunction for the Hecke operators discussed in §4. As a ﬁrst application, we can now determine the ring structure of M∗(Γ1) Proposition 4. The ring M∗(Γ1) is freely generated by the modular forms E4 and E6. Corollary. The inequality (7) for the dimension of Mk(Γ1) is an equality for all even k ≥0. Proof. The essential point is to show that the modular forms E4(z) and E6(z) are algebraically independent. To see this, we ﬁrst note that the forms E4(z)3 and E6(z)2 of weight 12 cannot be proportional. Indeed, if we had E6(z)2 = λE4(z)3 for some (necessarily non-zero) constant λ, then the meromorphic modular form f(z) = E6(z)/E4(z) of weight 2 would satisfy f 2 = λE4 (and also f 3 = λ−1E6) and would hence be holomorphic (a function whose square is holomorphic cannot have poles), contradicting the inequal-ity dim M2(Γ1) ≤0 of Corollary 1 of Proposition 2. But any two modular forms f1 and f2 of the same weight which are not proportional are necessarily algebraically independent. Indeed, if P(X, Y ) is any polynomial in C[X, Y ] such that P(f1(z), f2(z)) ≡0, then by considering the weights we see that Pd(f1, f2) has to vanish identically for each homogeneous component Pd of P. But Pd(f1, f2)/f d 2 = p(f1/f2) for some polynomial p(t) in one variable, and since p has only ﬁnitely many roots we can only have Pd(f1, f2) ≡0 if f1/f2 is a constant. It follows that E3 4 and E2 6, and hence also E4 and E6, are alge-braically independent. But then an easy calculation shows that the dimension of the weight k part of the subring of M∗(Γ1) which they generate equals the right-hand side of the inequality (7), so that the proposition and corollary follow from this inequality. 2.2 Fourier Expansions of Eisenstein Series Recall from (3) that any modular form on Γ1 has a Fourier expansion of the form ∞ n=0 anqn, where q = e2πiz. The coeﬃcients an often contain interesting arithmetic information, and it is this that makes modular forms important for classical number theory. For the Eisenstein series, normalized by (12), the coeﬃcients are given by: 16 D. Zagier Proposition 5. The Fourier expansion of the Eisenstein series Gk(z) (k even, k > 2) is Gk(z) = −Bk 2k + ∞ n=1 σk−1(n) qn , (13) where Bk is the kth Bernoulli number and where σk−1(n) for n ∈N denotes the sum of the (k −1)st powers of the positive divisors of n. We recall that the Bernoulli numbers are deﬁned by the generating function ∞ k=0 Bkxk/k! = x/(ex −1) and that the ﬁrst values of Bk (k > 0 even) are given by B2 = 1 6, B4 = −1 30, B6 = 1 42, B8 = −1 30, B10 = 5 66, B12 = −691 2730, and B14 = 7 6 . Proof. A well known and easily proved identity of Euler states that n∈Z 1 z + n = π tan πz z ∈C \ Z , (14) where the sum on the left, which is not absolutely convergent, is to be inter-preted as a Cauchy principal value ( = lim N −M where M, N tend to inﬁnity with M −N bounded). The function on the right is periodic of period 1 and its Fourier expansion for z ∈H is given by π tan πz = π cos πz sin πz = πi eπiz + e−πiz eπiz −e−πiz = −πi 1 + q 1 −q = −2πi 1 2 + ∞ r=1 qr , where q = e2πiz. Substitute this into (14), diﬀerentiate k −1 times and divide by (−1)k−1(k −1)! to get n∈Z 1 (z + n)k = (−1)k−1 (k −1)! dk−1 dzk−1 π tan πz = (−2πi)k (k −1)! ∞ r=1 rk−1 qr (k ≥2, z ∈H) , an identity known as Lipschitz’s formula. Now the Fourier expansion of Gk (k > 2 even) is obtained immediately by splitting up the sum in (10) into the terms with m = 0 and those with m ̸= 0: Gk(z) = 1 2 n∈Z n̸=0 1 nk + 1 2 m, n∈Z m̸=0 1 (mz + n)k = ∞ n=1 1 nk + ∞ m=1 ∞ n=−∞ 1 (mz + n)k = ζ(k) + (2πi)k (k −1)! ∞ m=1 ∞ r=1 rk−1 qmr = (2πi)k (k −1)! −Bk 2k + ∞ n=1 σk−1(n) qn , where in the last line we have used Euler’s evaluation of ζ(k) (k > 0 even) in terms of Bernoulli numbers. The result follows. Elliptic Modular Forms and Their Applications 17 The ﬁrst three examples of Proposition 5 are the expansions G4(z) = 1 240 + q + 9q2 + 28q3 + 73q4 + 126q5 + 252q6 + · · · , G6(z) = −1 504 + q + 33q2 + 244q3 + 1057q4 + · · · , G8(z) = 1 480 + q + 129q2 + 2188q3 + · · · . The other two normalizations of these functions are given by G4(z) = 16 π4 3! G4(z) = π4 90 E4(z) , E4(z) = 1 + 240q + 2160q2 + · · · , G6(z) = −64 π6 5! G6(z) = π6 945 E6(z) , E6(z) = 1 −504q −16632q2 −· · · , G8(z) = 256 π8 7! G8(z) = π8 9450 E8(z) , E8(z) = 1 + 480q + 61920q2 + · · · . Remark. We have discussed only Eisenstein series on the full modular group in detail, but there are also various kinds of Eisenstein series for subgroups Γ ⊂Γ1 . We give one example. Recall that a Dirichlet character modulo N ∈N is a homomorphism χ : (Z/NZ)∗→C∗, extended to a map χ : Z →C (traditionally denoted by the same letter) by setting χ(n) equal to χ(n mod N) if (n, N) = 1 and to 0 otherwise. If χ is a non-trivial Dirichlet character and k a positive integer with χ(−1) = (−1)k, then there is an Eisenstein series having the Fourier expansion Gk,χ(z) = ck(χ) + ∞ n=1 d\|n χ(d) dk−1 qn which is a “modular form of weight k and character χ on Γ0(N).” (This means that Gk,χ( az+b cz+d) = χ(a)(cz + d)kGk,χ(z) for any z ∈H and any a b c d ∈ SL(2, Z) with c ≡0 (mod N).) Here ck(χ) ∈Q is a suitable constant, given explicitly by ck(χ) = 1 2L(1 −k, χ), where L(s, χ) is the analytic continuation of the Dirichlet series ∞ n=1 χ(n)n−s . The simplest example, for N = 4 and χ = χ−4 the Dirichlet character modulo 4 given by χ−4(n) = ⎧ ⎪ ⎨ ⎪ ⎩ +1 if n ≡1 (mod 4) , −1 if n ≡3 (mod 4) , 0 if n is even (15) and k = 1, is the series G1,χ−4(z) = c1(χ−4)+ ∞ n=1 ⎛ ⎝ d\|n χ−4(d) ⎞ ⎠qn = 1 4 +q +q2 +q4 +2q5 +q8 +· · · . (16) 18 D. Zagier (The fact that L(0, χ−4) = 2c1(χ−4) = 1 2 is equivalent via the functional equation of L(s, χ−4) to Leibnitz’s famous formula L(1, χ−4) = 1 −1 3 + 1 5 − · · · = π 4 .) We will see this function again in §3.1. ♠ Identities Involving Sums of Powers of Divisors We now have our ﬁrst explicit examples of modular forms and their Fourier expansions and can immediately deduce non-trivial number-theoretic identi-ties. For instance, each of the spaces M4(Γ1), M6(Γ1), M8(Γ1), M10(Γ1) and M14(Γ1) has dimension exactly 1 by the corollary to Proposition 2, and is therefore spanned by the Eisenstein series Ek(z) with leading coeﬃcient 1, so we immediately get the identities E4(z)2 = E8(z) , E4(z)E6(z) = E10(z) , E6(z)E8(z) = E4(z)E10(z) = E14(z) . Each of these can be combined with the Fourier expansion given in Proposit-ion 5 to give an identity involving the sums-of-powers-of-divisors functions σk−1(n), the ﬁrst and the last of these being n−1 m=1 σ3(m)σ3(n −m) = σ7(n) −σ3(n) 120 , n−1 m=1 σ3(m)σ9(n −m) = σ13(n) −11σ9(n) + 10σ3(n) 2640 . Of course similar identities can be obtained from modular forms in higher weights, even though the dimension of Mk(Γ1) is no longer equal to 1. For instance, the fact that M12(Γ1) is 2-dimensional and contains the three modu-lar forms E4E8, E2 6 and E12 implies that the three functions are linearly dependent, and by looking at the ﬁrst two terms of the Fourier expansions we ﬁnd that the relation between them is given by 441E4E8 + 250E2 6 = 691E12, a formula which the reader can write out explicitly as an identity among sums-of-powers-of-divisors functions if he or she is so inclined. It is not easy to obtain any of these identities by direct number-theoretical reasoning (although in fact it can be done). ♥ 2.3 The Eisenstein Series of Weight 2 In §2.1 and §2.2 we restricted ourselves to the case when k > 2, since then the series (9) and (10) are absolutely convergent and therefore deﬁne modular forms of weight k. But the ﬁnal formula (13) for the Fourier expansion of Gk(z) converges rapidly and deﬁnes a holomorphic function of z also for k = 2, so Elliptic Modular Forms and Their Applications 19 in this weight we can simply deﬁne the Eisenstein series G2, G2 and E2 by equations (13), (12), and (11), respectively, i.e., G2(z) = −1 24 + ∞ n=1 σ1(n) qn = −1 24 + q + 3q2 + 4q3 + 7q4 + 6q5 + · · · , G2(z) = −4π2 G2(z) , E2(z) = 6 π2 G2(z) = 1 −24q −72q2 −· · · . (17) Moreover, the same proof as for Proposition 5 still shows that G2(z) is given by the expression (10), if we agree to carry out the summation over n ﬁrst and then over m : G2(z) = 1 2 n̸=0 1 n2 + 1 2 m̸=0 n∈Z 1 (mz + n)2 . (18) The only diﬀerence is that, because of the non-absolute convergence of the double series, we can no longer interchange the order of summation to get the modular transformation equation G2(−1/z) = z2G2(z). (The equation G2(z + 1) = G2(z), of course, still holds just as for higher weights.) Never-theless, the function G2(z) and its multiples E2(z) and G2(z) do have some modular properties and, as we will see later, these are important for many applications. Proposition 6. For z ∈H and a b c d ∈SL(2, Z) we have G2 az + b cz + d = (cz + d)2 G2(z) −πic(cz + d) . (19) Proof. There are many ways to prove this. We sketch one, due to Hecke, since the method is useful in many other situations. The series (10) for k = 2 does not converge absolutely, but it is just at the edge of convergence, since m,n \|mz + n\|−λ converges for any real number λ > 2. We therefore modify the sum slightly by introducing G2,ε(z) = 1 2 m, n ′ 1 (mz + n)2 \|mz + n\|2ε (z ∈H, ε > 0) . (20) (Here ′ means that the value (m, n) = (0, 0) is to be omitted from the summation.) The new series converges absolutely and transforms by G2,ε az+b cz+d = (cz + d)2\|cz + d\|2εG2,ε(z). We claim that lim ε→0 G2,ε(z) exists and equals G2(z) −π/2y, where y = I(z). It follows that each of the three non-holomorphic functions G∗ 2(z) = G2(z) −π 2y , E∗ 2(z) = E2(z) −3 πy , G∗ 2(z) = G2(z) + 1 8πy (21) transforms like a modular form of weight 2, and from this one easily deduces the transformation equation (19) and its analogues for E2 and G2. To prove 20 D. Zagier the claim, we deﬁne a function Iε by Iε(z) = ∞ −∞ dt (z + t)2 \|z + t\|2ε z ∈H, ε > −1 2 . Then for ε > 0 we can write G2,ε − ∞ m=1 Iε(mz) = ∞ n=1 1 n2+2ε + ∞ m=1 ∞ n=−∞ 1 (mz + n)2 \|mz + n\|2ε − n+1 n dt (mz + t)2\|mz + t\|2ε . Both sums on the right converge absolutely and locally uniformly for ε > −1 2 (the second one because the expression in square brackets is O \|mz+n\|−3−2ε by the mean-value theorem, which tells us that f(t) −f(n) for any diﬀeren-tiable function f is bounded in n ≤t ≤n + 1 by maxn≤u≤n+1 \|f ′(u)\|), so the limit of the expression on the right as ε →0 exists and can be obtained simply by putting ε = 0 in each term, where it reduces to G2(z) by (18). On the other hand, for ε > −1 2 we have Iε(x + iy) = ∞ −∞ dt (x + t + iy)2 ((x + t)2 + y2)ε = ∞ −∞ dt (t + iy)2 (t2 + y2)ε = I(ε) y1+2ε , where I(ε) = ∞ −∞(t+i)−2(t2+1)−εdt , so ∞ m=1 Iε(mz) = I(ε)ζ(1+2ε)/y1+2ε for ε > 0. Finally, we have I(0) = 0 (obvious), I′(0) = − ∞ −∞ log(t2 + 1) (t + i)2 dt = 1 + log(t2 + 1) t + i −tan−1 t ∞ −∞ = −π , and ζ(1+2ε) = 1 2ε +O(1), so the product I(ε)ζ(1+2ε)/y1+2ε tends to −π/2y as ε →0. The claim follows. Remark. The transformation equation (18) says that G2 is an example of what is called a quasimodular form, while the functions G∗ 2, E∗ 2 and G∗ 2 deﬁned in (21) are so-called almost holomorphic modular forms of weight 2. We will return to this topic in Section 5. 2.4 The Discriminant Function and Cusp Forms For z ∈H we deﬁne the discriminant function Δ(z) by the formula Δ(z) = e2πiz ∞ n=1 1 −e2πinz24 . (22) Elliptic Modular Forms and Their Applications 21 (The name comes from the connection with the discriminant of the elliptic curve Ez = C/(Z.z + Z.1), but we will not discuss this here.) Since \|e2πiz\| < 1 for z ∈H, the terms of the inﬁnite product are all non-zero and tend exponen-tially rapidly to 1, so the product converges everywhere and deﬁnes a holomor-phic and everywhere non-zero function in the upper half-plane. This function turns out to be a modular form and plays a special role in the entire theory. Proposition 7. The function Δ(z) is a modular form of weight 12 on SL(2, Z). Proof. Since Δ(z) ̸= 0, we can consider its logarithmic derivative. We ﬁnd 1 2πi d dz log Δ(z) = 1−24 ∞ n=1 n e2πinz 1 −e2πinz = 1−24 ∞ m=1 σ1(m) e2πimz = E2(z) , where the second equality follows by expanding e2πinz 1 −e2πinz as a geometric series ∞ r=1 e2πirnz and interchanging the order of summation, and the third equality from the deﬁnition of E2(z) in (17). Now from the transformation equation for E2 (obtained by comparing (19) and(11)) we ﬁnd 1 2πi d dz log Δ az+b cz+d (cz + d)12Δ(z) = 1 (cz + d)2 E2 az + b cz + d −12 2πi c cz + d −E2(z) = 0 . In other words, (Δ\|12γ)(z) = C(γ) Δ(z) for all z ∈H and all γ ∈Γ1, where C(γ) is a non-zero complex number depending only on γ, and where Δ\|12γ is deﬁned as in (8). It remains to show that C(γ) = 1 for all γ. But C : Γ1 →C∗ is a homomorphism because Δ →Δ\|12γ is a group action, so it suﬃces to check this for the generators T = 1 1 0 1 and S = 0 −1 1 0 of Γ1. The ﬁrst is obvious since Δ(z) is a power series in e2πiz and hence periodic of period 1, while the second follows by substituting z = i into the equation Δ(−1/z) = C(S) z12Δ(z) and noting that Δ(i) ̸= 0. Let us look at this function Δ(z) more carefully. We know from Corollary 1 to Proposition 2 that the space M12(Γ1) has dimension at most 2, so Δ(z) must be a linear combination of the two functions E4(z)3 and E6(z)2. From the Fourier expansions E3 4 = 1 + 720q + · · · , E6(z)2 = 1 −1008q + · · · and Δ(z) = q + · · · we see that this relation is given by Δ(z) = 1 1728 E4(z)3 −E6(z)2 . (23) This identity permits us to give another, more explicit, version of the fact that every modular form on Γ1 is a polynomial in E4 and E6 (Proposition 4). In-deed, let f(z) be a modular form of arbitrary even weight k ≥4, with Fourier expansion as in (3). Choose integers a, b ≥0 with 4a + 6b = k (this is always 22 D. Zagier possible) and set h(z) = f(z)−a0E4(z)aE6(z)b /Δ(z). This function is holo-morphic in H (because Δ(z) ̸= 0) and also at inﬁnity (because f −a0Ea 4Eb 6 has a Fourier expansion with no constant term and the Fourier expansion of Δ begins with q), so it is a modular form of weight k −12. By induction on the weight, h is a polynomial in E4 and E6, and then from f = a0Ea 4Eb 6 + Δh and (23) we see that f also is. In the above argument we used that Δ(z) has a Fourier expansion begin-ning q+O(q2) and that Δ(z) is never zero in the upper half-plane. We deduced both facts from the product expansion (22), but it is perhaps worth noting that this is not necessary: if we were simply to deﬁne Δ(z) by equation (23), then the fact that its Fourier expansion begins with q would follow from the knowledge of the ﬁrst two Fourier coeﬃcients of E4 and E6, and the fact that it never vanishes in H would then follow from Proposition 2 because the total number k/12 = 1 of Γ1-inequivalent zeros of Δ is completely accounted for by the ﬁrst-order zero at inﬁnity. We can now make the concrete normalization of the isomorphism between Γ1\H and P1(C) mentioned after Corollary 2 of Proposition 2. In the notation of that proposition, choose f(z) = E4(z)3 and g(z) = Δ(z). Their quotient is then the modular function j(z) = E4(z)3 Δ(z) = q−1 + 744 + 196884 q + 21493760 q2 + · · · , called the modular invariant. Since Δ(z) ̸= 0 for z ∈H, this function is ﬁnite in H and deﬁnes an isomorphism from Γ1\H to C as well as from Γ1\H to P1(C). The next (and most interesting) remarks about Δ(z) concern its Fourier expansion. By multiplying out the product in (22) we obtain the expansion Δ(z) = q ∞ n=1 1 −qn)24 = ∞ n=1 τ(n) qn (24) where q = e2πiz as usual (this is the last time we will repeat this!) and the coeﬃcients τ(n) are certain integers, the ﬁrst values being given by the table n 1 2 3 4 5 6 7 8 9 10 τ(n) 1 −24 252 −1472 4830 −6048 −16744 84480 −113643 −115920 Ramanujan calculated the ﬁrst 30 values of τ(n) in 1915 and observed several remarkable properties, notably the multiplicativity property that τ(pq) = τ(p)τ(q) if p and q are distinct primes (e.g., −6048 = −24 · 252 for p = 2, q = 3) and τ(p2) = τ(p)2 −p11 if p is prime (e.g., −1472 = (−24)2 −2048 for p = 2). This was proved by Mordell the next year and later generalized by Hecke to the theory of Hecke operators, which we will discuss in §4. Ramanujan also observed that \|τ(p)\| was bounded by 2p5√p for primes p < 30, and conjectured that this holds for all p. This property turned out Elliptic Modular Forms and Their Applications 23 to be immeasurably deeper than the assertion about multiplicativity and was only proved in 1974 by Deligne as a consequence of his proof of the famous Weil conjectures (and of his previous, also very deep, proof that these conjectures implied Ramanujan’s). However, the weaker inequality \|τ(p)\| ≤Cp6 with some eﬀective constant C > 0 is much easier and was proved in the 1930’s by Hecke. We reproduce Hecke’s proof, since it is simple. In fact, the proof applies to a much more general class of modular forms. Let us call a modular form on Γ1 a cusp form if the constant term a0 in the Fourier expansion (3) is zero. Since the constant term of the Eisenstein series Gk(z) is non-zero, any modular form can be written uniquely as a linear combination of an Eisenstein series and a cusp form of the same weight. For the former the Fourier coeﬃcients are given by (13) and grow like nk−1 (since nk−1 ≤σk−1(n) < ζ(k −1)nk−1 ). For the latter, we have: Proposition 8. Let f(z) be a cusp form of weight k on Γ1 with Fourier expan-sion ∞ n=1 anqn. Then \|an\| ≤Cnk/2 for all n, for some constant C depending only on f. Proof. From equations (1) and (2) we see that the function z →yk/2\|f(z)\| on H is Γ1-invariant. This function tends rapidly to 0 as y = I(z) →∞ (because f(z) = O(q) by assumption and \|q\| = e−2πy), so from the form of the fundamental domain of Γ1 as given in Proposition 1 it is clearly bounded. Thus we have the estimate \|f(z)\| ≤c y−k/2 (z = x + iy ∈H) (25) for some c > 0 depending only on f. Now the integral representation an = e2πny 1 0 f(x + iy) e−2πinx dx for an, valid for any y > 0, show that \|an\| ≤cy−k/2e2πny. Taking y = 1/n (or, optimally, y = k/4πn) gives the estimate of the proposition with C = c e2π (or, optimally, C = c (4πe/k)k/2 ). Remark. The deﬁnition of cusp forms given above is actually valid only for the full modular group Γ1 or for other groups having only one cusp. In general one must require the vanishing of the constant term of the Fourier expansion of f, suitably deﬁned, at every cusp of the group Γ, in which case it again follows that f can be estimated as in (25). Actually, it is easier to simply deﬁne cusp forms of weight k as modular forms for which yk/2f(x + iy) is bounded, a deﬁnition which is equivalent but does not require the explicit knowledge of the Fourier expansion of the form at every cusp. ♠Congruences for τ(n) As a mini-application of the calculations of this and the preceding sections we prove two simple congruences for the Ramanujan tau-function deﬁned by 24 D. Zagier equation (24). First of all, let us check directly that the coeﬃcient τ(n) of qn of the function deﬁned by (23) is integral for all n. (This fact is, of course, obvious from equation (22).) We have Δ = (1 + 240A)3 −(1 −504B)2 1728 = 5 A −B 12 +B +100A2−147B2+8000A3 (26) with A = ∞ n=1 σ3(n)qn and B = ∞ n=1 σ5(n)qn. But σ5(n)−σ3(n) is divisib-le by 12 for every n (because 12 divides d5 −d3 for every d), so (A −B)/12 has integral coeﬃcients. This gives the integrality of τ(n), and even a congru-ence modulo 2. Indeed, we actually have σ5(n) ≡σ3(n) (mod 24), because d3(d2 −1) is divisible by 24 for every d, so (A −B)/12 has even coeﬃ-cients and (26) gives Δ ≡B + B2 (mod 2) or, recalling that ( anqn)2 ≡ anq2n (mod 2) for every power series anqn with integral coeﬃcients, τ(n) ≡σ5(n) + σ5(n/2) (mod 2), where σ5(n/2) is deﬁned as 0 if 2 ∤n. But σ5(n), for any integer n, is congruent modulo 2 to the sum of the odd divisors of n, and this is odd if and only if n is a square or twice a square, as one sees by writing n = 2sn0 with n0 odd and pairing the complementary divisors of n0. It follows that σ5(n) + σ5(n/2) is odd if and only if n is an odd square, so we get the congruence: τ(n) ≡ 1 (mod 2) if n is an odd square , 0 (mod 2) otherwise . (27) In a diﬀerent direction, from dim M12(Γ1) = 2 we immediately deduce the linear relation G12(z) = Δ(z) + 691 156 E4(z)3 720 + E6(z)3 1008 and from this a famous congruence of Ramanujan, τ(n) ≡σ11(n) (mod 691) ∀n ≥1 , (28) where the “691” comes from the numerator of the constant term −B12/24 of G12. ♥ 3 Theta Series If Q is a positive deﬁnite integer-valued quadratic form in m variables, then there is an associated modular form of weight m/2, called the theta series of Q, whose nth Fourier coeﬃcient for every integer n ≥0 is the number of representations of n by Q. This provides at the same time one of the main constructions of modular forms and one of the most important sources of applications of the theory. In 3.1 we consider unary theta series (m = 1), while Elliptic Modular Forms and Their Applications 25 the general case is discussed in 3.2. The unary case is the most classical, going back to Jacobi, and already has many applications. It is also the basis of the general theory, because any quadratic form can be diagonalized over Q (i.e., by passing to a suitable sublattice it becomes the direct sum of m quadratic forms in one variable). 3.1 Jacobi’s Theta Series The simplest theta series, corresponding to the unary (one-variable) quadratic form x →x2, is Jacobi’s theta function θ(z) = n∈Z qn2 = 1 + 2q + 2q4 + 2q9 + · · · , (29) where z ∈H and q = e2πiz as usual. Its modular transformation properties are given as follows. Proposition 9. The function θ(z) satisﬁes the two functional equations θ(z + 1) = θ(z) , θ −1 4z = 2z i θ(z) (z ∈H) . (30) Proof. The ﬁrst equation in (30) is obvious since θ(z) depends only on q. For the second, we use the Poisson transformation formula. Recall that this formula says that for any function f : R →C which is smooth and small at inﬁnity, we have n∈Z f(n) = n∈Z f(n), where f(y) = ∞ −∞e2πixy f(x) dx is the Fourier transform of f. (Proof : the sum n∈Z f(n + x) is convergent and deﬁnes a function g(x) which is periodic of period 1 and hence has a Fourier expansion g(x) = n∈Z cne2πinx with cn = 1 0 g(x)e−2πinxdx = f(−n), so n f(n) = g(0) = n cn = n f(−n) = n f(n).) Applying this to the function f(x) = e−πtx2, where t is a positive real number, and noting that f(y) = ∞ −∞ e−πtx2+2πixy dx = e−πy2/t √ t ∞ −∞ e−πu2 du = e−πy2/t √ t (substitution u = √ t (x −iy/t) followed by a shift of the path of integration), we obtain ∞ n=−∞ e−πn2t = 1 √ t ∞ n=−∞ e−πn2/t (t > 0) . This proves the second equation in (30) for z = it/2 lying on the posi-tive imaginary axis, and the general case then follows by analytic continu-ation. 26 D. Zagier The point is now that the two transformations z →z + 1 and z →−1/4z generate a subgroup of SL(2, R) which is commensurable with SL(2, Z), so (30) implies that the function θ(z) is a modular form of weight 1/2. (We have not deﬁned modular forms of half-integral weight and will not discuss their theory in these notes, but the reader can simply interpret this statement as saying that θ(z)2 is a modular form of weight 1.) More speciﬁcally, for every N ∈N we have the “congruence subgroup” Γ0(N) ⊆Γ1 = SL(2, Z), consisting of matrices a b c d ∈Γ1 with c divisible by N, and the larger group Γ + 0 (N) = ⟨Γ0(N), WN⟩= Γ0(N) ∪Γ0(N)WN, where WN = 1 √ N 0 −1 N 0 (“Fricke involution”) is an element of SL(2, R) of order 2 which normalizes Γ0(N). The group Γ + 0 (N) contains the elements T = 1 1 0 1 and WN for any N. In general they generate a subgroup of inﬁnite index, so that to check the modularity of a given function it does not suﬃce to verify its behavior just for z →z + 1 and z →−1/Nz, but for N = 4 (like for N = 1 !) they generate the full group and this is suﬃcient. The proof is simple. Since W 2 N = −1, it is suﬃcient to show that the two matrices T and T = W4T W −1 4 = 1 0 4 1 generate the image of Γ0(4) in PSL(2, R), i.e., that any element γ = a b c d ∈Γ0(4) is, up to sign, a word in T and T. Now a is odd, so \|a\| ̸= 2\|b\|. If \|a\| < 2\|b\|, then either b+a or b−a is smaller than b in absolute value, so replacing γ by γ·T ±1 decreases a2 + b2. If \|a\| > 2\|b\| ̸= 0, then either a + 4b or a −4b is smaller than a in absolute value, so replacing γ by γ · T ±1 decreases a2 + b2. Thus we can keep multiplying γ on the right by powers of T and T until b = 0, at which point ±γ is a power of T. Now, by the principle “a ﬁnite number of q-coeﬃcients suﬃce” formu-lated at the end of Section 1, the mere fact that θ(z) is a modular form is already enough to let one prove non-trivial identities. (We enunciated the principle only in the case of forms of integral weight, but even without know-ing the details of the theory it is clear that it then also applies to half-integral weight, since a space of modular forms of half-integral weight can be mapped injectively into a space of modular forms of the next higher in-tegral weight by multiplying by θ(z).) And indeed, with almost no eﬀort we obtain proofs of two of the most famous results of number theory of the 17th and 18th centuries, the theorems of Fermat and Lagrange about sums of squares. ♠Sums of Two and Four Squares Let r2(n) = # (a, b) ∈Z2 \| a2 + b2 = n be the number of repre-sentations of an integer n ≥0 as a sum of two squares. Since θ(z)2 = a∈Z qa2 b∈Z qb2 , we see that r2(n) is simply the coeﬃcient of qn in θ(z)2. From Proposition 9 and the just-proved fact that Γ0(4) is generated by −Id2, T and T, we ﬁnd that the function θ(z)2 is a “modular form of weight 1 and character χ−4 on Γ0(4)” in the sense explained in the paragraph preceding equation (15), where χ−4 is the Dirichlet character modulo 4 deﬁned by (15). Elliptic Modular Forms and Their Applications 27 Since the Eisenstein series G1,χ−4 in (16) is also such a modular form, and since the space of all such forms has dimension at most 1 by Proposition 3 (because Γ0(4) has index 6 in SL(2, Z) and hence volume 2π), these two func-tions must be proportional. The proportionality factor is obviously 4, and we obtain: Proposition 10. Let n be a positive integer. Then the number of represen-tations of n as a sum of two squares is 4 times the sum of (−1)(d−1)/2, where d runs over the positive odd divisors of n . Corollary (Theorem of Fermat). Every prime number p ≡1 (mod 4) is a sum of two squares. Proof of Corollary. We have r2(p) = 4 1 + (−1)(p−1)/4 = 8 ̸= 0. The same reasoning applies to other powers of θ. In particular, the number r4(n) of representations of an integer n as a sum of four squares is the coef-ﬁcient of qn in the modular form θ(z)4 of weight 2 on Γ0(4), and the space of all such modular forms is at most two-dimensional by Proposition 3. To ﬁnd a basis for it, we use the functions G2(z) and G∗ 2(z) deﬁned in equa-tions (17) and (21). We showed in §2.3 that the latter function transforms with respect to SL(2, Z) like a modular form of weight 2, and it follows easily that the three functions G∗ 2(z), G∗ 2(2z) and G∗ 2(4z) transform like modular forms of weight 2 on Γ0(4) (exercise!). Of course these three functions are not holomorphic, but since G∗ 2(z) diﬀers from the holomorphic function G2(z) by 1/8πy, we see that the linear combinations G∗ 2(z)−2G∗ 2(2z) = G2(z)−2G2(2z) and G∗ 2(2z) −2G∗ 2(4z) = G2(2z) −2G2(4z) are holomorphic, and since they are also linearly independent, they provide the desired basis for M2(Γ0(4)). Looking at the ﬁrst two Fourier coeﬃcients of θ(z)4 = 1 + 8q + · · · , we ﬁnd that θ(z)4 equals 8 (G2(z)−2G2(2z))+16 (G2(2z)−2G2(4z)). Now comparing coeﬃcients of qn gives: Proposition 11. Let n be a positive integer. Then the number of representa-tions of n as a sum of four squares is 8 times the sum of the positive divisors of n which are not multiples of 4. Corollary (Theorem of Lagrange). Every positive integer is a sum of four squares. ♥ For another simple application of the q-expansion principle, we introduce two variants θM(z) and θF (z) (“M” and “F” for “male” and “female” or “minus sign” and “fermionic”) of the function θ(z) by inserting signs or by shifting the indices by 1/2 in its deﬁnition: θM(z) = n∈Z (−1)n qn2 = 1 −2q + 2q4 −2q9 + · · · , θF (z) = n∈Z+1/2 qn2 = 2q1/4 + 2q9/4 + 2q25/4 + · · · . 28 D. Zagier These are again modular forms of weight 1/2 on Γ0(4). With a little experi-mentation, we discover the identity θ(z)4 = θM(z)4 + θF (z)4 (31) due to Jacobi, and by the q-expansion principle all we have to do to prove it is to verify the equality of a ﬁnite number of coeﬃcients (here just one). In this particular example, though, there is also an easy combinatorial proof, left as an exercise to the reader. The three theta series θ, θM and θF , in a slightly diﬀerent guise and slightly diﬀerent notation, play a role in many contexts, so we say a little more about them. As well as the subgroup Γ0(N) of Γ1, one also has the principal congruence subgroup Γ(N) = {γ ∈Γ1 \| γ ≡Id2 (mod N)} for every integer N ∈N, which is more basic than Γ0(N) because it is a normal subgroup (the kernel of the map Γ1 →SL(2, Z/NZ) given by reduction modulo N). Exceptionally, the group Γ0(4) is isomorphic to Γ(2), simply by conjugation by 2 0 0 1 ∈GL(2, R), so that there is a bijection between modular forms on Γ0(4) of any weight and modular forms on Γ(2) of the same weight given by f(z) →f(z/2). In particular, our three theta functions correspond to three new theta functions θ3(z) = θ(z/2) , θ4(z) = θM(z/2) , θ2(z) = θF (z/2) (32) on Γ(2), and the relation (31) becomes θ4 2 + θ4 4 = θ4 3 . (Here the index “1” is missing because the fourth member of the quartet, θ1(z) = (−1)nq(n+1/2)2/2 is identically zero, as one sees by sending n to −n −1. It may look odd that one keeps a whole notation for the zero function. But in fact the functions θi(z) for 1 ≤i ≤4 are just the “Thetanullwerte” or “theta zero-values” of the two-variable series θi(z, u) = εn qn2/2 e2πinu, where the sum is over Z or Z + 1 2 and εn is either 1 or (−1)n, none of which vanishes identically. The functions θi(z, u) play a basic role in the theory of elliptic functions and are also the simplest example of Jacobi forms, a theory which is related to many of the themes treated in these notes and is also important in connection with Siegel modular forms of degree 2 as discussed in Part III of this book.) The quotient group Γ1/Γ(2) ∼ = SL(2, Z/2Z), which has order 6 and is isomorphic to the symmetric group S3 on three symbols, acts as the latter on the modular forms θi(z)8, while the fourth powers transform by Θ(z) := ⎛ ⎝ θ2(z)4 −θ3(z)4 θ4(z)4 ⎞ ⎠ ⇒ Θ(z + 1) = − ⎛ ⎝ 1 0 0 0 0 1 0 1 0 ⎞ ⎠Θ(z), z−2Θ −1 z = − ⎛ ⎝ 0 0 1 0 1 0 1 0 0 ⎞ ⎠Θ(z) . Elliptic Modular Forms and Their Applications 29 This illustrates the general principle that a modular form on a subgroup of ﬁnite index of the modular group Γ1 can also be seen as a component of a vector-valued modular form on Γ1 itself. The full ring M∗(Γ(2)) of modu-lar forms on Γ(2) is generated by the three components of Θ(z) (or by any two of them, since their sum is zero), while the subring M∗(Γ(2))S3 = M∗(Γ1) is generated by the modular forms θ2(z)8 + θ3(z)8 + θ4(z)8 and θ2(z)4 + θ3(z)4 θ3(z)4 + θ4(z)4 θ4(z)4 −θ2(z)4 of weights 4 and 6 (which are then equal to 2E4(z) and 2E6(z), respectively). Finally, we see that 1 256θ2(z)8θ3(z)8θ4(z)8 is a cusp form of weight 12 on Γ1 and is, in fact, equal to Δ(z). This last identity has an interesting consequence. Since Δ(z) is non-zero for all z ∈H, it follows that each of the three theta-functions θi(z) has the same property. (One can also see this by noting that the “visible” zero of θ2(z) at inﬁnity accounts for all the zeros allowed by the formula discussed in §1.3, so that this function has no zeros at ﬁnite points, and then the same holds for θ3(z) and θ4(z) because they are related to θ2 by a modular transformation.) This suggests that these three functions, or equivalently their Γ0(4)-versions θ, θM and θF , might have a product expansion similar to that of the function Δ(z), and indeed this is the case: we have the three identities θ(z) = η(2z)5 η(z)2η(4z)2 , θM(z) = η(z)2 η(2z) , θF (z) = 2 η(4z)2 η(2z) , (33) where η(z) is the “Dedekind eta-function” η(z) = Δ(z)1/24 = q1/24 ∞ n=1 1 −qn . (34) The proof of (33) is immediate by the usual q-expansion principle: we multiply the identities out (writing, e.g., the ﬁrst as θ(z)η(z)2η(4z)2 = η(2z)5) and then verify the equality of enough coeﬃcients to account for all possible zeros of a modular form of the corresponding weight. More eﬃciently, we can use our knowledge of the transformation behavior of Δ(z) and hence of η(z) under Γ1 to see that the quotients on the right in (33) are ﬁnite at every cusp and hence, since they also have no poles in the upper half-plane, are holomorphic modular forms of weight 1/2, after which the equality with the theta-functions on the left follows directly. More generally, one can ask when a quotient of products of eta-functions is a holomorphic modular form. Since η(z) is non-zero in H, such a quotient never has ﬁnite zeros, and the only issue is whether the numerator vanishes to at least the same order as the denominator at each cusp. Based on extensive numerical calculations, I formulated a general conjecture saying that there are essentially only ﬁnitely many such products of any given weight, and a second explicit conjecture giving the complete list for weight 1/2 (i.e., when the number of η’s in the numerator is one bigger than in the denominator). Both conjectures were proved by Gerd Mersmann in a brilliant Master’s thesis. 30 D. Zagier For the weight 1/2 result, the meaning of “essentially” is that the product should be primitive, i.e., it should have the form η(niz)ai where the ni are positive integers with no common factor. (Otherwise one would obtain inﬁnitely many examples by rescaling, e.g., one would have both θM(z) = η(z)2/η(2z) and θM(2z) = η(2z)2/η(4z) on the list.) The classiﬁcation is then as follows: Theorem (Mersmann). There are precisely 14 primitive eta-products which are holomorphic modular forms of weight 1/2 : η(z) , η(z)2 η(2z) , η(2z)2 η(z) , η(z) η(4z) η(2z) , η(2z)3 η(z) η(4z) , η(2z)5 η(z)2η(4z)2 , η(z)2η(6z) η(2z) η(3z) , η(2z)2η(3z) η(z) η(6z) , η(2z) η(3z)2 η(z) η(6z) , η(z) η(6z)2 η(2z) η(3z) , η(z) η(4z) η(6z)2 η(2z) η(3z) η(12z) , η(2z)2η(3z) η(12z) η(z) η(4z) η(6z) , η(2z)5η(3z) η(12z) η(z)2η(4z)2η(6z)2 , η(z) η(4z) η(6z)5 η(2z)2η(3z)2η(12z)2 . Finally, we mention that η(z) itself has the theta-series representation η(z) = ∞ n=1 χ12(n) qn2/24 = q1/24 −q25/24 −q49/24 + q121/24 + · · · where χ12(12m ± 1) = 1, χ12(12m ± 5) = −1, and χ12(n) = 0 if n is divisible by 2 or 3. This identity was discovered numerically by Euler (in the simpler-looking but less enlightening version ∞ n=1(1 −qn) = ∞ n=1(−1)n q(3n2+n)/2) and proved by him only after several years of eﬀort. From a modern point of view, his theorem is no longer surprising because one now knows the following beautiful general result, proved by J-P. Serre and H. Stark in 1976: Theorem (Serre–Stark). Every modular form of weight 1/2 is a linear combination of unary theta series. Explicitly, this means that every modular form of weight 1/2 with respect to any subgroup of ﬁnite index of SL(2, Z) is a linear combination of sums of the form n∈Z qa(n+c)2 with a ∈Q>0 and c ∈Q. Euler’s formula for η(z) is a typical case of this, and of course each of the other products given in Mers-mann’s theorem must also have a representation as a theta series. For instance, the last function on the list, η(z)η(4z)η(6z)5/η(2z)2η(3z)2η(12z)2, has the ex-pansion n>0, (n,6)=1 χ8(n)qn2/24, where χ8(n) equals +1 for n ≡±1 (mod 8) and −1 for n ≡±3 (mod 8). We end this subsection by mentioning one more application of the Jacobi theta series. Elliptic Modular Forms and Their Applications 31 ♠The Kac–Wakimoto Conjecture For any two natural numbers m and n, denote by Δm(n) the number of representations of n as a sum of m triangular numbers (numbers of the form a(a −1)/2 with a integral). Since 8a(a −1)/2 + 1 = (2a −1)2, this can also be written as the number rodd m (8n+m) of representations of 8n+m as a sum of m odd squares. As part of an investigation in the theory of aﬃne superalgebras, Kac and Wakimoto were led to conjecture the formula Δ4s2(n) = r1, a1, ..., rs, as ∈Nodd r1a1+···+rsas = 2n+s2 Ps(a1, . . . , as) (35) for m of the form 4s2 (and a similar formula for m of the form 4s(s + 1) ), where Nodd = {1, 3, 5, . . .} and Ps is the polynomial Ps(a1, . . . , as) = i ai · i 0 for x ̸= 0, the matrix A must be positive deﬁnite. This implies that det A > 0. Hence A is non-singular and A−1 exists and belongs to Mm(Q). The level of Q is then deﬁned as the smallest positive integer N = NQ such that NA−1 is again an even integral matrix. We also have the discriminant Δ = ΔQ of A, deﬁned as (−1)m det A. It is always congruent to 0 or 1 modulo 4, so there is an associated character (Kronecker symbol) χΔ, which is the unique Dirichlet character modulo N satisyﬁng χΔ(p) = Δ p (Legendre symbol) for any odd prime p ∤N. (The character χΔ in the special cases Δ = −4, 12 and 8 already occurred in §2.2 (eq. (15)) and §3.1.) The precise description of the modular behavior of ΘQ for m ∈2Z is then: Theorem (Hecke, Schoenberg). Let Q : Z2k →Z be a positive deﬁnite integer-valued form in 2k variables of level N and discriminant Δ. Then ΘQ is a modular form on Γ0(N) of weight k and character χΔ, i.e., we have ΘQ az+b cz+d = χΔ(a) (cz + d)k ΘQ(z) for all z ∈H and a b c d ∈Γ0(N). The proof, as in the unary case, relies essentially on the Poisson sum-mation formula, which gives the identity ΘQ(−1/Nz) = N k/2(z/i)k ΘQ∗(z), where Q∗(x) is the quadratic form associated to NA−1, but ﬁnding the pre-cise modular behavior requires quite a lot of work. One can also in principle reduce the higher rank case to the one-variable case by using the fact that every quadratic form is diagonalizable over Q, so that the sum in (36) can be broken up into ﬁnitely many sub-sums over sublattices or translated sub-lattices of Zm on which Q(x1, . . . , xm) can be written as a linear combination of m squares. There is another language for quadratic forms which is often more conve-nient, the language of lattices. From this point of view, a quadratic form is no longer a homogeneous quadratic polynomial in m variables, but a function Q Elliptic Modular Forms and Their Applications 33 from a free Z-module Λ of rank m to Z such that the associated scalar prod-uct (x, y) = Q(x + y) −Q(x) −Q(y) (x, y ∈Λ) is bilinear. Of course we can always choose a Z-basis of Λ, in which case Λ is identiﬁed with Zm and Q is described in terms of a symmetric matrix A as in (37), the scalar product being given by (x, y) = xtAy, but often the basis-free language is more conve-nient. In terms of the scalar product, we have a length function ∥x∥2 = (x, x) (actually this is the square of the length, but one often says simply “length” for convenience) and Q(x) = 1 2∥x∥2, so that the integer-valued case we are considering corresponds to lattices in which all vectors have even length. One often chooses the lattice Λ inside the euclidean space Rm with its standard length function (x, x) = ∥x∥2 = x2 1 + · · · + x2 m; in this case the square root of det A is equal to the volume of the quotient Rm/Λ, i.e., to the volume of a fundamental domain for the action by translation of the lattice Λ on Rm. In the case when this volume is 1, i.e., when Λ ∈Rm has the same covolume as Zm, the lattice is called unimodular. Let us look at this case in more detail. ♠Invariants of Even Unimodular Lattices If the matrix A in (37) is even and unimodular, then the above theorem tells us that the theta series ΘQ associated to Q is a modular form on the full modular group. This has many consequences. Proposition 12. Let Q : Zm →Z be a positive deﬁnite even unimodular quadratic form in m variables. Then (i) the rank m is divisible by 8, and (ii)the number of representations of n ∈N by Q is given for large n by the formula RQ(n) = −2k Bk σk−1(n) + O nk/2 (n →∞) , (38) where m = 2k and Bk denotes the kth Bernoulli number. Proof. For the ﬁrst part it is enough to show that m cannot be an odd multiple of 4, since if m is either odd or twice an odd number then 4m or 2m is an odd multiple of 4 and we can apply this special case to the quadratic form Q ⊕Q ⊕Q ⊕Q or Q ⊕Q, respectively. So we can assume that m = 2k with k even and must show that k is divisible by 4 and that (38) holds. By the theorem above, the theta series ΘQ is a modular form of weight k on the full modular group Γ1 = SL(2, Z) (necessarily with trivial character, since there are no non-trivial Dirichlet characters modulo 1). By the results of Section 2, this modular form is a linear combination of Gk(z) and a cusp form of weight k, and from the Fourier expansion (13) we see that the coeﬃcient of Gk in this decomposition equals −2k/Bk, since the constant term RQ(0) of ΘQ equals 1. (The only vector of length 0 is the zero vector.) Now Proposition 8 implies the 34 D. Zagier asymptotic formula (38), and the fact that k must be divisible by 4 also follows because if k ≡2 (mod 4) then Bk is positive and therefore the right-hand side of (38) tends to −∞as k →∞, contradicting RQ(n) ≥0. The ﬁrst statement of Proposition 12 is purely algebraic, and purely alge-braic proofs are known, but they are not as simple or as elegant as the modu-lar proof just given. No non-modular proof of the asymptotic formula (38) is known. Before continuing with the theory, we look at some examples, starting in rank 8. Deﬁne the lattice Λ8 ⊂R8 to be the set of vectors belonging to either Z8 or (Z+ 1 2)8 for which the sum of the coordinates is even. This is unimodular because the lattice Z8 ∪(Z + 1 2)8 contains both it and Z8 with the same index 2, and is even because x2 i ≡xi (mod 2) for xi ∈Z and x2 i ≡1 4 (mod 2) for xi ∈Z + 1 2. The lattice Λ8 is sometimes denoted E8 because, if we choose the Z-basis ui = ei −ei+1 (1 ≤i ≤6), u7 = e6 + e7, u8 = −1 2(e1 + · · · + e8) of Λ8, then every ui has length 2 and (ui, uj) for i ̸= j equals −1 or 0 according whether the ith and jth vertices (in a standard numbering) of the “E8” Dynkin diagram in the theory of Lie algebras are adjacent or not. The theta series of Λ8 is a modular form of weight 4 on SL(2, Z) whose Fourier expansion begins with 1, so it is necessarily equal to E4(z), and we get “for free” the information that for every integer n ≥1 there are exactly 240 σ3(n) vectors x in the E8 lattice with (x, x) = 2n. From the uniqueness of the modular form E4 ∈M4(Γ1) we in fact get that rQ(n) = 240σ3(n) for any even unimodular quadratic form or lattice of rank 8, but here this is not so interesting because the known classiﬁcation in this rank says that Λ8 is, in fact, the only such lattice up to isomorphism. However, in rank 16 one knows that there are two non-equivalent lattices: the direct sum Λ8 ⊕Λ8 and a second lattice Λ16 which is not decomposable. Since the theta series of both lattices are modular forms of weight 8 on the full modular group with Fourier expansions beginning with 1, they are both equal to the Eisen-stein series E8(z), so we have rΛ8⊕Λ8(n) = rΛ16(n) = 480 σ7(n) for all n ≥1, even though the two lattices in question are distinct. (Their distinctness, and a great deal of further information about the relative positions of vectors of various lengths in these or in any other lattices, can be obtained by using the theory of Jacobi forms which was mentioned brieﬂy in §3.1 rather than just the theory of modular forms.) In rank 24, things become more interesting, because now dim M12(Γ1) = 2 and we no longer have uniqueness. The even unimodular lattices of this rank were classiﬁed completely by Niemeyer in 1973. There are exactly 24 of them up to isomorphism. Some of them have the same theta series and hence the same number of vectors of any given length (an obvious such pair of lattices being Λ8⊕Λ8⊕Λ8 and Λ8⊕Λ16), but not all of them do. In particular, exactly one of the 24 lattices has the property that it has no vectors of length 2. This is the famous Leech lattice (famous among other reasons because it has a huge group of automorphisms, closely related to the monster group and Elliptic Modular Forms and Their Applications 35 other sporadic simple groups). Its theta series is the unique modular form of weight 12 on Γ1 with Fourier expansion starting 1 + 0q + · · · , so it must equal E12(z) −21736 691 Δ(z), i.e., the number rLeech(n) of vectors of length 2n in the Leech lattice equals 21736 691 σ11(n) −τ(n) for every positive integer n. This gives another proof and an interpretation of Ramanujan’s congruence (28). In rank 32, things become even more interesting: here the complete clas-siﬁcation is not known, and we know that we cannot expect it very soon, because there are more than 80 million isomorphism classes! This, too, is a consequence of the theory of modular forms, but of a much more sophisti-cated part than we are presenting here. Speciﬁcally, there is a fundamental theorem of Siegel saying that the average value of the theta series associated to the quadratic forms in a single genus (we omit the deﬁnition) is always an Eisenstein series. Specialized to the particular case of even unimodular forms of rank m = 2k ≡0 (mod 8), which form a single genus, this theorem says that there are only ﬁnitely many such forms up to equivalence for each k and that, if we number them Q1, . . . , QI, then we have the relation I i=1 1 wi ΘQi(z) = mk Ek(z) , (39) where wi is the number of automorphisms of the form Qi (i.e., the number of matrices γ ∈SL(m, Z) such that Qi(γx) = Qi(x) for all x ∈Zm) and mk is the positive rational number given by the formula mk = Bk 2k B2 4 B4 8 · · · B2k−2 4k −4 , where Bi denotes the ith Bernoulli number. In particular, by comparing the constant terms on the left- and right-hand sides of (39), we see that I i=1 1/wi = mk, the Minkowski-Siegel mass formula. The numbers m4 ≈ 1.44 × 10−9, m8 ≈2.49 × 10−18 and m12 ≈7, 94 × 10−15 are small, but m16 ≈4, 03 × 107 (the next two values are m20 ≈4.39 × 1051 and m24 ≈ 1.53 × 10121), and since wi ≥2 for every i (one has at the very least the automorphisms ± Idm ), this shows that I > 80000000 for m = 32 as as-serted. A further consequence of the fact that ΘQ ∈Mk(Γ1) for Q even and unimodular of rank m = 2k is that the minimal value of Q(x) for non-zero x ∈Λ is bounded by r = dim Mk(Γ1) = [k/12] + 1. The lattice L is called extremal if this bound is attained. The three lattices of rank 8 and 16 are extremal for trivial reasons. (Here r = 1.) For m = 24 we have r = 2 and the only extremal lattice is the Leech lattice. Extremal unimodular lattices are also known to exist for m = 32, 40, 48, 56, 64 and 80, while the case m = 72 is open. Surprisingly, however, there are no examples of large rank: Theorem (Mallows–Odlyzko–Sloane). There are only ﬁnitely many non-isomorphic extremal even unimodular lattices. 36 D. Zagier We sketch the proof, which, not surprisingly, is completely modular. Since there are only ﬁnitely many non-isomorphic even unimodular lattices of any given rank, the theorem is equivalent to saying that there is an absolute bound on the value of the rank m for extremal lattices. For simplicity, let us suppose that m = 24n. (The cases m = 24n + 8 and m = 24n + 16 are similar.) The theta series of any extremal unimodular lattice of this rank must be equal to the unique modular form fn ∈M12n(SL(2, Z)) whose q-development has the form 1 + O(qn+1). By an elementary argument which we omit but which the reader may want to look for, we ﬁnd that this q-development has the form fn(z) = 1 + n an qn+1 + nbn 2 −24 n (n + 31) an qn+2 + · · · where an and bn are the coeﬃcients of Δ(z)n in the modular functions j(z) and j(z)2, respectively, when these are expressed (locally, for small q) as Lau-rent series in the modular form Δ(z) = q −24q2 + 252q3 −· · · . It is not hard to show that an has the asymptotic behavior an ∼An−3/2Cn for some constants A = 225153.793389 · · · and C = 1/Δ(z0) = 69.1164201716 · · ·, where z0 = 0.52352170017992 · · ·i is the unique zero on the imaginary axis of the function E2(z) deﬁned in (17) (this is because E2(z) is the logarithmic derivative of Δ(z)), while bn has a similar expansion but with A replaced by 2λA with λ = j(z0) −720 = 163067.793145 · · ·. It follows that the coeﬃcient 1 2nbn −24n(n+31)an of qn+2 in fn is negative for n larger than roughly 6800, corresponding to m ≈163000, and that therefore extremal lattices of rank larger than this cannot exist. ♥ ♠Drums Whose Shape One Cannot Hear Marc Kac asked a famous question, “Can one hear the shape of a drum?” Ex-pressed more mathematically, this means: can there be two riemannian mani-folds (in the case of real “drums” these would presumably be two-dimensional manifolds with boundary) which are not isometric but have the same spectra of eigenvalues of their Laplace operators? The ﬁrst example of such a pair of manifolds to be found was given by Milnor, and involved 16-dimensional closed “drums.” More drum-like examples consisting of domains in R2 with polygonal boundary are now also known, but they are diﬃcult to construct, whereas Milnor’s example is very easy. It goes as follows. As we already men-tioned, there are two non-isomorphic even unimodular lattices Λ1 = Γ8 ⊕Γ8 and Λ2 = Γ16 in dimension 16. The fact that they are non-isomorphic means that the two Riemannian manifolds M1 = R16/Λ1 and M2 = R16/Λ2, which are topologically both just tori (S1)16, are not isometric to each other. But the spectrum of the Laplace operator on any torus Rn/Λ is just the set of norms ∥λ∥2 (λ ∈Λ), counted with multiplicities, and these spectra agree for M1 and M2 because the theta series λ∈Λ1 q∥λ∥2 and λ∈Λ2 q∥λ∥2 coincide. ♥ Elliptic Modular Forms and Their Applications 37 We should not leave this section without mentioning at least brieﬂy that there is an important generalization of the theta series (36) in which each term qQ(x1,...,xm) is weighted by a polynomial P(x1, . . . , xm). If this polynomial is homogeneous of degree d and is spherical with respect to Q (this means that ΔP = 0, where Δ is the Laplace operator with respect to a system of coordinates in which Q(x1, . . . , xm) is simply x2 1 + · · · + x2 m), then the theta series ΘQ,P (z) = x P(x)qQ(x) is a modular form of weight m/2 + d (on the same group and with respect to the same character as in the case P = 1), and is a cusp form if d is strictly positive. The possibility of putting non-trivial weights into theta series in this way considerably enlarges their range of applications, both in coding theory and elsewhere. 4 Hecke Eigenforms and L-series In this section we give a brief sketch of Hecke’s fundamental discoveries that the space of modular forms is spanned by modular forms with multiplicative Fourier coeﬃcients and that one can associate to these forms Dirichlet series which have Euler products and functional equations. These facts are at the basis of most of the higher developments of the theory: the relations of modular forms to arithmetic algebraic geometry and to the theory of motives, and the adelic theory of automorphic forms. The last two subsections describe some basic examples of these higher connections. 4.1 Hecke Theory For each integer m ≥1 there is a linear operator Tm, the mth Hecke operator, acting on modular forms of any given weight k. In terms of the description of modular forms as homogeneous functions on lattices which was given in §1.1, the deﬁnition of Tm is very simple: it sends a homogeneous function F of degree −k on lattices Λ ⊂C to the function TmF deﬁned (up to a suitable normalizing constant) by TmF(Λ) = F(Λ′), where the sum runs over all sublattices Λ′ ⊂Λ of index m. The sum is ﬁnite and obviously still homoge-neous in Λ of the same degree −k. Translating from the language of lattices to that of functions in the upper half-plane by the usual formula f(z) = F(Λz), we ﬁnd that the action of Tm is given by Tmf(z) = mk−1 a b c d ∈Γ1\Mm (cz + d)−k f az + b cz + d (z ∈H) , (40) where Mm denotes the set of 2 × 2 integral matrices of determinant m and where the normalizing constant mk−1 has been introduced for later conve-nience (Tm normalized in this way will send forms with integral Fourier co-eﬃcients to forms with integral Fourier coeﬃcients). The sum makes sense 38 D. Zagier because the transformation law (2) of f implies that the summand associated to a matrix M = a b c d ∈Mm is indeed unchanged if M is replaced by γM with γ ∈Γ1, and from (40) one also easily sees that Tmf is holomorphic in H and satisﬁes the same transformation law and growth properties as f, so Tm indeed maps Mk(Γ1) to Mk(Γ1). Finally, to calculate the eﬀect of Tm on Fourier developments, we note that a set of representatives of Γ1\Mm is given by the upper triangular matrices a b 0 d with ad = m and 0 ≤b < d (this is an easy exercise), so Tmf(z) = mk−1 ad=m a, d>0 1 dk b (mod d) f az + b d . (41) If f(z) has the Fourier development (3), then a further calculation with (41), again left to the reader, shows that the function Tmf(z) has the Fourier ex-pansion Tmf(z) = d\|m d>0 (m/d)k−1 n≥0 d\|n an qmn/d2 = n≥0 r\|(m,n) r>0 rk−1 amn/r2 qn . (42) An easy but important consequence of this formula is that the operators Tm (m ∈N) all commute. Let us consider some examples. The expansion (42) begins σk−1(m)a0 + amq + · · · , so if f is a cusp form (i.e., a0 = 0), then so is Tmf. In particular, since the space S12(Γ1) of cusp forms of weight 12 is 1-dimensional, spanned by Δ(z), it follows that TmΔ is a multiple of Δ for every m ≥1. Since the Fourier expansion of Δ begins q + · · · and that of TmΔ begins τ(m)q + · · · , the eigenvalue is necessarily τ(m), so TmΔ = τ(m)Δ and (42) gives τ(m) τ(n) = r\|(m,n) r11 τ mn r2 for all m, n ≥1 , proving Ramanujan’s multiplicativity observations mentioned in §2.4. By the same argument, if f ∈Mk(Γ1) is any simultaneous eigenfunction of all of the Tm, with eigenvalues λm, then am = λma1 for all m. We therefore have a1 ̸= 0 if f is not identically 0, and if we normalize f by a1 = 1 (such an f is called a normalized Hecke eigenform, or Hecke form for short) then we have Tmf = am f , am an = r\|(m,n) rk−1 amn/r2 (m, n ≥1) . (43) Examples of this besides Δ(z) are the unique normalized cusp forms f(z) = Δ(z)Ek−12(z) in the ﬁve further weights where dim Sk(Γ1) = 1 (viz. k = 16, 18, 20, 22 and 26) and the function Gk(z) for all k ≥4, for which we have TmGk = σk−1(m)Gk, σk−1(m)σk−1(n) = r\|(m,n) rk−1σk−1(mn/r2). (This Elliptic Modular Forms and Their Applications 39 was the reason for the normalization of Gk chosen in §2.2.) In fact, a theorem of Hecke asserts that Mk(Γ1) has a basis of normalized simultaneous eigen-forms for all k, and that this basis is unique. We omit the proof, though it is not diﬃcult (one introduces a scalar product on the space of cusp forms of weight k, shows that the Tm are self-adjoint with respect to this scalar product, and appeals to a general result of linear algebra saying that com-muting self-adjoint operators can always be simultaneously diagonalized), and content ourselves instead with one further example, also due to Hecke. Con-sider k = 24, the ﬁrst weight where dim Sk(Γ1) is greater than 1. Here Sk is 2-dimensional, spanned by ΔE3 4 = q + 696q2 + · · · and Δ2 = q2 −48q3 + . . . . Computing the ﬁrst two Fourier expansions of the images under T2 of these two functions by (42), we ﬁnd that T2(ΔE3 4) = 696 ΔE3 4 + 20736000 Δ2 and T2(Δ2) = ΔE3 4 + 384 Δ2. The matrix 696 20736000 1 384 has distinct eigen-values λ1 = 540 + 12 √ 144169 and λ2 = 540 −12 √ 144169, so there are precisely two normalized eigenfunctions of T2 in S24(Γ1), namely the func-tions f1 = ΔE3 4 −(156 −12 √ 144169)Δ2 = q + λ1q2 + · · · and f2 = ΔE3 4 −(156 + 12 √ 144169)Δ2 = q + λ2q2 + · · · , with T2fi = λifi for i = 1, 2. The uniqueness of these eigenfunctions and the fact that Tm commutes with T2 for all m ≥1 then implies that Tmfi is a multiple of fi for all m ≥1, so G24, f1 and f2 give the desired unique basis of M24(Γ1) consisting of normalized Hecke eigenforms. Finally, we mention without giving any details that Hecke’s theory gen-eralizes to congruence groups of SL(2, Z) like the group Γ0(N) of matrices a b c d ∈Γ1 with c ≡0 (mod N), the main diﬀerences being that the deﬁ-nition of Tm must be modiﬁed somewhat if m and N are not coprime and that the statement about the existence of a unique base of Hecke forms be-comes more complicated: the space Mk(Γ0(N)) is the direct sum of the space spanned by all functions f(dz) where f ∈Mk(Γ0(N ′)) for some proper di-visor N ′ of N and d divides N/N ′ (the so-called “old forms”) and a space of “new forms” which is again uniquely spanned by normalized eigenforms of all Hecke operators Tm with (m, N) = 1. The details can be found in any standard textbook. 4.2 L-series of Eigenforms Let us return to the full modular group. We have seen that Mk(Γ1) contains, and is in fact spanned by, normalized Hecke eigenforms f = amqm satisfy-ing (43). Specializing this equation to the two cases when m and n are coprime and when m = pν and n = p for some prime p gives the two equations (which together are equivalent to (43)) amn = am an if (m, n) = 1 , apν+1 = ap apν −pk−1 apν−1 (p prime, ν ≥1) . The ﬁrst says that the coeﬃcients an are multiplicative and hence that the Dirichlet series L(f, s) = ∞ n=1 an ns , called the Hecke L-series of f, has an Eu-40 D. Zagier ler product L(f, s) = p prime 1 + ap ps + ap2 p2s + · · · , and the second tells us that the power series ∞ ν=0 apνxν for p prime equals 1/(1 −apx + pk−1x2). Combining these two statements gives Hecke’s fundamental Euler product development L(f, s) = p prime 1 1 −ap p−s + pk−1−2s (44) for the L-series of a normalized Hecke eigenform f ∈Mk(Γ1), a simple ex-ample being given by L(Gk, s) = p 1 1 −(pk−1 + 1)p−s + pk−1−2s = ζ(s) ζ(s −k + 1) . For eigenforms on Γ0(N) there is a similar result except that the Euler factors for p\|N have to be modiﬁed suitably. The L-series have another fundamental property, also discovered by Hecke, which is that they can be analytically continued in s and then satisfy func-tional equations. We again restrict to Γ = Γ1 and also, for convenience, to cusp forms, though not any more just to eigenforms. (The method of proof ex-tends to non-cusp forms but is messier there since L(f, s) then has poles, and since Mk is spanned by cusp forms and by Gk, whose L-series is completely known, there is no loss in making the latter restriction.) From the estimate an = O(nk/2) proved in §2.4 we know that L(f, s) converges absolutely in the half-plane ℜ(s) > 1 + k/2. Take s in that half-plane and consider the Euler gamma function Γ(s) = ∞ 0 ts−1 e−t dt . Replacing t by λt in this integral gives Γ(s) = λs ∞ 0 ts−1 e−λt dt or λ−s = Γ(s)−1 ∞ 0 ts−1 e−λt dt for any λ > 0. Applying this to λ = 2πn, multiplying by an, and summing over n, we obtain (2π)−s Γ(s) L(f, s) = ∞ n=1 an ∞ 0 ts−1 e−2πnt dt = ∞ 0 ts−1 f(it) dt ℜ(s) > k 2 + 1 , where the interchange of integration and summation is justiﬁed by the abso-lute convergence. Now the fact that f(it) is exponentially small for t →∞ (because f is a cusp form) and for t →0 (because f(−1/z) = zkf(z) ) im-plies that the integral converges absolutely for all s ∈C and hence that the function L∗(f, s) := (2π)−s Γ(s) L(f, s) = (2π)−s Γ(s) ∞ n=1 an ns (45) Elliptic Modular Forms and Their Applications 41 extends holomorphically from the half-plane ℜ(s) > 1 + k/2 to the entire complex plane. The substitution t →1/t together with the transformation equation f(i/t) = (it)kf(it) of f then gives the functional equation L∗(f, k −s) = (−1)k/2 L∗(f, s) (46) of L∗(f, s). We have proved: Proposition 13. Let f = ∞ n=1 anqn be a cusp form of weight k on the full modular group. Then the L-series L(f, s) extends to an entire function of s and satisﬁes the functional equation (46), where L∗(f, s) is deﬁned by equation (45). It is perhaps worth mentioning that, as Hecke also proved, the converse of Proposition 13 holds as well: if an (n ≥1) are complex numbers of polynomial growth and the function L∗(f, s) deﬁned by (45) continues analytically to the whole complex plane and satisﬁes the functional equation (46), then f(z) = ∞ n=1 ane2πinz is a cusp form of weight k on Γ1 . 4.3 Modular Forms and Algebraic Number Theory In §3, we used the theta series θ(z)2 to determine the number of represen-tations of any integer n as a sum of two squares. More generally, we can study the number r(Q, n) of representations of n by a positive deﬁnite binary quadratic Q(x, y) = ax2 + bxy + cy2 with integer coeﬃcients by considering the weight 1 theta series ΘQ(z) = x,y∈Z qQ(x,y) = ∞ n=0 r(Q, n)qn. This theta series depends only on the class [Q] of Q up to equivalences Q ∼Q ◦γ with γ ∈Γ1. We showed in §1.2 that for any D < 0 the number h(D) of Γ1-equivalence classes [Q] of binary quadratic forms of discriminant b2 −4ac = D is ﬁnite. If D is a fundamental discriminant (i.e., not representable as D′r2 with D′ congruent to 0 or 1 mod 4 and r > 1), then h(D) equals the class number of the imaginary quadratic ﬁeld K = Q( √ D) of discriminant D and there is a well-known bijection between the Γ1-equivalence classes of binary quadratic forms of discriminant D and the ideal classes of K such that r(Q, n) for any form Q equals w times the number r(A, n) of integral ideals a of K of norm n belonging to the corresponding ideal class A, where w is the num-ber of roots of unity in K (= 6 or 4 if D = −3 or D = −4 and 2 oth-erwise). The L-series L(ΘQ, s) of ΘQ is therefore w times the “partial zeta-function” ζK,A(s) = a∈A N(a)−s. The ideal classes of K form an abelian group. If χ is a homomorphism from this group to C∗, then the L-series LK(s, χ) = a χ(a)/N(a)s (sum over all integral ideals of K) can be writ-ten as A χ(A) ζK,A(s) (sum over all ideal classes of K) and hence is the L-series of the weight 1 modular form fχ(z) = w−1 A χ(A)ΘA(z). On the other hand, from the unique prime decomposition of ideals in K it follows that LK(s, χ) has an Euler product. Hence fχ is a Hecke eigenform. If χ = χ0 is the trivial character, then LK(s, χ) = ζK(s), the Dedekind zeta function 42 D. Zagier of K, which factors as ζ(s)L(s, εD), the product of the Riemann zeta-function and the Dirichlet L-series of the character εD(n) = D n (Kronecker sym-bol). Therefore in this case we get [Q] r(Q, n) = w d\|n εD(d) (an identity known to Gauss) and correspondingly fχ0(z) = 1 w [Q] ΘQ(z) = h(D) 2 + ∞ n=1 d\|n D d qn , an Eisenstein series of weight 1. If the character χ has order 2, then it is a so-called genus character and one knows that LK(s, χ) factors as L(s, εD1)L(s, εD2) where D1 and D2 are two other discriminants with prod-uct D. In this case, too, fχ(z) is an Eisenstein series. But in all other cases, fχ is a cusp form and the theory of modular forms gives us non-trivial information about representations of numbers by quadratic forms. ♠Binary Quadratic Forms of Discriminant −23 We discuss an explicit example, taken from a short and pretty article written by van der Blij in 1952. The class number of the discriminant D = −23 is 3, with the SL(2, Z)-equivalence classes of binary quadratic forms of this discriminant being represented by the three forms Q0(x, y) = x2 + xy + 6y2 , Q1(x, y) = 2x2 + xy + 3y2 , Q2(x, y) = 2x2 −xy + 3y2 . Since Q1 and Q2 represent the same integers, we get only two distinct theta series ΘQ0(z) = 1 + 2q + 2q4 + 4q6 + 4q8 + · · · , ΘQ1(z) = 1 + 2q2 + 2q3 + 2q4 + 2q6 + · · · . The linear combination corresponding to the trivial character is the Eisenstein series fχ0 = 1 2 ΘQ0 + 2ΘQ1 = 3 2 + ∞ n=1 d\|n −23 d qn = 3 2 + q + 2q2 + 2q3 + 3q4 + · · · , in accordance with the general identity w−1 [Q] r(Q, n) = d\|n εD(d) men-tioned above. If χ is one of the two non-trivial characters, with values e±2πi/3 = 1 2(−1 ± i √ 3) on Q1 and Q2, we have Elliptic Modular Forms and Their Applications 43 fχ = 1 2 ΘQ0 −ΘQ1 = q −q2 −q3 + q6 + · · · . This is a Hecke eigenform in the space S1(Γ0(23), ε−23). Its L-series has the form L(fχ, s) = p 1 1 −ap p−s + ε−23(p) p−2s where ε−23(p) equals the Legendre symbol (p/23) by quadratic reciprocity and ap = ⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩ 1 if p = 23 , 0 if (p/23) = −1 , 2 if (p/23) = 1 and p is representable as x2 + xy + 6y2 , −1 if (p/23) = 1 and p is representable as 2x2 + xy + 3y2 . (47) On the other hand, the space S1(Γ0(23), ε−23) is one-dimensional, spanned by the function η(z) η(23z) = q ∞ n=1 1 −qn 1 −q23n . (48) We therefore obtain the explicit “reciprocity law” Proposition 14 (van der Blij). Let p be a prime. Then the number ap deﬁned in (47) is equal to the coeﬃcient of qp in the product (48). As an application of this, we observe that the q-expansion on the right-hand side of (48) is congruent modulo 23 to Δ(z) = q (1 −qn)24 and hence that τ(p) is congruent modulo 23 to the number ap deﬁned in (47) for every prime number p, a congruence for the Ramanujan function τ(n) of a somewhat diﬀerent type than those already given in (27) and (28). ♥ Proposition 14 gives a concrete example showing how the coeﬃcients of a modular form – here η(z)η(23z) – can answer a question of number theory – here, the question whether a given prime number which splits in Q(√−23) splits into principal or non-principal ideals. But actually the connection goes much deeper. By elementary algebraic number theory we have that the L-series L(s) = LK(s, χ) is the quotient ζF (s)/ζ(s) of the Dedekind function of F by the Riemann zeta function, where F = Q(α) (α3 −α −1 = 0) is the cubic ﬁeld of discriminant −23. (The composite K ·F is the Hilbert class ﬁeld of K.) Hence the four cases in (47) also describe the splitting of p in F: 23 is ramiﬁed, quadratic non-residues of 23 split as p = p1p2 with N(pi) = pi, and quadratic residues of 23 are either split completely (as products of three prime ideals of norm p) or are inert (remain prime) in F, according whether they are represented by Q0 or Q1. Thus the modular form η(z)η(23z) describes not only the algebraic number theory of the quadratic ﬁeld K, but also the splitting of primes in the higher degree ﬁeld F. This is the ﬁrst non-trivial example of the connection found by Weil–Langlands and Deligne–Serre which relates modular forms of weight one to the arithmetic of number ﬁelds whose Galois groups admit non-trivial two-dimensional representations. 44 D. Zagier 4.4 Modular Forms Associated to Elliptic Curves and Other Varieties If X is a smooth projective algebraic variety deﬁned over Q, then for al-most all primes p the equations deﬁning X can be reduced modulo p to deﬁne a smooth variety Xp over the ﬁeld Z/pZ = Fp. We can then count the number of points in Xp over the ﬁnite ﬁeld Fpr for all r ≥1 and, putting all this information together, deﬁne a “local zeta function” Z(Xp, s) = exp ∞ r=1 \|Xp(Fpr)\| p−rs/r (ℜ(s) ≫0) and a “global zeta function” (Hasse– Weil zeta function) Z(X/Q, s) = p Z(Xp, s), where the product is over all primes. (The factors Zp(X, s) for the “bad” primes p, where the equations deﬁning X yield a singular variety over Fp, are deﬁned in a more complicated but completely explicit way and are again power series in p−s.) Thanks to the work of Weil, Grothendieck, Dwork, Deligne and others, a great deal is known about the local zeta functions – in particular, that they are rational functions of p−s and have all of their zeros and poles on the vertical lines ℜ(s) = 0, 1 2, . . . , n −1 2, n where n is the dimension of X – but the global zeta function remains mysterious. In particular, the general conjecture that Z(X/Q, s) can be meromorphically continued to all s is known only for very special classes of varieties. In the case where X = E is an elliptic curve, given, say, by a Weierstrass equation y2 = x3 + Ax + B (A, B ∈Z, Δ := −4A3 −27B2 ̸= 0) , (49) the local factors can be made completely explicit and we ﬁnd that Z(E/Q, s) = ζ(s)ζ(s −1) L(E/Q, s) where the L-series L(E/Q, s) is given for ℜ(s) ≫0 by an Euler product of the form L(E/Q, s) = p∤Δ 1 1−ap(E) p−s+p1−2s · p\|Δ 1 (polynomial of degree ≤2 in p−s) (50) with ap(E) deﬁned for p ∤Δ as p − (x, y) ∈(Z/pZ)2 \| y2 = x3 + Ax + B . In the mid-1950’s, Taniyama noticed the striking formal similarity between this Euler product expansion and the one in (44) when k = 2 and asked whether there might be cases of overlap between the two, i.e., cases where the L-function of an elliptic curve agrees with that of a Hecke eigenform of weight 2 having eigenvalues ap ∈Z (a necessary condition if they are to agree with the integers ap(E)). Numerical examples show that this at least sometimes happens. The sim-plest elliptic curve (if we order elliptic curves by their “conductor,” an invariant in N which is divisible only by primes dividing the discriminant Δ in (49)) is the curve Y 2 −Y = X3 −X2 of conductor 11. (This can be put into the form (49) by setting y = 216Y −108, x = 36X −12, giving A = −432, Elliptic Modular Forms and Their Applications 45 B = 8208, Δ = −28 ·312 ·11, but the equation in X and Y , the so-called “min-imal model,” has much smaller coeﬃcients.) We can compute the numbers ap by counting solutions of Y 2 −Y = X3 −X2 in (Z/pZ)2. (For p > 3 this is equivalent to the recipe given above because the equations relating (x, y) and (X, Y ) are invertible in characteristic p, and for p = 2 or 3 the minimal model gives the correct answer.) For example, we have a5 = 5 −4 = 1 because the equation Y 2 −Y = X3 −X2 has the 4 solutions (0,0), (0,1), (1,0) and (1,1) in (Z/5Z)2. Then we have L(E/Q, s) = 1 + 2 2s + 2 22s −1 1 + 1 3s + 3 32s −1 1 −1 5s + 5 52s −1 · · · = 1 1s −2 2s −1 3s + 2 4s + 1 5s + · · · = L(f, s) , where f ∈S2(Γ0(11)) is the modular form f(z) = η(z)2η(11z)2 = q ∞ n=1 1−qn2 1−q11n2 = x−2q2−q3+2q4+q5+ · · · . In the 1960’s, one direction of the connection suggested by Taniyama was proved by M. Eichler and G. Shimura, whose work establishes the following theorem. Theorem (Eichler–Shimura). Let f(z) be a Hecke eigenform in S2(Γ0(N)) for some N ∈N with integral Fourier coeﬃcients. Then there exists an elliptic curve E/Q such that L(E/Q, s) = L(f, s). Explicitly, this means that ap(E) = ap(f) for all primes p, where an(f) is the coeﬃcient of qn in the Fourier expansion of f. The proof of the theorem is in a sense quite explicit. The quotient of the upper half-plane by Γ0(N), compactiﬁed appropriately by adding a ﬁnite number of points (cusps), is a complex curve (Riemann surface), traditionally denoted by X0(N), such that the space of holomorphic 1-forms on X0(N) can be identiﬁed canoni-cally (via f(z) →f(z)dz) with the space of cusp forms S2(Γ0(N)). One can also associate to X0(N) an abelian variety, called its Jacobian, whose tangent space at any point can be identiﬁed canonically with S2(Γ0(N)). The Hecke operators Tp introduced in 4.1 act not only on S2(Γ0(N)), but on the Jaco-bian variety itself, and if the Fourier coeﬃcients ap = ap(f) are in Z, then so do the diﬀerences Tp −ap · Id . The subvariety of the Jacobian annihilated by all of these diﬀerences (i.e., the set of points x in the Jacobian whose image under Tp equals ap times x; this makes sense because an abelian variety has the structure of a group, so that we can multiply points by integers) is then precisely the sought-for elliptic curve E. Moreover, this construction shows that we have an even more intimate relationship between the curve E and the form f than the L-series equality L(E/Q, s) = L(f, s), namely, that there is an actual map from the modular curve X0(N) to the elliptic curve E which 46 D. Zagier is induced by f. Speciﬁcally, if we deﬁne φ(z) = ∞ n=1 an(f) n e2πinz, so that φ′(z) = 2πif(z)dz, then the fact that f is modular of weight 2 implies that the diﬀerence φ(γ(z)) −φ(z) has zero derivative and hence is constant for all γ ∈Γ0(N), say φ(γ(z)) −φ(z) = C(γ). It is then easy to see that the map C : Γ0(N) →C is a homomorphism, and in our case (f an eigenform, eigenval-ues in Z), it turns out that its image is a lattice Λ ⊂C, and the quotient map C/Λ is isomorphic to the elliptic curve E. The fact that φ(γ(z)) −φ(z) ∈Λ then implies that the composite map H φ − →C pr − →C/Λ factors through the projection H →Γ0(N)\H, i.e., φ induces a map (over C) from X0(N) to E. This map is in fact deﬁned over Q, i.e., there are modular functions X(z) and Y (z) with rational Fourier coeﬃcients which are invariant under Γ0(N) and which identically satisfy the equation Y (z)2 = X(z)3+AX(z)+B (so that the map from X0(N) to E in its Weierstrass form is simply z →(X(z), Y (z) ) ) as well as the equation X′(z)/2Y (z) = 2πif(z). (Here we are simplifying a little.) Gradually the idea arose that perhaps the answer to Taniyama’s orig-inal question might be yes in all cases, not just sometimes. The results of Eichler and Shimura showed this in one direction, and strong evidence in the other direction was provided by a theorem proved by A. Weil in 1967 which said that if the L-series of an elliptic curve E/Q and certain “twists” of it satisﬁed the conjectured analytic properties (holomorphic con-tinuation and functional equation), then E really did correspond to a modu-lar form in the above way. The conjecture that every E over Q is modu-lar became famous (and was called according to taste by various subsets of the names Taniyama, Weil and Shimura, although none of these three people had ever stated the conjecture explicitly in print). It was ﬁnally proved at the end of the 1990’s by Andrew Wiles and his collaborators and followers: Theorem (Wiles–Taylor, Breuil–Conrad–Diamond–Taylor). Every el-liptic curve over Q can be parametrized by modular functions. The proof, which is extremely diﬃcult and builds on almost the entire apparatus built up during the previous decades in algebraic geometry, repre-sentation theory and the theory of automorphic forms, is one of the pinnacles of mathematical achievement in the 20th century. ♠Fermat’s Last Theorem In the 1970’s, Y. Hellegouarch was led to consider the elliptic curve (49) in the special case when the roots of the cubic polynomial on the right were nth powers of rational integers for some prime number n > 2, i.e., if this cubic factors as (x −an)(x −bn)(x −cn) where a, b, c satisfy the Fermat equation an + bn + cn = 0. A decade later, G. Frey studied the same el-liptic curve and discovered that the associated Galois representation (we Elliptic Modular Forms and Their Applications 47 do not explain this here) had properties which contradicted the properties which Galois representations of elliptic curves were expected to satisfy. Pre-cise conjectures about the modularity of certain Galois representations were then made by Serre which would fail for the representations attached to the Hellegouarch-Frey curve, so that the correctness of these conjectures would imply the insolubility of Fermat’s equation. (Very roughly, the conjectures imply that, if the Galois representation associated to the above curve E is modular at all, then the corresponding cusp form would have to be congru-ent modulo n to a cusp form of weight 2 and level 1 or 2, and there aren’t any.) In 1990, K. Ribet proved a special case of Serre’s conjectures (the gen-eral case is now also known, thanks to recent work of Khare, Wintenberger, Dieulefait and Kisin) which was suﬃcient to yield the same implication. The proof by Wiles and Taylor of the Taniyama-Weil conjecture (still with some minor restrictions on E which were later lifted by the other authors cited above, but in suﬃcient generality to make Ribet’s result applicable) thus suf-ﬁced to give the proof of the following theorem, ﬁrst claimed by Fermat in 1637: Theorem (Ribet, Wiles–Taylor). If n > 2, there are no positive integers with an + bn = cn. ♥ Finally, we should mention that the connection between modularity and algebraic geometry does not apply only to elliptic curves. Without going into detail, we mention only that the Hasse–Weil zeta function Z(X/Q, s) of an arbitrary smooth projective variety X over Q splits into factors corre-sponding to the various cohomology groups of X, and that if any of these cohomology groups (or any piece of them under some canonical decompo-sition, say with respect to the action of a ﬁnite group of automorphisms of X) is two-dimensional, then the corresponding piece of the zeta func-tion is conjectured to be the L-series of a Hecke eigenform of weight i + 1, where the cohomology group in question is in degree i. This of course in-cludes the case when X = E and i = 1, since the ﬁrst cohomology group of a curve of genus 1 is 2-dimensional, but it also applies to many higher-dimensional varieties. Many examples are now known, an early one, due to R. Livné, being given by the cubic hypersurface x3 1 + · · · + x3 10 = 0 in the projective space {x ∈P9 \| x1 + · · · + x10 = 0}, whose zeta-function equals 7 j=0 ζ(s −i)mi · L(s −2, f)−1 where (m0, . . . m7) = (1, 1, 1, −83, 43, 1, 1, 1) and f = q + 2q2 −8q3 + 4q4 + 5q5 + · · · is the unique new form of weight 4 on Γ0(10). Other examples arise from so-called “rigid Calabi-Yau 3-folds,” which have been studied intensively in recent years in connection with the phenomenon, ﬁrst discovered by mathematical physicists, called “mirror sym-metry.” We skip all further discussion, referring to the survey paper and book cited in the references at the end of these notes. 48 D. Zagier 5 Modular Forms and Diﬀerential Operators The starting point for this section is the observation that the derivative of a modular form is not modular, but nearly is. Speciﬁcally, if f is a modular form of weight k with the Fourier expansion (3), then by diﬀerentiating (2) we see that the derivative Df = f ′ := 1 2πi d f dz = q d f dq = ∞ n=1 n an qn (51) (where the factor 2πi has been included in order to preserve the rationality properties of the Fourier coeﬃcients) satisﬁes f ′ az + b cz + d = (cz + d)k+2 f ′(z) + k 2πi c (cz + d)k+1f(z) . (52) If we had only the ﬁrst term, then f ′ would be a modular form of weight k+2. The presence of the second term, far from being a problem, makes the theory much richer. To deal with it, we will: • modify the diﬀerentiation operator so that it preserves modularity; • make combinations of derivatives of modular forms which are again modu-lar; • relax the notion of modularity to include functions satisfying equations like (52); • diﬀerentiate with respect to t(z) rather than z itself, where t(z) is a modu-lar function. These four approaches will be discussed in the four subsections 5.1–5.4, re-spectively. 5.1 Derivatives of Modular Forms As already stated, the ﬁrst approach is to introduce modiﬁcations of the op-erator D which do preserve modularity. There are two ways to do this, one holomorphic and one not. We begin with the holomorphic one. Comparing the transformation equation (52) with equations (19) and (17), we ﬁnd that for any modular form f ∈Mk(Γ1) the function ϑkf := f ′ −k 12 E2 f , (53) sometimes called the Serre derivative, belongs to Mk+2(Γ1). (We will often drop the subscript k, since it must always be the weight of the form to which the operator is applied.) A ﬁrst consequence of this basic fact is the following. We introduce the ring M∗(Γ1) := M∗(Γ1)[E2] = C[E2, E4, E6], called the ring of quasimodular forms on SL(2, Z). (An intrinsic deﬁnition of the elements of this ring, and a deﬁnition for other groups Γ ⊂G, will be given in the next subsection.) Then we have: Elliptic Modular Forms and Their Applications 49 Proposition 15. The ring M∗(Γ1) is closed under diﬀerentiation. Speciﬁ-cally, we have E′ 2 = E2 2 −E4 12 , E′ 4 = E2E4 −E6 3 , E′ 6 = E2E6 −E2 4 2 . (54) Proof. Clearly ϑE4 and ϑE6, being holomorphic modular forms of weight 6 and 8 on Γ1, respectively, must be proportional to E6 and E2 4, and by looking at the ﬁrst terms in their Fourier expansion we ﬁnd that the factors are −1/3 and −1/2. Similarly, by diﬀerentiating (19) we ﬁnd the analogue of (53) for E2, namely that the function E′ 2 −1 12 E2 2 belongs to M4(Γ). It must therefore be a multiple of E4, and by looking at the ﬁrst term in the Fourier expansion one sees that the factor is −1/12 . Proposition 15, ﬁrst discovered by Ramanujan, has many applications. We de-scribe two of them here. Another, in transcendence theory, will be mentioned in Section 6. ♠Modular Forms Satisfy Non-Linear Diﬀerential Equations An immediate consequence of Proposition 15 is the following: Proposition 16. Any modular form or quasi-modular form on Γ1 satisﬁes a non-linear third order diﬀerential equation with constant coeﬃcients. Proof. Since the ring M∗(Γ1) has transcendence degree 3 and is closed under diﬀerentiation, the four functions f, f ′, f ′′ and f ′′′ are algebraically dependent for any f ∈ M∗(Γ1). As an example, by applying (54) several times we ﬁnd that the function E2 satisﬁes the non-linear diﬀerential equation f ′′′ −ff ′′ + 3 2 f ′2 = 0 . This is called the Chazy equation and plays a role in the theory of Painlevé equations. We can now use modular/quasimodular ideas to describe a full set of solu-tions of this equation. First, deﬁne a “modiﬁed slash operator” f →f∥2 g by (f∥2 g)(z) = (cz +d)−2f az+b cz+d + π 12 c cz+d for g = a b c d . This is not linear in f (it is only aﬃne), but it is nevertheless a group operation, as one checks easily, and, at least locally, it makes sense for any matrix a b c d ∈SL(2, C). Now one checks by a direct, though tedious, computation that Ch ! f∥2 g " = Ch[f]\|8 g (where deﬁned) for any g ∈SL(2, C), where Ch[f] = f ′′′ −ff ′′ + 3 2 f ′2. (Again this is surprising, because the operator Ch is not linear.) Since E2 is a so-lution of Ch[f] = 0, it follows that E2∥2 g is a solution of the same equation (in g−1H ⊂P1(C)) for every g ∈SL(2, C), and since E2∥2 γ = E2 for γ ∈Γ1 and ∥2 is a group operation, it follows that this function depends only on the class of g in Γ1\SL(2, C). But Γ1\SL(2, C) is 3-dimensional and a third-order diﬀerential equation generically has a 3-dimensional solution space (the values of f, f ′ and f ′′ at a point determine all higher derivatives recursively 50 D. Zagier and hence ﬁx the function uniquely in a neighborhood of any point where it is holomorphic), so that, at least generically, this describes all solutions of the non-linear diﬀerential equation Ch[f] = 0 in modular terms. ♥ Our second “application” of the ring M∗(Γ1) describes an unexpected ap-pearance of this ring in an elementary and apparently unrelated context. ♠Moments of Periodic Functions This very pretty application of the modular forms E2, E4, E6 is due to P. Gallagher. Denote by P the space of periodic real-valued functions on R, i.e., functions f : R →R satisfying f(x + 2π) = f(x). For each ∞-tuple n = (n0, n1, . . . ) of non-negative integers (all but ﬁnitely many equal to 0) we deﬁne a “coordinate” In on the inﬁnite-dimensional space P by In[f] = 2π 0 f(x)n0f ′(x)n1 · · · dx (higher moments). Apart from the relations among these coming from integration by parts, like f ′′(x)dx = 0 or f ′(x)2dx = − f(x)f ′′(x)dx, we also have various inequalities. The general problem, cer-tainly too hard to be solved completely, would be to describe all equalities and inequalities among the In[f]. As a special case we can ask for the complete list of inequalities satisﬁed by the four moments A[f], B[f], C[f], D[f] := 2π 0 f, 2π 0 f 2, 2π 0 f 3, 2π 0 f ′2 as f ranges over P. Surprisingly enough, the answer involves quasimodular forms on SL(2, Z). First, by making a lin-ear shift f →λf + μ with λ, μ ∈R we can suppose that A[f] = 0 and D[f] = 1. The problem is then to describe the subset X ⊂R2 of pairs B[f], C[f] = 2π 0 f(x)2dx, 2π 0 f(x)3dx where f ranges over functions in P satisfying 2π 0 f(x)dx = 0 and 2π 0 f ′(x)2 dx = 1. Theorem (Gallagher). We have X = (B, C) ∈R2 \| 0 < B ≤1, C2 ≤ Φ(B) where the function Φ : (0, 1] →R≥0 is given parametrically by Φ G′ 2(it) G′ 4(it) = (G′ 4(it) −G′′ 2(it))2 2 G′ 4(it)3 0 < t ≤∞ . The idea of the proof is as follows. First, from A = 0 and D = 1 we deduce 0 < B[f] ≤1 by an inequality of Wirtinger (just look at the Fourier expansion of f). Now let f ∈P be a function – but one must prove that it exists! – which maximizes C = C[f] for given values of A, B and D. By a standard calculus-of-variations-type argument (replace f by f + εg where g ∈P is orthogonal to 1, f and f ′′, so that A, B and D do not change to ﬁrst order, and then use that C also cannot change to ﬁrst order since otherwise its value could not be extremal, so that g must also be orthogonal to f 2), we show that the four functions 1, f, f 2 and f ′′ are linearly dependent. From this it follows by integrating once that the ﬁve functions 1, f, f 2, f 3 and f ′2 are also linearly dependent. After a rescaling f →λf + μ, we can write this dependency as f ′(x)2 = 4f(x)3 −g2f(x) −g3 for some constants g2 and g3. But this is the Elliptic Modular Forms and Their Applications 51 famous diﬀerential equation of the Weierstrass ℘-function, so f(2πx) is the restriction to R of the function ℘(x, Zτ + Z) for some τ ∈H, necessarily of the form τ = it with t > 0 because everything is real. Now the coeﬃcients g2 and g3, by the classical Weierstrass theory, are simple multiples of G4(it) and G6(it), and for this function f the value of A(f) = 2π 0 f(x)dx is known to be a multiple of G2(it). Working out all the details, and then rescaling f again to get A = 0 and D = 1, one ﬁnds the result stated in the theorem. ♥ We now turn to the second modiﬁcation of the diﬀerentiation operator which preserves modularity, this time, however, at the expense of sacriﬁcing holomorphy. For f ∈Mk(Γ) (we now no longer require that Γ be the full modular group Γ1) we deﬁne ∂kf(z) = f ′(z) − k 4πy f(z) , (55) where y denotes the imaginary part of z. Clearly this is no longer holomorphic, but from the calculation 1 I(γz) = \|cz + d\|2 y = (cz + d)2 y −2ic(cz+d) γ = a b c d ∈SL(2, R) and (52) one easily sees that it transforms like a modular form of weight k+2, i.e., that (∂kf)\|k+2γ = ∂kf for all γ ∈Γ. Moreover, this remains true even if f is modular but not holomorphic, if we interpret f ′ as 1 2πi ∂f ∂z . This means that we can apply ∂= ∂k repeatedly to get non-holomorphic modular forms ∂nf of weight k + 2n for all n ≥0. (Here, as with ϑk, we can drop the subscript k because ∂k will only be applied to forms of weight k; this is convenient because we can then write ∂nf instead of the more correct ∂k+2n−2 · · · ∂k+2∂kf .) For example, for f ∈Mk(Γ) we ﬁnd ∂2f = 1 2πi ∂ ∂z −k + 2 4πy f ′ − k 4πy f = f ′′ − k 4πy f ′ − k 16π2y2 f −k + 2 4πy f ′ + k(k + 2) 16π2y2 f = f ′′ −k + 1 2πy f ′ + k(k + 1) 16π2y2 f and more generally, as one sees by an easy induction, ∂nf = n r=0 (−1)n−r n r (k + r)n−r (4πy)n−r Drf , (56) where (a)m = a(a+1) · · · (a+m−1) is the Pochhammer symbol. The inversion of (56) is 52 D. Zagier Dnf = n r=0 n r (k + r)n−r (4πy)n−r ∂rf , (57) and describes the decomposition of the holomorphic but non-modular form f (n) = Dnf into non-holomorphic but modular pieces: the function yr−n∂rf is multiplied by (cz + d)k+n+r(c¯ z + d)n−r when z is replaced by az+b cz+d with a b c d ∈Γ. Formula (56) has a consequence which will be important in §6. The usual way to write down modular forms is via their Fourier expansions, i.e., as power series in the quantity q = e2πiz which is a local coordinate at inﬁn-ity for the modular curve Γ\H. But since modular forms are holomorphic functions in the upper half-plane, they also have Taylor series expansions in the neighborhood of any point z = x + iy ∈H. The “straight” Taylor se-ries expansion, giving f(z + w) as a power series in w, converges only in the disk \|w\| < y centered at z and tangent to the real line, which is un-natural since the domain of holomorphy of f is the whole upper half-plane, not just this disk. Instead, we should remember that we can map H iso-morphically to the unit disk, with z mapping to 0, by sending z′ ∈H to w = z′−z z′−¯ z . The inverse of this map is given by z′ = z−¯ zw 1−w , and then if f is a modular form of weight k we should also include the automorphy factor (1 −w)−k corresponding to this fractional linear transformation (even though it belongs to PSL(2, C) and not Γ). The most natural way to study f near z is therefore to expand (1 −w)−kf z−¯ zw 1−w in powers of w. The following proposition describes the coeﬃcients of this expansion in terms of the opera-tor (55). Proposition 17. Let f be a modular form of weight k and z = x+iy a point of H. Then 1 −w −k f z −¯ zw 1 −w = ∞ n=0 ∂nf(z) (4πyw)n n! (\|w\| < 1) . (58) Proof. From the usual Taylor expansion, we ﬁnd 1 −w −kf z −¯ zw 1 −w = 1 −w −kf z + 2iyw 1 −w = 1 −w −k ∞ r=0 Drf(z) r! −4πyw 1 −w r , and now expanding (1 −w)−k−r by the binomial theorem and using (56) we obtain (58). Proposition 17 is useful because, as we will see in §6, the expansion (58), after some renormalizing, often has algebraic coeﬃcients that contain interesting arithmetic information. Elliptic Modular Forms and Their Applications 53 5.2 Rankin–Cohen Brackets and Cohen–Kuznetsov Series Let us return to equation (52) describing the near-modularity of the deriva-tive of a modular form f ∈Mk(Γ). If g ∈Mℓ(Γ) is a second modu-lar form on the same group, of weight ℓ, then this formula shows that the non-modularity of f ′(z)g(z) is given by an additive correction term (2πi)−1kc (cz + d)k+ℓ+1f(z) g(z). This correction term, multiplied by ℓ, is symmetric in f and g, so the diﬀerence [f, g] = kfg′ −ℓf ′g is a modular form of weight k + ℓ+ 2 on Γ. One checks easily that the bracket [ · , · ] deﬁned in this way is anti-symmetric and satisﬁes the Jacobi identity, making M∗(Γ) into a graded Lie algebra (with grading given by the weight + 2). Further-more, the bracket g →[f, g] with a ﬁxed modular form f is a derivation with respect to the usual multiplication, so that M∗(Γ) even acquires the structure of a so-called Poisson algebra. We can continue this construction to ﬁnd combinations of higher deriva-tives of f and g which are modular, setting [f, g]0 = fg, [f, g]1 = [f, g] = kfg′ −ℓf ′g, [f, g]2 = k(k + 1) 2 fg′′ −(k + 1)(ℓ+ 1)f ′g′ + ℓ(ℓ+ 1) 2 f ′′g , and in general [f, g]n = r, s≥0 r+s=n (−1)r k + n −1 s ℓ+ n −1 r Drf Dsg (n ≥0) , (59) the nth Rankin–Cohen bracket of f and g. Proposition 18. For f ∈Mk(Γ) and g ∈Mℓ(Γ) and for every n ≥0, the function [f, g]n deﬁned by (59) belongs to Mk+ℓ+2n(Γ). There are several ways to prove this. We will do it using Cohen–Kuznetsov series. If f ∈Mk(Γ), then the Cohen–Kuznetsov series of f is deﬁned by fD(z, X) = ∞ n=0 Dnf(z) n! (k)n Xn ∈Hol0(H) , (60) where (k)n = (k+n−1)!/(k−1)! = k(k+1) · · · (k+n−1) is the Pochhammer symbol already used above and Hol0(H) denotes the space of holomorphic functions in the upper half-plane of subexponential growth (see §1.1). This series converges for all X ∈C (although for our purposes only its properties as a formal power series in X will be needed). Its key property is given by: Proposition 19. If f ∈Mk(Γ), then the Cohen–Kuznetsov series deﬁned by (60) satisﬁes the modular transformation equation fD az + b cz + d , X (cz + d)2 = (cz + d)k exp c cz + d X 2πi fD(z, X) . (61) for all z ∈H, X ∈C, and γ = a b c d ∈Γ . 54 D. Zagier Proof. This can be proved in several diﬀerent ways. One way is direct: one shows by induction on n that the derivative Dnf(z) transforms under Γ by Dnf az + b cz + d = n r=0 n r (k + r)n−r (2πi)n−r cn−r(cz + d)k+n+r Drf(z) for all n ≥0 (equation (52) is the case n = 1 of this), from which the claim follows easily. Another, more elegant, method is to use formula (56) or (57) to establish the relationship fD(z, X) = eX/4πy f∂(z, X) (z = x + iy ∈H, X ∈C) (62) between fD(z, X) and the modiﬁed Cohen–Kuznetsov series f∂(z, X) = ∞ n=0 ∂nf(z) n! (k)n Xn ∈Hol0(H) . (63) The fact that each function ∂nf(z) transforms like a modular form of weight k + 2n on Γ implies that f∂(z, X) is multiplied by (cz + d)k when z and X are replaced by az+b cz+d and X (cz+d)2 , and using (62) one easily de-duces from this the transformation formula (61). Yet a third way is to observe that fD(z, X) is the unique solution of the diﬀerential equation X ∂2 ∂X2 + k ∂ ∂X −D fD = 0 with the initial condition fD(z, 0) = f(z) and that (cz + d)−ke−cX/2πi(cz+d) fD az+b cz+d, X (cz+d)2 satisﬁes the same diﬀerential equation with the same initial condition. Now to deduce Proposition 18 we simply look at the product of fD(z, −X) with gD(z, X). Proposition 19 implies that this product is multiplied by (cz + d)k+ℓwhen z and X are replaced by az+b cz+d and X (cz+d)2 (the factors involving an exponential in X cancel), and this means that the coeﬃcient of Xn in the product, which is equal to [f,g]n (k)n(ℓ)n , is modular of weight k + ℓ+ 2n for every n ≥0. Rankin–Cohen brackets have many applications in the theory of modu-lar forms. We will describe two – one very straightforward and one more subtle – at the end of this subsection, and another one in §5.4. First, how-ever, we make a further comment about the Cohen–Kuznetsov series at-tached to a modular form. We have already introduced two such series: the series fD(z, X) deﬁned by (60), with coeﬃcients proportional to Dnf(z), and the series f∂(z, X) deﬁned by (63), with coeﬃcients proportional to ∂nf(z). But, at least when Γ is the full modular group Γ1, we had de-ﬁned a third diﬀerentiation operator besides D and ∂, namely the opera-tor ϑ deﬁned in (53), and it is natural to ask whether there is a correspond-ing Cohen–Kuznetsov series fϑ here also. The answer is yes, but this series Elliptic Modular Forms and Their Applications 55 is not simply given by n≥0 ϑnf(z) Xn/n! (k)n . Instead, for f ∈Mk(Γ1), we deﬁne a sequence of modiﬁed derivatives ϑ[n]f ∈Mk+2n(Γ1) for n ≥0 by ϑf = f, ϑ f = ϑf, ϑ[r+1]f = ϑ ϑ[r]f −r(k+r−1) E4 144ϑ[r−1]f for r ≥1 (64) (the last formula also holds for r = 0 with f [−1] deﬁned as 0 or in any other way), and set fϑ(z, X) = ∞ n=0 ϑ[n]f(z) n! (k)n Xn . Using the ﬁrst equation in (54) we ﬁnd by induction on n that Dnf = n r=0 n r (k + r)n−r E2 12 n−r ϑ[r]f (n = 0, 1, . . . ) , (65) (together with similar formulas for ∂nf in terms of ϑ[n]f and for ϑ[n]f in terms of Dnf or ∂nf), the explicit version of the expansion of Dnf as a polynomial in E2 with modular coeﬃcients whose existence is guaran-teed by Proposition 15. This formula together with (62) gives us the rela-tions fϑ(z, X) = e−XE2(z)/12 fD(z, X) = e−XE∗ 2 (z)/12 f∂(z, X) (66) between the new series and the old ones, where E∗ 2(z) is the non-holomorphic Eisenstein series deﬁned in (21), which transforms like a modular form of weight 2 on Γ1. More generally, if we are on any discrete subgroup Γ of SL(2, R), we choose a holomorphic or meromorphic function φ in H such that the function φ∗(z) = φ(z) − 1 4πy transforms like a modular form of weight 2 on Γ, or equivalently such that φ az+b cz+d = (cz + d)2φ(z) + 1 2πic(cz + d) for all a b c d ∈Γ. (Such a φ always exists, and if Γ is commensurable with Γ1 is simply the sum of 1 12E2(z) and a holomorphic or meromorphic modular form of weight 2 on Γ ∩Γ1.) Then, just as in the special case φ = E2/12, the operator ϑφ deﬁned by ϑφf := Df −kφf for f ∈Mk(Γ) sends Mk(Γ) to Mk+2(Γ), the function ω := φ′ −φ2 belongs to M4(Γ) (gen-eralizing the ﬁrst equation in (54)), and if we generalize the above deﬁni-tion by introducing operators ϑ[n] φ : Mk(Γ) →Mk+2n(Γ) (n = 0, 1, . . . ) by ϑ φ f = f , ϑ[r+1] φ f = ϑ[r] φ f + r(k + r −1) ω ϑ[r−1] φ f for r ≥0 , (67) then (65) holds with 1 12E2 replaced by φ, and (66) is replaced by fϑφ(z, X) := ∞ n=0 ϑ[n] φ f(z) Xn n! (k)n = e−φ(z)X fD(z, X) = e−φ∗(z)X f∂(z, X) . (68) 56 D. Zagier These formulas will be used again in §§6.3–6.4. We now give the two promised applications of Rankin–Cohen brackets. ♠Further Identities for Sums of Powers of Divisors In §2.2 we gave identities among the divisor power sums σν(n) as one of our ﬁrst applications of modular forms and of identities like E2 4 = E8. By including the quasimodular form E2 we get many more identities of the same type, e.g., the relationship E2 2 = E4 + 12E′ 2 gives the identity n−1 m=1 σ1(m)σ1(n −m) = 1 12 5σ3(n) −(6n −1)σ1(n) and similarly for the other two formulas in (54). Using the Rankin–Cohen brackets we get yet more. For instance, the ﬁrst Rankin–Cohen bracket of E4(z) and E6(z) is a cusp form of weight 12 on Γ1, so it must be a multiple of Δ(z), and this leads to the for-mula τ(n) = n( 7 12σ5(n) + 5 12σ3(n)) −70 n−1 m=1(5m −2n)σ3(m)σ5(n −m) for the coeﬃcient τ(n) of qn in Δ. (This can also be expressed completely in terms of elementary functions, without mentioning Δ(z) or τ(n), by writing Δ as a linear combination of any two of the three functions E12, E4E8 and E2 6.) As in the case of the identities mentioned in §2.2, all of these identities also have combinatorial proofs, but these involve much more work and more thought than the (quasi)modular ones. ♥ ♠Exotic Multiplications of Modular Forms A construction which is familiar both in symplectic geometrys and in quantum theory (Moyal brackets) is that of the deformation of the multiplication in an algebra. If A is an algebra over some ﬁeld k, say with a commutative and asso-ciative multiplication μ : A ⊗A →A, then we can look at deformations με of μ given by formal power series με(x, y) = μ0(x, y)+μ1(x, y)ε+μ2(x, y)ε2+· · · with μ0 = μ which are still associative but are no longer necessarily commu-tative. (To be more precise, με is a multiplication on A if there is a topol-ogy and the series above is convergent and otherwise, after being extended k-linearly, a multiplication on A.) The linear term μ1(x, y) in the ex-pansion of με is then anti-symmetric and satisﬁes the Jacobi identity, making A (or A) into a Lie algebra, and the μ1-product with a ﬁxed element of A is a derivation with respect to the original multiplication μ, giving the structure of a Poisson algebra. All of this is very reminiscent of the zeroth and ﬁrst Rankin–Cohen brackets, so one can ask whether these two brack-ets arise as the beginning of the expansion of some deformation of the or-dinary multiplication of modular forms. Surprisingly, this is not only true, but there is in fact a two-parameter family of such deformed multiplica-tions: Elliptic Modular Forms and Their Applications 57 Theorem. Let u and v be two formal variables and Γ any subgroup of SL(2, R). Then the multiplication μu,v on ∞ k=0 Mk(Γ) deﬁned by μu,v(f, g) = ∞ n=0 tn(k, ℓ; u, v) [f, g]n (f ∈Mk(Γ), g ∈Mℓ(Γ)) , (69) where the coeﬃcients tn(k, ℓ; u, v) ∈Q(k, l)[u, v] are given by tn(k, ℓ; u, v) = vn 0≤j≤n/2 n 2j −1 2 j u v −1 j −u v j −k −1 2 j −ℓ−1 2 j n + k + l −3 2 j , (70) is associative. Of course the multiplication given by (u, v) = (0, 0) is just the usual mul-tiplication of modular forms, and the multiplications associated to (u, v) and (λu, λv) are isomorphic by rescaling f →λkf for f ∈Mk, so the set of new multiplications obtained this way is parametrized by a projective line. Two of the multiplications obtained are noteworthy: the one corresponding to (u : v) = (1 : 0) because it is the only commutative one, and the one cor-responding to (u, v) = (0, 1) because it is so simple, being given just by f ∗g = n[f, g]n. These deformed multiplications were found, at approximately the same time, in two independent investigations and as consequences of two quite diﬀerent theories. On the one hand, Y. Manin, P. Cohen and I studied Γ-invariant and twisted Γ-invariant pseudodiﬀerential operators in the up-per half-plane. These are formal power series Ψ(z) = n≥h fn(z)D−n, with D as in (52), transforming under Γ by Ψ az+b cz+d = Ψ(z) or by Ψ az+b cz+d = (cz + d)κΨ(z)(cz + d)−κ, respectively, where κ is some complex parameter. (Notice that the second formula is diﬀerent from the ﬁrst because multiplica-tion by a non-constant function of z does not commute with the diﬀerentiation operator D, and is well-deﬁned even for non-integral κ because the ambiguity of argument involved in choosing a branch of (cz + d)κ cancels out when one divides by (cz + d)κ on the other side.) Using that D transforms under the action of a b c d to (cz + d)2D, one ﬁnds that the leading coeﬃcient fh of F is then a modular form on Γ of weight 2h and that the higher coeﬃcients fh+j (j ≥0) are speciﬁc linear combinations (depending on h, j and κ) of Djg0, Dj−1g1, . . . , gj for some modular forms gi ∈M2h+2i(G), so that (assuming that Γ contains −1 and hence has only modular forms of even weight) we can canonically identify the space of invariant or twisted invariant pseudodif-ferential operators with ∞ k=0 Mk(Γ). On the other hand, pseudo-diﬀerential operators can be multiplied just by multiplying out the formal series deﬁning them using Leibnitz’s rule, this multiplication clearly being associative, and this then leads to the family of non-trivial multiplications of modular forms given in the theorem, with u/v = κ−1/2. The other paper in which the same 58 D. Zagier multiplications arise is one by A. and J. Untenberger which is based on a cer-tain realization of modular forms as Hilbert-Schmidt operators on L2(R). In both constructions, the coeﬃcients tn(k, ℓ; u, v) come out in a diﬀerent and less symmetric form than (70); the equivalence with the form given above is a very complicated combinatorial identity. ♥ 5.3 Quasimodular Forms We now turn to the deﬁnition of the ring M∗(Γ) of quasimodular forms on Γ. In §5.1 we deﬁned this ring for the case Γ = Γ1 as C[E2, E4, E6]. We now give an intrinsic deﬁnition which applies also to other discrete groups Γ. We start by deﬁning the ring of almost holomorphic modular forms on Γ. By deﬁnition, such a form is a function in H which transforms like a modular form but, instead of being holomorphic, is a polynomial in 1/y (with y = I(z) as usual) with holomorphic coeﬃcients, the motivating examples being the non-holomorphic Eisenstein series E∗ 2(z) and the non-holomorphic derivative ∂f(z) of a holomorphic modular form as deﬁned in equations (21) and (55), respectively. More precisely, an almost holomorphic modular form of weight k and depth ≤p on Γ is a function of the form F(z) = p r=0 fr(z)(−4πy)−r with each fr ∈Hol0(H) (holomorphic functions of moderate growth) satisfying F\|kγ = F for all γ ∈Γ. We denote by # M (≤p) k = # M (≤p) k (Γ) the space of such forms and by # M∗= k # Mk, # Mk = ∪p # M (≤p) k the graded and ﬁltered ring of all almost holomorphic modular forms, usually omitting Γ from the notations. For the two basic examples E∗ 2 ∈# M (≤1) 2 (Γ1) and ∂kf ∈# M (≤1) k+2 (Γ) (where f ∈Mk(Γ)) we have f0 = E2, f1 = 12 and f0 = Df, f1 = kf, respectively. We now deﬁne the space M (≤p) k = M (≤p) k (Γ) of quasimodular forms of weight k and depth ≤p on Γ as the space of “constant terms” f0(z) of F(z) as F runs over # M (≤p) k . It is not hard to see that the almost holomorphic modular form F is uniquely determined by its constant term f0, so the ring M∗= k Mk ( Mk = ∪p M (≤p) k ) of quasimodular forms on Γ is canonically isomorphic to the ring # M∗of almost holomorphic modular forms on Γ. One can also deﬁne quasimodular forms directly, as was pointed out to me by W. Nahm: a quasimodular form of weight k and depth ≤p on Γ is a function f ∈Hol0(H) such that, for ﬁxed z ∈H and variable γ = a b c d ∈Γ, the function f\|kγ (z) is a polynomial of degree ≤p in c cz+d. Indeed, if f(z) = f0(z) ∈ Mk corresponds to F(z) = r fr(z) (−4πy)−r ∈# Mk, then the modularity of F implies the identity f\|kγ (z) = r fr(z) c cz+d r with the same coeﬃcients fr(z), and conversely. The basic facts about quasimodular forms are summarized in the following proposition, in which Γ is a non-cocompact discrete subgroup of SL(2, R) and φ ∈ M2(Γ) is a quasimodular form of weight 2 on Γ which is not modular, e.g., φ = E2 if Γ is a subgroup of Γ1. For Γ = Γ1 part (i) of the proposition reduces to Proposition 15 above, while part (ii) shows that the general deﬁ-Elliptic Modular Forms and Their Applications 59 nition of quasimodular forms given here agrees with the ad hoc one given in §5.1 in this case. Proposition 20. (i) The space of quasimodular forms on Γ is closed un-der diﬀerentiation. More precisely, we have D M (≤p) k ⊆ M (≤p+1) k+2 for all k, p ≥0. (ii) Every quasimodular form on Γ is a polynomial in φ with modular co-eﬃcients. More precisely, we have M (≤p) k (Γ) = p r=0 Mk−2r(Γ) · φr for all k, p ≥0. (iii) Every quasimodular form on Γ can be written uniquely as a linear com-bination of derivatives of modular forms and of φ. More precisely, for all k, p ≥0 we have M (≤p) k (Γ) = ⎧ ⎨ ⎩ p r=0 Dr Mk−2r(Γ) if p < k/2 , k/2−1 r=0 Dr Mk−2r(Γ) ⊕C · Dk/2−1φ if p ≥k/2 . Proof. Let F = fr(−4πy)−r ∈# Mk correspond to f = f0 ∈ Mk. The almost holomorphic form ∂kF ∈ # Mk+2 then has the expansion ∂kF = ! D(fr) + (k −r + 1)fr−1 " (−4πy)−r, with constant term Df. This proves the ﬁrst statement. (One can also prove it in terms of the direct deﬁnition of quasimodular forms by diﬀerentiating the formula expressing the transfor-mation behavior of f under Γ.) Next, one checks easily that if F belongs to # M (≤p) k , then the last coeﬃcient fp(z) in its expansion is a modular form of weight k −2p. It follows that p ≤k/2 (if fp ̸= 0, i.e., if F has depth exactly p) and also, since the almost holomorphic modular form φ∗corresponding to φ is the sum of φ and a non-zero multiple of 1/y, that F is a linear combi-nation of fpφ∗p and an almost holomorphic modular form of depth strictly smaller than p, from which statement (ii) for almost holomorphic modular forms (and therefore also for quasimodular forms) follows by induction on p. Statement (iii) is proved exactly the same way, by subtracting from F a mul-tiple of ∂pfp if p < k/2 and a multiple of ∂k/2−1φ if p = k/2 to prove by induction on p the corresponding statement with “quasimodular” and D re-placed by “almost holomorphic modular” and ∂, and then again using the isomorphism between # M∗and M∗. There is one more important element of the structure of the ring of quasi-modular (or almost holomorphic modular) forms. Let F = fr(−4πy)−r and f = f0 be an almost holomorphic modular form of weight k and depth ≤p and the quasimodular form which corresponds to it. One sees easily using the properties above that each coeﬃcient fr is quasimodular (of weight k−2r and depth ≤p−r) and that, if δ : M∗→ M∗is the map which sends f to f1, then fr = δrf/r! for all r ≥0 (and δrf = 0 for r > p), so that the expansion of F(z) in powers of −1/4πy is a kind of Taylor expansion formula. This gives us three operators from M∗to itself: the diﬀerentiation operator D, the operator E 60 D. Zagier which multiplies a quasimodular form of weight k by k, and the operator δ. Each of these operators is a derivation on the ring of quasimodular forms, and they satisfy the three commutation relations [E, D] = 2 D , [D, δ] = −2 δ , [D, δ] = E (of which the ﬁrst two just say that D and δ raise and lower the weight of a quasimodular form by 2, respectively), giving to M∗the structure of an sl2C)-module. Of course the ring # M∗, being isomorphic to M∗, also be-comes an sl2C)-module, the corresponding operators being ϑ (= ϑk on Mk), E (= multiplication by k on # Mk) and δ∗(= “derivation with respect to −1/4πy”). From this point of view, the subspace M∗of M∗or # M∗appears sim-ply as the kernel of the lowering operator δ (or δ∗). Using this sl2C)-module structure makes many calculations with quasimodular or almost holomorphic modular forms simpler and more transparent. ♠Counting Ramiﬁed Coverings of the Torns We end this subsection by describing very brieﬂy a beautiful and unexpected context in which quasimodular forms occur. Deﬁne Θ(X, z, ζ) = n>0 (1 −qn) n>0 n odd 1 −en2X/8 qn/2 ζ 1 −e−n2X/8 qn/2 ζ−1 , expand Θ(X, z, ζ) as a Laurent series n∈Z Θn(X, z) ζn, and expand Θ0(X, z) as a Taylor series ∞ r=0 Ar(z) X2r. Then a somewhat intricate calculation involving Eisenstein series, theta series and quasimodular forms on Γ(2) shows that each Ar is a quasimodular form of weight 6r on SL(2, Z), i.e., a weighted homogeneous polynomial in E2, E4, and E6. In particular, A0(z) = 1 (this is a consequence of the Jacobi triple product identity, which says that Θn(0, z) = (−1)nqn2/2), so we can also expand log Θ0(X, z) as ∞ r=1 Fr(z) X2r and Fr(z) is again quasimodular of weight 6r. These functions arose in the study of the “one-dimensional analogue of mirror symmetry”: the coeﬃcient of qm in Fr counts the generically ramiﬁed coverings of degree m of a curve of genus 1 by a curve of genus r + 1. (“Generic” means that each point has ≥m −1 preimages.) We thus obtain: Theorem. The generating function of generically ramiﬁed coverings of a torus by a surface of genus g > 1 is a quasimodular form of weight 6g−6 on SL(2, Z). As an example, we have F1 = A1 = 1 103680(10E3 2 −6E2E4 −4E6) = q2 + 8q3 + 30q4 + 80q5 + 180q6 + · · · . In this case (but for no higher genus), the function F1 = 1 1440(E′ 4 + 10E′′ 2 ) is also a linear combination of derivatives of Eisenstein series and we get a simple explicit formula n(σ3(n)−nσ1(n))/6 for the (correctly counted) number of degree n coverings of a torus by a surface of genus 2. ♥ Elliptic Modular Forms and Their Applications 61 5.4 Linear Diﬀerential Equations and Modular Forms The statement of Proposition 16 is very simple, but not terribly useful, because it is diﬃcult to derive properties of a function from a non-linear diﬀerential equation. We now prove a much more useful fact: if we express a modular form as a function, not of z, but of a modular function of z (i.e., a meromorphic modular form of weight zero), then it always satisﬁes a linear diﬀerentiable equation of ﬁnite order with algebraic coeﬃcients. Of course, a modular form cannot be written as a single-valued function of a modular function, since the latter is invariant under modular transformations and the former transforms with a non-trivial automorphy factor, but in can be expressed locally as such a function, and the global non-uniqueness then simply corresponds to the monodromy of the diﬀerential equation. The fact that we have mentioned is by no means new – it is at the heart of the original discovery of modular forms by Gauss and of the later work of Fricke and Klein and others, and appears in modern literature as the theory of Picard–Fuchs diﬀerential equations or of the Gauss–Manin connection – but it is not nearly as well known as it ought to be. Here is a precise statement. Proposition 21. Let f(z) be a (holomorphic or meromorphic) modular form of positive weight k on some group Γ and t(z) a modular function with re-spect to Γ. Express f(z) (locally) as Φ(t(z)). Then the function Φ(t) satisﬁes a linear diﬀerential equation of order k + 1 with algebraic coeﬃcients, or with polynomial coeﬃcients if Γ\H has genus 0 and t(z) generates the ﬁeld of modular functions on Γ. This proposition is perhaps the single most important source of applications of modular forms in other branches of mathematics, so with no apology we sketch three diﬀerent proofs, each one giving us diﬀerent information about the diﬀerential equation in question. Proof 1: We want to ﬁnd a linear relation among the derivatives of f with respect to t. Since f(z) is not deﬁned in Γ\H, we must replace d/dt by the operator Dt = t′(z)−1d/dz, which makes sense in H. We wish to show that the functions Dn t f (n = 0, 1, . . . , k + 1) are linearly dependent over the ﬁeld of modular functions on Γ, since such functions are algebraic functions of t(z) in general and rational functions in the special cases when Γ\H has genus 0 and t(z) is a “Hauptmodul” (i.e., t : Γ\H →P1(C) is an isomorphism). The diﬃculty is that, as seen in §5.1, diﬀerentiating the transformation equation (2) produces undesired extra terms as in (52). This is because the automorphy factor (cz+d)k in (2) is non-constant and hence contributes non-trivially when we diﬀerentiate. To get around this, we replace f(z) by the vector-valued function F : H →Ck+1 whose mth component (0 ≤m ≤k) is Fm(z) = zk−mf(z). Then Fm az + b cz + d = (az + b)k−m (cz + d)m f(z) = m n=0 Mmn Fn(z) , (71) 62 D. Zagier for all γ = a b c d ∈Γ, where M = Sk(γ) ∈SL(k + 1, Z) denotes the kth symmetric power of γ. In other words, we have F(γz) = MF(z) where the new “automorphy factor” M, although it is more complicated than the automorphy factor in (2) because it is a matrix rather than a scalar, is now independent of z, so that when we diﬀerentiate we get simply (cz +d)−2F ′(γz) = MF ′(z). Of course, this equation contains (cz +d)−2, so diﬀerentiating again with respect to z would again produce unwanted terms. But the Γ-invariance of t implies that t′ az+b cz+d = (cz+d)2t′(z), so the factor (cz+d)−2 is cancelled if we replace d/dz by Dt = d/dt. Thus (71) implies DtF(γz) = MDtF(z), and by induction Dr t F(γz) = MDr t F(z) for all r ≥0. Now consider the (k + 2)× (k + 2) matrix f Dtf · · · Dk+1 t f F DtF · · · Dk+1 t F . The top and bottom rows of this matrix are identical, so its determinant is 0. Expanding by the top row, we ﬁnd 0 = k+1 n=0(−1)n det(An(z)) Dn t f(z), where An(z) is the (k + 1) × (k + 1) matrix F DtF · · · Dn t F · · · Dk+1 t F . From Dr t F(γz) = MDr t F(z) (∀r) we get An(γz) = MAn(z) and hence, since det(M) = 1, that An(z) is a Γ-invariant function. Since An(z) is also mero-morphic (including at the cusps), it is a modular function on Γ and hence an algebraic or rational function of t(z), as desired. The advantage of this proof is that it gives us all k+1 linearly independent solutions of the diﬀerential equa-tion satisﬁed by f(z): they are simply the functions f(z), zf(z), . . ., zkf(z). Proof 2 (following a suggestion of Ouled Azaiez): We again use the diﬀer-entiation operator Dt, but this time work with quasimodular rather than vector-valued modular forms. As in §5.2, choose a quasimodular form φ(z) of weight 2 on Γ with δ(φ) = 1 (e.g., 1 12E2 if Γ = Γ1), and write each Dn t f(z) (n = 0, 1, 2, . . . ) as a polynomial in φ(z) with (meromorphic) modular coeﬃ-cients. For instance, for n = 1 we ﬁnd Dtf = k f t′ φ + ϑφf t′ where ϑφ denotes the Serre derivative with respect to φ. Using that φ′ −φ2 ∈M4, one ﬁnds by induction that each Dn t f is the sum of k(k −1) · · · (k −n + 1) (φ/t′)n f and a polynomial of degree < n in φ. It follows that the k + 2 func-tions Dn t (f)/f 0≤n≤k+1 are linear combinations of the k + 1 functions {(φ/t′)n}0≤n≤k with modular functions as coeﬃcients, and hence that they are linearly dependent over the ﬁeld of modular functions on Γ. Proof 3: The third proof will give an explicit diﬀerential equation satisﬁed by f. Consider ﬁrst the case when f has weight 1. The funcion t′ = D(t) is a (meromorphic) modular form of weight 2, so we can form the Rankin–Cohen brackets [f, t′]1 and [f, f]2 of weights 5 and 6, respectively, and the quotients A = [f,t′]1 ft′2 and B = −[f,f]2 2f 2t′2 of weight 0. Then D2 t f + A Dtf + Bf = 1 t′ f ′ t′ ′ + ft′′ −2f ′t′ ft′2 f ′ t′ −ff ′′ −2f ′2 t′2f 2 f = 0 . Elliptic Modular Forms and Their Applications 63 Since A and B are modular functions, they are rational (if t is a Hauptmodul) or algebraic (in any case) functions of t, say A(z) = a(t(z)) and B(z) = b(t(z)), and then the function Φ(t) deﬁned locally by f(z) = Φ(t(z)) satisﬁes Φ′′(t) + a(t)Φ′(t) + b(t)Φ(t) = 0. Now let f have arbitrary (integral) weight k > 0 and apply the above construction to the function h = f 1/k, which is formally a modular form of weight 1. Of course h is not really a modular form, since in general it is not even a well-deﬁned function in the upper half-plane (it changes by a kth root of unity when we go around a zero of f). But it is deﬁned locally, and the functions A = [h,t′]1 ht′2 and B = −[h,h]2 2h2t′2 are well-deﬁned, because they are both homogeneous of degree 0 in h, so that the kth roots of unity occur-ring when we go around a zero of f cancel out. (In fact, a short calculation shows that they can be written directly in terms of Rankin–Cohen brack-ets of f and t′ , namely A = [f,t′]1 kft′2 and B = − [f,f]2 k2(k+1)f 2t′2 .) Now just as before A(z) = a(t(z)), B(z) = b(t(z)) for some rational or algebraic func-tions a(t) and b(t). Then Φ(t)1/k is annihilated by the second order diﬀer-ential operator L = d2/dt2 + a(t)d/dt + b(t) and Φ(t) itself is annihilated by the (k + 1)st order diﬀerential operator Symk(L) whose solutions are the kth powers or k-fold products of solutions of the diﬀerential equation LΨ = 0. The coeﬃcients of this operator can be given by explicit expres-sions in terms of a and b and their derivatives (for instance, for k = 2 we ﬁnd Sym2(L) = d3/dt3 + 3a d2/dt2 + (a′ + 2a2 + 4b) d/dt + 2(b′ + 2ab) ), and these in turn can be written as weight 0 quotients of appropriate Rankin– Cohen brackets. Here are two classical examples of Proposition 21. For the ﬁrst, we take Γ = Γ(2), f(z) = θ3(z)2 (with θ3(z) = qn2/2 as in (32)) and t(z) = λ(z), where λ(z) is the Legendre modular function λ(z) = 16 η(z/2)8 η(2z)16 η(z)24 = 1 −η(z/2)16 η(2z)8 η(z)24 = θ2(z) θ3(z) 4 , (72) which is known to be a Hauptmodul for the group Γ(2). Then θ3(z)2 = ∞ n=0 2n n λ(z) 16 n = F 1 2, 1 2; 1; λ(z) , (73) where F(a, b; c; x) = ∞ n=0 (a)n(b)n n! (c)n xn with (a)n as in eq. (56) denotes the Gauss hypergeometric function, which satisﬁes the second order diﬀerential equation x(x −1) y′′ + (a + b + 1)x −c y′ + aby = 0. For the second example we take Γ = Γ1, t(z) = 1728/j(z), and f(z) = E4(z). Since f is a modular form of weight 4, it should satisfy a ﬁfth order linear diﬀerential equation with respect to t(z), but by the third proof above one should even have that the 64 D. Zagier fourth root of f satisﬁes a second order diﬀerential equation, and indeed one ﬁnds 4 E4(z) = F 1 12, 5 12; 1; t(z) = 1 + 1 · 5 1 · 1 12 j(z) + 1 · 5 · 13 · 17 1 · 1 · 2 · 2 122 j(z)2 + · · · , (74) a classical identity which can be found in the works of Fricke and Klein. ♠The Irrationality of ζ(3) In 1978, Roger Apéry created a sensation by proving: Theorem. The number ζ(3) = n≥1 n−3 is irrational. What he actually showed was that, if we deﬁne two sequences {an} = {1, 5, 73, 1445, . . . } and {bn} = {0, 6, 351/4, 62531/36, . . . } as the solutions of the recursion (n + 1)3 un+1 = (34n3 + 51n2 + 27n + 5) un −n3 un−1 (n ≥1) (75) with initial conditions a0 = 1, a1 = 5, b0 = 0, b1 = 6, then we have an ∈Z (∀n ≥0) , D3 n bn ∈Z (∀n ≥0) , lim n→∞ bn an = ζ(3) , (76) where Dn denotes the least common multiple of 1, 2, . . ., n. These three as-sertions together imply the theorem. Indeed, both an and bn grow like Cn by the recursion, where C = 33.97 . . . is the larger root of C2 −34C + 1 = 0. On the other hand, an−1bn −anbn−1 = 6/n3 by the recursion and induction, so the diﬀerence between bn/an and its limiting value ζ(3) decreases like C−2n. Hence the quantity xn = D3 n(bn−anζ(3)) grows like by D3 n/(C+o(1))n, which tends to 0 as n →∞since C > e3 and D3 n = (e3 + o(1))n by the prime num-ber theorem. (Chebyshev’s weaker elementary estimate of Dn would suﬃce here.) But if ζ(3) were rational then the ﬁrst two statements in (76) would imply that the xn are rational numbers with bounded denominators, and this is a contradiction. Apéry’s own proof of the three properties (76), which involved complicated explicit formulas for the numbers an and bn as sums of binomial coeﬃcients, was very ingenious but did not give any feeling for why any of these three properties hold. Subsequently, two more enlightening proofs were found by Frits Beukers, one using representations of an and bn as multiple integrals involving Legendre polynomials and the other based on modularity. We give a brief sketch of the latter one. Let Γ = Γ + 0 (6) as in §3.1 be the group Γ0(6)∪Γ0(6)W, where W = W6 = 1 √ 6 0 −1 6 0 . This group has genus 0 and the Hauptmodul t(z) = η(z) η(6z) η(2z) η(3z) 12 = q −12 q2 + 66 q3 −220 q4 + . . . , Elliptic Modular Forms and Their Applications 65 where η(z) is the Dedekind eta-function deﬁned in (34). For f(z) we take the function f(z) = η(2z) η(3z) 7 η(z) η(6z) 5 = 1 + 5 q + 13 q2 + 23 q3 + 29 q4 + . . . , which is a modular form of weight 2 on Γ (and in fact an Eisenstein series, namely 5G2(z)−2G2(2z)+3G2(3z)−30G2(6z) ). Proposition 21 then implies that if we expand f(z) as a power series 1 + 5 t(z) + 73 t(z)2+ 1445 t(z)3 + · · · in t(z), then the coeﬃcients of the expansion satisfy a linear recursion with polynomial coeﬃcients (this is equivalent to the statement that the power series itself satisﬁes a diﬀerential equation with polynomial coeﬃcients), and if we go through the proof of the proposition to calculate the diﬀerential equation explicitly, we ﬁnd that the recursion is exactly the one deﬁning an. Hence f(z) = ∞ n=0 an t(z)n, and the integrality of the coeﬃcients an follows immediately since f(z) and t(z) have integral coeﬃcients and t has leading coeﬃcient 1. To get the properties of the second sequence {bn} is a little harder. Deﬁne g(z) = G4(z) −28 G4(2z) + 63 G4(3z) −36 G4(6z) = q −14 q2 + 91 q3 −179 q4 + · · · , an Eisenstein series of weight 4 on Γ. Write the Fourier expansion as g(z) = ∞ n=1 cn qn and deﬁne ˜ g(z) = ∞ n=1 n−3cn qn, so that ˜ g′′′ = g. This is the so-called Eichler integral associated to g and inherits certain modular properties from the modularity of g. (Speciﬁcally, the diﬀerence ˜ g\|−2 γ−˜ g is a polynomial of degree ≤2 for every γ ∈Γ, where \|−2 is the slash operator deﬁned in (8).) Using this (we skip the details), one ﬁnds by an argument analogous to the proof of Proposition 21 that if we expand the product f(z)˜ g(z) as a power series t(z) + 117 8 t(z)2 + 62531 216 t(z)3 + · · · in t(z), then this power series again satisﬁes a diﬀerential equation. This equation turns out to be the same one as the one satisﬁed by f (but with right-hand side 1 instead of 0), so the coeﬃcients of the new expansion satisfy the same recursion and hence (since they begin 0, 1) must be one-sixth of Apéry’s coeﬃcients bn, i.e., we have 6f(z)˜ g(z) = ∞ n=0 bn t(z)n. The integrality of D3 nbn (indeed, even of D3 nbn/6) follows immediately: the coeﬃcients cn are integral, so D3 n is a common de-nominator for the ﬁrst n terms of f˜ g as a power series in q and hence also as a power series in t(z). Finally, from the deﬁnition (13) of G4(z) we ﬁnd that the Fourier coeﬃcients of g are given by the Dirichlet series identity ∞ n=1 cn ns = 1 −28 2s + 63 3s −36 6s ζ(s) ζ(s −3) , so the limiting value of bn/an (which must exist because {an} and {bn} satisfy the same recursion) is given by 66 D. Zagier 1 6 lim n→∞ bn an = ˜ g(z) t(z)=1/C = ∞ n=1 cn n3 qn q=1 = ∞ n=1 cn ns s=3 = 1 6 ζ(3) , proving also the last assertion in (76). ♥ ♠An Example Coming from Percolation Theory Imagine a rectangle R of width r > 0 and height 1 on which has been drawn a ﬁne square grid (of size roughly rN × N for some integer N going to in-ﬁnity). Each edge of the grid is colored black or white with probability 1/2 (the critical probability for this problem). The (horizontal) crossing probability Π(r) is then deﬁned as the limiting value, as N →∞, of the probability that there exists a path of black edges connecting the left and right sides of the rectangle. A simple combinatorial argument shows that there is always either a vertex-to-vertex path from left to right passing only through black edges or else a square-to-square path from top to bottom passing only through white edges, but never both. This implies that Π(r) + Π(1/r) = 1. The hypothesis of conformality which, though not proved, is universally believed and is at the basis of the modern theory of percolation, says that the corresponding prob-lem, with R replaced by any (nice) open domain in the plane, is unchanged under conformal (biholomorphic) mappings, and this can be used to compute the crossing probability as the solution of a diﬀerential equation. The result, due to J. Cardy, is the formula Π(r) = 2π √ 3 Γ( 1 3)−3 t1/3 F 1 3, 2 3; 4 3; t), where t is the cross-ratio of the images of the four vertices of the rectangle R when it is mapped biholomorphically onto the unit disk by the Riemann uniformiza-tion theorem. This cross-ratio is known to be given by t = λ(ir) with λ(z) as in (72). In modular terms, using Proposition 21, we ﬁnd that this translates to the formula Π(r) = −27/3 3−1/2 π2 Γ( 1 3)−3 ∞ r η(iy)4dy ; i.e., the deriva-tive of Π(r) is essentially the restriction to the imaginary axis of the modular form η(z)4 of weight 2. Conversely, an easy argument using modular forms on SL(2, Z) shows that Cardy’s function is the unique function satisfying the functional equation Π(r) + Π(1/r) = 1 and having an expansion of the form e−2παr times a power series in e−2πr for some α ∈R (which is then given by α = 1/6). Unfortunately, there seems to be no physical argument implying a priori that the crossing probability has the latter property, so one cannot (yet?) use this very simple modular characterization to obtain a new proof of Cardy’s famous formula. ♥ 6 Singular Moduli and Complex Multiplication The theory of complex multiplication, the last topic which we will treat in detail in these notes, is by any standards one of the most beautiful chapters in all of number theory. To describe it fully one needs to combine themes relating to elliptic curves, modular forms, and algebraic number theory. Given Elliptic Modular Forms and Their Applications 67 the emphasis of these notes, we will discuss mostly the modular forms side, but in this introduction we brieﬂy explain the notion of complex multiplication in the language of elliptic curves. An elliptic curve over C, as discussed in §1, can be represented by a quo-tient E = C/Λ, where Λ is a lattice in C. If E′ = C/Λ′ is another curve and λ a complex number with λΛ ⊆Λ′, then multiplication by λ induces an algebraic map from E to E′. In particular, if λΛ ⊆Λ, then we get a map from E to itself. Of course, we can always achieve this with λ ∈Z, since Λ is a Z-module. These are the only possible real values of λ, and for generic lattices also the only possible complex values. Elliptic curves E = C/Λ where λΛ ⊆Λ for some non-real value of λ are said to admit complex multiplication. As we have seen in these notes, there are two completely diﬀerent ways in which elliptic curves are related to modular forms. On the one hand, the moduli space of elliptic curves is precisely the domain of deﬁnition Γ1\H of modular functions on the full modular group, via the map Γ1z ↔ ! C/Λz " , where Λz = Zz + Z ⊂C. On the other hand, elliptic curves over Q are sup-posed (and now ﬁnally known) to have parametrizations by modular functions and to have Hasse-Weil L-functions that coincide with the Hecke L-series of certain cusp forms of weight 2. The elliptic curves with complex multiplication are of special interest from both points of view. If we think of H as parametriz-ing elliptic curves, then the points in H corresponding to elliptic curves with complex multiplication (usually called CM points for short) are simply the numbers z ∈H which satisfy a quadratic equation over Z. The basic fact is that the value of j(z) (or of any other modular function with algebraic coef-ﬁcients evaluated at z) is then an algebraic number; this says that an elliptic curve with complex multiplication is always deﬁned over Q (i.e., has a Weier-strass equation with algebraic coeﬃcients). Moreover, these special algebraic numbers j(z), classically called singular moduli, have remarkable properties: they give explicit generators of the class ﬁelds of imaginary quadratic ﬁelds, the diﬀerences between them factor into small prime factors, their traces are themselves the coeﬃcients of modular forms, etc. This will be the theme of the ﬁrst two subsections. If on the other hand we consider the L-function of a CM elliptic curve and the associated cusp form, then again both have very special properties: the former belongs to two important classes of number-theoretical Dirichlet series (Epstein zeta functions and L-series of grossencharacters) and the latter is a theta series with spherical coeﬃcients associated to a binary quadratic form. This will lead to several applications which are treated in the ﬁnal two subsections. 6.1 Algebraicity of Singular Moduli In this subsection we will discuss the proof, reﬁnements, and applications of the following basic statement: Proposition 22. Let z ∈H be a CM point. Then j(z) is an algebraic number. 68 D. Zagier Proof. By deﬁnition, z satisﬁes a quadratic equation over Z, say Az2 + Bz + C = 0. There is then always a matrix M ∈M(2, Z), not proportional to the identity, which ﬁxes z. (For instance, we can take M = B C −A 0 .) This matrix has a positive determinant, so it acts on the upper half-plane in the usual way. The two functions j(z) and j(Mz) are both modular functions on the subgroup Γ1 ∩M −1Γ1M of ﬁnite index in Γ1, so they are algebraically dependent, i.e., there is a non-zero polynomial P(X, Y ) in two variables such that P(j(Mz), j(z)) vanishes identically. By looking at the Fourier expansion at ∞we can see that the polynomial P can be chosen to have coeﬃcients in Q. (We omit the details, since we will prove a more precise statement below.) We can also assume that the polynomial P(X, X) is not identically zero. (If some power of X −Y divides P then we can remove it without aﬀecting the validity of the relation P(j(Mz), j(z)) ≡0, since j(Mz) is not identically equal to j(z).) The number j(z) is then a root of the non-trivial polynomial P(X, X) ∈Q[X], so it is an algebraic number. More generally, if f(z) is any modular function (say, with respect to a sub-group of ﬁnite index of SL(2, Z)) with algebraic Fourier coeﬃcients in its q-expansion at inﬁnity, then f(z) ∈Q, as one can see either by showing that f(Mz) and f(z) are algebraically dependent over Q for any M ∈M(2, Z) with det M > 0 or by showing that f(z) and j(z) are algebraically dependent over Q and using Proposition 22. The full theory of complex multiplication describes precisely the number ﬁeld in which these numbers f(z) lie and the way that Gal(Q/Q) acts on them. (Roughly speaking, any Galois conjugate of any f(z) has the form f ∗(z∗) for some other modular form with algebraic coeﬃcients and CM point z∗, and there is a recipe to compute both.) We will not explain anything about this in these notes, except for a few words in the third application below, but refer the reader to the texts listed in the references. We now return to the j-function and the proof above. The key point was the algebraic relation between j(z) and j(Mz), where M was a matrix in M(2, Z) of positive determinant m ﬁxing the point z. We claim that the poly-nomial P relating j(Mz) and j(z) can be chosen to depend only on m. More precisely, we have: Proposition 23. For each m ∈N there is a polynomial Ψm(X, Y ) ∈Z[X, Y ], symmetric up to sign in its two arguments and of degree σ1(m) with respect to either one, such that Ψm(j(Mz), j(z)) ≡0 for every matrix M ∈M(2, Z) of determinant m. Proof. Denote by Mm the set of matrices in M(2, Z) of determinant m. The group Γ1 acts on Mm by right and left multiplication, with ﬁnitely many orbits. More precisely, an easy and standard argument shows that the ﬁnite set M∗ m = $a b 0 d a, b, d ∈Z, ad = m, 0 ≤b < d % ⊂Mm (77) Elliptic Modular Forms and Their Applications 69 is a full set of representatives for Γ1\Mm; in particular, we have Γ1\Mm = M∗ m = ad=m d = σ1(m) . (78) We claim that we have an identity M∈Γ1\Mm X −j(Mz) = Ψm(X, j(z)) (z ∈H, X ∈C) (79) for some polynomial Ψm(X, Y ). Indeed, the left-hand side of (79) is well-deﬁned because j(Mz) depends only on the class of M in Γ1\Mm, and it is Γ1-invariant because Mm is invariant under right multiplication by elements of Γ1. Furthermore, it is a polynomial in X (of degree σ1(m), by (78)) each of whose coeﬃcients is a holomorphic function of z and of at most exponential growth at inﬁnity, since each j(Mz) has these properties and each coeﬃcient of the product is a polynomial in the j(Mz). But a Γ1-invariant holomorphic function in the upper half-plane with at most exponential growth at inﬁnity is a polynomial in j(z), so that we indeed have (79) for some polynomial Ψm(X, Y ) ∈C[X, Y ]. To see that the coeﬃcients of Ψm are in Z, we use the set of representatives (77) and the Fourier expansion of j, which has the form j(z) = ∞ n=−1 cnqn with cn ∈Z (c−1 = 1, c0 = 744, c1 = 196884 etc.). Thus Ψm(X, j(z)) = ad=m d>0 d−1 b=0 X −j az + b d = ad=m d>0 b (mod d) X − ∞ n=−1 cn ζbn d qan/d , where qα for α ∈Q denotes e2πiαz and ζd = e2πi/d. The expression in parentheses belongs to the ring Z[ζd][X][q−1/d, q1/d]] of Laurent series in q1/d with coeﬃcients in Z[ζd], but applying a Galois conjugation ζd →ζr d with r ∈(Z/dZ)∗just replaces b in the inner product by br, which runs over the same set Z/dZ, so the inner product has coeﬃcients in Z. The fractional pow-ers of q go away at the same time (because the product over b is invariant under z →z +1), so each inner product, and hence Ψm(X, j(z)) itself, belongs to Z[X][q−1, q]]. Now the fact that it is a polynomial in j(z) and that j(z) has a Fourier expansion with integral coeﬃcients and leading coeﬃcient q−1 imlies that Ψm(X, j(z)) ∈Z[X, j(z)]. Finally, the symmetry of Ψm(X, Y ) up to sign follows because z′ = Mz with M = a b c d ∈Mm is equivalent to z = M ′z′ with M ′ = d −b −c a ∈Mm. 70 D. Zagier An example will make all of this clearer. For m = 2 we have M∈Γ1\Mm X −j(z) = X −j z 2 X −j z + 1 2 X −j 2z by (77). Write this as X3 −A(z)X2 + B(z)X −C(z). Then A(z) = j z 2 + j z + 1 2 + j 2z = q−1/2 + 744 + o(q) + −q−1/2 + 744 + o(q) + q−2 + 744 + o(q) = q−2 + 0 q−1 + 2232 + o(1) = j(z)2 −1488 j(z) + 16200 + o(1) as z →i∞, and since A(z) is holomorphic and Γ1-invariant this implies that A = j2 −1488j+16200. A similar calculation gives B = 1488j2+40773375j+ 8748000000 and C = −j3 + 162000j2 −8748000000j + 157464000000000, so Ψ2(X, Y ) = −X2Y 2 + X3 + 1488X2Y + 1488XY 2 + Y 3 −162000X2 + 40773375XY −162000Y 2 + 8748000000X + 8748000000Y −157464000000000 . (80) Remark. The polynomial Ψm(X, Y ) is not in general irreducible: if m is not square-free, then it factors into the product of all Φm/r2(X, Y ) with r ∈N such that r2\|m, where Φm(X, Y ) is deﬁned exactly like Ψm(X, Y ) but with Mm replaced by the set M0 m of primitive matrices of determinant m. The polynomial Φm(X, Y ) is always irreducible. To obtain the algebraicity of j(z), we used that this value was a root of P(X, X). We therefore should look at the restriction of the polynomial Ψm(X, Y ) to the diagonal X = Y . In the example (80) just considered, two properties of this restriction are noteworthy. First, it is (up to sign) monic, of degree 4. Second, it has a striking factorization: Ψ2(X, X) = −(X −8000) · (X + 3375)2 · (X −1728) . (81) We consider each of these properties in turn. We assume that m is not a square, since otherwise Ψm(X, X) is identically zero because Ψm(X, Y ) contains the factor Ψ1(X, Y ) = X −Y . Proposition 24. For m not a perfect square, the polynomial Ψm(X, X) is, up to sign, monic of degree σ+ 1 (m) := d\|m max(d, m/d). Elliptic Modular Forms and Their Applications 71 Proof. Using the identity b (mod d)(x −ζb dy) = xd −yd we ﬁnd Ψm(j(z), j(z)) = ad=m b (mod d) j(z) −j az + b d = ad=m b (mod d) q−1 −ζ−b d q−a/d + o(1) = ad=m q−d −q−a + (lower order terms) ∼± q−σ+ 1 (m) as I(z) →∞, and this proves the proposition since j(z) ∼q−1. Corollary. Singular moduli are algebraic integers. □ Now we consider the factors of Ψm(X, X). First let us identify the three roots in the factorization for m = 2 just given. The three CM points i, (1 + i √ 7)/2 and i √ 2 are ﬁxed, respectively, by the three matrices 0 −1 1 0 , 1 1 −1 1 and 0 −1 2 0 of determinant 2, so each of the corresponding j-values must be a root of the polynomial (81). Computing these values numerically to low accuracy to see which is which, we ﬁnd j(i) = 1728 , j 1 + i √ 7 2 = −3375 , j(i √ 2) = 8000 . (Another way to distinguish the roots, without doing any transcendental cal-culations, would be to observe, for instance, that i and i √ 2 are ﬁxed by matrices of determinant 3 but (1 + i √ 7)/2 is not, and that X + 3375 does not occur as a factor of Ψ3(X, X) but both X −1728 and X −8000 do.) The same method can be used for any other CM point. Here is a table of the ﬁrst few values of j(zD), where zD equals 1 2 √ D for D even and 1 2(1+ √ D) for D odd: D −3 −4 −7 −8 −11 −12 −15 −16 −19 j(zD) 0 1728 −3375 8000 −32768 54000 −191025+85995 √ 5 2 287496 −884736 We can make this more precise. For each discriminant (integer congruent to 0 or 1 mod 4) D < 0 we consider, as at the end of §1.2, the set QD of prim-itive positive deﬁnite binary quadratic forms of discriminant D, i.e., functions Q(x, y) = Ax2 + Bxy + Cy2 with A, B, C ∈Z, A > 0, gcd(A, B, C) = 1 and B2 −4AC = D. To each Q ∈QD we associate the root zQ of Q(z, 1) = 0 in H. This gives a Γ1-equivariant bijection between QD and the set ZD ⊂H of CM points of discriminant D. (The discriminant of a CM point is the smallest discriminant of a quadratic polynomial over Z of which it is a root.) In particular, the cardinality of Γ1\ZD is h(D), the class number of D. We choose a set of representatives {zD,i}1≤i≤h(D) for Γ1\ZD (e.g., the points of 72 D. Zagier ZD ∩ F1, where F1 is the fundamental domain constructed in §1.2, corre-sponding to the set Qred D in (5)), with zD,1 = zD. We now form the class polynomial HD(X) = z∈Γ1\ZD X −j(z) = 1≤i≤h(D) X −j(zD,i) . (82) Proposition 25. The polynomial HD(X) belongs to Z[X] and is irreducible. In particular, the number j(zD) is algebraic of degree exactly h(D) over Q, with conjugates j(zD,i) (1 ≤i ≤h(D)). Proof. We indicate only the main ideas of the proof. We have already proved that j(z) for any CM point z is a root of the equation Ψm(X, X) = 0 whenever m is the determinant of a matrix M ∈M(2, Z) ﬁxing z. The main point is that the set of these m’s depends only on the discriminant D of z, not on z itself, so that the diﬀerent numbers j(zD,i) are roots of the same equations and hence are conjugates. Let Az2 + Bz + C = 0 (A > 0) be the minimal equation of z and suppose that M = a b c d ∈Mm ﬁxes z. Then since cz2 + (d −a)z −b = 0 we must have (c, d −a, −b) = u(A, B, C) for some u ∈Z. This gives M = 1 2(t −Bu) −Cu Au 1 2(t + Bu) , det M = t2 −Du2 4 , (83) where t = tr M. Convesely, if t and u are any integers with t2 −Du2 = 4m, then (83) gives a matrix M ∈Mm ﬁxing z. Thus the set of integers m = det M with Mz = z is the set of numbers 1 4(t2 −Du2) with t ≡Du (mod 2) or, more invariantly, the set of norms of elements of the quadratic order OD = Z[zD] = $t + u √ D 2 t, u ∈Z, t ≡Du (mod 2) % (84) of discriminant D, and this indeed depends only on D, not on z. We can then obtain HD(X), or at least its square, as the g.c.d. of ﬁnitely many polymials Ψm(X, X), just as we obtained H−7(X)2 = (X + 3375)2 as the g.c.d. of Ψ2(X, X) and Ψ3(X, X) in the example above. (Start, for example, with a prime m1 which is the norm of an element of OD – there are known to be inﬁnitely many – and then choose ﬁnitely many further m’s which are also norms of elements in OD but not any of the ﬁnitely many other quadratic orders in which m1 is a norm.) This argument only shows that HD(X) has rational coeﬃcients, not that it is irreducible. The latter fact is proved most naturally by studying the arithmetic of the corresponding elliptic curves with complex multiplication (roughly speaking, the condition of having complex multiplication by a given order OD is purely algebraic and hence is preserved by Galois conjugation), but since the emphasis in these notes is on modular methods and their appli-cations, we omit the details. Elliptic Modular Forms and Their Applications 73 The proof just given actually yields the formula Ψm(X, X) = ± D<0 HD(X)rD(m)/w(D) (m ∈N, m ̸= square) , (85) due to Kronecker. Here rD(m) = { ( t, u ) ∈Z2 \| t2 −Du2 = 4m } = {λ ∈OD \| N(λ) = m} and w(D) is the number of units in OD, which is equal to 6 or 4 for D = −3 or −4, respectively, and to 2 otherwise. The product in (85) is ﬁnite since rD(m) ̸= 0 ⇒4m = t2 −u2D ≥\|D\| because u can’t be equal to 0 if m is not a square. There is another form of formula (85) which will be used in the second ap-plication below. As well as the usual class number h(D), one has the Hurwitz class number h∗(D) (the traditional notation is H(\|D\|)), deﬁned as the num-ber of all Γ1-equivalence classes of positive deﬁnite binary quadratic forms of discriminant D, not just the primitive ones, counted with multiplicity equal to one over the order of their stabilizer in Γ1 (which is 2 or 3 if the corresponding point in the fundamental domain for Γ1\H is at i or ρ and is 1 otherwise). In formulas, h∗(D) = r2\|D h′(D/r2), where the sum is over all r ∈N for which r2\|D (and for which D/r2 is still congruent to 0 or 1 mod 4, since otherwise h′(D/r2) will be 0) and h′(D) = h(D)/ 1 2w(D) with w(D) as above. Simi-larly, we can deﬁne a modiﬁed class “polynomial” H∗ D(X), of “degree” h∗(D), by H∗ D(X) = r2\|D HD/r2(X)2/w(D) , e.g., H∗ −12(X) = X1/3(X −54000). (These are actual polynomials unless \|D\| or 3\|D\| is a square.) Then (85) can be written in the following considerably simpler form: Ψm(X, X) = ± t2<4m H∗ t2−4m(X) (m ∈N, m ̸= square) . (86) This completes our long discussion of the algebraicity of singular moduli. We now describe some of the many applications. ♠Strange Approximations to π We start with an application that is more fun than serious. The discriminant D = −163 has class number one (and is in fact known to be the smallest such discriminant), so Proposition 25 implies that j(zD) is a rational integer. Moreover, it is large (in absolute value) because j(z) ≈q−1 and the q = e2πiz corresponding to z = z163 is roughly −4 × 10−18. But then from j(z) = q−1 + 744 + O(q) we ﬁnd that q−1 is extremely close to an integer, giving the formula eπ √ 163 = 262537412640768743.999999999999250072597 · · · 74 D. Zagier which would be very startling if one did not know about complex multiplica-tion. By taking logarithms, one gets an extremely good approximation to π: π = 1 √ 163 log(262537412640768744) −(2.237 · · · × 10−31) . (A poorer but simpler approximation, based on the fact that j(z163) is a perfect cube, is π ≈ 3 √ 163 log(640320), with an error of about 10−16.) Many further identities of this type were found, still using the theory of complex multipli-cation, by Ramanujan and later by Dan Shanks, the most spectactular one being π − 6 √ 3502 log ! 2 ε(x1) ε(x2) ε(x3) ε(x4) " ≈7.4 × 10−82 where x1 = 429 + 304 √ 2, x2 = 627 2 + 221 √ 2, x3 = 1071 2 + 92 √ 34, x4 = 1553 2 + 133 √ 34, and ε(x) = x + √ x2 −1. Of course these formulas are more curiosities than useful ways to compute π, since the logarithms are no easier to compute numerically than π is and in any case, if we allow complex numbers, then Euler’s formula π = 1 √−1 log(−1) is an exact formula of the same kind! ♥ ♠Computing Class Numbers By comparing the degrees on both sides of (85) or (86) and using Proposit-ion 24, we obtain the famous Hurwitz-Kronecker class number relations σ+ 1 (m) = D<0 h(D) w(D) rD(m) = t2<4m h∗(t2 −4m) (m ∈N, m ̸= square) . (87) (In fact the equality of the ﬁrst and last terms is true also for m square if we replace the summation condition by t2 ≤4m and deﬁne h∗(0) = −1 12, as one shows by a small modiﬁcation of the proof given here.) We mention that this formula has a geometric interpretation in terms of intersection numbers. Let X denote the modular curve Γ1\H. For each m ≥1 there is a curve Tm ⊂X × X, the Hecke correspondence, corresponding to the mth Hecke operator Tm introduced in §4.1. (The preimage of this curve in H× H consists of all pairs (z, Mz) with z ∈H and M ∈Mm.) For m = 1, this curve is just the diagonal. Now the middle or right-hand term of (87) counts the “physical” intersection points of Tm and T1 in X × X (with appropriate multiplicities if the intersections are not transversal), while the left-hand term computes the same number homologically, by ﬁrst compactifying X to ¯ X = X ∪{∞} (which is isomorphic to P1(C) via z →j(z)) and Tm and T1 to their closures ¯ Tm and ¯ T1 in ¯ X × ¯ X and then computing the intersection number of the homology classes of ¯ Tm and ¯ T1 in H2( ¯ X × ¯ X) ∼ = Z2 and correcting this by the contribution coming from intersections at inﬁnity of the compactiﬁed curves. Elliptic Modular Forms and Their Applications 75 Equation (87) gives a formula for h∗(−4m) in terms of h∗(D) with \|D\| < 4m. This does not quite suﬃce to calculate class numbers recursively since only half of all discriminants are multiples of 4. But by a quite similar type of argument (cf. the discussion in §6.2 below) one can prove a second class number relation, namely t2≤4m (m −t2) h∗(t2 −4m) = d\|m min(d, m/d)3 (m ∈N) (88) (again with the convention that h∗(0) = −1 12), and this together with (87) does suﬃce to determine h∗(D) for all D recursively, since together they ex-press h∗(−4m) + 2h∗(1 −4m) and mh∗(1 −4m) + 2(m −1)h∗(1 −4m) as linear combinations of h∗(D) with \|D\| < 4m −1, and every negative discrim-inant has the form −4m or 1 −4m for some m ∈N. This method is quite reasonable computationally if one wants to compute a table of class numbers h∗(D) for −X < D < 0, with about the same running time (viz., O(X3/2) operations) as the more direct method of counting all reduced quadratic forms with discriminant in this range. ♥ ♠Explicit Class Field Theory for Imaginary Quadratic Fields Class ﬁeld theory, which is the pinnacle of classical algebraic number theory, gives a complete classiﬁcation of all abelian extensions of a given number ﬁeld K. In particular, it says that the unramiﬁed abelian extensions (we omit all deﬁnitions) are the subﬁelds of a certain ﬁnite extension H/K, the Hilbert class ﬁeld, whose degree over K is equal to the class number of K and whose Galois group over K is canonically isomorphic to the class group of K, while the ramiﬁed abelian extensions have a similar description in terms of more complicated partititons of the ideals of OK into ﬁnitely many classes. However, this theory, beautiful though it is, gives no method to actually construct the abelian extensions, and in fact it is only known how to do this explicitly in two cases: Q and imaginary quadratic ﬁelds. If K = Q then the Hilbert class ﬁeld is trivial, since the class number is 1, and the ramiﬁed abelian extensions are just the subﬁelds of Q(e2πi/N) (N ∈N) by the Kronecker-Weber theorem. For imaginary quadratic ﬁelds, the result (in the unramiﬁed case) is as follows. Theorem. Let K be an imaginary quadratic ﬁeld, with discriminant D and Hilbert class ﬁeld H. Then the h(D) singular moduli j(zD,i) are conjugate to one another over K (not just over Q), any one of them generates H over K, and the Galois group of H over K permutes them transitively. More precisely, if we label the CM points of discriminant D by the ideal classes of K and identify Gal(H/K) with the class group of K by the fundamental isomorphism of class ﬁeld theory, then for any two ideal classes A, B of K the element σA of Gal(H/K) sends j(zB) to j(zAB). The ramiﬁed abelian extensions can also be described completely by com-plex multiplication theory, but then one has to use the values of other modu-lar functions (not just of j(z)) evaluated at all points of K ∩H (not just 76 D. Zagier those of discriminant D). Actually, even for the Hilbert class ﬁelds it can be advantageous to use modular functions other than j(z). For instance, for D = −23, the ﬁrst discriminant with class number not a power of 2 (and therefore the ﬁrst non-trivial example, since Gauss’s theory of genera de-scribes the Hilbert class ﬁeld of D when the class group has exponent 2 as a composite quadratic extension, e.g., H for K = Q(√−15) is Q(√−3, √ 5) ), the Hilbert class ﬁeld is generated over K by the real root α of the polynomial X3 −X −1. The singular modulus j(zD), which also generates this ﬁeld, is equal to −53α12(2α −1)3(3α + 2)3 and is a root of the much more compli-cated irreducible polynomial H−23(X) = X3 + 3491750X2 −5151296875X + 12771880859375, but using more detailed results from the theory of com-plex multiplication one can show that the number 2−1/2eπi/24η(zD)/η(2zD) also generates H, and this number turns out to be α itself! The improve-ment is even more dramatic for larger values of \|D\| and is important in situations where one actually wants to compute a class ﬁeld explicitly, as in the applications to factorization and primality testing mentioned in §6.4 below. ♥ ♠Solutions of Diophantine Equations In §4.4 we discussed that one can parametrize an elliptic curve E over Q by modular functions X(z), Y (z), i.e., functions which are invariant under Γ0(N) for some N (the conductor of E) and identically satisfy the Weier-strass equation Y 2 = X3+AX +B deﬁning E. If K is an imaginary quadratic ﬁeld with discriminant D prime to N and congruent to a square modulo 4N (this is equivalent to requiring that OK = OD contains an ideal n with OD/n ∼ = Z/NZ), then there is a canonical way to lift the h(D) points of Γ1\H of discriminant D to h(D) points in the covering Γ0(N)\H, the so-called Heegner points. (The way is quite simple: if D ≡r2 (mod 4N), then one looks at points zQ with Q(x, y) = Ax2 + Bxy + Cy2 ∈QD satisfying A ≡0 (mod N) and B ≡r (mod 2N); there is exactly one Γ0(N)-equivalence class of such points in each Γ1-equivalence class of points of ZD.) One can then show that the values of X(z) and Y (z) at the Heegner points be-have exactly like the values of j(z) at the points of ZD, viz., these values all lie in the Hilbert class ﬁeld H of K and are permuted simply transi-tively by the Galois group of H over K. It follows that we get h(D) points Pi = (X(zi), Y (zi)) in E with coordinates in H which are permuted by Gal(H/K), and therefore that the sum PK = P1 + · · · + Ph(D) has coor-dinates in K (and in many cases even in Q). This method of constructing potentially non-trivial rational solutions of Diophantine equations of genus 1 (the question of their actual non-triviality will be discussed brieﬂy in §6.2) was invented by Heegner as part of his proof of the fact, mentioned above, that −163 is the smallest quadratic discriminant with class number 1 (his proof, which was rather sketchy at many points, was not accepted at the time by the mathematical community, but was later shown by Stark to Elliptic Modular Forms and Their Applications 77 be correct in all essentials) and has been used to construct non-trivial so-lutions of several classical Diophantine equations, e.g., to show that every prime number congruent to 5 or 7 (mod 8) is a “congruent number” (= the area of a right triangle with rational sides) and that every prime number congruent to 4 or 7 (mod 9) is a sum of two rational cubes (Sylvester’s problem). ♥ 6.2 Norms and Traces of Singular Moduli A striking property of singular moduli is that they always are highly factoriz-able numbers. For instance, the value of j(z−163) used in the ﬁrst application in §6.1 is −6403203 = −2183353233293, and the value of j(z−11) = −32768 is the 15th power of −2. As part of our investigation about the heights of Heegner points (see below), B. Gross and I were led to a formula which ex-plains and generalizes this phenomenon. It turns out, in fact, that the right numbers to look at are not the values of the singular moduli themselves, but their diﬀerences. This is actually quite natural, since the deﬁnition of the modular function j(z) involves an arbitrary choice of additive constant: any function j(z) + C with C ∈Z would have the same analytic and arithmetic properties as j(z). The factorization of j(z), however, would obviously change completely if we replaced j(z) by, say, j(z) + 1 or j(z) −744 = q−1 + O(q) (the so-called “normalized Hauptmodul” of Γ1), but this replacement would have no eﬀect on the diﬀerence j(z1) −j(z2) of two singular moduli. That the original singular moduli j(z) do nevertheless have nice factorizations is then due to the accidental fact that 0 itself is a singular modulus, namely j(z−3), so that we can write them as diﬀerences j(z) −j(z−3). (And the fact that the values of j(z) tend to be perfect cubes is then related to the fact that z−3 is a ﬁxed-point of order 3 of the action of Γ1 on H.) Secondly, since the singu-lar moduli are in general algebraic rather than rational integers, we should not speak only of their diﬀerences, but of the norms of their diﬀerences, and these norms will then also be highly factored. (For instance, the norms of the singular moduli j(z−15) and j(z−23), which are algebraic integers of de-gree 2 and 3 whose values were given in §6.1, are −3653113 and −59113173, respectively.) If we restrict ourselves for simplicity to the case that the discriminants D1 and D2 of z1 and z2 are coprime, then the norm of j(z1) −j(z2) depends only on D1 and D2 and is given by J(D1, D2) = z1∈Γ1\ZD1 z2∈Γ1\ZD2 j(z1) −j(z2) . (89) (If h(D1) = 1 then this formula reduces simply to HD2(zD1), while in general J(D1, D2) is equal, up to sign, to the resultant of the two irreducible polyno-mials HD1(X) and HD2(X).) These are therefore the numbers which we want to study. We then have: 78 D. Zagier Theorem. Let D1 and D2 be coprime negative discriminants. Then all prime factors of J(D1, D2) are ≤ 1 4D1D2. More precisely, any prime factors of J(D1, D2) must divide 1 4(D1D2 −x2) for some x ∈Z with \|x\| < √D1D2 and x2 ≡D1D2 (mod 4). There are in fact two proofs of this theorem, one analytic and one arith-metic. We give some brief indications of what their ingredients are, without deﬁning all the terms occurring. In the analytic proof, one looks at the Hilbert modular group SL(2, OF ) (see the notes by J. Bruinier in this volume) asso-ciated to the real quadratic ﬁeld F = Q(√D1D2) and constructs a certain Eisenstein series for this group, of weight 1 and with respect to the “genus character” associated to the decomposition of the discriminant of F as D1·D2. Then one restricts this form to the diagonal z1 = z2 (the story here is actually more complicated: the Eisenstein series in question is non-holomorphic and one has to take the holomorphic projection of its restriction) and makes use of the fact that there are no holomorphic modular forms of weight 2 on Γ1 . In the arithmetic proof, one uses that p\|J(D1, D2) if and only the CM ellip-tic curves with j-invariants j(zD1) and j(zD2) become isomorphic over Fp . Now it is known that the ring of endomorphisms of any elliptic curve over Fp is isomorphic either to an order in a quadratic ﬁeld or to an order in the (unique) quaternion algebra Bp,∞over Q ramiﬁed at p and at inﬁnity. For the elliptic curve E which is the common reduction of the curves with com-plex multiplication by OD1 and OD2 the ﬁrst alternative cannot occur, since a quadratic order cannot contain two quadratic orders coming from diﬀer-ent quadratic ﬁelds, so there must be an order in Bp,∞which contains two elements α1 and α2 with square D1 and D2, respectively. Then the element α = α1α2 also belongs to this order, and if x is its trace then x ∈Z (because α is in an order and hence integral), x2 < N(α) = D1D2 (because Bp,∞is ramiﬁed at inﬁnity), and x2 ≡D1D2 (mod p) (because Bp,∞is ramiﬁed at p). This proves the theorem, except that we have lost a factor “4” because the elements αi actually belong to the smaller orders O4Di and we should have worked with the elements αi/2 or (1 + αi)/2 (depending on the parity of Di) in ODi instead. The theorem stated above is actually only the qualitative version of the full result, which gives a complete formula for the prime factorization of J(D1, D2). Assume for simplicity that D1 and D2 are fundamental. For each positive integer n of the form 1 4(D1D2 −x2), we deﬁne a function ε from the set of divisors of n to {±1} by the requirement that ε is com-pletely multiplicative (i.e., ε pr1 1 · · · prs s = ε p1 r1 · · · ε ps rs for any divisor pr1 1 · · · prs s of n) and is given on primes p\|n by ε(p) = χD1(p) if p ∤D1 and by ε(p) = χD2(p) if p ∤D2, where χD is the Dirichlet character modulo D introduced at the beginning of §3.2. Notice that this makes sense: at least one of the two alternatives must hold, since (D1, D2) = 1 and p is prime, and if they both hold then the two deﬁnitions agree because D1D2 is then congruent to a non-zero square modulo p if p is odd and to an odd square modulo 8 if p = 2, so χD1(p)χD2(p) = 1. We then deﬁne F(n) (still for n of the form Elliptic Modular Forms and Their Applications 79 1 4(D1D2 −x2)) by F(n) = d\|n dε(n/d) . This number, which is a priori only rational since ε(n/d) can be positive or negative, is actually integral and in fact is always a power of a sin-gle prime number: one can show easily that ε(n) = −1, so n contains an odd number of primes p with ε(p) = −1 and 2 ∤ordp(n), and if we write n = p2α1+1 1 · · · p2αr+1 r p2β1 r+1 · · · p2βs r+sqγ1 1 · · · qγt t with r odd and ε(pi) = −1, ε(qj) = +1, αi, βi, γj ≥0, then F(n) = p(α1+1)(γ1+1)···(γt+1) if r = 1, p1 = p , 1 if r ≥3 . (90) The complete formula for J(D1, D2) is then J(D1, D2)8/w(D1)w(D2) = x2 1, then in general (namely, whenever X0(N) has positive genus) the Jacobian is non-trivial and the heights do not have to vanish identically. The famous conjecture of Birch and Swinnerton-Dyer, one of the seven million-dollar Clay Millennium Problems, says that the heights of points on an abelian variety are related to the values or derivatives of a certain L-function; more concretely, in the case of an elliptic curve E/Q, the conjecture predicts that the order to which the L-series L(E, s) vanishes at s = 1 is equal to the rank of the Mordell-Weil group E(Q) and that the value of the ﬁrst non-zero derivative of L(E, s) at s = 1 is equal to a certain explicit expression involving the height pairings of a system of generators of E(Q) with one another. The same kind of calculations as in the case N = 1 permitted a veriﬁcation of this prediction in the case of Heegner points, the relevant derivative of the L-series being the ﬁrst one: Theorem. Let E be an elliptic curve deﬁned over Q whose L-series vanishes at s = 1. Then the height of any Heegner point is an explicit (and in general non-zero) multiple of the derivative L′(E/Q, 1). Elliptic Modular Forms and Their Applications 81 The phrase “in general non-zero” in this theorem means that for any given elliptic curve E with L(E, 1) = 0 but L′(E, 1) ̸= 0 there are Heegner points whose height is non-zero and which therefore have inﬁnite order in the Mordell-Weil group E(Q). We thus get the following (very) partial statement in the direction of the full BSD conjecture: Corollary. If E/Q is an elliptic curve over Q whose L-series has a simple zero at s = 1, then the rank of E(Q) is at least one. (Thanks to subsequent work of Kolyvagin using his method of “Euler systems” we in fact know that the rank is exactly equal to one in this case.) In the opposite direction, there are elliptic curves E/Q and Heegner points P in E(Q) for which we know that the multiple occurring in the theorem is non-zero but where P can be checked directly to be a torsion point. In that case the theorem says that L′(E/Q, 1) must vanish and hence, if the L-series is known to have a functional equation with a minus sign (the sign of the functional equation can be checked algorithmically), that L(E/Q, s) has a zero of order at least 3 at s = 1. This is important because it is exactly the hypothesis needed to apply an earlier theorem of Goldfeld which, assuming that such a curve is known, proves that the class numbers of h(D) go to inﬁnity in an eﬀective way as D →−∞. Thus modular methods and the theory of Heegner points suﬃce to solve the nearly 200-year problem, due to Gauss, of showing that the set of discriminants D < 0 with a given class number is ﬁnite and can be determined explicitly, just as in Heegner’s hands they had already suﬃced to solve the special case when the given value of the class number was one. ♥ So far in this subsection we have discussed the norms of singular moduli (or more generally, the norms of their diﬀerences), but algebraic numbers also have traces, and we can consider these too. Now the normalization of j does matter; it turns out to be best to choose the normalized Hauptmodul j0(z) = j(z) −744. For every discriminant D < 0 we therefore deﬁne T (D) ∈Z to be the trace of j0(zD). This is the sum of the h(D) singular moduli j0(zD,i), but just as in the case of the Hurwitz class numbers it is better to use the modiﬁed trace T ∗(d) which is deﬁned as the sum of the values of j0(z) at all points z ∈ Γ1\H satisfying a quadratic equation, primitive or not, of discriminant D (and with the points i and ρ, if they occur at all, being counted with multiplicity 1/2 or 1/3 as usual), i.e., T ∗(D) = r2\|D T (D/r2)/( 1 2w(D/r2)) with the same conventions as in the deﬁnition of h∗(D). The result, quite diﬀerent from (and much easier to prove than) the formula for the norms, is that these numbers T ∗(D) are the coeﬃcients of a modular form. Speciﬁcally, if we denote by g(z) the meromorphic modular form θM(z)E4(4z)/η(4z)6 of weight 3/2, where θM(z) is the Jacobi theta function n∈Z(−1)nqn2 as in §3.1, then we have: Theorem. The Fourier expansion of g(z) is given by g(z) = q−1 −2 − d>0 T ∗(−d) qd . (92) 82 D. Zagier We can check the ﬁrst few coeﬃcients of this by hand: the Fourier ex-pansion of g begins q−1 −2 + 248 q3 −492 q4 + 4119 q7 −· · · and indeed we have T ∗(−3) = 1 3(0 −744) = −248, T ∗(−4) = 1 2(1728 −744) = 492, and T ∗(−7) = −3375 −744 = −4119. The proof of this theorem, though somewhat too long to be given here, is fairly elementary and is essentially a reﬁnement of the method used to prove the Hurwitz-Kronecker class number relations (87) and (88). More pre-cisely, (87) was proved by comparing the degrees on both sides of (86), and these in turn were computed by looking at the most negative exponent oc-curring in the q-expansion of the two sides when X is replaced by j(z); in particular, the number σ+ 1 (m) came from the calculation in Proposition 24 of the most negative power of q in Ψm(j(z), j(z)). If we look instead at the next coeﬃcient (i.e., that of q−σ+ 1 (m)+1), then we ﬁnd − t2<4m T ∗(t2 −4m) for the right-hand side of (86), because the modular functions HD(j(z)) and H∗ D(j(z)) have Fourier expansions beginning q−h(D)(1 −T (D)q + O(q2)) and q−h∗(D)(1 −T ∗(D)q + O(q2)), respectively, while by looking carefully at the “lower order terms” in the proof of Proposition 24 we ﬁnd that the corre-sponding coeﬃcient for the left-hand side of (86) vanishes for all m unless m or 4m+ 1 is a square, when it equals +4 or −2 instead. The resulting identity can then be stated uniformly for all m as t∈Z T ∗(t2 −4m) = 0 (m ≥0) , (93) where we have artiﬁcially deﬁned T ∗(0) = 2, T ∗(1) = −1, and T ∗(D) = 0 for D > 1 ( a deﬁnition made plausible by the result in the theorem we want to prove). By a somewhat more complicated argument involving modular forms of higher weight, we prove a second relation, analogous to (88): t∈Z (m −t2)T ∗(t2 −4m) = 1 if m = 0 , 240σ3(n) if m ≥1 . (94) (Of course, the factor m −t2 here could be replaced simply by −t2, in view of (93), but (94) is more natural, both from the proof and by anal-ogy with (88).) Now, just as in the discussion of the Hurwitz-Kronecker class number relations, the two formulas (93) and (94) suﬃce to determine all T ∗(D) by recursion. But in fact we can solve these equations directly, rather than recursively (though this was not done in the original paper). Write the right-hand side of (92) as t0(z) + t1(z) where t0(z) = ∞ m=0 T ∗(−4m) q4m and t1(z) = ∞ m=0 T ∗(1 −4m) q4m−1, and deﬁne two unary theta series θ0(z) and θ1(z) (= θ(4z) and θF (4z) in the notation of §3.1) as the sums qt2 with t ranging over the even or odd integers, respectively. Then (93) says that t0θ0 + t1θ1 = 0 and (94) says that [t0, θ0] + [t1, θ1] = 4E4(4z), where [ti, θi] = 3 2tiθ′ i −1 2t′ iθi is the ﬁrst Rankin–Cohen bracket of ti (in weight 3/2) and θi. By taking a linear combination of the second equation with the deriva-Elliptic Modular Forms and Their Applications 83 tive of the ﬁrst, we deduce that θ0(z) θ1(z) θ′ 0(z) θ′ 1(z) t0(z) t1(z) = 2 0 E4(4z) , and from this we can immediately solve for t0(z) and t1(z) and hence for their sum, proving the theorem. ♠The Borcherds Product Formula This is certainly not an “application” of the above theorem in any reasonable sense, since Borcherds’s product formula is much deeper and more general and was proved earlier, but it turns out that there is a very close link and that one can even use this to give an elementary proof of Borcherds’s formula in a special case. This special case is the beautiful product expansion H∗ D(j(z)) = q−h∗(D) ∞ n=1 1 −qnAD(n2) , (95) where AD(m) denotes the mth Fourier coeﬃcient of a certain meromor-phic modular form fD(z) (with Fourier expansion beginning qD + O(q)) of weight 1/2. The link is that one can prove in an elementary way a “duality formula” AD(m) = −Bm(\|D\|), where Bm(n) is the coeﬃcient of qn in the Fourier expansion of a certain other meromorphic modular form gm(z) (with Fourier expansion beginning q−m + O(1)) of weight 3/2. For m = 1 the func-tion gm coincides with the g of the theorem above, and since (95) immediately implies that T ∗(D) = AD(1) this and the duality imply (92). Conversely, by applying Hecke operators (in half-integral weight) in a suitable way, one can give a generalization of (92) to all functions gn2, and this together with the duality formula gives the complete formula (95), not just its subleading coef-ﬁcient. ♥ 6.3 Periods and Taylor Expansions of Modular Forms In §6.1 we showed that the value of any modular function (with rational or algebraic Fourier coeﬃcients; we will not always repeat this) at a CM point z is algebraic. This is equivalent to saying that for any modular form f(z), of weight k, the value of f(z) is an algebraic multiple of Ωk z , where Ωz depends on z only, not on f or on k. Indeed, the second statement implies the ﬁrst by specializing to k = 0, and the ﬁrst implies the second by observing that if f ∈Mk and g ∈Mℓthen f ℓ/gk has weight 0 and is therefore algebraic at z, so that f(z)1/k and g(z)1/ℓare algebraically proportional. Furthermore, the number Ωz is unchanged (at least up to an algebraic number, but it is only deﬁned up to an algebraic number) if we replace z by Mz for any M ∈M(2, Z) with positive determinant, because f(Mz)/f(z) is a modular function, and since any two CM points which generate the same imaginary quadratic ﬁeld are related in this way, this proves: 84 D. Zagier Proposition 26. For each imaginary quadratic ﬁeld K there is a number ΩK ∈C∗such that f(z) ∈Q · Ωk K for all z ∈K ∩H, all k ∈Z, and all modular forms f of weight k with algebraic Fourier coeﬃcients. □ To ﬁnd ΩK, we should compute f(z) for some special modular form f (of non-zero weight!) and some point or points z ∈K ∩H. A natural choice for the modular form is Δ(z), since it never vanishes. Even better, to achieve weight 1, is its 12th root η(z)2, and better yet is the function Φ(z) = I(z)\| η(z)\|4 (which at z ∈K ∩H is an algebraic multiple of Ω2 K), since it is Γ1-invariant. As for the choice of z, we can look at the CM points of discriminant D (= discriminant of K), but since there are h(D) of them and none should be preferred over the others (their j-invariants are conjugate algebraic numbers), the only rea-sonable choice is to multiply them all together and take the h(D)-th root – or rather the h′(D)-th root (where h′(D) as previously denotes h(D)/ 1 2w(D), i.e., h′(D) = 1 3, 1 2 or h(K) for D = −3, D = −4, or D < −4), because the elliptic ﬁxed points ρ and i of Γ1\H are always to be counted with multiplic-ity 1 3 and 1 2, respectively. Surprisingly enough, the product of the invariants Φ(zD,i) (i = 1, . . . , h(D)) can be evaluated in closed form: Theorem. Let K be an imaginary quadratic ﬁeld of discriminant D. Then z∈Γ1\ZD 4π \|D\| Φ(z) 2/w(D) = \|D\|−1 j=1 Γ j/\|D\| χD(j) , (96) where χD is the quadratic character associated to K and Γ(x) is the Euler gamma function. Corollary. The number ΩK in Proposition 26 can be chosen to be ΩK = 1 2π\|D\| ⎛ ⎝ \|D\|−1 j=1 Γ j \|D\| χD(j) ⎞ ⎠ 1/2h′(D) . (97) Formula (96), usually called the Chowla–Selberg formula, is contained in a paper published by S. Chowla and A. Selberg in 1949, but it was later no-ticed that it already appears in a paper of Lerch from 1897. We cannot give the complete proof here, but we describe the main idea, which is quite sim-ple. The Dedekind zeta function ζK(s) = N(a)−s (sum over all non-zero integral ideals of K) has two decompositions: an additive one as A ζK,A(s), where A runs over the ideal classes of K and ζK,A(s) is the associated “par-tial zeta function” (= N(a)−s with a running over the ideals in A), and a multiplicative one as ζ(s)L(s, χD), where ζ(s) denotes the Riemann zeta function and L(s, χD) = ∞ n=1 χD(n)n−s. Using these two decompositions, one can compute in two diﬀerent ways the two leading terms of the Laurent expansion ζK(s) = A s−1 + B + O(s −1) as s →1. The residue at s = 1 of ζK,A(s) is independent of A and equals π/ 1 2w(D) \|D\|, leading to Dirichlet’s Elliptic Modular Forms and Their Applications 85 class number formula L(1, χD) = π h′(D)/ \|D\|. (This is, of course, precisely the method Dirichlet used.) The constant term in the Laurent expansion of ζK,A(s) at s = 1 is given by the famous Kronecker limit formula and is (up to some normalizing constants) simply the value of log(Φ(z)), where z ∈K ∩H is the CM point corresponding to the ideal class A. The Riemann zeta function has the expansion (s −1)−1 −γ + O(s −1) near s = 1 (γ = Euler’s constant), and L′(1, χD) can be computed by a relatively elementary analytic argument and turns out to be a simple multiple of j χD(j) log Γ(j/\|D\|). Combining everything, one obtains (96). As our ﬁrst “application,” we mention two famous problems of transcend-ence theory which were solved by modular methods, one using the Chowla– Selberg formula and one using quasimodular forms. ♠Two Transcendence Results In 1976, G.V. Chudnovsky proved that for any z ∈H, at least two of the three numbers E2(z), E4(z) and E6(z) are algebraically independent. (Equivalently, the ﬁeld generated by all f(z) with f ∈ M∗(Γ1)Q = Q[E2, E4, E6] has tran-scendence degree at least 2.) Applying this to z = i, for which E2(z) = 3/π, E4(z) = 3 Γ( 1 4)8/(2π)6 and E6(z) = 0, one deduces immediately that Γ( 1 4) is transcendental (and in fact algebraically independent of π). Twenty years later, Nesterenko, building on earlier work of Barré-Sirieix, Diaz, Gramain and Philibert, improved this result dramatically by showing that for any z ∈H at least three of the four numbers e2πiz, E2(z), E4(z) and E6(z) are algebraically independent. His proof used crucially the basic properties of the ring M∗(Γ1)Q discussed in §5, namely, that it is closed under diﬀerentiation and that each of its elements is a power series in q = e2πiz with rational coeﬃcients of bounded denominator and polynomial growth. Specialized to z = i, Nesterenko’s result implies that the three numbers π, eπ and Γ( 1 4) are algebraically indepen-dent. The algebraic independence of π and eπ (even without Γ( 1 4)) had been a famous open problem. ♥ ♠Hurwitz Numbers Euler’s famous result of 1734 that ζ(2r)/π2r is rational for every r ≥1 can be restated in the form n∈Z n̸=0 1 nk = (rational number) · πk for all k ≥2 , (98) where the “rational number” of course vanishes for k odd since then the con-tributions of n and −n cancel. This result can be obtained, for instance, by looking at the Laurent expansion of cot x near the origin. Hurwitz asked the corresponding question if one replaces Z in (98) by the ring Z[i] of Gaussian 86 D. Zagier integers and, by using an elliptic function instead of a trigonometric one, was able to prove the corresponding assertion λ∈Z[i] λ̸=0 1 λk = Hk k! ωk for all k ≥3 , (99) for certain rational numbers H4 = 1 10, H8 = 3 10, H12 = 567 130, . . . (here Hk = 0 for 4 ∤k since then the contributions of λ and −λ or of λ and iλ cancel), where ω = 4 1 0 dx √ 1 −x4 = Γ( 1 4)2 √ 2π = 5.24411 · · · . Instead of using the theory of elliptic functions, we can see this result as a spe-cial case of Proposition 26, since the sum on the left of (99) is just the special value of the modular form 2Gk(z) deﬁned in (10), which is related by (12) to the modular form Gk(z) with rational Fourier coeﬃcients, at z = i and ω is 2π √ 2 times the Chowla–Selberg period ΩQ(i). Similar considerations, of course, apply to the sum λ−k with λ running over the non-zero elements of the ring of integers (or of any other ideal) in any imaginary quadratic ﬁeld, and more generally to the special values of L-series of Hecke “grossencharacters” which we will consider shortly. ♥ We now turn to the second topic of this subsection: the Taylor expansions (as opposed to simply the values) of modular forms at CM points. As we al-ready explained in the paragraph preceding equation (58), the “right” Taylor expansion for a modular form f at a point z ∈H is the one occurring on the left-hand side of that equation, rather than the straight Taylor expansion of f. The beautiful fact is that, if z is a CM point, then after a renormal-ization by dividing by suitable powers of the period Ωz, each coeﬃcient of this expansion is an algebraic number (and in many cases even rational). This follows from the following proposition, which was apparently ﬁrst observed by Ramanujan. Proposition 27. The value of E∗ 2(z) at a CM point z ∈K ∩H is an algebraic multiple of Ω2 K. Corollary. The value of ∂nf(z), for any modular form f with algebraic Fourier coeﬃcients, any integer n ≥0 and any CM point z ∈K ∩H, is an algebraic multiple of Ωk+2n K , where k is the weight of f. Proof. We will give only a sketch, since the proof is similar to that al-ready given for j(z). For M = a b c d ∈Mm we deﬁne (E∗ 2\|2M)(z) = m(cz + d)−2E∗ 2(Mz) ( = the usual slash operator for the matrix m−1/2M ∈ SL(2, R)). From formula (1) and the fact that E∗ 2(z) is a linear combination of 1/y and a holomorphic function, we deduce immediately that the diﬀerence E∗ 2 −E∗ 2\|2M is a holomorphic modular form of weight 2 (on the subgroup of Elliptic Modular Forms and Their Applications 87 ﬁnite index M −1Γ1M ∩Γ1 of Γ1). It then follows by an argument similar to that in the proof of Proposition 23 that the function Pm(z, X) := M∈Γ1\Mm X + E∗ 2(z) −(E∗ 2\|2M)(z) (z ∈H, X ∈C) is a polynomial in X (of degree σ1(m)) whose coeﬃcients are modular forms of appropriate weights on Γ1 with rational coeﬃcients. (For example, P2(z, X) = X3 − 3 4 E4(z) X + 1 4 E6(z).) The algebraicity of E∗ 2(z)/Ω2 K for a CM point z now follows from Proposition 26, since if M ̸∈Z·Id2 ﬁxes z then E∗ 2(z) −(E∗ 2\|2M)(z) is a non-zero algebraic multiple of E∗ 2(z). The corollary follows from (58) and the fact that each non-holomorphic derivative ∂nf(z) is a polynomial in E∗ 2(z) with coeﬃcients that are holomorphic modular forms with algebraic Fourier coeﬃcients, as one sees from equation (66). Propositions 26 and 27 are illustrated for z = zD and f = E∗ 2, E4, E6 and Δ in the following table, in which α in the penultimate row is the real root of α3 −α −1 = 0. Observe that the numbers in the ﬁnal column of this table are all units; this is part of the general theory. D \|D\|1/2E∗ 2(zD) Ω2 D E4(zD) Ω4 D E6(zD) \|D\|1/2Ω6 D Δ(zD) Ω12 D −3 0 0 24 −1 −4 0 12 0 1 −7 3 15 27 −1 −8 4 20 28 1 −11 8 32 56 −1 −15 6 + 3 √ 5 15 + 12 √ 5 42 + 63 √ 5 3− √ 5 2 −19 24 96 216 −1 −20 12 + 4 √ 5 40 + 12 √ 5 72 + 112 √ 5 √ 5 −2 −23 7+11α+12α2 a1/3 5α1/3(6 + 4α + α2) 469+1176α+504α2 23 −α−8 −24 12 + 12 √ 2 60 + 24 √ 2 84 + 72 √ 2 3 −2 √ 2 In the paragraph preceding Proposition 17 we explained that the non-holomorphic derivatives ∂nf of a holomorphic modular form f(z) are more natural and more fundamental than the holomorphic derivatives Dnf, because the Taylor series Dnf(z) tn/n! represents f only in the disk \|z′ −z\| < I(z) whereas the series (58) represents f everywhere. The above corollary gives a second reason to prefer the ∂nf : at a CM point z they are monomials in the period Ωz with algebraic coeﬃcients, whereas the derivatives Dnf(z) are polynomials in Ωz by equation (57) (in which there can be no cancellation since Ωz is known to be transcendental). If we set cn = cn(f, z, Ω) = ∂nf(z) Ωk+2n (n = 0, 1, 2, . . . ) , (100) where Ω is a suitably chosen algebraic multiple of ΩK, then the cn (up to a possible common denominator which can be removed by multiplying f by 88 D. Zagier a suitable integer) are even algebraic integers, and they still can be considered to be “Taylor coeﬃcients of f at z” since by (58) the series ∞ n=0 cntn/n! equals (Ct + D)−kf At+B Ct+D where A B C D = −¯ z z −1 1 1/4πyΩ 0 0 Ω ∈GL(2, C). These normalized Taylor coeﬃcients have several interesting number-theoretical ap-plications, as we will see below, and from many points of view are actually better number-theoretic invariants than the more familiar coeﬃcients an of the Fourier expansion f = anqn. Surprisingly enough, they are also much easier to calculate: unlike the an, which are mysterious numbers (think of the Ramanujan function τ(n)) and have to be calculated anew for each modular form, the Taylor coeﬃcients ∂nf or cn are always given by a simple recursive procedure. We illustrate the procedure by calculating the numbers ∂nE4(i), but the method is completely general and works the same way for all modular forms (even of half-integral weight, as we will see in §6.4). Proposition 28. We have ∂nE4(i) = pn(0)E4(i)1+n/2, with pn(t) ∈Z[ 1 6][t] deﬁned recursively by p0(t) = 1, pn+1(t) = t2 −1 2 p′ n(t)−n + 2 6 tpn(t)−n(n + 3) 144 pn−1(t) (n ≥0) . Proof. Since E∗ 2(i) = 0, equation (66) implies that f∂(i, X) = fϑ(i, X) for any f ∈Mk(Γ1), i.e., we have ∂nf(i) = ϑ[n]f(i) for all n ≥0, where {ϑ[n]f}n=0,1,2,... is deﬁned by (64). We use Ramanujan’s notations Q and R for the modular forms E4(z) and E6(z). By (54), the derivation ϑ sends Q and R to −1 3R and −1 2Q2, respectively, so ϑ acts on M∗(Γ1) = C[Q, R] as −R 3 ∂ ∂Q −Q2 2 ∂ ∂R . Hence ϑ[n]Q = Pn(Q, R) for all n, where the polynomials Pn(Q, R) ∈Q[Q, R] are given recursively by P0 = Q , Pn+1 = −R 3 ∂Pn ∂Q −Q2 2 ∂Pn ∂R −n(n + 3) QPn−1 144 . Since Pn(Q, R) is weighted homogeneous of weight 2n + 4, where Q and R have weight 4 and 6, we can write Pn(Q, R) as Q1+n/2pn(R/Q3/2) where pn is a polynomial in one variable. The recursion for Pn then translates into the recursion for pn given in the proposition. The ﬁrst few polynomials pn are p1 = −1 3t, p2 = 5 36, p3 = −5 72t, p4 = 5 216t2 + 5 288, . . . , giving ∂2E4(i) = 5 36E4(i)2, ∂4E4(i) = 5 288E4(i)3, etc. (The values for n odd vanish because i is a ﬁxed point of the element S ∈Γ1 of order 2.) With the same method we ﬁnd ∂nf(i) = qn(0)E4(i)(k+2n)/4 for any modular form f ∈Mk(Γ1), with polynomials qn satisfying the same recursion as pn but with n(n + 3) replaced by n(n + k −1) (and of course with a diﬀerent initial value q0(t)). If we consider a CM point z other than i, then the method and result are similar but we have to use (68) instead of (66), where φ is a quasimodular form diﬀering from E2 by a (meromorphic) modular form of weight 2 and chosen so that φ∗(z) vanishes. If we replace Γ1 by some other group Γ, then Elliptic Modular Forms and Their Applications 89 the same method works in principle but we need an explicit description of the ring M∗(Γ) to replace the description of M∗(Γ1) as C[Q, R] used above, and if the genus of the group is larger than 0 then the polynomials pn(t) have to be replaced by elements qn of some ﬁxed ﬁnite algebraic extension of Q(t), again satisfying a recursion of the form qn+1 = Aq′ n+(nB+C)qn+n(n+k−1)Dqn−1 with A, B, C and D independent of n. The general result says that the values of the non-holomorphic derivatives ∂nf(z0) of any modular form at any point z0 ∈H are given “quasi-recursively” as special values of a sequence of algebraic functions in one variable which satisfy a diﬀerential recursion. ♠Generalized Hurwitz Numbers In our last “application” we studied the numbers Gk(i) whose values, up to a factor of 2, are given by the Hurwitz formula (99). We now discuss the meaning of the non-holomorphic derivatives ∂nGk(i). From (55) we ﬁnd ∂k 1 (mz + n)k = k 2πi(z −¯ z) · m¯ z + n (mz + n)k+1 and more generally ∂r k 1 (mz + n)k = (k)r (2πi(z −¯ z))r · (m¯ z + n)r (mz + n)k+r for all r ≥0, where ∂r k = ∂k+2r−2◦· · ·◦∂k+2◦∂k and (k)r = k(k+1) · · · (k+r−1) as in §5. Thus ∂nGk(i) = (k)n 2 (−4π)n λ∈Z[i] λ̸=0 ¯ λn λk+n (n = 0, 1, 2, . . . ) (and similarly for ∂nGk(z) for any CM point z, with Z[i] replaced by Zz+Z and −4π by −4πI(z)). If we observe that the class number of the ﬁeld K = Q(i) is 1 and that any integral ideal of K can be written as a principal ideal (λ) for exactly four numbers λ ∈Z[i], then we can write λ∈Z[i] λ̸=0 ¯ λn λk+n = λ∈Z[i] λ̸=0 ¯ λk+2n (λ¯ λ)k+n = 4 a ψk+2n(a) N(a)k+n where the sum runs over the integral ideals a of Z[i] and ψk+2n(a) is deﬁned as ¯ λk+2n, where λ is any generator of a. (This is independent of λ if k+2n is divis-ible by 4, and in the contrary case the sum vanishes.) The functions ψk+2n are called Hecke “grossencharacters” (the German original of this semi-anglicized word is “Größencharaktere”, with ﬁve diﬀerences of spelling, and means liter-ally “characters of size,” referring to the fact that these characters, unlike the 90 D. Zagier usual ideal class characters of ﬁnite order, depend on the size of the generator of a principal ideal) and their L-series LK(s, ψk+2n) = a ψk+2n(a)N(a)−s are an important class of L-functions with known analytic continuation and functional equations. The above calculation shows that ∂nGk(i) (or more gen-erally the non-holomorphic derivatives of any holomorphic Eisenstein series at any CM point) are simple multiples of special values of these L-series at integral arguments, and Proposition 28 and its generalizations give us an al-gorithmic way to compute these values in closed form. ♥ 6.4 CM Elliptic Curves and CM Modular Forms In the introduction to §6, we deﬁned elliptic curves with complex multiplica-tion as quotients E = C/Λ where λΛ ⊆Λ for some non-real complex num-ber λ. In that case, as we have seen, the lattice Λ is homothetic to Zz + Z for some CM point z ∈H and the singular modulus j(E) = j(z) is algebraic, so E has a model over Q. The map from E to E induced by multiplication by λ is also algebraic and deﬁned over Q. For simplicity, we concentrate on those z whose discriminant is one of the 13 values −3, −4, −7, . . . , −163 with class number 1, so that j(z) ∈Q and E (but not the complex multiplication) can be deﬁned over Q. For instance, the three elliptic curves y2 = x3 + x , y2 = x3 + 1 , y2 = x3 −35x −98 (101) have j-invariants 1728, 0 and −3375, corresponding to multiplication by the orders Z[i], Z ! 1+√−3 2 " , and Z ! 1+√−7 2 " , respectively. For the ﬁrst curve (or more generally any curve of the form y2 = x3 + Ax with A ∈Z) the mul-tiplication by i corresponds to the obvious endomorphism (x, y) →(−x, iy) of the curve, and similarly for the second curve (or any curve of the form y2 = x3 + B) we have the equally obvious endomorphism (x, y) →(ωx, y) of order 3, where ω is a non-trivial cube root of 1. For the third curve (or any of its “twists” E : Cy2 = x3 −35x −98) the existence of a non-trivial endomorphism is less obvious. One checks that the map φ : (x, y) → γ2 x + β2 x + α , γ3 y 1 − β2 (x + α)2 , where α = (7 + √−7)/2, β = (7 + 3√−7)/2, and γ = (1 + √−7)/4, maps E to itself and satisﬁes φ(φ(P)) −φ(P) + 2P = 0 for any point P on E, where the addition is with respect to the group law on the curve, so that we have a map from O−7 to the endomorphisms of E sending λ = m 1+√−7 2 + n to the endomorphism P →mφ(P) + nP. The key point about elliptic curves with complex multiplication is that the number of their points over ﬁnite ﬁelds is given by a simple formula. For the three curves above this looks as follows. Recall that the number of points over Fp of an elliptic curve E/Q given by a Weierstrass equation Elliptic Modular Forms and Their Applications 91 y2 = F(x) = x3 + Ax + B equals p + 1 −ap where ap (for p odd and not di-viding the discriminant of F) is given by − x (mod p) F (x) p . For the three curves in (101) we have x (mod p) x3 + x p = 0 if p ≡3 (mod 4) , −2a if p = a2 + 4b2 , a ≡1 (mod 4) (102a) x (mod p) x3 + 1 p = 0 if p ≡2 (mod 3) , −2a if p = a2 + 3b2 , a ≡1 (mod 3) (102b) x (mod p) x3 −35x −98 p = 0 if (p/7) = −1 , −2a if p = a2 + 7b2 , (a/7) = 1 . (102c) (For other D with h(D) = 1 we would get a formula for ap(E) as ±A where 4p = A2 + \|D\|B2.) The proofs of these assertions, and of the more general statements needed when h(D) > 1, will be omitted since they would take us too far aﬁeld, but we give the proof of the ﬁrst (due to Gauss), since it is elementary and quite pretty. We prove a slightly more general but less precise statement. Proposition 29. Let p be an odd prime and A an integer not divisible by p. Then x (mod p) x3 + Ax p = ⎧ ⎪ ⎨ ⎪ ⎩ 0 if p ≡3 (mod 4) , ±2a if p ≡1 (mod 4) and (A/p) = 1 , ±4b if p ≡1 (mod 4) and (A/p) = −1 , (103) where \|a\| and \|b\| in the second and third lines are deﬁned by p = a2 + 4b2. Proof. The ﬁrst statement is trivial since if p ≡3 (mod 4) then (−1/p) = −1 and the terms for x and −x in the sum cancel, so we can suppose that p ≡1 (mod 4). Denote the sum on the left-hand side of (103) by sp(A). Replacing x by rx with r ̸≡0 (mod p) shows that sp(r2A) = r p sp(A), so the number sp(A) takes on only four values, say ±2α for A = g4i or A = g4i+2 and ±2β for A = g4i+1 or A = g4i+3, where g is a primitive root modulo p. (That sp(A) is always even is obvious by replacing x by −x.) Now we take the sum of the squares of sp(A) as A ranges over all integers modulo p, noting that sp(0) = 0. This gives 2(p −1)(α2 + β2) = A, x, y ∈Fp x3 + Ax p y3 + Ay p = x, y ∈Fp xy p A ∈Fp (x2 + A)(y2 + A) p = x, y ∈Fp xy p −1 + p δx2,y2 = 2p(p −1) , 92 D. Zagier where in the last line we have used the easy fact that z∈Fp z(z+r) p equals p −1 for r ≡0 (mod p) and −1 otherwise. (Proof: the ﬁrst statement is obvious; the substitution z →rz shows that the sum is independent of r if r ̸≡0; and the sum of the values for all integers r mod p clearly vanishes.) Hence α2 + β2 = p. It is also obvious that α is odd (for (A/p) = +1 there are p−1 2 −1 values of x between 0 and p/2 with x3+Ax p non-zero and hence odd) and that β is even (for (A/p) = −1 all p−1 2 values of x3+Ax p with 0 < x < p/2 are odd), so α = ±a, β = ±2b as asserted. An almost exactly similar proof works for equation (102b): here one deﬁnes sp(B) = x x3+B p and observes that sp(r3B) = r p sp(B), so that sp(B) takes on six values ±α, ±β and ±γ depending on the class of i (mod 6), where p ≡1 (mod 6) (otherwise all sp(B) vanish) and B = gi with g a primitive root modulo p. Summing sp(B) over all quadratic residues B (mod p) shows that α+β+γ = 0, and summing sp(B)2 over all B gives α2+β2+γ2 = 6p. These two equations imply that α ≡β ≡γ ̸≡0 (mod 3) and that 4p = α2+3((β−γ)/3)2, which gives (102b) since it is easily seen that α is even and β and γ are odd. I do not know of any elementary proof of this sort for equation (102c) or for the similar identities corresponding to the other imaginary quadratic ﬁelds of class number 1. The reader is urged to try to ﬁnd such a proof. ♠Factorization, Primality Testing, and Cryptography We mention brieﬂy one “practical” application of complex multiplication the-ory. Many methods of modern cryptography depend on being able to identify very large prime numbers quickly or on being able to factor (or being relatively sure that no one else will be able to factor) very large composite numbers. Sev-eral methods involve the arithmetic of elliptic curves over ﬁnite ﬁelds, which yield ﬁnite groups in which certain operations are easily performed but not easily inverted. The diﬃculty of the calculations, and hence the security of the method, depends on the structure of the group of points of the curve over the ﬁnite ﬁelds, so one would like to be able to construct, say, examples of elliptic curves E over Q whose reduction modulo p for some very large prime p has an order which itself contains a very large prime factor. Since counting the points on E(Fp) directly is impractical when p is very large, it is essential here to know curves E for which the number ap, and hence the cardinality of E(Fp), is known a priori, and the existence of closed formulas like the ones in (102) implies that the curves with complex multiplication are suitable. In practice one wants the complex multiplication to be by an order in a quadratic ﬁeld which is not too big but also not too small. For this purpose one needs eﬀective ways to construct the Hilbert class ﬁelds (which is where the needed singular moduli will lie) eﬃciently, and here again the methods mentioned in 6.1, and in particular the simpliﬁcations arising by replacing the modular function j(z) by better modular functions, become relevant. For more information, see the bibliography. ♥ Elliptic Modular Forms and Their Applications 93 We now turn from elliptic curves with complex multiplication to an impor-tant and related topic, the so-called CM modular forms. Formula (102a) says that the coeﬃcient ap of the L-series L(E, s) = ann−s of the elliptic curve E : y2 = x3 + x for a prime p is given by ap = Tr(λ) = λ + ¯ λ, where λ is one of the two numbers of norm p in Z[i] of the form a + bi with a ≡1 (mod 4), b ≡0 (mod 2) (or is zero if there is no such λ). Now using the multiplicative properties of the an we ﬁnd that the full L-series is given by L(E, s) = LK(s, ψ1) , (104) where LK(s, ψ1) = 0̸=a⊆Z[i] ψ1(a)N(a)−s is the L-series attached as in §6.3 to the ﬁeld K = Q(i) and the grossencharacter ψ1 deﬁned by ψ1(a) = 0 if 2 \| N(a) , λ if 2 ∤N(a) , a = (λ), λ ∈4Z + 1 + 2Z i . (105) In fact, the L-series of the elliptic curve E belongs to three important classes of L-series: it is an L-function coming from algebraic geometry (by deﬁnition), the L-series of a generalized character of an algebraic number ﬁeld (by (104)), and also the L-series of a modular form, namely L(E, s) = L(fE, s) where fE(z) = 0̸=a⊆Z[i] ψ1(a) qN(a) = a≡1 (mod 4) b≡0 (mod 2) a qa2+b2 (106) which is a theta series of the type mentioned in the ﬁnal paragraph of §3, associated to the binary quadratic form Q(a, b) = a2 + b2 and the spherical polynomial P(a, b) = a (or a+ib) of degree 1. The same applies, with suitable modiﬁcations, for any elliptic curve with complex multiplication, so that the Taniyama-Weil conjecture, which says that the L-series of an elliptic curve over Q should coincide with the L-series of a modular form of weight 2, can be seen explicitly for this class of curves (and hence was known for them long before the general case was proved). The modular form fE deﬁned in (106) can also be denoted θψ1, where ψ1 is the grossencharacter (105). More generally, the L-series of the powers ψd = ψ d 1 of ψ1 are the L-series of modular forms, namely LK(ψd, s) = L(θψd, s) where θψd(z) = 0̸=a⊆Z[i] ψd(a) qN(a) = a≡1 (mod 4) b≡0 (mod 2) (a + ib)d qa2+b2 , (107) which is a modular form of weight d + 1. Modular forms constructed in this way – i.e., theta series associated to a binary quadratic form Q(a, b) and a spherical polynomial P(a, b) of arbitrary degree d – are called CM modular forms, and have several remarkable properties. First of all, they are always linear combinations of the theta series θψ(z) = a ψ(α) qN(a) associated to some grossencharacter ψ of an imaginary quadratic ﬁeld. (This is because any 94 D. Zagier positive deﬁnite binary quadratic form over Q is equivalent to the norm form λ →λ¯ λ on an ideal in an imaginary quadratic ﬁeld, and since the Laplace operator associated to this form is ∂2 ∂λ ∂λ, the only spherical polynomials of degree d are linear combinations of λd and ¯ λd.) Secondly, since the L-series L(θψ, s) = LK(ψ, s) has an Euler product, the modular forms θψ attached to grossencharacters are always Hecke eigenforms, and this is the only inﬁnite family of Hecke eigenforms, apart from Eisenstein series, which are known explicitly. (Other eigenforms do not appear to have any systematic rule of construction, and even the number ﬁelds in which their Fourier coeﬃcients lie are totally mysterious, an example being the ﬁeld Q( √ 144169) of coeﬃ-cients of the cusp form of level 1 and weight 24 mentioned in §4.1.) Thirdly, sometimes modular forms constructed by other methods turn out to be of CM type, leading to new identities. For instance, in four cases the modular form deﬁned in (107) is a product of eta-functions: θψ0(z) = η(8z)4/η(4z)2, θψ1(z) = η(8z)8/η(4z)2η(16z)2, θψ2(z) = η(4z)6, θψ4(z) = η(4z)14/η(8z)4. More generally, we can ask which eta-products are of CM type. Here I do not know the answer, but for pure powers there is a complete result, due to Serre. The fact that both η(z) = n≡1 (mod 6) (−1)(n−1)/6qn2/24 and η(z)3 = n≡1 (mod 4) n qn2/8 are unary theta series implies that each of the functions η2, η4 and η6 is a binary theta series and hence a modular form of CM type (the function η(z)6 equals θψ2(z/4), as we just saw, and η(z)4 equals fE(z/6), where E is the second curve in (101)), but there are other, less obvious, examples, and these can be completely classiﬁed: Theorem (Serre). The function η(z)n (n even) is a CM modular form for n = 2, 4, 6, 8, 10, 14 or 26 and for no other value of n. Finally, the CM forms have another property called “lacunarity” which is not shared by any other modular forms. If we look at (106) or (107), then we see that the only exponents which occur are sums of two squares. By the theorem of Fermat proved in §3.1, only half of all primes (namely, those congruent to 1 modulo 4) have this property, and by a famous theorem of Landau, only O(x/(log x)1/2) of the integers ≤x do. The same applies to any other CM form and shows that 50% of the coeﬃcients ap (p prime) vanish if the form is a Hecke eigenform and that 100% of the coeﬃcients an (n ∈N) vanish for any CM form, eigenform or not. Another diﬃcult theorem, again due to Serre, gives the converse statements: Theorem (Serre). Let f = anqn be a modular form of integral weight ≥2. Then: 1. If f is a Hecke eigenform, then the density of primes p for which ap ̸= 0 is equal to 1/2 if f corresponds to a grossencharacter of an imaginary quadratic ﬁeld, and to 1 otherwise. 2. The number of integers n ≤x for which an ̸= 0 is O(x/√log x) as x →∞ if f is of CM type and is larger than a positive multiple of x otherwise. Elliptic Modular Forms and Their Applications 95 ♠Central Values of Hecke L-Series We just saw above that the Hecke L-series associated to grossencharacters of degree d are at the same time the Hecke L-series of CM modular forms of weight d + 1, and also, if d = 1, sometimes the L-series of elliptic curves over Q. On the other hand, the Birch–Swinnerton-Dyer conjecture predicts that the value of L(E, 1) for any elliptic curve E over Q (not just one with CM) is related as follows to the arithmetic of E : it vanishes if and only if E has a rational point of inﬁnite order, and otherwise is (essentially) a certain period of E multiplied by the order of a mysterious group Ш, the Tate–Shafarevich group of E. Thanks to the work of Kolyvagin, one knows that this group is indeed ﬁnite when L(E, 1) ̸= 0, and it is then a stan-dard fact (because Ш admits a skew-symmetric non-degenerate pairing with values in Q/Z) that the order of Ш is a perfect square. In summary, the value of L(E, 1), normalized by dividing by an appropriate period, is al-ways a perfect square. This suggests looking at the central point ( = point of symmetry with respect to the functional equation) of other types of L-series, and in particular of L-series attached to grossencharacters of higher weights, since these can be normalized in a nice way using the Chowla– Selberg period (97), to see whether these numbers are perhaps also always squares. An experiment to test this idea was carried out over 25 years ago by B. Gross and myself and conﬁrmed this expectation. Let K be the ﬁeld Q(√−7) and for each d ≥1 let ψd = ψ d 1 be the grossencharacter of K which sends an ideal a ⊆OK to λd if a is prime to p7 = (√−7) and to 0 otherwise, where λ in the former case is the generator of a which is con-gruent to a square modulo p7. For d = 1, the L-series of ψd coincides with the L-series of the third elliptic curve in (101), while for general odd values of d it is the L-series of a modular form of weight d + 1 and trivial charac-ter on Γ0(49). The L-series L(ψ2m−1, s) has a functional equation sending s to 2m −s, so the point of symmetry is s = m. The central value has the form L(ψ2m−1, m) = 2Am (m −1)! 2π √ 7 m Ω2m−1 K (108) where ΩK = 4 √ 7 η 1 + √−7 2 2 = Γ( 1 7)Γ( 2 7)Γ( 4 7) 4π2 is the Chowla–Selberg period attached to K and (it turns out) Am ∈Z for all m > 1. (We have A1 = 1 4.) The numbers Am vanish for m even because the functional equation of L(ψ2m−1, s) has a minus sign in that case, but the nu-merical computation suggested that the others were indeed all perfect squares: A3 = A5 = 12, A7 = 32, A9 = 72, . . . , A33 = 447622863272552. Many years later, in a paper with Fernando Rodriguez Villegas, we were able to conﬁrm this prediction: 96 D. Zagier Theorem. The integer Am is a square for all m > 1. More precisely, we have A2n+1 = bn(0)2 for all n ≥0, where the polynomials bn(x) ∈Q[x] are deﬁned recursively by b0(x) = 1 2, b1(x) = 1 and 21 bn+1(x) = −(x−7)(64x−7)b′ n(x)+ (32nx −56n + 42)bn(x) −2n(2n −1)(11x + 7)bn−1(x) for n ≥1. The proof is too complicated to give here, but we can indicate the main idea. The ﬁrst point is that the numbers Am themselves can be computed by the method explained in §6.3. There we saw (for the full modular group, but the method works for higher level) that the value at s = k + n of the L-series of a grossencharacter of degree d = k + 2n is essentially equal to the nth non-holomorphic derivative of an Eisenstein series of weight k at a CM point. Here we want d = 2m −1 and s = m, so k = 1, n = m −1. More precisely, one ﬁnds that L(ψ2m−1, m) = (2π/ √ 7)m (m−1)! ∂m−1G1,ε(z7), where G1,ε is the Eisenstein series G1,ε = 1 2 + ∞ n=1 d\|n ε(n) qn = 1 2 + q + 2q2 + 3q4 + q7 + · · · of weight 1 associated to the character ε(n) = n 7 and z7 is the CM point 1 2 1 + i √ 7 . These coeﬃcients can be obtained by a quasi-recursion like the one in Proposition 28 (though it is more complicated here because the analogues of the polynomials pn(t) are now elements in a quadratic extension of Q[t]), and therefore are very easy to compute, but this does not explain why they are squares. To see this, we ﬁrst observe that the Eisenstein series G1,ε is one-half of the binary theta series Θ(z) = m, n∈Z qm2+mn+2n2 . In his thesis, Villegas proved a beautiful formula expressing certain linear combinations of values of binary theta series at CM points as the squares of linear combinations of values of unary theta series at other CM points. The same turns out to be true for the higher non-holomorphic derivatives, and in the case at hand we ﬁnd the remarkable formula ∂2nΘ(z7) = 22n72n+1/4 ∂nθ2(z∗ 7) 2 for all n ≥0 , (109) where θ2(z) = n∈Z q(n+1/2)2/2 is the Jacobi theta-series deﬁned in (32) and z∗ 7 = 1 2(1 + i √ 7). Now the values of the non-holomorphic derivatives ∂nθ2(z∗ 7) can be computed quasi-recursively by the method explained in 6.3. The result is the formula given in the theorem. Remarks 1. By the same method, using that every Eisenstein series of weight 1 can be written as a linear combination of binary theta series, one can show that the correctly normalized central values of all L-series of grossen-characters of odd degree are perfect squares. 2. Identities like (109) seem very surprising. I do not know of any other case in mathematics where the Taylor coeﬃcients of an analytic function at some point are in a non-trivial way the squares of the Taylor coeﬃcients of another analytic function at another point. Elliptic Modular Forms and Their Applications 97 3. Equation (109) is a special case of a yet more general identity, proved in the same paper, which expresses the non-holomorphic derivative ∂2nθψ(z0) of the CM modular form associated to a grossencharacter ψ of degree d = 2r as a simple multiple of ∂nθ2(z1) ∂n+rθ2(z2), where z0, z1 and z2 are CM points belonging to the quadratic ﬁeld associated to ψ. ♥ We end with an application of these ideas to a classical Diophantine equa-tion. ♠Which Primes are Sums of Two Cubes? In §3 we gave a modular proof of Fermat’s theorem that a prime number can be written as a sum of two squares if and only if it is congruent to 1 modulo 4. The corresponding question for cubes was studied by Sylvester in the 19th century. For squares the answer is the same whether one considers integer or rational squares (though in the former case the representation, when it exists, is unique and in the latter case there are inﬁnitely many), but for cubes one only considers the problem over the rational numbers because there seems to be no rule for deciding which numbers have a decomposition into integral cubes. The question therefore is equivalent to asking whether the Mordell-Weil group Ep(Q) of the elliptic curve Ep : x3 + y3 = p is non-trivial. Except for p = 2, which has the unique decomposition 13 + 13, the group Ep(Q) is torsion-free, so that if there is even one rational solution there are inﬁnitely many. An equivalent question is therefore: for which primes p is the rank rp of Ep(Q) greater than 0 ? Sylvester’s problem was already mentioned at the end of §6.1 in connection with Heegner points, which can be used to construct non-trivial solutions if p ≡4 or 7 mod 9. Here we consider instead an approach based on the Birch– Swinnerton-Dyer conjecture, according to which rp > 0 if and only if the L-series of Ep vanishes at s = 1. By a famous theorem of Coates and Wiles, one direction of this conjecture is known: if L(Ep, 1) ̸= 0 then Ep(Q) has rank 0. The question we want to study is therefore: when does L(Ep, 1) vanish? For ﬁve of the six possible congruence classes for p (mod 9) (we assume that p > 3, since r2 = r3 = 0) the answer is known. If p ≡4, 7 or 8 (mod 9), then the functional equation of L(Ep, s) has a minus sign, so L(Ep, 1) = 0 and rp is expected (and, in the ﬁrst two cases, known) to be ≥1; it is also known by an “inﬁnite descent” argument to be ≤1 in these cases. If p ≡2 or 5 (mod 9), then the functional equation has a plus sign and L(Ep, 1) divided by a suitable period is ≡1 (mod 3) and hence ̸= 0, so by the Coates-Wiles theorem these primes can never be sums of two cubes, a result which can also be proved in an elementary way by descent. In the remaining case p ≡1 (mod 9), however, the answer can vary: here the functional equation has a plus sign, so ords=1L(Ep, s) is even and rp is also expected to be even, and descent gives rp ≤2, but both cases rp = 0 and rp = 2 occur. The following result, again proved jointly with F. Villegas, gives a criterion for these primes. 98 D. Zagier Theorem. Deﬁne a sequence of numbers c0 = 1, c1 = 2, c2 = −152, c3 = 6848, . . . by cn = sn(0) where s0(x) = 1, s1(x) = 3x2 and sn+1(x) = (1−8x3)s′ n(x)+(16n+3)x2sn(x)−4n(2n−1)xsn−1(x) for n ≥1. If p = 9k+1 is prime, then L(Ep, 1) = 0 if and only if p\|ck. For instance, the numbers c2 = −152 and c4 = −8103296 are divis-ible by p = 19 and p = 37, respectively, so L(Ep, 1) vanishes for these two primes (and indeed 19 = 33 + (−2)3, 37 = 43 + (−3)3), whereas c8 = 532650564250569441280 is not divisible by 73, which is therefore not a sum of two rational cubes. Note that the numbers cn grow very quickly (roughly like n3n), but to apply the criterion for a given prime number p = 9k + 1 one need only compute the polynomials sn(x) modulo p for 0 ≤n ≤k, so that no large numbers are required. We again do not give the proof of this theorem, but only indicate the main ingredients. The central value of L(Ep, s) for p ≡1 (mod 9) is given by L(Ep, 1) = 3 Ω 3 √p Sp where Ω = Γ( 1 3)3 2π √ 3 and Sp is an integer which is supposed to be a perfect square (namely the order of the Tate-Shafarevich group of Ep if rp = 0 and 0 if rp > 0). Using the methods from Villegas’s thesis, one can show that this is true and that both Sp and its square root can be expressed as the traces of certain algebraic numbers deﬁned as special values of modu-lar functions at CM points: Sp = Tr(αp) = Tr(βp)2, where αp = 3 √p 54 Θ(pz0) Θ(z0) and βp = 6 √p √±12 η(pz1) η(z1/p) with Θ(z) = m,n qm2+mn+n2, z0 = 1 2 + i 6 √ 3 and z1 = r + √−3 2 (r ∈Z, r2 ≡−3 (mod 4p)). This gives an explicit formula for L(Ep, 1), but it is not very easy to compute since the numbers αp and βp have large degree (18k and 6k, respectively if p = 9k + 1) and lie in a diﬀerent number ﬁeld for each prime p. To obtain a formula in which everything takes place over Q, one observes that the L-series of Ep is the L-series of a cubic twist of a grossencharacter ψ1 of K = Q(√−3) which is independent of p. More precisely, L(Ep, s) = L(χpψ1, s) where ψ1 is deﬁned (just like the grossenchar-acters for Q(i) and Q(√−7) deﬁned in (105) and in the previous “application”) by ψ1(a) = λ if a = (λ) with λ ≡1 (mod 3) and ψ1(α) = 0 if 3\|N(a), and χp is the cubic character which sends a = (λ) to the unique cube root of unity in K which is congruent to (¯ λ/λ)(p−1)/3 modulo p. This means that formally the L-value L(Ep, 1) for p = 9k + 1 is congruent modulo p to the central value L(ψ12k+1, 6k + 1). Of course both of these numbers are transcendental, but the theory of p-adic L-functions shows that their “algebraic parts” are in fact congruent modulo p: we have L(ψ12k+1 1 , 6k + 1) = 39k−1 Ω12k+1 (2π)6k Ck (6k)! for some integer Ck ∈Z, and Sp ≡Ck (mod p). Now the calculation of Ck proceeds exactly like that of the number Am deﬁned in (108): Ck is, up to nor-malizing constants, equal to the value at z = z0 of the 6k-th non-holomorphic Elliptic Modular Forms and Their Applications 99 derivative of Θ(z), and can be computed quasi-recursively, and Ck is also equal to c2 k where ck is, again up to normalizing constants, the value of the 3k-th non-holomorphic derivative of η(z) at z = 1 2(−1+√−3) and is given by the formula in the theorem. Finally, an estimate of the size of L(Ep, 1) shows that \|Sp\| < p, so that Sp vanishes if and only if ck ≡0 (mod p), as claimed. The numbers ck can also be described by a generating function rather than a quasi-recursion, using the relations between modular forms and diﬀerential equations discussed in §5, namely 1 −x)1/24 F 1 3, 1 3; 2 3; x 1/2 = ∞ n=0 cn (3n)! x F( 2 3, 2 3; 4 3; x)3 8 F( 1 3, 1 3; 2 3; x)3 n , where F(a, b; c; x) denotes Gauss’s hypergeometric function. ♥ References and Further Reading There are several other elementary texts on the theory of modular forms which the reader can consult for a more detailed introduction to the ﬁeld. Four fairly short introductions are the classical book by Gunning (Lectures on Modular Forms, Annals of Math. Studies 48, Princeton, 1962), the last chapter of Serre’s Cours d’Arithmétique (Presses Universitaires de France, 1970; English translation A Course in Arithmetic, Graduate Texts in Math-ematics 7, Springer 1973), Ogg’s book Modular Forms and Dirichlet Series (Benjamin, 1969; especially for the material covered in §§3–4 of these notes), and my own chapter in the book From Number Theory to Physics (Springer 1992). The chapter on elliptic curves by Henri Cohen in the last-named book is also a highly recommended and compact introduction to a ﬁeld which is intimately related to modular forms and which is touched on many times in these notes. Here one can also recommend N. Koblitz’s Introduction to Elliptic Curves and Modular forms (Springer Graduate Texts 97, 1984). An excellent book-length Introduction to Modular Forms is Serge Lang’s book of that title (Springer Grundlehren 222, 1976). Three books of a more classical na-ture are B. Schoeneberg’s Elliptic Modular Functions (Springer Grundlehren 203, 1974), R.A. Rankin’s Modular Forms and Modular Functions (Cam-bridge, 1977) and (in German) Elliptische Funktionen und Modulformen by M. Koecher and A. Krieg (Springer, 1998). The point of view in these notes leans towards the analytic, with as many results as possible (like the algebraicity of j(z) when z is a CM point) be-ing derived purely in terms of the theory of modular forms over the complex numbers, an approach which was suﬃcient – and usually simpler – for the type of applications which I had in mind. The books listed above also belong to this category. But for many other applications, including the deepest ones in Diophantine equations and arithmetic algebraic geometry, a more arith-metic and more advanced approach is required. Here the basic reference is 100 D. Zagier Shimura’s classic Introduction to the Arithmetic Theory of Automorphic Func-tions (Princeton 1971), while two later books that can also be recommended are Miyake’s Modular Forms (Springer 1989) and the very recent book A First Course in Modular Forms by Diamond and Shurman (Springer 2005). We now give, section by section, some references (not intended to be in any sense complete) for various of the speciﬁc topics and examples treated in these notes. 1–2. The material here is all standard and can be found in the books listed above, except for the statement in the ﬁnal section of §1.2 that the class numbers of negative discriminants are the Fourier coeﬃcients of some kind of modular form of weight 3/2, which was proved in my paper in CRAS Paris (1975), 883–886. 3.1. Proposition 11 on the number of representations of integers as sums of four squares was proved by Jacobi in the Fundamenta Nova Theoriae Ellip-ticorum, 1829. We do not give references for the earlier theorems of Fermat and Lagrange. For more information about sums of squares, one can consult the book Representations of Integers as Sums of Squares (Springer, 1985) by E. Grosswald. The theory of Jacobi forms mentioned in connection with the two-variable theta functions θi(z, u) was developed in the book The Theory of Jacobi Forms (Birkhäuser, 1985) by M. Eichler and myself. Mersmann’s theorem is proved in his Bonn Diplomarbeit, “Holomorphe η-Produkte und nichtverschwindende ganze Modulformen für Γ0(N)” (Bonn, 1991), unfortu-nately never published in a journal. The theorem of Serre and Stark is given in their paper “Modular forms of weight 1/2 ” in Modular Forms of One Vari-able VI (Springer Lecture Notes 627, 1977, editors J-P. Serre and myself; this is the sixth volume of the proceedings of two big international conferences on modular forms, held in Antwerp in 1972 and in Bonn in 1976, which con-tain a wealth of further material on the theory). The conjecture of Kac and Wakimoto appeared in their article in Lie Theory and Geometry in Honor of Bertram Kostant (Birkhäuser, 1994) and the solutions by Milne and myself in the Ramanujan Journal and Mathematical Research Letters, respectively, in 2000. 3.2. The detailed proof and references for the theorem of Hecke and Schoen-berg can be found in Ogg’s book cited above. Niemeier’s classiﬁcation of unimodular lattices of rank 24 is given in his paper “Deﬁnite quadratische Formen der Dimension 24 und Diskriminante 1” (J. Number Theory 5, 1973). Siegel’s mass formula was presented in his paper “Über die analytische Theo-rie der quadratischen Formen” in the Annals of Mathematics, 1935 (No. 20 of his Gesammelte Abhandlungen, Springer, 1966). A recent paper by M. King (Math. Comp., 2003) improves by a factor of more than 14 the lower bound on the number of inequivalent unimodular even lattices of dimension 32. The paper of Mallows-Odlyzko-Sloane on extremal theta series appeared in J. Al-gebra in 1975. The standard general reference for the theory of lattices, which contains an immense amount of further material, is the book by Conway and Elliptic Modular Forms and Their Applications 101 Sloane (Sphere Packings, Lattices and Groups, Springer 1998). Milnor’s ex-ample of 16-dimensional tori as non-isometric isospectral manifolds was given in 1964, and 2-dimensional examples (using modular groups!) were found by Vignéras in 1980. Examples of pairs of truly drum-like manifolds – i.e., do-mains in the ﬂat plane – with diﬀerent spectra were ﬁnally constructed by Gordon, Webb and Wolpert in 1992. For references and more on the history of this problem, we refer to the survey paper in the book What’s Happening in the Mathematical Sciences (AMS, 1993). 4.1–4.3. For a general introduction to Hecke theory we refer the reader to the books of Ogg and Lang mentioned above or, of course, to the beautifully written papers of Hecke himself (if you can read German). Van der Blij’s example is given in his paper “Binary quadratic forms of discriminant −23” in Indagationes Math., 1952. A good exposition of the connection between Galois representations and modular forms of weight one can be found in Serre’s article in Algebraic Number Fields: L-Functions and Galois Properties (Academic Press 1977). 4.4. Book-length expositions of the Taniyama–Weil conjecture and its proof using modular forms are given in Modular Forms and Fermat’s Last The-orem (G. Cornell, G. Stevens and J. Silverman, eds., Springer 1997) and, at a much more elementary level, Invitation to the Mathematics of Fermat-Wiles (Y. Hellegouarch, Academic Press 2001), which the reader can con-sult for more details concerning the history of the problem and its solution and for further references. An excellent survey of the content and status of Serre’s conjecture can be found in the book Lectures on Serre’s Conjec-tures by Ribet and Stein ( For the ﬁnal proof of the conjecture and references to all earlier work, see “Modularity of 2-adic Barsotti-Tate representations” by M. Kisin ( Livné’s example ap-peared in his paper “Cubic exponential sums and Galois representations” (Contemp. Math. 67, 1987). For an exposition of the conjectural and known examples of higher-dimensional varieties with modular zeta functions, in par-ticular of those coming from mirror symmetry, we refer to the recent paper “Modularity of Calabi-Yau varieties” by Hulek, Kloosterman and Schütt in Global Aspects of Complex Geometry (Springer, 2006) and to the monograph Modular Calabi-Yau Threefolds by C. Meyer (Fields Institute, 2005). However, the proof of Serre’s conjectures means that the modularity is now known in many more cases than indicated in these surveys. 5.1. Proposition 15, as mentioned in the text, is due to Ramanujan (eq. (30) in “On certain Arithmetical Functions,” Trans. Cambridge Phil. Soc., 1916). For a good discussion of the Chazy equation and its relation to the “Painlevé property” and to SL(2, C), see the article “Symmetry and the Chazy equation” by P. Clarkson and P. Olver (J. Diﬀ. Eq. 124, 1996). The result by Gallagher on means of periodic functions which we describe as our second application is 102 D. Zagier described in his very nice paper “Arithmetic of means of squares and cubes” in Internat. Math. Res. Notices, 1994. 5.2. Rankin–Cohen brackets were deﬁned in two stages: the general conditions needed for a polynomial in the derivatives of a modular form to itself be modular were described by R. Rankin in “The construction of automorphic forms from the derivatives of a modular form” (J. Indian Math. Soc., 1956), and then the speciﬁc bilinear operators [ · , · ]n satisfying these conditions were given by H. Cohen as a lemma in “Sums involving the values at negative integers of L functions of quadratic characters” (Math. Annalen, 1977). The Cohen–Kuznetsov series were deﬁned in the latter paper and in the paper “A new class of identities for the Fourier coeﬃcients of a modular form” (in Russian) by N.V. Kuznetsov in Acta Arith., 1975. The algebraic theory of “Rankin–Cohen algebras” was developed in my paper “Modular forms and diﬀerential operators” in Proc. Ind. Acad. Sciences, 1994, while the papers by Manin, P. Cohen and myself and by the Untenbergers discussed in the second application in this subsection appeared in the book Algebraic Aspects of Integrable Systems: In Memory of Irene Dorfman (Birkhäuser 1997) and in the J. Anal. Math., 1996, respectively. 5.3. The name and general deﬁnition of quasimodular forms were given in the paper “A generalized Jacobi theta function and quasimodular forms” by M. Kaneko and myself in the book The Moduli Space of Curves (Birkhäuser 1995), immediately following R. Dijkgraaf’s article “Mirror symmetry and el-liptic curves” in which the problem of counting ramiﬁed coverings of the torus is presented and solved. 5.4. The relation between modular forms and linear diﬀerential equations was at the center of research on automorphic forms at the turn of the (previous) century and is treated in detail in the classical works of Fricke, Klein and Poincaré and in Weber’s Lehrbuch der Algebra. A discussion in a modern lan-guage can be found in §5 of P. Stiller’s paper in the Memoirs of the AMS 299, 1984. Beukers’s modular proof of the Apéry identities implying the irrational-ity of ζ(2) and ζ(3) can be found in his article in Astérisque 147–148 (1987), which also contains references to Apéry’s original paper and other related work. My paper with Kleban on a connection between percolation theory and modular forms appeared in J. Statist. Phys. 113 (2003). 6.1. There are several references for the theory of complex multiplication. A nice book giving an introduction to the theory at an accessible level is Primes of the form x2 + ny2 by David Cox (Wiley, 1989), while a more ad-vanced account is given in the Springer Lecture Notes Volume 21 by Borel, Chowla, Herz, Iwasawa and Serre. Shanks’s approximation to π is given in a paper in J. Number Theory in 1982. Heegner’s original paper attacking the class number one problem by complex multiplication methods appeared in Math. Zeitschrift in 1952. The result quoted about congruent numbers was proved by Paul Monsky in “Mock Heegner points and congruent numbers” Elliptic Modular Forms and Their Applications 103 (Math. Zeitschrift 1990) and the result about Sylvester’s problem was an-nounced by Noam Elkies, in “Heegner point computations” (Springer Lecture Notes in Computing Science, 1994). See also the recent preprint “Some Dio-phantine applications of Heegner points” by S. Dasgupta and J. Voight. 6.2. The formula for the norms of diﬀerences of singular moduli was proved in my joint paper “Singular moduli” (J. reine Angew. Math. 355, 1985) with B. Gross, while our more general result concerning heights of Heegner points appeared in “Heegner points and derivatives of L-series” (Invent. Math. 85, 1986). The formula describing traces of singular moduli is proved in my pa-per of the same name in the book Motives, Polylogarithms and Hodge Theory (International Press, 2002). Borcherds’s result on product expansions of auto-morphic forms was published in a celebrated paper in Invent. Math. in 1995. 6.3. The Chowla–Selberg formula is discussed, among many other places, in the last chapter of Weil’s book Elliptic Functions According to Eisenstein and Kronecker (Springer Ergebnisse 88, 1976), which contains much other beauti-ful historical and mathematical material. Chudnovsky’s result about the tran-scendence of Γ( 1 4) is given in his paper for the 1978 (Helskinki) International Congress of Mathematicians, while Nesterenko’s generalization giving the al-gebraic independence of π and eπ is proved in his paper “Modular functions and transcendence questions” (in Russian) in Mat. Sbornik, 1996. A good summary of this work, with further references, can be found in the “featured review” of the latter paper in the 1997 Mathematical Reviews. The algorith-mic way of computing Taylor expansions of modular forms at CM points is described in the ﬁrst of the two joint papers with Villegas cited below, in connection with the calculation of central values of L-series. 6.4. A discussion of formulas like (102) can be found in any of the general ref-erences for the theory of complex multiplication listed above. The applications of such formulas to questions of primality testing, factorization and cryptogra-phy is treated in a number of papers. See for instance “Eﬃcient construction of cryptographically strong elliptic curves” by H. Baier and J. Buchmann (Springer Lecture Notes in Computer Science 1977, 2001) and its bibliogra-phy. Serre’s results on powers of the eta-function and on lacunarity of modu-lar forms are contained in his papers “Sur la lacunarité des puissances de η” (Glasgow Math. J., 1985) and “Quelques applications du théorème de densité de Chebotarev” (Publ. IHES, 1981), respectively. The numerical experiments concerning the numbers deﬁned in (108) were given in a note by B. Gross and myself in the memoires of the French mathematical society (1980), and the two papers with F. Villegas on central values of Hecke L-series and their applications to Sylvester’s problems appeared in the proceedings of the third and fourth conferences of the Canadian Number Theory Association in 1993 and 1995, respectively. Hilbert Modular Forms and Their Applications Jan Hendrik Bruinier Fachbereich Mathematik, Technische Universität Darmstadt, Schloßgartenstraße 7, 64289 Darmstadt, Germany E-mail: bruinier@mathematik.tu-darmstadt.de Introduction The present notes contain the material of the lectures given by the author at the summer school on “Modular Forms and their Applications” at the Sophus Lie Conference Center in the summer of 2004. We give an introduction to the theory of Hilbert modular forms and some geometric and arithmetic applications. We tried to keep the informal style of the lectures. In particular, we often do not work in greatest possible generality, but rather consider a reasonable special case, in which the main ideas of the theory become clear. For a more comprehensive account to Hilbert modular varieties, we refer to the books by Freitag [Fr], Garrett [Ga], van der Geer [Ge1], and Goren [Go]. We hope that the present text will be a useful addition to these references. Hilbert modular surfaces can also be realized as modular varieties corre-sponding to the orthogonal group of a rational quadratic space of type (2, 2). This viewpoint leads to several interesting features of these surfaces. For in-stance, they come with a natural family of divisors arising from embeddings of “smaller” orthogonal groups, the so-called Hirzebruch–Zagier divisors. Their study led to important discoveries and triggered generalizations in various di-rections. Moreover, the theta correspondence provides a source of automorphic forms related to the geometry of Hirzebruch–Zagier divisors. A more recent development is the regularized theta lifting due to Borcherds, Harvey and Moore, which yields to automorphic products and automorphic Green functions. The focus of the present text is on these topics, highlight-ing the role of the orthogonal group. We added some background material on quadratic spaces and orthogonal groups, to make the connection explicit. I thank G. van der Geer and D. Zagier for several interesting conversations during the summer school at the Sophus Lie Conference Center. Moreover, I thank J. Funke and T. Yang for their helpful comments on earlier versions of this manuscript. 106 J. H. Bruinier 1 Hilbert Modular Surfaces In this section we give a brief introduction to Hilbert modular surfaces asso-ciated to real quadratic ﬁelds. For details we refer to [Fr], [Ga], [Ge1], [Go]. 1.1 The Hilbert Modular Group Let d > 1 be a squarefree integer. Then F = Q( √ d) is a real quadratic ﬁeld, which we view as a subﬁeld of R. The discriminant of F is D = d, if d ≡1 (mod 4) , 4d, if d ≡2, 3 (mod 4) . (1.1) We write OF for the ring of integers in F, so OF = Z + 1+ √ d 2 Z, if d ≡1 (mod 4) , Z + √ dZ, if d ≡2, 3 (mod 4) . (1.2) The ring OF is a Dedekind domain, that is, it is a noetherian integrally closed integral domain in which every non-zero prime ideal is maximal. We denote by O∗ F the group of units in OF . By the Dirichlet unit theorem there is a unique unit ε0 > 1 such that O∗ F = {±1} × {εn 0; n ∈Z}. It is called the fundamental unit of F. We write x →x′ for the conjugation, N(x) = xx′ for the norm in F, and tr(x) = x + x′ for the trace in F. The diﬀerent of F is denoted by dF . Note that dF = ( √ D). Recall that an (integral) ideal of OF is a OF -submodule of OF . A fractional ideal of F is a ﬁnitely generated OF -submodule of F. Fractional ideals form a group together with the ideal multiplication. The neutral element is OF and the inverse of a fractional ideal a ⊂F is a−1 = {x ∈F; xa ⊂OF } . Since F is a quadratic extension of Q, we have the useful formula a−1 = 1 N(a)a′, where a′ is the conjugate of a. Two fractional ideals a, b are called equivalent, if there is a r ∈F such that a = rb. The group of equivalence classes Cl(F) is called the ideal class group of F. It is a ﬁnite abelian group. Two fractional ideals a, b are called equivalent in the narrow sense, if there is a totally positive r ∈F such that a = rb. The group of equivalence classes Cl+(F) is called the narrow ideal class group of F. It is equal to Cl(F), if and only if ε0 has norm −1. Otherwise it is an extension of degree 2 of Cl(F). The (narrow) class number of F is the order of the (narrow) ideal class group. It measures how far OF is from being a principal ideal domain. If the class number of F is greater than 1, there are ideals which cannot be generated by a single element. However, we have the following fact, which holds in any Dedekind ring. Hilbert Modular Forms 107 Remark 1.1. If a ⊂F is a fractional ideal, then there exist α, β ∈F such that a = αOF + βOF . □ The group SL2(F) is embedded into SL2(R) × SL2(R) by the two real embeddings of F. It acts on H × H, where H = {τ ∈C; I(τ) > 0} is the complex upper half plane, via fractional linear transformations, a b c d z = az1 + b cz1 + d, a′z2 + b′ c′z2 + d′ . (1.3) Here and throughout we use z = (z1, z2) as a standard variable on H2. If z ∈H2 and a b c d ∈SL2(F), we write N(cz + d) = (cz1 + d)(c′z2 + d′) . (1.4) Lemma 1.2. For z ∈H2 and γ = a b c d ∈SL2(F) we have ℑ(γz) = ℑ(z) \| N(cz + d)\|2 . Proof. This follows immediately from the analogous assertion in the one-dimensional case. □ If a is a fractional ideal of F, we write Γ(OF ⊕a) = $a b c d ∈SL2(F); a, d ∈OF , b ∈a−1, c ∈a % (1.5) for the Hilbert modular group corresponding to a. Moreover, we write ΓF = Γ(OF ⊕OF ) = SL2(OF ) . (1.6) Let Γ ⊂SL2(F) be a subgroup which is commensurable with ΓF , i.e., Γ ∩ΓF has ﬁnite index in both, Γ and ΓF . Then Γ acts properly discontinuously on H2, i.e., if W ⊂H2 is compact, then {γ ∈Γ; γW ∩W ̸= ∅} is ﬁnite (see Corol-lary 1.17). In particular, for any a ∈H2, the stabilizer Γa = {γ ∈Γ; γa = a} is a ﬁnite subgroup of Γ. Let ¯ Γa be the image of Γa in PSL2(F) = SL2(F)/{±1}. If # ¯ Γa > 1 then a is called an elliptic ﬁxed point for Γ and # ¯ Γa is called the order of a. The order of a only depends of the Γ-class. Moreover, there are only ﬁnitely many Γ-classes of elliptic ﬁxed points. It can be shown that Γ always has a ﬁnite index subgroup which has no elliptic ﬁxed points. The quotient Y (Γ) = Γ\H2 (1.7) is a normal complex surface. The singularities are given by the elliptic ﬁxed points. They are ﬁnite quotient singularities. 108 J. H. Bruinier The surface Y (Γ) is non-compact. It can be compactiﬁed by adding a ﬁnite number of points, the cusps of Γ. They can be described as follows. The group SL2(F) also acts on P1(F) = F ∪{∞} by a b c d α β = a α β + b c α β + d = aα + bβ cα + dβ . Notice that, since a b c d ∞= a c , the action of SL2(F) is transitive. The Γ-classes of P1(F) are called the cusps of Γ. Lemma 1.3. The map ϕ : ΓF \P1(F) Cl(F), (α : β) αOF + βOF , is bijective. Proof. We begin by showing that ϕ is well-deﬁned: It is clear that ϕ(α : β) = ϕ(rα : rβ). Now let a b c d ∈ΓF , and let γ δ = a b c d α β . We need to show that ϕ(γ : δ) = ϕ(α : β). We have ϕ (γ : δ) = γOF + δOF = (aα + bβ)OF + (cα + dβ)OF ⊂ϕ (α : β) . Interchanging the roles of (γ : δ) and (α : β), we see ϕ (α : β) = (dγ −bδ)OF + (−cγ + aδ)OF ⊂ϕ (γ : δ) . Consequently, ϕ(γ : δ) = ϕ(α : β). The surjectivity of ϕ follows from Remark 1.1. Finally, we show that ϕ is injective. Let a = ϕ(α : β) = ϕ(γ : δ). Then 1 ∈ OF = aa−1 = αa−1 + βa−1. So there exist ˜ α, ˜ β ∈a−1 such that 1 = α˜ β −β ˜ α. We ﬁnd that M1 := α ˜ α β ˜ β ∈ a a−1 a a−1 ∩SL2(F) , and M1∞= (α : β). In the same way we ﬁnd M2 := γ ˜ γ δ ˜ δ ∈ a a−1 a a−1 ∩SL2(F) such that M2∞= (γ : δ). Therefore we have M2M −1 1 ∈ a a−1 a a−1 a−1 a−1 a a = OF OF OF OF . Hence, M2M −1 1 ∈ΓF and M2M −1 1 (α : β) = (γ : δ). This concludes the proof of the Lemma. □ Hilbert Modular Forms 109 Corollary 1.4. The number of cusps of ΓF is equal to the class number h(F) of F. A subgroup Γ ⊂SL2(F) which is commensurable with ΓF has ﬁnitely many cusps. Remark 1.5. Let Γ∞⊂Γ be the stabilizer of ∞. Then there is a Z-module M ⊂F of rank 2 and a ﬁnite index subgroup V ⊂O∗ F acting on M such that the group G(M, V ) = $ε μ 0 ε−1 ; μ ∈M and ε ∈V % (1.8) is contained in Γ∞with ﬁnite index. Example 1.6. If a ⊂F is a fractional ideal, then Γ(OF ⊕a)∞= $ε μ 0 ε−1 ; μ ∈a−1, ε ∈O∗ F % . 1.2 The Baily–Borel Compactiﬁcation We embed P1(F) into P1(R) × P1(R) via the two real embeddings of F. Then we may view P1(F) as the set of rational boundary points of H2 in the same way as P1(Q) is viewed as the set of rational boundary points of H. Here we consider (H2)∗= H2 ∪P1(F) . (1.9) By introducing a suitable topology on (H2)∗, the quotient Γ(H2)∗can be made into a compact Hausdorﬀspace. This leads to the Baily–Borel com-pactiﬁcation of Y (Γ). Proposition 1.7. On (H2)∗there is a unique topology with the following properties: (i) The induced topology on H2 agrees with the usual one. (ii) H2 is open and dense in (H2)∗. (iii) The sets UC ∪∞, where UC = (z1, z2) ∈H2; I(z1)I(z2) > C for C > 0, form a base of open neighborhoods of the cusp ∞. (iv) If κ ∈P1(F) and ρ ∈SL2(F) with ρ∞= κ, then the sets ρ(UC ∪∞) (C > 0) form a base of open neighborhoods of the cusp κ. □ 110 J. H. Bruinier Remark 1.8. The system of open neighborhoods of κ deﬁned by (iv) does not depend on the choice of ρ. The stabilizer Γ∞of ∞acts on UC. If γ = ε μ 0 ε−1 ∈ Γ∞, then γz = ε2z1 + εμ, ε′2z2 + ε′μ′ . We consider the quotient X(Γ) = Γ(H2)∗. (1.10) Theorem 1.9. The quotient space X(Γ), together with the quotient topology, is a compact Hausdorﬀspace. □ Proposition 1.10. For C > 0 suﬃciently large, the canonical map Γ∞\UC ∪∞ Γ(H2)∗ is an open embedding. □ The group SL2(F) acts by topological automorphisms on (H2)∗. Hence, for ρ ∈SL2(F), the natural map X(Γ) − →X(ρ−1Γρ), z →ρ−1z is topological. If ρ∞= κ, it takes the cusp κ of Γ to the cusp ∞of ρ−1Γρ. In that way, local considerations near the cusps can often be reduced to con-siderations at the cusp ∞(for a conjugate group), for which one can use Proposition 1.10. We deﬁne a complex structure on X(Γ) as follows. For an open subset V ⊂ X(Γ) we let U ⊂(H2)∗be the inverse image under the canonical projection pr : (H2)∗→X(Γ), and let U ′ be the inverse image in H2. We have the diagram H2 (H2)∗ pr X(Γ) U ′ U V . We deﬁne OX(Γ )(V ) to be the ring of continuous functions f : V →C such that pr∗(f) is holomorphic on U ′. This deﬁnes a sheaf OX(Γ ) of rings on X(Γ), and the pair (X(Γ), OX(Γ )) is a locally ringed space. Theorem 1.11 (Baily–Borel). The space (X(Γ), OX(Γ )) is a normal com-plex space. □ The proof is based on a criterion of Baily and Cartan for the continuation of complex structures, see [Fr] p. 112. In contrast to the case of modular curves the resulting normal complex space X(Γ) is not regular. There are ﬁnite quotient singularities at the elliptic Hilbert Modular Forms 111 ﬁxed points, and more seriously, the cusps are highly singular points. By the theory of Hironaka the singularities can be resolved [Hi]. A weak form of Hironaka’s result states that there exists a desingularization π : X(Γ) − →X(Γ) , (1.11) where X(Γ) is a non-singular connected projective variety such that π induces a biholomorphic map π−1(X(Γ)reg) →X(Γ)reg. Here X(Γ)reg is the regular locus of X(Γ). One can further require that the complement of π−1(X(Γ)reg) is a divisor with normal crossings. The minimal resolution of singularities was constructed by Hirzebruch [Hz]. It can be shown that there is an ample line bundle on X(Γ), the line bundle of modular forms (in suﬃciently divisible weight). Consequently, the space X(Γ) carries the structure of a projective algebraic variety over C. The surface Y (Γ) is a Zariski-open subvariety and therefore quasi-projective. Remark 1.12. The Hilbert modular surfaces Y (Γ) often have a moduli inter-pretation, analogously to the fact that SL2(Z)\H parametrizes isomorphism classes of elliptic curves over C. It can be used to construct integral models of Hilbert modular surfaces. For instance, Y (Γ(OF ⊕a)) is the coarse moduli space for isomorphism classes of triples (A, ı, m), where A is an abelian surface over C, and ı : OF →End(A) is an embedding of rings (real multiplication), and m is an isomorphism from the polarization module of A to (adF )−1 re-specting the positivities, cf. [Go] Theorem 2.17. The variety Y (Γ(OF ⊕a)) can be interpreted as the complex points of a moduli stack over Z. One can also construct toroidal compactiﬁcations and Baily–Borel compactiﬁcations over Z, cf. [Rap], [DePa], [Ch]. Siegel Domains Here we recall the properties of Siegel domains for Hilbert modular surfaces. They are nice substitutes for fundamental domains. We write (x1, x2) for the real part and (y1, y2) for the imaginary part of (z1, z2). The top degree diﬀerential form dμ = dx1 dy1 y2 1 dx2 dy2 y2 2 (1.12) on H2 is invariant under the action of SL2(R)2. It deﬁnes an invariant measure on H2, which is induced by the Haar measure on SL2(R). Deﬁnition 1.13. A subset S ⊂H2 is called a fundamental set for Γ, if H2 = & γ∈Γ γ(S) . 112 J. H. Bruinier Deﬁnition 1.14. A fundamental set S for Γ is called a fundamental domain for Γ, if (i) S is measurable. (ii) There is a subset N ⊂S of measure zero, such that for all z, w ∈S\N we have z ∼Γ w ⇒z = w . Remark 1.15. It can be shown that every measurable fundamental set contains a fundamental domain. Nice fundamental sets for the action of Γ on H2 are given by Siegel do-mains: For a positive real number t we put St = z ∈H2; \| xi\| < t and yi > 1 t for i = 1, 2 . (1.13) Proposition 1.16. For any ﬁxed t ∈R>0 there exist only ﬁnitely many γ ∈Γ such that γSt ∩St ̸= ∅. (1.14) Proof. It is clear that there are only ﬁnitely many γ = a b c d ∈Γ with c = 0 satisfying condition (1.14). On the other hand, assume that γ ∈Γ with c ̸= 0, and assume that there is a z ∈γSt ∩St. Then we have y1 \|cz1 + d\|2 > 1 t , (1.15) y2 \|c′z2 + d′\|2 > 1 t . (1.16) The ﬁrst inequality implies that y1 > 1 t (cx1 + d)2 + c2y2 1 ≥1 t c2y2 1 > 1 t2 c2y1 , (1.17) and therefore \|c\| < t. In the same way, inequality (1.16) implies that \|c′\| < t. Hence there are only ﬁnitely many possibilities for c. For these, by (1.17) and its analogue for y2, the imaginary part (y1, y2) is bounded, and there are also just ﬁnitely many possibilities for d. Moreover, replacing γ by γ−1 in the above argument, we ﬁnd that there are only ﬁnitely many possibilities for a. But a, c, d determine γ. □ Corollary 1.17. The action of Γ on H2 is properly discontinuous, that is, if W ⊂H2 is compact, then {γ ∈Γ; γW ∩W ̸= ∅} is ﬁnite. Proof. This follows from Proposition 1.16 and the fact that ' t∈R>0 St = H2. □ Hilbert Modular Forms 113 Theorem 1.18. Let κ1, . . . , κr ∈P1(F) be a set of representatives for the cusps of Γ, and let ρ1, . . . , ρr ∈SL2(F) such that ρj ∞= κj. There is a t > 0 such that S = r & j=1 ρjSt is a measurable fundamental set for Γ. Proof. See [Ga], Chapter 1.6. □ 1.3 Hilbert Modular Forms Let Γ ⊂SL2(F) be a subgroup which is commensurable with ΓF , and let (k1, k2) ∈Z2. Deﬁnition 1.19. A meromorphic function f : H2 →C is called a meromor-phic Hilbert modular form of weight (k1, k2) for Γ, if f(γz) = (cz1 + d)k1(c′z2 + d′)k2f(z) (1.18) for all γ = a b c d ∈Γ. If k1 = k2 =: k, then f is said to have parallel weight, and is simply called a meromorphic Hilbert modular form of weight k. If f is holomorphic on H2, then it is called a holomorphic Hilbert modular form. Finally, a Hilbert modular form f is called symmetric, if f(z1, z2) = f(z2, z1) . For a function f : H2 →C and γ = a b c d ∈SL2(F) we deﬁne the Petersson slash operator by (f \|k1,k2 γ)(z) = (cz1 + d)−k1(c′z2 + d′)−k2f(γz) . The assignment f →f \|k1,k2 γ deﬁnes a right action of SL2(F) on complex valued functions on H2. Using it, we may rewrite condition (1.18) as f \|k1,k2 γ = f, γ ∈Γ . If k1 = k2 =: k, we simply write f \|k γ instead of f \|k1,k2 γ. If f is a holomorphic Hilbert modular form for Γ, it has a Fourier expansion at the cusp ∞of the following form. Let M ⊂F be a Z-module of rank 2 and let V ⊂O∗ F be a ﬁnite index subgroup acting on M such that the group G(M, V ) = $ε μ 0 ε−1 ; μ ∈M and ε ∈V % (1.19) is contained in Γ∞with ﬁnite index. The transformation law (1.18) for γ ∈ G(M, V ) implies that f(z + μ) = f(z) 114 J. H. Bruinier for all μ ∈M. Therefore, f has a normally convergent Fourier expansion f = ν∈M∨ aν e(tr(νz)) , (1.20) where e(w) = e2πiw, tr(νz) = νz1 + ν′z2, and M ∨= {λ ∈F; tr(μλ) ∈Z for all μ ∈M} (1.21) is the dual lattice of M with respect to the trace form on F. The Fourier coeﬃcients aν are given by aν = 1 vol(R2/M) R2/M f(z)e(−tr(νz)) dx1 dx2 . (1.22) More generally, if κ ∈P1(F) is any cusp of Γ, we take ρ ∈SL2(F) such that ρ∞= κ, and consider f \|k1,k2 ρ at ∞. The Fourier expansion of f at κ is the expansion of f \|k1,k2 ρ at ∞(it depends on the choice of ρ). It is a striking fact that, in contrast to the one-dimensional situation, a holomorphic Hilbert modular form is automatically holomorphic at the cusps by the Götzky–Koecher principle. Theorem 1.20 (Götzky–Koecher principle). Let f : H2 →C be a holo-morphic function satisfying f \|k1,k2 γ = f for all γ ∈G(M, V ) as in (1.19). Then (i) aε2ν = εk1ε′k2aν for all ν ∈M ∨and ε ∈V , (ii) aν ̸= 0 ⇒ν = 0 or ν ≫0. Proof. (i) For ε ∈V we have ε−1 0 0 ε ∈G(M, V ). The transformation law implies that ε−k1ε′−k2 ν∈M∨ aν e(tr(νε−2z)) = ν∈M∨ aν e(tr(νz)) . Comparing Fourier coeﬃcients, we obtain the ﬁrst assertion. (ii) Suppose that there is a ν ∈M ∨such that aν ̸= 0 and such that ν < 0 or ν′ < 0. Without loss of generality we assume ν < 0. There is an ε ∈V with ε > 1 and 0 < ε′ < 1 such that tr(εν) < 0. Then tr(ε2nν) goes to −∞for n →∞. The series n≥1 aνε2n e(i tr(νε2n)) is a subseries of the Fourier expansion of f(z) at z = (i, i) and therefore converges absolutely. But by (i) we have n≥1 \|aνε2n e(i tr(νε2n))\| = \|aν\| n≥1 εk1nε′k2ne−2π tr(νε2n) →∞, contradicting the convergence. □ Hilbert Modular Forms 115 Corollary 1.21. A holomorphic Hilbert modular form for the group Γ has a Fourier expansion at the cusp ∞of the form f(z) = a0 + ν∈M∨ ν≫0 aν e(tr(νz)) . (1.23) The constant term a0 is called the value of f at ∞. We write f(∞) = a0. More generally, if κ ∈P1(F) is any cusp of Γ, we take ρ ∈SL2(F) such that ρ∞= κ. We put f(κ) = (f \|k1,k2 ρ)(∞). If (k1, k2) ̸= (0, 0), the value f(κ) of f at κ depends on the choice of ρ (by a non-zero factor). Deﬁnition 1.22. A holomorphic Hilbert modular form f is called a cusp form, if it vanishes at all cusps of Γ. Proposition 1.23. Let f be a holomorphic modular form of weight (k1, k2) for Γ. If k1 ̸= k2, then f is a cusp form. Proof. This follows from Theorem 1.20 (i), applied to the constant terms at the cusps. □ Proposition 1.24. Let f be a modular form of weight (k1, k2) for Γ. Then the function h(z) = \|f(z)yk1/2 1 yk2/2 2 \| is Γ-invariant. Proof. This follows from Lemma 1.2. □ Proposition 1.25. Let f be a holomorphic modular form of weight (k1, k2) for Γ, and let h(z) = \|f(z)yk1/2 1 yk2/2 2 \| be the Γ-invariant function of Proposit-ion 1.24. (i) If f has parallel negative weight k = k1 = k2, then h attains a maximum on H2. (ii) If f is a cusp form, then h vanishes at the cusps and attains a maximum on H2. Proof. We only prove the ﬁrst statement, the second is similar. By Proposit-ion 1.24, it suﬃces to consider h on a fundamental set for Γ. In view of The-orem 1.18, it suﬃces to show that for any ρ ∈SL2(F) and any t ∈R>0, the function h(ρz) is bounded and attains its maximum on the Siegel domain St. Using the Fourier expansion of f at the cusp ρ∞, we see that h(ρz) = (y1y2)k/2a0 + (y1y2)k/2 ν∈M∨ ν≫0 aν e(tr(νz)) for a suitable rank 2 lattice M ⊂F. Since the weight k is negative, we ﬁnd that limy1y2→∞h(ρz) = 0 on St. Consequently, h(ρz) is bounded and attains a maximum on St. □ 116 J. H. Bruinier Proposition 1.26. Let f be a holomorphic modular form of weight (k1, k2) for Γ. Then f vanishes identically unless k1, k2 are both positive or k1 = k2 = 0. In the latter case f is constant. Proof. Let us ﬁrst consider the case that k1 = 0 and k2 ̸= 0. By Proposit-ion 1.23, f is a cusp form. The function h(z) = yk2/2 2 f(z) is holomorphic in z1, and, by Proposition 1.25, \|h\| attains a maximum on H2. According to the maximum principle, h must be constant as a function of z1. Hence, h(z1, z2) = h(z1, γ′z2) for all γ ∈Γ. Since {γz2; γ ∈Γ} is dense in H, the function h must be constant on H2. Because h vanishes at the cusps, it must vanish identically. In the same way we see that f = 0, if k2 = 0 and k1 ̸= 0. Let us now consider the case that k1 = k2 = 0. If f is a cusp form, then Proposition 1.25 implies that \|f\| attains a maximum on H2. Hence, by the maximum principle, f must be constant. Since f vanishes at ∞, we obtain that f ≡0. If f is holomorphic (but not necessarily cuspidal), we consider the cusp form g(z) := κ∈Γ \P1(F ) (f(z) −f(κ)) . We ﬁnd that g ≡0, and therefore f is constant. Finally, assume that some ki, say k1, is negative. If k1 ̸= k2, then, by Proposition 1.23, f is a cusp form. If k1 = k2, then f has parallel negative weight. In both cases, Proposition 1.25 implies that h(z) = \|f(z)yk1/2 1 yk2/2 2 \| is bounded by a constant C > 0 on H2. We consider the Fourier expansion of f at the cusp ∞as in (1.20). The coeﬃcients aν are given by (1.22). We ﬁnd that \|aν\| ≤ 1 vol(R2/M) R2/M \|f(z)e(−tr(νz))\| dx1 dx2 ≤Cy−k1/2 1 y−k2/2 2 e−2π tr(νy) . Taking the limit y1 →0, we see that aν vanishes for all ν ∈M ∨, and therefore f ≡0. □ Corollary 1.27. Let X(Γ) →X(Γ) be a desingularization as in (1.11). Then any holomorphic 1-form on X(Γ) vanishes identically. Proof. Let ω be a holomorphic 1-form on X(Γ) and denote by η its pullback to the regular locus of X(Γ). Viewing η as a Γ-invariant holomorphic 1-form on H2, we may write η = f1(z) dz1 + f2(z) dz2 , where f1 and f2 are holomorphic Hilbert modular forms of weights (2, 0) and (0, 2), respectively. Hence, by Proposition 1.26, η vanishes identically. □ Hilbert Modular Forms 117 In the same way one sees that any antiholomorphic 1-form on X(Γ) van-ishes identically. Consequently, H1( X(Γ), O e X(Γ )) = 0 , that is, the surface X(Γ) is regular. Using Hodge theory we see that the ﬁrst cohomolgy group H1( X(Γ), C) vanishes. In particular, the interesting part of the cohomology of a Hilbert modular surface is in degree 2. It can be shown that the Hilbert modular surfaces X(Γ(OF ⊕a)) corres-ponding to the groups Γ(OF ⊕a) are simply connected. This also implies the vanishing of the ﬁrst Betti number. However, there are also examples of Hilbert modular surfaces which are not simply connected. See [Ge1], Chapter IV.6. (Recall that the fundamental group of a complex surface is a birational invariant.) For the rest of these notes we will only be considering Hilbert modular forms of parallel weight k. Notation 1.28. Let k ∈Z. We write Mk(Γ) for the C-vector space of holo-morphic Hilbert modular forms of weight k for the group Γ, and denote by Sk(Γ) the subspace of cusp forms. The codimension of Sk(Γ) in Mk(Γ) is clearly bounded by the number of cusps of Γ. Proposition 1.29. Let X(Γ) →X(Γ) be a desingularization. (i) Meromorphic Hilbert modular forms of weight 0 for Γ, can be identiﬁed with meromorphic functions on X(Γ). (ii) Meromorphic Hilbert modular forms of weight 2 for Γ, can be identiﬁed with meromorphic 2-forms on X(Γ). (iii) Hilbert cusp forms of weight 2 for Γ, can be identiﬁed with holomorphic 2-forms on X(Γ). Proof. The ﬁrst two assertions are easy. For the third assertion we refer to [Fr], Chapter II.4. □ In particular, the arithmetic genus of the surface X(Γ), that is, the Euler characteristic of the structure sheaf, is given by χ(O e X(Γ )) = 2 p=0 (−1)p dim Hp( X(Γ), O e X(Γ )) = 1 + dim(S2(Γ)) . Holomorphic Hilbert modular forms can be interpreted as sections of the sheaf Mk(Γ) of modular forms, which can be deﬁned as follows: If we write pr : H2 →Y (Γ) for the canonical projection, then the sections over an open subset U ⊂Γ\H2 are holomorphic functions on pr−1(U), satisfying the trans-formation law (1.18). By the Koecher principle, this sheaf on Y (Γ) extends to X(Γ). It is a coherent OX(Γ )-module. 118 J. H. Bruinier Let n(Γ) denote the least common multiple of the orders of all elliptic ﬁxed points for Γ. When n(Γ) \| k, then Mk(Γ) is a line bundle. One can show that this line bundle is ample and thereby prove that X(Γ) is algebraic. Notice that Mnk(Γ) = Mk(Γ)⊗n for any positive integer n. 1.4 Mk(Γ ) is Finite Dimensional In this section we show that Mk(Γ) is ﬁnite dimensional. The argument is based on the comparison of two diﬀerent norms on the space of cusp forms. It is a rather general principle and works in a much more general setting (cf. [Fr], Chapter I.6). We begin by deﬁning the Petersson scalar product on Mk(Γ). The top degree diﬀerential form dμ = dx1 dy1 y2 1 dx2 dy2 y2 2 on H2 is invariant under the action of SL2(R)2. Deﬁnition 1.30. Let f, g ∈Mk(Γ) such that the product fg is a cusp form. We deﬁne the Petersson scalar product of f and g by ⟨f, g⟩= F f(z)g(z)(y1y2)k dμ , where F is a fundamental domain for Γ. Lemma 1.31. For f, g as above the Petersson scalar product converges abso-lutely and is independent of the choice of the fundamental domain. Proof. Arguing as in Proposition 1.25, we see that f(z)g(z) (y1y2)k is invari-ant under Γ and bounded on H2. Hence, that the integral does not depend on the choice of F follows from the absolute convergence using the theorem on dominated convergence for the Lebesgue integral. To prove the absolute convergence, it suﬃces to show that F dμ < ∞. In view of Proposition 1.18, it suﬃces to show that St dμ < ∞ for all t > 0. This follows from the fact that ∞ 1/t dy y2 < ∞. □ In particular, the Petersson scalar product deﬁnes a hermitian scalar prod-uct on Sk(Γ). We denote the corresponding L2-norm on Sk(Γ) by ∥f∥2 := ⟨f, f⟩. (1.24) On the other hand we have the maximum norm on Sk(Γ) which is deﬁned by ∥f∥∞= max z∈F \|f(z)\|(y1y2)k/2 . (1.25) Hilbert Modular Forms 119 Lemma 1.32. There is a constant A = A(Γ, k) > 0 such that ∥f∥∞≤A · ∥f∥2 for all f ∈Sk(Γ). Proof. The L2-norm can be estimated by considering the Fourier expansions of f at the cusps of Γ and using Siegel domains (Proposition 1.18). See [Fr], Lemma 6.2 for details. □ Theorem 1.33. The vector space Mk(Γ) is ﬁnite dimensional. Proof. It suﬃces to show that dim Sk(Γ) < ∞. Let f1, . . . , fm be an orthonor-mal set with respect to the Petersson scalar product, that is, ⟨fi, fj⟩= δij. For an arbitrary linear combination f = m j=1 cjfj with coeﬃcients cj ∈C, we have ∥f∥∞≤A∥f∥2 by Lemma 1.32. Hence, for all z ∈H2 we have m j=1 cjfj(z)(y1y2)k/2 ≤A ⎛ ⎝ m j=1 \|cj\|2 ⎞ ⎠ 1/2 . We consider the inequality for cj = fj(z). We ﬁnd m j=1 \|fj(z)\|2(y1y2)k/2 ≤A ⎛ ⎝ m j=1 \|fj(z)\|2 ⎞ ⎠ 1/2 . Dividing by the sum on the right hand side and taking the square we obtain m j=1 \| fj(z)\|2(y1 y2)k ≤A2 . Integrating over F we ﬁnd that m ≤A2 vol(Γ\H2) . This concludes the proof of the theorem. □ 1.5 Eisenstein Series Here we deﬁne Eisenstein series for Hilbert modular groups. For simplicity we only consider the full Hilbert modular group ΓF = SL2(OF ) of the real 120 J. H. Bruinier quadratic ﬁeld F. We write N(x) for the norm of x ∈F, and N(a) for the norm of a fractional ideal a ⊂F. Let B ∈Cl(F) be an ideal class of F. There is a zeta function associated with B which is deﬁned by ζB(s) = c∈B c⊂OF N(c)−s . (1.26) Here s is a complex variable, and the sum runs through all integral ideals in the ideal class B. The series converges for ℜ(s) > 1. It has a meromorphic continuation to the full complex plane and the completed zeta function ΛB(s) = Ds/2π−sΓ(s/2)2ζB(s) (1.27) satisﬁes the functional equation ΛB(s) = ΛdB−1(1 −s) (1.28) (see e.g. [Ne], Chapter VII.5). Here, d = dF = √ DOF is the diﬀerent of F. The Dedekind zeta function ζF (s) of F is given by ζF (s) = B∈Cl(F ) ζB(s) = c⊂OF integral ideal N(c)−s . (1.29) Let b be a fractional ideal in the ideal class B. The group of units O∗ F acts on b × b by (c, d) →(εc, εd) for ε ∈O∗ F . Recall that for a b c d ∈SL2(F) and z = (z1, z2) ∈H2, we write N(cz + d) = (cz1 + d)(c′z2 + d′). Deﬁnition 1.34. Let k > 2 be an even integer. We deﬁne the Eisenstein series of weight k associated to B by1 Gk,B(z) = N(b)k ′ (c,d)∈O∗ F \b×b N(cz + d)−k . The Eisenstein series Gk,B does not depend on the choice of the representative b of the ideal class B and converges uniformly absolutely in every Siegel domain St (t > 0). Consequently, it deﬁnes an element of Mk(ΓF ). The value at the cusp ∞is given by Gk,B(∞) = lim z∈St y1y2→∞ Gk,B(z) = N(b)k ′ d∈O∗ F \b N(d)−k = ′ d∈O∗ F \b N(db−1)−k = ζB−1(k) . 1 The superscript at the summation sign means that the zero summand is omitted. Hilbert Modular Forms 121 If κ ∈P1(F) is any cusp and ρ = α β γ δ ∈SL2(F) with ρ∞= κ, then Gk,B(κ) = lim z∈St y1y2→∞ (Gk,B \|k ρ)(z) = N(a)kζ[a]B−1(k) , where a = OF α + OF γ. Theorem 1.35. Let k > 2 be an even integer. The Eisenstein series Gk,B ∈ Mk(ΓF ), where B ∈Cl(F), are linearly independent. The space Mk(ΓF ) can be decomposed as a direct sum Mk(ΓF ) = Sk(ΓF ) ⊕ ( B∈Cl(F ) CGk,B . Proof. See [Ge1], p. 21. □ Remark 1.36. One can deﬁne Gk,B for k > 2 odd in the same way. In this case it is easily seen that Gk,B ≡0 if OF contains a unit of negative norm. Moreover, the theorem also holds for k = 2. In this case one can deﬁne G2,B by analytic continuation using the “Hecke-trick”. In turns out that all Eisenstein series of weight 2 are holomorphic (in contrast to the case of elliptic modular forms, where the constant term is sometimes non-holomorphic). The Fourier expansion of the Eisenstein series can be computed in the same way as in the case of elliptic modular forms. Theorem 1.37. Let k ≥2 even. The Eisenstein series Gk,B has the Fourier expansion Gk,B(z) = ζB−1(k) + (2πi)2k (k −1)!2 D1/2−k ν∈d−1 ν≫0 σk−1,dB(dν) e(tr(νz)) . Here, for A ∈Cl(F) and an integral ideal l ⊂OF , σs,A(l) denotes the divisor sum σs,A(l) = c ∈A integral c\|l N(c)s . □ Using the functional equation of ζB(s), we may write Gk,B(z) = (2πi)2k (k −1)!2 D1/2−k 1 4ζdB(1 −k) + ν∈d−1 ν≫0 σk−1,dB(dν) e(tr(νz)) . (1.30) Recall that a Hilbert modular form f of weight k is called symmetric, if f(z1, z2) = f(z2, z1). It is easily seen that the Eisenstein series Gk,B are symmetric. 122 J. H. Bruinier Restriction to the Diagonal If f ∈Mk(ΓF ) is a Hilbert modular form, we consider its restriction to the diagonal g(τ) = f(τ, τ). Since the elliptic modular group SL2(Z) is embed-ded diagonally into ΓF = SL2(OF ), the function g has the transformation behavior g(γτ) = f(γτ, γτ) = (cτ + d)2kf(τ, τ) for γ = a b c d ∈SL2(Z). Therefore g(τ) is an elliptic modular form for SL2(Z) of weight 2k. If f has the Fourier expansion f(z) = a0 + ν∈d−1 ν≫0 aν e(tr(νz)) at the cusp ∞, then g has the expansion g(τ) = a0 + n≥1 ν∈d−1 ν≫0 tr(ν)=n aν e(nτ) . (1.31) The geometric interpretation is the following. The diagonal embedding H → H2, τ →(τ, τ) induces a morphism ϕ : SL2(Z)\H →Y (ΓF ), which is birational onto its image. If we view f as a section of the line bundle of modular forms of weight k over Y (ΓF ), then g = ϕ∗(f) is the pull-back. We now consider the restriction of the Eisenstein series Gk,B. Using the Fourier expansion (1.30), we see that 1 4ζdB(1 −k) + n≥1 ν∈d−1 ν≫0 tr(ν)=n σk−1,dB(dν) e(nτ) (1.32) is a modular form for SL2(Z) of weight 2k. As a ﬁrst consequence we see that the special values ζdB(1 −k) must be rational numbers. This follows from the fact that the divisor sums σk−1,dB(dν) are rational (integers), and the fact that the spaces of elliptic modular forms for SL2(Z) have bases with rational Fourier coeﬃcients. If k = 2, 4, then (1.32) must be a multiple of the elliptic Eisenstein series E2k(τ) = −B2k 4k + n≥1 σ2k−1(n) e(nτ) . Here B2k is the usual Bernoulli number and σs(n) = d\|n ds. Comparing the ﬁrst Fourier coeﬃcients we obtain a formula for ζdB(1 −k) due to Siegel. Hilbert Modular Forms 123 Theorem 1.38 (Siegel). If k = 2, 4, then ζB(1 −k) = −B2k k ν∈d−1 ν≫0 tr(ν)=1 σk−1,B(dν) . □ By means of (1.29), and using B4 = B8 = −1/30, we obtain: Corollary 1.39. The special values of the Dedekind zeta function of F at the arguments −1, −3 are given by ζF (−1) = 1 60 x∈Z x2 0 such that \|aν\| ≤C R2/M (y1y2)−k/2e−2π tr(νy) dx1 dx2 for all y ∈(R>0)2. Choosing y = ( 1 ν , 1 ν′ ), we see that \|aν\| ≤C vol(R2/M) N(ν)k/2 . This proves the proposition. □ 126 J. H. Bruinier For the rest of this section we only consider the full Hilbert modular group ΓF . Let f ∈Mk(ΓF ), and denote the Fourier expansion by f = a0 + ν∈d−1 ν≫0 aν e(tr(νz)) . Let O∗,+ F be the group of totally positive units of OF , and let U = {ε2; ε ∈ O∗,+ F }. Then U has index 2 in the cyclic group O∗,+ F . We have aν = aεν for all ν ∈d−1 and all ε ∈U. Deﬁnition 1.43. We deﬁne an L-series associated to f by L(f, s) = ν∈d−1/U ν≫0 aν N(νd)−s . Example 1.44. For the Eisenstein series Gk,B (see Deﬁnition 1.34) one easily checks that L(Gk,B, s) = 2 (2πi)2k (k −1)!2 D1/2−kζd−1B−1(s) ζdB(s + 1 −k) . The functional equation (1.28) of the partial Dedekind zeta function ζB(s) implies that L(Gk,B, s) has a meromorphic continuation and satisﬁes a func-tional equation relating s and k−s. Therefore it is reasonable to expect similar properties for the L-functions of cusp forms as well. Theorem 1.45. Let f ∈Sk(ΓF ). The completed L-function Λ(f, s) = Ds(2π)−2sΓ(s)2L(f, s) has a holomorphic continuation to C, is entire and bounded in vertical strips, and satisﬁes the functional equation Λ(f, s) = (−1)kΛ(f, k −s) . Proof. Using the Euler integral for the Gamma function, we see that (2π)−2sΓ(s)2 N(ν)−s = ∞ 0 ∞ 0 e−2π tr(νy)(y1y2)s dy1 y1 dy2 y2 . Hence, by unfolding we ﬁnd that Λ(f, s) = (R>0)2/U f(iy)(y1y2)s dy1 y1 dy2 y2 . We split up the integral into an integral over y1y2 > 1 and a second in-tegral over y1y2 < 1. The modularity of f implies that f i 1 y1 , 1 y2 = Hilbert Modular Forms 127 (−1)k(y1y2)kf(iy). Hence the second integral can be transformed into an in-tegral over y1y2 > 1 as well. We ﬁnd that Λ(f, s) = (R>0)2/U y1y2>1 f(iy)(y1y2)s dy1 y1 dy2 y2 + (−1)k (R>0)2/U y1y2>1 f(iy)(y1y2)k−s dy1 y1 dy2 y2 . This integral representation converges for all s ∈C and deﬁnes the holomor-phic continuation of Λ(f, s). Moreover, the functional equation is obvious now. □ We now suppose that OF contains a unit of negative norm. Then Mk(ΓF ) contains no non-zero element for k odd. So we further suppose that k is even. Then the Fourier coeﬃcients of f ∈Mk(ΓF ) satisfy aν = aεν for all ν ∈d−1 and all ε ∈O∗,+ F . Thus, aν only depends on the ideal (νd) ⊂OF , and we write a((νd)) = aν. Then we may rewrite the L-function of f in the form L(f, s) = a⊂OF principal ideal a(a) N(a)−s . So this L-function is analogous to the zeta function ζB(s) associated to an ideal class B of F (here the unit class). It is natural to associate more general L-functions to f, for instance, an L-function where one sums over all integral ideals of F analogous to the full Dedekind zeta function of F. To this end it is more convenient to view Hilbert modular forms as automorphic functions on (ResF/Q SL2)(A), where ResF/Q denotes the Weil restriction of scalars, and A denotes the ring of adeles of Q (cf. [Ga]). 2 The Orthogonal Group O(2, n) An important property of Hilbert modular surfaces is that they can also be regarded as modular varieties associated to the orthogonal group of a suitable rational quadratic space V of type (2, 2). There is an accidental isomorphism ResF/Q SL2 ∼ = SpinV of algebraic groups over Q. Modular varieties for orthog-onal groups O(2, n) come with natural families of special algebraic cycles on them arising from embeddings of “smaller” orthogonal groups. They provide a rich source of extra structure and can be used to study geometric questions. In the O(2, 2)-case of Hilbert modular surfaces these special cycles lead to Hirzebruch–Zagier divisors (codimension 1) and certain CM-points (codimen-sion 2). To put things in the right context, in this section we study quadratic spaces and modular forms for orthogonal groups in slightly greater generality than needed for the application to Hilbert modular surfaces. However, we hope that this will rather clarify things than complicate them. For a detailed account of the theory of quadratic forms and orthogonal groups we refer to [Ki], [Scha]. 128 J. H. Bruinier 2.1 Quadratic Forms Let R be a commutative ring with unity 1. We write R∗for the group of units in R. Let M be a ﬁnitely generated R-module. A quadratic form on M is a mapping Q : M →R such that (i) Q(rx) = r2Q(x) for all r ∈R and all x ∈M, (ii) B(x, y) := Q(x + y) −Q(x) −Q(y) is a bilinear form. The ﬁrst condition follows from the second if 2 is invertible in R. Then we have Q(x) = 1 2B(x, x). The pair (M, Q) is called a quadratic module over R. If R is a ﬁeld, we frequently say space instead of module. Two elements x, y ∈M are called orthogonal if B(x, y) = 0. If A ⊂M is a subset, we denote the orthogonal complement by A⊥= {x ∈M; B(x, y) = 0 for all y ∈A} . (2.1) It is a submodule of M. For x ∈M we brieﬂy write x⊥instead of {x}⊥. The quadratic module M is called non-degenerate, if M ⊥= {0}. A non-zero vector x ∈M is called isotropic if Q(x) = 0, and anisotropic, if Q(x) ̸= 0, respectively. Let (M, Q) and (M ′, Q′) be quadratic modules over R. An R-linear map σ : M →M ′ is called an isometry, if σ is injective and Q′(σ(x)) = Q(x) for all x ∈M. If σ is also surjective then M and N are called isometric. The orthogonal group of M, OM = {σ ∈Aut(M); σ isometry} , (2.2) is the group of all isometries from M onto itself. The special orthogonal group is the subgroup SOM = {σ ∈OM; det(σ) = 1} . (2.3) Important examples of isometries are given by reﬂections. For an element x ∈M with Q(x) ∈R∗we deﬁne τx : M →M by τx(y) = y −B(y, x)Q(x)−1x, y ∈M . (2.4) Then τx is an isometry and satisﬁes τx(x) = −x τx(y) = y , for y ∈x⊥, τ2 x = id . So τx is the reﬂection in the hyperplane x⊥. Hilbert Modular Forms 129 Further examples of isometries are given by Eichler elements. Let u ∈M be isotropic and v ∈M with B(u, v) = 0. We deﬁne Eu,v : M →M by Eu,v(y) = y + B(y, u)v −B(y, v)u −B(y, u)Q(v)u, y ∈M . (2.5) One easily checks that Eu,v is an isometry and Eu,v(u) = u, Eu,v(v) = v −2Q(v)u, Eu,v1Eu,v2 = Eu,v1+v2 for v1, v2 ∈u⊥. If M is free, and v1, . . . , vn is a basis of M, we have the corresponding Gram matrix S = (B(vi, vj))i,j. The class of det(S) in R∗/(R∗)2 is independent of the choice of the basis. It is called the discriminant of M and is denoted by d(M). Note that if v1, . . . , vn is an orthogonal basis of M, we have d(M) = 2nQ(v1) · · · Q(vn) . (2.6) For the rest of this subsection, let M be a quadratic space of dimension n over a ﬁeld k of characteristic ̸= 2. The space is non-degenerate, if and only if d(M) ̸= 0. If M is non-degenerate and v1 ∈M is an anisotropic vector, there exist anisotropic vectors v2, . . . , vn ∈M such that v1, . . . , vn is an or-thogonal basis of M. Consequently, the reﬂection corresponding to v1 satisﬁes det(τv1) = −1. Theorem 2.1. Let M be a regular quadratic space over a ﬁeld k of character-istic ̸= 2. Then OM is generated by reﬂections. Moreover, SOM is the subgroup of elements of OM which can be written as a product of an even number of reﬂections. □ Example 2.2. Let p, q be non-negative integers. We denote by Rp,q the quadratic space over R given by Rp+q with the quadratic form Q(x) = x2 1 + · · · + x2 p −x2 p+1 · · · −x2 p+q . If V is a ﬁnite dimensional quadratic space over R, then there exist non-negative integers p, q such that V is isometric to Rp,q. The pair (p, q) is uniquely determined by V and is called the type of V . Moreover, p −q is called the signature of V . The orthogonal group of Rp,q is also denoted by O(p, q). 2.2 The Cliﬀord Algebra As before, let R be a commutative ring with unity 1, and let (V, Q) be a ﬁnitely generated quadratic module over R. If A is any R-algebra, we write Z(A) for the center of A. 130 J. H. Bruinier We consider the tensor algebra TV = ∞ ( m=0 V ⊗m = R ⊕V ⊕(V ⊗R V ) ⊕. . . of V . Let IV ⊂TV be the two-sided ideal generated by v ⊗v −Q(v) for v ∈V . The Cliﬀord algebra of V is deﬁned by CV = TV /IV . (2.7) Observe that R and V are embedded into CV via the canonical maps. For simplicity, the element of CV represented by v1 ⊗· · · ⊗vm (where vi ∈V ) is denoted by v1 · · · vm. By deﬁnition, we have for u, v ∈V ⊂CV : v2 = Q(v), uv + vu = B(u, v) . In particular, uv = −vu if and only if u and v are orthogonal. The Cliﬀord algebra has the following universal property. Proposition 2.3. Let f : V →A be an R-linear map to an R-algebra A with unity 1A such that f(v)2 = Q(v)1A for all v ∈V . Then there exists a unique R-algebra homomorphism CV →A such that the following diagram commutes: V CV A . The universal property implies that an isometry ϕ : V →V ′ of quadratic spaces over R induces a unique R-algebra homomorphism ˜ ϕ : CV →CV ′ compatible with the natural inclusions. Therefore, the assignment V →CV deﬁnes a functor from the category of quadratic spaces over R with isometries as morphisms to the category of associative R-algebras with unity. Moreover, if we ﬁx a quadratic space (V, Q) over R, for any commutative R-algebra S with unity we can consider the extension of scalars V (S) = V ⊗R S, which is a quadratic module over S in a natural way (the quadratic form being deﬁned by Q(v ⊗s) = s2Q(v)). In the same way, we consider CV (S) = CV ⊗R S. One easily checks that CV (S) = CV (S). So taking the Cliﬀord algebra commutes with extension of scalars. Example 2.4. We denote by Cp,q the Cliﬀord algebra of the real quadratic space Rp,q of Example 2.2. For small p, q we have C0,0 = R, C1,0 = R ⊕R, C0,1 = C, C2,0 = M2(R), C1,1 = M2(R), C0,2 = H . Here H denotes the Hamilton quaternion algebra. (This follows from Ex-amples 2.7 and 2.8 below). Hilbert Modular Forms 131 Now assume that V is free. If v1, . . . , vn is a basis of V , then these vectors generate CV as an R-algebra. The elements vi1 · · · vir (1 ≤i1 < · · · < ir ≤n and 0 ≤r ≤n) form a basis of CV . In particular, CV is a free R-module of rank 2n. Observe that for the trivial quadratic form Q ≡0 the Cliﬀord algebra CV is simply the Grassmann algebra of V . We write C0 V for the R-subalgebra of CV generated by products of an even number of basis vectors of V , and C1 V for the R-submodule of CV generated by products of an odd number of basis vectors of V . This deﬁnition is meaningful, since the deﬁning relations of CV only involve products of an even number of basis vectors. We obtain a decomposition CV = C0 V ⊕C1 V , which is a Z/2Z-grading on CV . The subalgebra C0 V is called the even Cliﬀord algebra of V (or the second Cliﬀord algebra of V ). Multiplication by −1 deﬁnes an isometry of V . By Proposition 2.3 it in-duces an algebra automorphism J : CV − →CV , (2.8) called the canonical automorphism. If 2 is invertible in R, then the even Clif-ford algebra can be characterized by C0 V = {v ∈CV ; J(v) = v} . There is a second involution on CV , which is an anti-automorphism. It is called the canonical involution on CV and is deﬁned by t : CV − →CV , (x1 ⊗· · · ⊗xm)t = xm ⊗· · · ⊗x1 . (2.9) It reduces to the identity on R ⊕V . It is used to deﬁne the Cliﬀord norm on CV by N : CV − →CV , N(x) = xtx . (2.10) For x ∈V we have N(x) = Q(x). So the norm map extends the quadratic form on V . Note that the Cliﬀord norm is in general not multiplicative on CV . For the rest of this subsection, let k be a ﬁeld of characteristic ̸=2, and let (V, Q) be a non-degenerate quadratic space over k of dimension n. Moreover, let v1, . . . , vn be an orthogonal basis of V . We put δ = v1 · · · vn ∈CV . 132 J. H. Bruinier Remark 2.5. When n is even, we have δvi = −viδ, δ2 = (−1)n/22−nd(V ) ∈k∗/(k∗)2 . When n is odd, we have δvi = viδ, δ2 = (−1)(n−1)/22−nd(V ) ∈k∗/(k∗)2 . □ Theorem 2.6. The center of CV is given by Z(CV ) = k if n is even , k + kδ if n is odd . The center of C0 V is given by Z(C0 V ) = k + kδ if n is even , k if n is odd . □ Let A be a ring with unity such that k ⊂Z(A). Recall that A is called a quaternion algebra over k, if it has a basis {1, x1, x2, x3} as a k-vector space such that x2 1 = α, x2 2 = β, x3 = x1x2 = −x2x1 for some α, β ∈k∗. Then it is denoted by (α, β). The parameters α, β deter-mine A up to k-algebra isomorphism. It is easily seen that k = Z(A). The conjugation in A is deﬁned by x = a0 + a1x1 + a2x2 + a3x3 →¯ x = a0 −a1x1 −a2x2 −a3x3 for x ∈A. The norm is deﬁned by N(x) = x¯ x ∈k. A quaternion algebra over k is either isomorphic to M2(k) or it is a division algebra. For more details we refer to [Ki] Chapter 1.5. We end this section by giving some examples of Cliﬀord algebras associated to low dimensional quadratic spaces (see [Ki], p. 28). Example 2.7. If n = 1 then CV = k + kδ and δ2 = d(V )/2. As a k-algebra, we have CV ∼ = k[X]/(X2 −d(V )/2) . When d(V )/2 is not a square in k, this is a quadratic ﬁeld extension of k. When d(V )/2 is a square in k, then CV ∼ = k ⊕k. Example 2.8. Suppose that n = 2 and that V has the orthogonal basis v1, v2 with Q(vi) = qi ∈k∗. Then CV = k ⊕kv1 ⊕kv2 ⊕kv1v2 is isomorphic to the quaternion algebra (q1, q2) over k. Moreover, C0 V ∼ = k[X]/(X2 + d(V )). Hilbert Modular Forms 133 Example 2.9. Suppose that n = 3 and that V has the orthogonal basis v1, v2, v3 with Q(vi) = qi ∈k∗. Then C0 V = k ⊕kv1v2 ⊕kv2v3 ⊕kv1v3 is isomorphic to the quaternion algebra (−q1q2, −q2q3) over k. The conjugation in the quaternion algebra is identiﬁed with the main involution of the Cliﬀord algebra, and the norm with the Cliﬀord norm. Example 2.10. Suppose that n = 4 and that the space V has the orthogonal basis v1, v2, v3, v4 with Q(vi) = qi ∈k∗. Then the center Z of the even Cliﬀord algebra C0 V is k + kδ, and we have C0 V = Z + Zv1v2 + Zv2v3 + Zv1v3 . Since (v1v2)2 = −q1q2, (v2v3)2 = −q2q3, and (v1v2)(v2v3) = q2(v1v3), the algebra C0 V is isomorphic to the quaternion algebra (−q1q2, −q2q3) over Z. The conjugation in the quaternion algebra is identiﬁed with the main involution of the Cliﬀord algebra, and the norm with the Cliﬀord norm. 2.3 The Spin Group As before, let R be a commutative ring with unity 1, and let (V, Q) be a ﬁnitely generated quadratic module over R. The Cliﬀord group CGV of V is deﬁned by CGV = {x ∈CV ; x invertible and xV J(x)−1 = V } . (2.11) By deﬁnition, every x ∈CGV deﬁnes an automorphism αx of V by αx(v) = xvJ(x)−1 . We obtain a linear representation α : CGV →AutR(V ), x →αx, called the vector representation. It is easily seen that the involution x →xt takes CGV to itself. Consequently, if x ∈CGV , then the Cliﬀord norm N(x) belongs to CGV as well. Lemma 2.11. Assume that R = k is a ﬁeld of characteristic ̸= 2. The kernel of the vector representation α : CGV →Autk(V ) is equal to k∗. The Cliﬀord norm induces a homomorphism CGV →k∗. Proof. It is clear that k∗is contained in ker(α). We now show that ker(α) ⊂k∗. Let x ∈ker(α) ⊂CGV . We write x = x0 + x1 with x0 ∈C0 V and x1 ∈C1 V . Then we have xvJ(x)−1 = v for all v ∈V . Hence x0v = vx0 , x1v = −vx1 . Since V generates CV as an algebra, we see that x0 ∈Z(CV )∗∩C0 V = k∗. Moreover, one can check that the second condition implies that x1 = 0. 134 J. H. Bruinier We know that N(x) acts trivially on V via α for x ∈CGV . Let v ∈V . Since w := αx(v) ∈V , we have w = −J(w)t. This implies xvJ(x)−1 = (xt)−1vJ(xt) , and therefore N(x)vJ(N(x))−1 = v. Hence N(x) ∈k∗. The multiplicativity of the norm is a direct consequence. Lemma 2.12. For x ∈CGV the automorphism αx ∈AutR(V ) is an isome-try. Proof. Let v ∈V . Since w = αx(v) ∈V , we have Q(w) = N(w) = J(x−1)tvtxtxvJ(x−1) = Q(v) . This shows that αx is an isometry. □ Consequently, the vector representation deﬁnes a homomorphism α : CGV →OV . (2.12) Moreover, if x ∈CGV ∩V , then Q(x) ∈R∗and αx is equal to the reﬂection τx in the hyperplane x⊥. Deﬁnition 2.13. We deﬁne the general Spin group GSpinV and the Spin group SpinV of V by GSpinV = CGV ∩C0 V , SpinV = {x ∈GSpinV ; N(x) = 1} . For the rest of this section we assume that R = k is a ﬁeld of characteristic ̸=2. We brieﬂy discuss the structure of the Cliﬀord and the Spin group. In this case, by Theorem 2.1, the vector representation (2.12) is surjective onto OV . Moreover, the kernel is given by k∗(see Lemma 2.11). Hence CGV and GSpinV are central extensions of OV and SOV , respectively, 1 k∗ CGV OV 1 , 1 k∗ GSpinV SOV 1 . According to Lemma 2.11 and Theorem 2.6, the Cliﬀord norm deﬁnes a homomorphism CGV →k∗. It induces a homomorphism θ : OV − →k∗/(k∗)2 , (2.13) called the spinor norm. It is characterized by the property that for the reﬂec-tion τv corresponding to an anisotropic vector v ∈V we have Hilbert Modular Forms 135 θ(τv) = Q(v) ∈k∗/(k∗)2 . We obtain the exact sequence 1 {±1} SpinV α SOV θ k∗/(k∗)2 . The groups CGV , GSpin, and SpinV can be viewed as the groups of k-valued points of an aﬃne algebraic group over k. If A is a commutative k-algebra with unity, then the group of A-valued points of CGV is CGV (A) = CGV (A), and analogously for the other groups. The following lemma will be useful in Section 2.7. Lemma 2.14. Assume that dim(V ) ≤4. Then GSpinV = {x ∈C0 V ; N(x) ∈k∗} , SpinV = {x ∈C0 V ; N(x) = 1} . Proof. It is clear that the left hand sides are contained in the right hand sides. Conversely, let x ∈C0 V with N(x) ∈k∗. Then x is invertible, because y = xt N(x)−1 ∈C0 V is inverse to x. Hence, it suﬃces to show that xV x−1 ⊂V . Let v ∈V . It is clear that w := xvx−1 ∈C1 V . The assumption dim(V ) ≤4 implies that V = {g ∈C1 V ; gt = g} . Therefore it suﬃces to show that w = wt. Since N(x) ∈k∗, we have N(x)v N(x)−1 = v. This implies that xvx−1 = (xt)−1vxt and therefore w = wt. □ Quadratic Spaces in Dimension Four We now consider the special cases that (V, Q) is a rational quadratic space of dimension 4 over the ﬁeld k. Let v1, v2, v3, v4 be an orthogonal basis of V and put qi = Q(vi) ∈k∗. By means of Example 2.10 and Lemma 2.14, we see that SpinV is the group of norm 1 elements in the quaternion algebra (−q1q2, −q2q3) over Z := Z(C0 V ) = k + kδ, where δ = v1v2v3v4. We would like to describe the vector representation of SpinV intrinsically in terms of C0 V . This can be done by identifying V with an isometric copy ˜ V in C0 V . (Note that by deﬁnition V ̸⊂C0 V .) The vector representation on V translates to a “twisted” vector representation on ˜ V . We partly follow [KR] §0. Lemma 2.15. Let v0 ∈V with q0 = Q(v0) ̸= 0, and denote by σ = Ad(v0) the adjoint automorphism of C0 V associated to v0, i.e., xσ = v0xv−1 0 for x ∈C0 V . Then δσ = −δ and the ﬁxed algebra of σ in C0 V is a quaternion algebra B0 over k such that C0 V = B0 ⊗k Z. 136 J. H. Bruinier Proof. See [KR] Lemma 0.2. □ In particular, on the center Z of C0 V , the automorphism σ agrees with the conjugation in Z/k. Let ˜ V = {x ∈C0 V ; xt = xσ} . (2.14) This is a quadratic space over k together with the quadratic form ˜ Q(x) = q0 · xσx = q0 · N(x) . (2.15) The group SpinV acts on ˜ V by x →˜ αg(x) := gxg−σ , (2.16) for x ∈˜ V and g ∈SpinV . The quadratic form ˜ Q is preserved under this action: ˜ Q(gxg−σ) = q0 · (gxg−σ)t(gxg−σ) = q0 · xtx = ˜ Q(x) . (2.17) Lemma 2.16. The assignment x →x · v0 deﬁnes an isometry of quadratic spaces ( ˜ V , ˜ Q) − →(V, Q) , which is compatible with the actions of SpinV . Proof. See [KR] Lemma 0.3. □ 2.4 Rational Quadratic Spaces of Type (2, n) Let (V, Q) be a non-degenerate quadratic space over Q. Then V (R) = V ⊗Q R is isometric to Rp,q for a unique pair of non-negative integers (p, q), called the type of V . If K ⊂OV (R) is a maximal compact subgroup, then OV (R)/K is a symmetric space. It is hermitian, i.e., has a complex structure, if and only if p = 2 or q = 2. Since this is the case of interest to us, throughout this subsection we assume that V has type (2, n). We discuss several realizations of the corresponding hermitian symmetric domain. We frequently write (·, ·) for the bilinear form B(·, ·). The Grassmannian Model We consider the Grassmannian of 2-dimensional subspaces of V (R) on which the quadratic form is positive deﬁnite, Gr(V ) = {v ⊂V (R); dim v = 2 and Q\|v > 0} . By Witt’s theorem, OV (R) acts transitively on Gr(V ). If v0 ∈Gr(V ) is ﬁxed, then the stabilizer K of v0 is a maximal compact subgroup of OV (R), and K ∼ = O(2) × O(n). Thus Gr(V ) ∼ = OV (R)/K is a realization of the hermitian symmetric space. The Grassmannian model has the advantage that it provides an easy description of OV (R)/K, but unfortunately we do not see the complex structure. Hilbert Modular Forms 137 The Projective Model We consider the complexiﬁcation V (C) of V and the corresponding projective space P(V (C)) = (V (C){0})/C∗. (2.18) The zero quadric N = {[Z] ∈P(V (C)); (Z, Z) = 0} (2.19) is a closed algebraic subvariety. The subset K = {[Z] ∈P(V (C)); (Z, Z) = 0 , (Z, ¯ Z) > 0} (2.20) of the zero quadric is a complex manifold of dimension n consisting of two connected components. The orthogonal group OV (R) acts transitively on K. The subgroup O+ V (R) of elements whose spinor norm equals the determinant preserves the components of K, whereas OV (R)\ O+ V (R) interchanges them. We choose one ﬁxed component of K and denote it by K+. If Z ∈V (C) we write Z = X + iY with X, Y ∈V (R) for the decomposition into real and imaginary part. Lemma 2.17. The assignment [Z] →v(Z) := RX + RY deﬁnes a real ana-lytic isomorphism K+ →Gr(V ). Proof. If Z ∈V (C), then the condition [Z] ∈K is equivalent to X ⊥Y, and (X, X) = (Y, Y ) > 0 . (2.21) But this means that X and Y span a two dimensional positive deﬁnite sub-space of V (R) and thereby deﬁne an element of Gr(V ). Conversely for a given v ∈Gr(V ) we may choose a (suitably oriented) orthogonal basis X, Y as in (2.21) and obtain a unique [Z] = [X + iY ] ∈K+. (Then [ ¯ Z] ∈K corresponds to the same v ∈Gr(V ).) We get a real analytic isomorphism between K+ and Gr(V ). □ The advantage of the projective model is that it comes with a natural complex structure. However, it is not the direct analogue of the upper half plane, the standard model for the hermitian symmetric space for SL2(R). The Tube Domain Model We may realize K+ as a tube domain in the following way. Suppose that e1 ∈V is a non-zero isotropic vector and e2 ∈V with (e1, e2) = 1. We deﬁne 138 J. H. Bruinier a rational quadratic subspace W ⊂V by W = V ∩e⊥ 1 ∩e⊥ 2 . Then W is Lorentzian, that is, has type (1, n −1) and V = W ⊕Qe2 ⊕Qe1 . If Z ∈V (C) and Z = z + ae2 + be1 with z ∈W(C) and a, b ∈C, we brieﬂy write Z = (z, a, b). We consider the tube domain H = {z ∈W(C); Q(I(z)) > 0} . (2.22) Lemma 2.18. The assignment z →ψ(z) := [(z, 1, −Q(z) −Q(e2))] (2.23) deﬁnes a biholomorphic map ψ : H →K. Proof. One easily checks that if z ∈H then ψ(z) ∈K. Conversely assume that [Z] ∈K with Z = X + iY . From the fact that X, Y span a two dimensional positive deﬁnite subspace of V (R) it follows that (Z, e1) ̸= 0. Thus [Z] has a unique representative of the form (z, 1, b). The condition Q(Z) = 0 implies that b = −Q(z) −Q(e2), and thereby [Z] = [(z, 1, −Q(z) −Q(e2))]. Moreover, from (Z, ¯ Z) > 0 one easily deduces Q(I(z)) > 0. We may infer that the map ψ is biholomorphic. □ The domain H ⊂W(C) ∼ = Cn has two components corresponding to the two cones of positive norm vectors in the Lorentzian space W(R). We denote by H+ the component which is mapped to K+ under the above isomorphism. It can be viewed as a generalized upper half plane. The group O+ V (R) acts transitively on it. In the O(2, 1) case the domain H+ can be identiﬁed with the usual upper half plane H. For O(2, 2) it can be identiﬁed with H2 as we shall see below. However, a disadvantage of the tube domain model is that the action of O+ V (R) is not linear anymore. Lattices As before, let (V, Q) be a non-degenerate quadratic space over Q of type (2, n). Deﬁnition 2.19. A lattice in V is a Z-module L ⊂V such that V = L ⊗Z Q. A lattice L ⊂V is called integral if the bilinear form is integral valued on L, that is, (x, y) ∈Z for all x, y ∈L. A lattice is called even if the quadratic form is integral valued on L, that is, Q(x) ∈Z for all x ∈L. So an even lattice is a free quadratic module over Z of ﬁnite rank. Clearly every even lattice is integral. The dual lattice L∨is deﬁned by L∨= {x ∈V ; (x, y) ∈Z for all y ∈L} . Hilbert Modular Forms 139 The lattice L is integral if and only if L ⊂L∨. In this case the quotient L∨/L is a ﬁnite abelian group, called the discriminant group. If S is the Gram matrix corresponding to a lattice basis of L, we have \|L∨/L\| = \| det(S)\| . For the rest of this section we assume that L ⊂V is an even lattice. The orthogonal group OL is a discrete subgroup of OV (R) ∼ = O(2, n). Let Γ ⊂OL ∩O+ V (R) (2.24) be a subgroup of ﬁnite index. Then Γ acts properly discontinuously on Gr(V ), K+, and H+. We consider the quotient Y (Γ) = Γ\H+ (2.25) similarly as in the construction of Hilbert modular surfaces in Section 1.1. It is a normal complex space, which is compact if and only if V is anisotropic. If Y (Γ) is non-compact, it can be compactiﬁed by adding rational bound-ary components (see e.g. [BrFr]). These boundary components are most easily described in the projective model K+. The boundary points of K+ in the zero quadric N correspond to non-trivial isotropic subspaces of V (R). Let F ⊂V (R) be an isotropic line. Then F represents a boundary point of K+. A boundary point of this type is called special, otherwise generic. A set consisting of one special boundary point is called a zero-dimensional boundary component. Let F ⊂V (R) be a two-dimensional isotropic subspace. The set of all generic boundary points of K+ which can be represented by an element of F(C) is called the one-dimensional boundary component attached to F. By a boundary component we understand a one- or two-dimensional boundary component. One can show (see [BrFr], Section 2): Lemma 2.20. There is a bijective correspondence between boundary compo-nents of K+ in the zero quadric N and non-zero isotropic subspaces F ⊂V (R) of the corresponding dimension. The boundary of K+ is the disjoint union of the boundary components. □ A boundary component is called rational if the corresponding isotropic space F is deﬁned over Q. The union of K+ with all rational boundary com-ponents is denoted by (K+)∗. The rational orthogonal group OV (Q) ∩O+ V (R) acts on (K+)∗. By the theory of Baily–Borel, the quotient X(Γ) = (K+)∗/Γ together with the Baily–Borel topology is a compact Hausdorﬀspace. There is a natural complex structure on X(Γ) as a normal complex space. More-over, using modular forms, one can construct an ample line bundle on X(Γ). Therefore, X(Γ) is projective algebraic. It is called the modular variety asso-ciated to Γ. Using the theory of canonical models, one can show that X(Γ) is actually deﬁned over a number ﬁeld (see [Ku2]). 140 J. H. Bruinier Heegner Divisors Let Γ be as above, see (2.24). In order to understand the geometry of X(Γ), we study special divisors on this variety, obtained from embeddings of modular varieties corresponding to quadratic subspaces of V . Let λ ∈L∨with Q(λ) < 0. Then the orthogonal complement Vλ = λ⊥⊂V is a rational quadratic space of type (2, n −1). Moreover, the orthogonal complement of λ in K+, Hλ = {[Z] ∈K+; (Z, λ) = 0} , is an analytic divisor. It is the hermitian symmetric domain corresponding to (Vλ, Q\|Vλ). Let us brieﬂy look at the description of Hλ in the tube domain model H+ using the above notation. We write λ = λW + ae2 + be1 with λW ∈W and a, b ∈Q. Then Hλ ∼ = {z ∈H+; aQ(z) −(z, λW ) −aq(e2) −b = 0} is given by a quadratic equation in the coordinates of H+. (Therefore it is sometimes called a rational quadratic divisor.) If β ∈L∨/L is ﬁxed and m is a ﬁxed negative rational number, then H(β, m) = λ∈β+L Q(λ)=m Hλ (2.26) deﬁnes an analytic divisor on K+ called the Heegner divisor of discriminant (β, m). If Γ acts trivially on L∨/L, then, by Chow’s lemma, this divisor de-scends to an algebraic divisor on Y (Γ) (denoted in the same way). By [Ku2], it is deﬁned over a number ﬁeld. Here we mainly consider the composite Heegner divisor H(m) = 1 2 β∈L∨/L H(β, m) = λ∈L∨/{±1} Q(λ)=m Hλ . (2.27) It is Γ-invariant and descends to an algebraic divisor on Y (Γ). Hence, Y (Γ) comes with a natural family of algebraic divisors indexed by negative rational numbers (with denominators bounded by the level of L). The existence of such a family is special for orthogonal and unitary groups. 2.5 Modular Forms for O(2, n) Let V , L, Γ be as above. We write K+ = {Z ∈V (C){0}; [Z] ∈K+} for the cone over K+. Hilbert Modular Forms 141 Deﬁnition 2.21. Let k ∈Z, and let χ be character of Γ. A meromorphic function F on K+ is called a meromorphic modular form of weight k and character χ for the group Γ, if (i) F is homogeneous of degree −k, i.e., F(cZ) = c−kF(Z) for any c ∈C{0}; (ii) F is invariant under Γ, i.e., F(gZ) = χ(g)F(Z) for any g ∈Γ; (iii) F is meromorphic at the boundary. If f is actually holomorphic on K+ and at the boundary, it is called a holo-morphic modular form. By the Koecher principle the boundary condition is automatically fulﬁlled if the Witt rank of V , that is, the dimension of a maximal isotropic subspace, is smaller than n. (Note that because of the signature the Witt rank of L is always ≤2.) 2.6 The Siegel Theta Function Examples of modular forms on orthogonal groups can be constructed using Eisenstein series similarly as in Section 1.5. However, we do not discuss this. Here we consider a rather diﬀerent source of modular forms, the so called theta lifting. The groups SL2(R) and O(2, n) form a dual reductive pair in the sense of Howe [Ho]. Hence, Howe duality gives rise to a correspondence between automorphic representations for the two groups. Often one can re-alize this correspondence as a lifting from automorphic forms on one group to the other, by integrating against certain kernel functions given by theta functions. Let V , L, Γ be as above and assume that n is even so that dim(V ) is even. Let N = min{a ∈Z>0 ; aQ(λ) ∈Z for all λ ∈L∨} be the level of L. We modify the discriminant of L by a sign and consider Δ = (−1) n+2 2 det S , where S is the Gram matrix for a lattice basis of L. One can show that Δ ≡1, 0 (mod 4). Therefore χΔ = Δ · is a quadratic Dirichlet character modulo N. For λ ∈V (R) and v ∈Gr(V ) we have a unique decomposition λ = λv + λv⊥, where λv and λv⊥are the orthogonal projections of λ to v and v⊥, respectively. The positive deﬁnite quadratic form Qv(λ) = Q(λv) −Q(λv⊥) on V is called the majorant associated to v. If Z ∈ K+, we brieﬂy write λZ and QZ instead of λv(Z) and Qv(Z), where v(Z) is the positive deﬁnite plane corresponding to Z via Lemma 2.17. 142 J. H. Bruinier Deﬁnition 2.22. Let r ∈Z≥0. The Siegel theta function of weight r of the lattice L is deﬁned by Θr(τ, Z) = vn/2 λ∈L∨ (λ, Z)r (Z, ¯ Z)r e Q(λZ)Nτ + Q(λZ⊥)N ¯ τ = vn/2 λ∈L∨ (λ, Z)r (Z, ¯ Z)r e Q(λ)Nu + QZ(λ)Niv , for τ = u + iv ∈H and Z ∈ K+ (see e.g. [Bo4], [Od1], [RS]). Here we denote e(w) = e2πiw as usual. Because of the rapid decay of the exponential term e(QZ(λ)Niv), the series converges normally on H× K+. It deﬁnes a real analytic function, which is non-holomorphic in both variables, τ and Z. Using the Poisson summation formula, or the theory of the Weil representation, one can show that as a function in τ, the Siegel theta function satisﬁes Θr(γτ, Z) = χΔ(d)(cτ + d)r+ 2−n 2 Θr(τ, Z) (2.28) for all γ = a b c d ∈Γ0(N), where Γ0(N) = $a b c d ∈SL2(Z); c ≡0 (mod N) % . (2.29) Moreover, in the variable Z, the function Θr(τ, Z) transforms as a modular form of weight r for Γ. This follows by direct inspection. We may use the Siegel theta function as an integral kernel to lift elliptic modular forms for Γ0(N) to modular forms on the orthogonal group. More precisely, let f ∈Sk(Γ0(N), χΔ) be a cusp form for Γ0(N) with character χΔ of weight k = r + 2−n 2 . We deﬁne the theta lift Φ0(Z, f) of f by the integral Φ0(Z, f) = F f(τ)Θr(τ, Z)vk du dv v2 , (2.30) where F denotes a fundamental domain for Γ0(N). Theorem 2.23. The theta lift Φ0(Z, f) of f is a holomorphic modular form of weight r = k −2−n 2 for the orthogonal group Γ. Proof. The transformation properties of the Siegel theta function immediately imply that Φ0(Z, f) transforms as a modular form of weight r for the group Γ. However, it is not clear at all, that Φ0(Z, f) is holomorphic. This can be proved by computing the Fourier expansion. For details we refer to e.g. [Bo4] Theorem 14.3, [Od1] Section 5, Theorem 2, or [RS]. □ Remark 2.24. The linear map f →Φ0(Z, f) often has a non-trivial kernel. The question when it vanishes is related to the vanishing of a special value of the standard L-function of f [Ral]. Therefore it can be rather diﬃcult. However, in many cases it is also possible to obtain non-vanishing results by computing the Fourier expansion of the lift. Hilbert Modular Forms 143 2.7 The Hilbert Modular Group as an Orthogonal Group In this section we discuss the accidental isomorphism relating the Hilbert modular group to an orthogonal group of type (2, 2) in detail. The Heegner divisors of the previous section give rise to certain algebraic divisors on Hilbert modular surfaces, known as Hirzebruch–Zagier divisors [HZ]. Let d ∈Q∗be not a square, and put F = Q( √ d). We consider the four dimensional Q-vector space V = Q ⊕Q ⊕F together with the quadratic form Q(a, b, ν) = νν′ −ab, where ν →ν′ denotes the conjugation in F. So (V, Q) is a rational quadratic space of type (2, 2) if d > 0 and of type (3, 1) if d < 0. We consider the orthogonal basis v1 = (1, 1, 0), v3 = (0, 0, 1), v2 = (1, −1, 0), v4 = (0, 0, √ d) . Then δ = v1v2v3v4 satisﬁes δ2 = d. According to Remark 2.5 and Theorem 2.6, the center Z(C0 V ) of the even Cliﬀord algebra of V is given by Z(C0 V ) = Q + Qδ ∼ = F. Moreover, in view of Example 2.10, C0 V = Z + Zv1v2 + Zv2v3 + Zv1v3 is isomorphic to the split quaternion algebra M2(F) over F. This isomorphism can be realized by the assignment 1 → 1 0 0 1 , v2v3 → 0 1 −1 0 , v1v2 → 1 0 0 −1 , v1v3 → 0 1 1 0 . The canonical involution on C0 V corresponds to the conjugation a b c d ∗ = d −b −c a in M2(F). The Cliﬀord norm corresponds to the determinant. Hence, by Lemma 2.14, SpinV can be identiﬁed with SL2(F). As algebraic groups over Q we have SpinV ∼ = ResF/Q SL2. Consequently, the group ΓF = SL2(OF ) and commensurable groups can be viewed as arithmetic subgroups of SpinV . For instance, using (2.32) below, it is easily seen that ΓF = SpinL, where L de-notes the lattice Z ⊕Z ⊕OF ⊂V . We now describe the vector representation explicitly using Lemmas 2.15 and 2.16. Let σ = Ad(v1) be the adjoint automorphism of C0 V associated to the basis vector v1, i.e., xσ = v1xv−1 1 for x ∈C0 V . Then δσ = −δ, and on the 144 J. H. Bruinier center F of C0 V , the automorphism σ agrees with the conjugation in F/Q. On M2(F) the action of σ is given by a b c d → a b c d σ = d′ −c′ −b′ a′ . As in (2.14) let ˜ V = {X ∈M2(F); X∗= Xσ} = X ∈M2(F); Xt = X′ = $a ν′ ν b ; a, b ∈Q and ν ∈F % . This is a rational quadratic space together with the quadratic form ˜ Q(X) = −Xσ · X = −det(X) . The corresponding bilinear form is ˜ B(X1, X2) = −tr(X1 · X∗ 2) , for X1, X2 ∈˜ V . The group SL2(F) ∼ = SpinV acts isometrically on ˜ V by x →g.X := gXg−σ = gXg′t , (2.31) for X ∈˜ V and g ∈SL2(F). A computation shows that in the present case the isometry of quadratic spaces ˜ V →V , X →X · v1, of Lemma 2.16 is given by a ν′ ν b →(a, b, ν) . (2.32) Throughout the rest of this section we work with ˜ V and the twisted vector representation (2.31). We assume that d is positive so that F is real quadratic. We now describe the hermitian symmetric space corresponding to O ˜ V as in Section 2.4. The two real embeddings x →(x, x′) ∈R2 induce an embedding ˜ V → M2(R). Hence we have ˜ V (C) = M2(C) and K = {[Z] ∈P(M2(C)); det(Z) = 0, −tr(Z ¯ Z∗) > 0} . We consider the isotropic vectors e1 = −1 0 0 0 and e2 = ( 0 0 0 1 ) in ˜ V , and the orthogonal complement W = ˜ V ∩e⊥ 1 ∩e⊥ 2 . For z = (z1, z2) ∈C2 ∼ = W(C) we put M(z) = z1z2 z1 z2 1 ∈M2(C) . Hilbert Modular Forms 145 Then [M(z)] lies in the zero quadric in P(M2(C)), and [M(z)] ∈K if and only if I(z1)I(z2) > 0. Consequently, we may identify H2 with H+. If we denote by K+ the corresponding component of K, we obtain a biholomorphic map H2 − →K+, z →[M(z)] . (2.33) It commutes with the actions of SL2(F), where the action on K+ is given by (2.31). More precisely, in the cone K+ we have γ.M(z) = N(cz + d)M(γz) (2.34) for γ = a b c d ∈SL2(F). This implies that modular forms of weight k in the sense of Deﬁnition 2.21 can be identiﬁed with Hilbert modular forms of parallel weight k in the sense of Deﬁnition 1.19. We consider in V the lattice L = Z ⊕Z ⊕OF ∼ = $ a ν′ ν b ∈˜ V ; a, b ∈Z and ν ∈OF % . (2.35) The dual lattice of L is L∨= Z ⊕Z ⊕d−1 F ∼ = $a ν′ ν b ∈˜ V ; a, b ∈Z and ν ∈d−1 F % . (2.36) The discriminant group is given by L∨/L ∼ = OF /dF . Proposition 2.25. Under the isomorphism SpinV ∼ = SL2(F), the subgroup SpinL is identiﬁed with ΓF . □ The map (2.33) induces an isomorphism of modular varieties Y (ΓF ) → Y (SpinL). Remark 2.26. More generally, let a be a fractional ideal of F and put A = N(a). We may consider the lattices L(a) = $a ν′ ν Ab ∈˜ V ; a, b ∈Z and ν ∈a % , L∨(a) = $ a ν′ ν Ab ∈˜ V ; a, b ∈Z and ν ∈ad−1 F % . Observe that L(a) is A-integral (that is, the bilinear form has values in AZ), and L∨(a) is the AZ-dual of L(a). The group Γ(OF ⊕a) ⊂SL2(F) deﬁned in (1.5) preserves these lattices. Hirzebruch–Zagier Divisors In view of the above discussion, the construction of Heegner divisors provides a natural family of algebraic divisors on a Hilbert modular surface, in this case known as Hirzebruch–Zagier divisors [HZ]. 146 J. H. Bruinier If A = a λ′ λ b ∈˜ V and z = (z1, z2) ∈H2, then (M(z), A) = −tr(M(z) · A∗) = −bz1z2 + λz1 + λ′z2 −a . The zero locus of the right hand side deﬁnes an analytic divisor on H2. Deﬁnition 2.27. Let m be a positive integer. The Hirzebruch–Zagier divisor Tm of discriminant m is deﬁned as the Heegner divisor H(−m/D) for the lattice L ⊂V , i.e., Tm = (a,b,λ)∈L∨/{±1} ab−λλ′=m/D (z1, z2) ∈H2; az1z2 + λz1 + λ′z2 + b = 0 . It deﬁnes an algebraic divisor on the Hilbert modular surface Y (ΓF ). Here the multiplicities of all irreducible components are 1. (There is no ramiﬁcation in codimension 1.) By taking the closure, we also obtain a divisor on X(ΓF ). We will denote these divisors by Tm as well, since it will be clear from the context where they are considered. It is well known that Tm is deﬁned over Q. Remark 2.28. When m is not a square modulo D, then Tm = ∅. Example 2.29. The divisor T1 on X(ΓF ) can be identiﬁed with the image of the modular curve X(1) = SL2(Z)\H under the diagonal embedding considered in Section 1.5. 3 Additive and Multiplicative Liftings Let F ⊂R be the real quadratic ﬁeld of discriminant D. Let (V, Q) be the corresponding rational quadratic space of type (2, 2) as in Section 2.7, and let L ⊂V be the even lattice (2.35). The corresponding Siegel theta function Θk(τ, z) in weight k is modular in both variables τ and z: As a function of τ, Θk(τ, z) is a non-holomorphic modular form of weight k for the group Γ0(D) with character χD = D · . As a function in z, Θk(τ, z) is a non-holomorphic modular form of weight k for the Hilbert modular group ΓF . For a cusp form f ∈Sk(D, χD) of weight k for Γ0(D) with character χD, we may consider the theta integral Φ0(z, f) as in (2.30). By means of Theorem 2.23 we ﬁnd that Φ0(z, f) deﬁnes a Hilbert cusp form of weight k for the group ΓF (which may vanish identically). Similar constructions can be done for the Hilbert modular groups Γ(OF ⊕a) and for their congruence subgroups. 3.1 The Doi–Naganuma Lift In the following we discuss the theta lift in more detail. To keep the exposition simple, we assume that D = p is a prime and F = Q(√p). We consider the full Hilbert modular group ΓF . Hilbert Modular Forms 147 For explicit computations it is convenient to modify the theta lifting a bit. Let Mk(p, χp) denote the space of holomorphic modular forms of weight k for Γ0(p) and χp. Since this space is trivial when k is odd, we assume that k is even. A function f ∈Mk(p, χp) has a Fourier expansion f(τ) = n≥0 c(n)qn , where q = e2πiτ as usual. We deﬁne the “plus” and “minus” subspace of Mk(p, χp) by M ± k (p, χp) = {f ∈Mk(p, χp); χp(n) = ∓1 ⇒c(n) = 0} , (3.1) and write S± k (p, χp) for the subspace of cusp forms. Examples of modular forms in M ± k (p, χp) can be constructed by means of Eisenstein series. Recall that there are the two Eisenstein series Gk(τ) = 1 + 2 L(1 −k, χp) ∞ n=1 d\|n dk−1χp(d)qn, (3.2) Hk(τ) = ∞ n=1 d\|n dk−1χp(n/d)qn (3.3) in Mk(p, χp) (cf. [He] Werke p. 818), the former corresponding to the cusp ∞, the latter corresponding to the cusp 0. The linear combination E± k = 1 + n≥1 B± k (n)qn = 1 + 2 L(1 −k, χp) n≥1 d\|n dk−1 (χp(d) ± χp(n/d)) qn (3.4) belongs to M ± k (p, χp). Proposition 3.30 (Hecke). The space Mk(p, χp) decomposes into the direct sum Mk(p, χp) = M + k (p, χp) ⊕M − k (p, χp) . Moreover, M ± k (p, χp) = CE± k ⊕S± k (p, χp) . □ Modular forms in the plus space behave in many ways similarly as modular forms on the full elliptic modular group SL2(Z). In fact, Theorem 5 of [BB] states that M ± k (p, χp) is isomorphic to the space of vector-valued modular forms of weight k for SL2(Z) transforming with the Weil representation of L∨/L. 148 J. H. Bruinier Notation 3.31. For a formal Laurent series c(n)qn ∈C((q)) we put ˜ c(n) = c(n), if p ∤n, 2c(n), if p \| n . (3.5) Proposition 3.32. Let f = c(n)qn ∈M ± k (p, χp) and g = b(n)qn ∈ M ± k′(p, χp). Then ⟨f, g⟩= n∈Z m∈Z ˜ c(m)b(pn −m)qn is a modular form of weight k + k′ for SL2(Z). The assignment (f, g) →⟨f, g⟩ deﬁnes a bilinear pairing. Proof. This can be proved by interpreting modular forms in the plus space as vector valued modular forms for SL2(Z), see [BB]. □ Remark 3.33. Proposition 3.32 implies some amusing identities of divisor sums arising from the equalities ⟨E+ k , E+ k ⟩= E2k for k = 2, 4. Here E2k denotes the Eisenstein series of weight 2k for SL2(Z) normalized such that the constant term is 1. Note that the statement of Proposition 3.32 does not depend on the holomorphicity of f. An analogous result holds for non-holomorphic modu-lar forms. For instance, the complex conjugate of the Siegel theta function Θk(τ, z) of the lattice L satisﬁes the plus space condition. This follows from Deﬁnition 2.22, since for (a, b, λ) ∈L∨we have −pQ(a, b, λ) = p(ab −λλ′) ≡□ (mod p) . Deﬁnition 3.34. For f ∈M + k (p, χp) we deﬁne the (modiﬁed) theta lift by the integral Φ(z, f) = SL2(Z)\H ⟨f(τ), Θk(τ, z)⟩vk du dv v2 . The integral converges absolutely if f is a cusp form. If f is not cuspidal, the integral has to be regularized (see [Bo4]). By computing the Fourier expan-sion of Φ(z, f), the following theorem can be proved (cf. [Za1], [Bo4] Theo-rem 14.3). Theorem 3.35. Let f = n c(n)qn ∈M + k (p, χp). The theta lift Φ(z, f) has the following properties. (i) Φ(z, f) is a Hilbert modular form of weight k for ΓF . Hilbert Modular Forms 149 (ii) It has the Fourier expansion Φ(z, f) = −Bk 2k ˜ c(0) + ν∈d−1 F ν≫0 d\|ν dk−1˜ c pνν′ d2 qν 1qν′ 2 , where Bk denotes the k-th Bernoulli number, and qj = e2πizj. (iii) The lift takes cusp forms to cusp forms. □ If we deﬁne in addition Φ(z, f) to be identically zero on M − k (p, χp), we obtain the Doi–Naganuma lift (see [DN], [Na]), DN : Mk(p, χp) − →Mk(ΓF ) . It is a fundamental fact that the Doi–Naganuma lift (and theta lifts in general) behave well with respect to the actions of the Hecke algebras. Theorem 3.36 (Doi–Naganuma, Zagier). The Doi–Naganuma lifting takes Hecke eigenforms to Hecke eigenforms. For a normalized Hecke eigen-form f = n c(n)qn ∈Mk(p, χp) we have L(DN(f), s) = L(f, s) · L(f ρ, s) , where L(f, s) denotes the Hecke L-function of f and f ρ = c(n)qn. □ Let Λ(f, s) = ps/2(2π)−sΓ(s)L(f, s) be the completed Hecke L-function of the eigenform f. It has the functional equation Λ(f, s) = C · Λ(f ρ, k −s) with a non-zero constant C ∈C. Therefore R(s) = ps (2π)−2s Γ(s)2L(f, s)L(f ρ, s) has the functional equation R(s) = R(k −s) , which looks as the functional equation of the L-function of a Hilbert modular form of weight k, see Theorem 1.45 in Section 1.6. Moreover, all further an-alytic properties of R(s) agree with those of L-functions of Hilbert modular forms. Hence, using a converse theorem (similar to Hecke’s converse theorem), one can infer that R(s) really comes from a Hilbert modular form. Originally, this argument led Doi and Naganuma to the discovery of the lifting. Using the converse theorem argument they were able to prove the existence of the lifting in the few cases where OF is euclidian. Employing a later result of Vaserstein (see [Ge1] Chapter IV.6) on generators of Hilbert modular groups, the proof can be generalized. The theta lifting approach came up later, and was suggested by Eichler and Shintani and worked out by Kudla, Oda, Vignerás, and others. 150 J. H. Bruinier 3.2 Borcherds Products Here we consider the Borcherds lift for Hilbert modular surfaces. It can be viewed as a multiplicative analogue of the Doi–Naganuma lift. It takes cer-tain weakly holomorphic elliptic modular forms of weight 0 to meromorphic Hilbert modular forms which have an inﬁnite product expansion resembling the Dedekind eta function. The zeros and poles of such Borcherds products are supported on Hirzebruch–Zagier divisors. Local Borcherds Products As a warm up, we study a local analogue of Borcherds products at the cusps of Hilbert modular surfaces. This is a special case of the more general results for O(2, n) of [BrFr]. We return to the setup of Section 1.1. In particular, F ⊂R is a real quadratic ﬁeld of discriminant D and ΓF = SL2(OF ) denotes the Hilbert modular group. We ask whether the Hirzebruch–Zagier divisors Tm on X(ΓF ) are Q-Cartier. Since the non-compact Hilbert modular surface Y (ΓF ) is non-singular except for the ﬁnite quotient singularities corresponding to the elliptic ﬁxed points, it is clear that Tm is Q-Cartier on Y (ΓF ). We only have to investigate the behavior at the cusps. Lemma 3.37. Let A = (a, b, λ) ∈L∨with ab −λλ′ > 0. The closure of the image in YF of (z1, z2) ∈H2; az1z2 + λz1 + λ′z2 + b = 0 goes through the cusp ∞if and only if a = 0. Proof. This is an immediate consequence of Proposition 1.7 (iii). □ Let m be a positive integer. We deﬁne the local Hirzebruch–Zagier divisor at ∞of discriminant m by T ∞ m = λ∈d−1 F /{±1} −λλ′=m/D b∈Z (z1, z2) ∈H2; λz1 + λ′z2 + b = 0 ⊂H2 . This divisor is invariant under the stabilizer ΓF,∞of ∞. Theorem 3.38. The Hirzebruch–Zagier divisor Tm on X(ΓF ) is Q-Cartier. Proof. We have to investigate the behavior at the cusps. Here we only consider the cusp ∞, the other cusps can be treated in the same way. We have to show that there is a small open neighborhood U ⊂X(ΓF ) of ∞and a holomorphic function f on U such that div(f) = r · Tm\|U ∈Div(U) Hilbert Modular Forms 151 for some positive integer r. Here Tm\|U denotes the restriction of Tm to U. In view of Proposition 1.10 and Lemma 3.37 it suﬃces to show that there exists a ΓF,∞-invariant holomorphic function ˜ f : H2 →C such that div( ˜ f) = r ·T ∞ m . This follows from Proposition 3.40 below. □ Remark 3.39. The statement of Theorem 3.38 does not generalize to Heegner divisors on O(2, n). For instance, for n > 3 there are obstructions to the Q-Cartier property at generic boundary points, which are related to theta series of even deﬁnite lattices of rank n −2 with harmonic polynomials of degree 2. (See [BrFr], [Lo].) The local Hirzebruch–Zagier divisor T ∞ m decomposes as a sum T ∞ m = λ∈d−1 F /O∗,2 F −λλ′=m/D λ>0 T ∞ λ , (3.6) where O∗,2 F denotes the subgroup of squares in the unit group O∗ F , and T ∞ λ = u∈O∗,2 F b∈Z (z1, z2) ∈H2; λuz1 + λ′u′z2 + b = 0 . (3.7) The divisor Tλ is invariant under ΓF,∞. In the following, we construct a holo-morphic function on H2/ΓF,∞whose divisor is T ∞ λ , using local Borcherds products [BrFr]. We start by introducing some notation. The subset S(m) = & λ∈d−1 F −λλ′=m/D {y ∈(R>0)2; λy1 + λ′y2 = 0} (3.8) of (R>0)2 is a union of hyperplanes. It is invariant under ΓF,∞. The comple-ment (R>0)2 \ S(m) is not connected. The connected components are called the Weyl chambers (of d−1 F ) of index m. Let W be a subset of a Weyl chamber of index m and λ ∈d−1 F with −λλ′ = m/D. Then λ is called positive with respect to W, if tr(λw) > 0 for all w ∈W (which is equivalent to requiring tr(λw0) > 0 for some w0 ∈W). In this case we write (λ, W) > 0 . Moreover, λ is called reduced with respect to W, if (uλ, W) < 0, and (λ, W) > 0 , for all u ∈O∗,2 F with u < 1. This condition is equivalent to (ε−2 0 λ, W) < 0, and (λ, W) > 0 . 152 J. H. Bruinier It implies that λ > 0. We denote by R(m, W) the set of all λ ∈d−1 F with −λλ′ = m/D which are reduced with respect to W. (Note that this deﬁnition slightly diﬀers from the one in [BB].) It is a ﬁnite set and {λ ∈d−1 F ; −λλ′ = m/D} = {±λu; λ ∈R(m, W) and u ∈O∗,2 F } . (3.9) Let W be a subset of a Weyl chamber of index m and λ ∈d−1 F with −λλ′ = m/D. We deﬁne a holomorphic function ψ∞ λ : H2 →C by ψ∞ λ (z) = u∈O∗,2 F ! 1 −e(σu tr(uλz)) " , where σu = +1, if (uλ, W) > 0 , −1, if (uλ, W) < 0 . The sign σu has to be inserted to obtain a convergent inﬁnite product. By construction we have ψ∞ λ = ψ∞ −λ and div(ψ∞ λ ) = T ∞ λ . Moreover, the product is invariant under translations 1 μ 0 1 ∈ΓF,∞. However, ψ∞ λ is not invariant under the full stabilizer of ∞. It deﬁnes an automorphy factor J(γ, z) = ψ∞ λ (γz)/ψ∞ λ (z) (3.10) of ΓF,∞acting on H2, that is, an element of H1(ΓF,∞, O(H2)∗). We need to show that this automorphy factor is trivial up to torsion. It suﬃces to con-sider what happens under the generator ε0 0 0 ε−1 0 of the subgroup of diagonal matrices in ΓF,∞. We have ψ∞ λ (ε2 0z) ψ∞ λ (z) = u∈O∗,2 F 1 −e(σu/ε2 0 tr(uλz)) 1 −e(σu tr(uλz)) . In this product only one factor is not equal to 1. If we assume that λ is reduced with respect to W, we obtain ψ∞ λ (ε2 0z) ψ∞ λ (z) = 1 −e(−tr(λz)) 1 −e(tr(λz)) = e(1/2 −tr(λz)) . On the other hand, we consider the invertible holomorphic function Iλ(z) = e tr λ ε2 0 −1z Hilbert Modular Forms 153 on H2. It satisﬁes Iλ(ε2 0z) Iλ(z) = e(tr(λz)) . Moreover, Iλ(z + μ) = Iλ(z) for all μ ∈(ε2 0 −1)OF . Therefore, up to torsion, the automorphy factor J(γ, z) in (3.10) can be trivialized with Iλ(z). The function Ψ ∞ λ (z) = Iλ(z) · ψ∞ λ (z) = e tr λ ε2 0 −1z u∈O∗,2 F ! 1 −e(σu tr(uλz)) " (3.11) is holomorphic on H2, has divisor T ∞ λ , and a power of it is invariant un-der ΓF,∞. Observe that Ψ ∞ λ does not depend on the choice of the Weyl chamber W, although the factors Iλ and ψ∞ λ do. Now it is easy to construct an analogous function for T ∞ m . We deﬁne the Weyl vector of index m for the Weyl chamber W by ρm,W = λ∈R(m,W) λ ε2 0 −1 . (3.12) Moreover, we deﬁne the local Borcherds product for T ∞ m by Ψ ∞ m (z) = λ∈d−1 F /O∗,2 F −λλ′=m/D λ>0 Ψ ∞ λ (z) = e tr(ρm,W z) λ∈d−1 F −λλ′=m/D (λ,W)>0 ! 1 −e(tr(λz)) " . (3.13) Proposition 3.40. The divisor of Ψ ∞ m is equal to T ∞ m . A power of Ψ ∞ m is invariant under ΓF,∞. □ Example 3.41. We compute Ψ ∞ 1 more explicitly. The point (1, ε0) ∈(R>0)2 does not belong to S(1). Hence it lies in a unique Weyl chamber W of index 1. The set of λ ∈d−1 F with −λλ′ = 1/D which are reduced with respect to W is given by R(1, W) = {ε2 0/ √ D}, if ε0ε′ 0 = −1 , {ε0/ √ D, ε2 0/ √ D}, if ε0ε′ 0 = +1 . The corresponding Weyl vector is equal to ρ1,W = ⎧ ⎨ ⎩ ε0 √ D 1 tr(ε0), if εε′ 0 = −1 , 1+ε0 tr( √ Dε0), if εε′ 0 = +1 . 154 J. H. Bruinier In the case εε′ 0 = −1, the point (ε−1 0 , ε0) lies in the same Weyl chamber W. It is often more convenient to work with this base point. If εε′ 0 = 1, then (ε−1 0 , ε0) ∈S(1). The Borcherds Lift For the material of the next two sections we also refer to [Br3]. The Doi– Naganuma lift of the Section 3.1 only deﬁnes a non-trivial map when k > 0. (For k = 0 we have Mk(D, χD) = 0.) It is natural to ask if one can also do something meaningful in the border case k = 0 where the Siegel theta func-tion (2.22) reduces to the theta function Θ0(τ, z) associated to the standard Gaussian on V (R). To get a feeling for this question, one can pretend that there is a non-trivial element f = n c(n)qn ∈M + 0 (p, χp) and formally write down its lifting according to Theorem 3.35. We ﬁnd that it has the Fourier expansion Φ(z, f) = −B0 2k ˜ c(0) + ν∈d−1 F ν≫0 d\|ν 1 d˜ c pνν′ d2 qν 1qν′ 2 . Reordering the summation, this can be written as Φ(z, f) = −B0 2k ˜ c(0) − ν∈d−1 F ν≫0 log 1 −qν 1qν′ 2 ˜ c(pνν′) . Hence, the lifting looks as the logarithm of a “modular” inﬁnite product, resem-bling the Dedekind eta function. The idea of Borcherds, Harvey and Moore was to drop the assumption on f being holomorphic and to replace it by something weaker [Bo1], [Bo2], [Bo4], [HM]. They consider a regularized theta lift for weakly holomorphic modular forms. It leads to meromorphic modular forms with inﬁnite product expansions (roughly of the above type). This construction works in greater generality for O(2, n). It yields a lift from weakly holomorphic modular forms of weight 1 −n/2 to meromorphic modular forms on O(2, n) with zeros and poles supported on Heegner divisors. Here we only consider the O(2, 2)-case of Hilbert modular surfaces. Moreover, to simplify the exposition, we assume that the real quadratic ﬁeld F has prime discriminant p. Let Γ be a subgroup of SL2(Q) which is commensurable with SL2(Z). Recall that a meromorphic modular form of weight k with respect to Γ is called weakly holomorphic if it is holomorphic outside the cusps. At the cusp ∞ such a modular form f has a Fourier expansion of the form f(τ) = n∈Z n≥N c(n)qn/h , Hilbert Modular Forms 155 where N ∈Z, and h ∈Z>0 is the width of the cusp ∞. By an elementary argument it can be proved that the Fourier coeﬃcients of f are bounded by c(n) = O eC√n , n →∞, (3.14) for some positive constant C > 0 depending on the order of the poles at the various cusps of Γ (see [BrFu1] Section 3). This estimate is also a consequence of the (much more precise) Hardy–Rademacher–Ramanujan asymptotic for the coeﬃcients of weakly holomorphic modular forms. Let Wk(p, χp) be the space of weakly holomorphic modular forms of weight k for the group Γ0(p) with character χp. Any modular form f in this space has a Fourier expansion of the form f = n≫−∞c(n)qn. Similarly as in (3.1) we denote by W + k (p, χp) the subspace of those f ∈Wk(p, χp), whose coeﬃcients c(n) satisfy the plus space condition, that is, c(n) = 0 whenever χp(n) = −1. Lemma 3.42. Let k ≤0. A weakly holomorphic modular form f = n c(n) · qn ∈W + k (p, χp) is uniquely determined by its principal part n<0 c(n)qn ∈C[q−1] . Proof. The diﬀerence of two elements of W + k (p, χp) with the same principal part is holomorphic at the cups ∞. Using the plus space condition (Lemma 3 of [BB]), one infers that the diﬀerence is also holomorphic at the cusp 0. Hence, it is a holomorphic modular form of weight k ≤0 with Nebentypus, and therefore vanishes identically. □ Corollary 3.43. Let k ≤0. Assume that f ∈W + k (p, χp) has principal part in Q[q−1]. Then all Fourier coeﬃcients of f are rational with bounded denomi-nators. Proof. This follows from Lemma 3.42 and the properties of the Galois action on Wk(p, χp). □ Let f = n c(n)qn ∈W + k (p, χp). Then (R>0)2 \ & m>0 c(−m)̸=0 S(m) is not connected. The connected components are called the Weyl chambers associated to f. If W ⊂(R>0)2 is such a Weyl chamber, then the Weyl vector corresponding to f and W is deﬁned by ρf,W = m>0 ˜ c(−m)ρm,W ∈F . (3.15) Here ρm,W is given by (3.12) and we have used the notation (3.5). We are now ready to state Borcherds’ theorem in a formulation that ﬁts nicely our setting (see [Bo4] Theorem 13.3 and [BB] Theorem 9). 156 J. H. Bruinier Theorem 3.44 (Borcherds). Let f = n≫−∞c(n)qn be a weakly holo-morphic modular form in W + 0 (p, χp) and assume that ˜ c(n) ∈Z for all n < 0. Then there exists a meromorphic Hilbert modular form Ψ(z, f) for ΓF (with some multiplier system of ﬁnite order) such that: (i) The weight of Ψ is equal to the constant term c(0) of f. (ii) The divisor Z(f) of Ψ is determined by the principal part of f at the cusp ∞. It equals Z(f) = n<0 ˜ c(n)T−n . (iii) Let W be a Weyl chamber associated to f and put N = min{n; c(n) ̸= 0}. The function Ψ has the Borcherds product expansion Ψ(z, f) = qρ 1qρ′ 2 ν∈d−1 F (ν,W)>0 1 −qν 1qν′ 2 ˜ c(pνν′) , which converges normally for all z with y1y2 > \|N\|/p outside the set of poles. Here ρ = ρf,W is the Weyl vector corresponding to f and W, and qν j = e2πiνzj for ν ∈F. Proof. We indicate the idea of the proof. We consider the theta lift (Sec-tion 2.6) for the lattice L in the quadratic space V = Q ⊕Q ⊕F (Section 2.7) and use the accidental isomorphism ΓF ∼ = SpinV . The corresponding Siegel theta function Θ0(τ, z) in weight 0 transforms as an element of M + 0 (p, χp) in the variable τ. As a function of z it is invariant under ΓF . The pairing ⟨f(τ), Θ0(τ, z)⟩(see Proposition 3.32) is a SL2(Z)-invariant function in τ. We consider the theta integral F ⟨f(τ), Θ0(τ, z)⟩du dv v2 , (3.16) where F = {τ ∈H; \|τ\| ≥1, \|u\| ≤1/2} denotes the standard fundamental domain for SL2(Z). Formally it deﬁnes a ΓF -invariant function on H2. Un-fortunately, because of the exponential growth of f at the cusps, the integral diverges. However, Harvey and Moore discovered that it can be regularized as follows [HM], [Bo4], [Kon]: If the constant term c(0) of f vanishes, one can regularize (3.16) by taking lim t→∞ Ft ⟨f(τ), Θ0(τ, z)⟩du dv v2 , (3.17) where Ft = {τ ∈F; \|v\| ≤t} denotes the truncated standard fundamental domain. So the regularization consists in prescribing the order of integration. We ﬁrst integrate over u and then over v. If the constant term of f does not Hilbert Modular Forms 157 vanish, the limit in (3.17) still diverges. It can be regularized by considering Φ(z, f, s) = lim t→∞ Ft ⟨f(τ), Θ0(τ, z)⟩v−s du dv v2 (3.18) for s ∈C. The limit exists for ℜ(s) large enough and has a meromorphic continuation to the whole complex plane. We deﬁne the regularized theta integral Φ(z, f) to be the constant term in the Laurent expansion of Φ(z, f, s) at s = 0. One can show that Φ(z, f) deﬁnes a ΓF -invariant real analytic function on H2\Z(f) with a logarithmic singularity2 along the divisor −4Z(f) ([Bo4] §6). The Fourier expansion of Φ(z, f) can be computed explicitly by applying some partial Poisson summation on the theta kernel. It turns out that Φ(z, f) = −4 log Ψ(z, f)(y1y2)c(0)/2 −2c(0) (log(2π) + Γ ′(1)) , in the domain of convergence of the inﬁnite product for Ψ(z, f). Using this identity and the properties of Φ(z, f), one can prove that the inﬁnite prod-uct has a meromorphic continuation to H2 satisfying the hypotheses of the theorem. □ Remark 3.45. The fact that Ψ(f, z) only converges in a suﬃciently small neigh-borhood of the cusp ∞is due to the rapid growth of the Fourier coeﬃcients of weakly holomorphic modular forms, see (3.14). Meromorphic Hilbert modular forms that arise as liftings of weakly holo-morphic modular forms by Theorem 3.44 are called Borcherds products. The following two propositions highlight the arithmetic nature of Borcherds products. Via the q-expansion principle (see [Rap], [Ch]) they imply that a suitable power of a Borcherds product deﬁnes a rational section of the line bundle of Hilbert modular forms over Z. Proposition 3.46. Any meromorphic Borcherds product is the quotient of two holomorphic Borcherds products. Proof. See [BBK] Proposition 4.5. □ Proposition 3.47. For any holomorphic Borcherds product Ψ there exists a positive integer n such that: (i) Ψ n is a Hilbert modular form for ΓF with trivial multiplier system. (ii) All Fourier coeﬃcients of Ψ n are contained in Z. (iii) The greatest common divisor of the Fourier coeﬃcients of Ψ n equals 1. Proof. The ﬁrst assertion is clear. The second and the third follow by Corol-lary 3.43 from the inﬁnite product expansion given in Theorem 3.44(iii). □ 2 If X is a normal complex space, D ⊂X a Cartier divisor, and f a smooth function on X \ supp(D), then f has a logarithmic singularity along D, if for any local equation g for D on an open subset U ⊂X, the function f −log \|g\| is smooth on U. 158 J. H. Bruinier Obstructions The Borcherds lift provides explicit relations among Hirzebruch–Zagier divi-sors on a Hilbert modular surface. It is natural to seek for a precise description of those linear combinations of Hirzebruch–Zagier divisors, which are divisors of Borcherds products. Since the divisor of a Borcherds product Ψ(z, f) is de-termined by the principal part of the weakly holomorphic modular form f, it suﬃces to understand which Fourier polynomials n<0 c(n)qn ∈C[q−1] can occur as principal parts of elements of W + 0 (p, χp). A necessary condition is easily obtained. If f ∈W + k (p, χp) with Fourier coeﬃcients c(n), and g ∈M + 2−k(p, χp) with Fourier coeﬃcients b(n), then the pairing ⟨f, g⟩is a weakly holomorphic modular form of weight 2 for SL2(Z). Thus ⟨f, g⟩dτ is a meromorphic diﬀerential on the Riemann sphere whose only pole is at the cusp ∞. By the residue theorem its residue has to vanish. But the residue is just the constant term in the Fourier expansion of ⟨f, g⟩. We ﬁnd that n≤0 ˜ c(n)b(−n) = 0 . (3.19) Applying this condition to the Eisenstein series E+ 2−k(τ), see (3.4), one gets a formula for the constant term of f. Proposition 3.48. Let k be a non-positive integer. Let f = n c(n)qn ∈ W + k (p, χp). Then c(0) = −1 2 n<0 ˜ c(n)B+ 2−k(−n) . □ Using Serre duality for vector bundles on Riemann surfaces, Borcherds showed that the necessary condition is also suﬃcient (see [Bo6] and [BB] Theorem 6). Theorem 3.49. There exists an f ∈W + k (p, χp) with prescribed principal part n<0 c(n)qn (where c(n) = 0 if χp(n) = −1), if and only if n<0 ˜ c(n)b(−n) = 0 for every cusp form g = m>0 b(m)qm in S+ 2−k(p, χp). □ Corollary 3.50. A formal power series m≥0 b(m)qm ∈C+ is the Fourier expansion of a modular form in M + 2−k(p, χp), if and only if n≤0 ˜ c(n)b(−n) = 0 for every f = n c(n)qn in W + k (p, χp). Hilbert Modular Forms 159 Proof. This follows immediately from Theorem 3.49, see [Br3], Corollary 4.2. □ If X is a regular projective algebraic variety, we write Div(X) for the group of divisors of X, and Rat(X) for the subgroup of divisors of rational functions on X. The ﬁrst Chow group of X is the quotient CH1(X) = Div(X)/ Rat(X) . Furthermore, we put CH1(X)Q = CH1(X) ⊗Z Q. Recall that CH1(X) is iso-morphic to the Picard group of X, the group of isomorphism classes of alge-braic line bundles on X. The isomorphism is given by mapping a line bundle L to the class c1(L) of the divisor of a rational section of L. The Chow group CH1(X) is an important invariant of X. It is ﬁnitely generated. Let π : X →X(ΓF ) be a desingularization. If k is a positive integer divisible by the order of all elliptic ﬁxed points of ΓF , then Mk := π∗Mk(ΓF ), the pullback of the line bundle of modular forms of weight k, deﬁnes an element of Pic( X). We consider its class in CH1( X). More generally, if k is any rational number, we chose an integer n such that nk is a positive integer divisible by n(ΓF ) and put c1(Mk) = 1 n c1(Mnk) ∈CH1( X)Q. The Hirzebruch–Zagier divisors are Q-Cartier on X(ΓF ). Their pullbacks deﬁne elements in CH1( X)Q. We want to describe their positions in this Chow group. To this end we consider the generating series A(τ) = c1(M−1/2) + m>0 π∗(Tm)qm ∈Q ⊗Q CH1( X)Q . (3.20) Combining Theorem 3.44 and Corollary 3.50 one obtains the following striking application. Theorem 3.51. The divisors π∗(Tm) generate a subspace of CH1( X)Q of di-mension ≤dim(M + 2 (p, χp)). The generating series A(τ) is a modular form in M + 2 (p, χp) with values in CH1( X)Q, i.e., an element of M + 2 (p, χp) ⊗Q CH1( X)Q. In other words, if λ is a linear functional on CH1( X)Q, then λ c1(M−1/2) + m>0 λ(π∗Tm)qm ∈M + 2 (p, χp) . A typical linear functional, one can take for λ, is given by the intersection pairing with a ﬁxed divisor on X. Theorem 3.51 was ﬁrst proved by Hirze-bruch and Zagier by computing intersection numbers of Hirzebruch–Zagier divisors with other such divisors and with the exceptional divisors coming from Hirzebruch’s resolution of the cusp singularities [HZ]. Their discovery triggered important investigations by several people, showing that more gen-erally periods of certain special cycles in arithmetic quotients of orthogonal 160 J. H. Bruinier or unitary type can be viewed as the coeﬃcients of Siegel modular forms. For instance, Oda considered cycles on quotients of O(2, n) given by embedded quotients of O(1, n) [Od1], and Kudla–Millson studied more general cycles on quotients of O(p, q) and U(p, q) using the Weil representation and theta func-tions with values in closed diﬀerential forms [KM1,KM2,KM3], see also [Fu] for the case of non-compact quotients. The relationship of the Kudla–Millson lift and the regularized theta lift is clariﬁed in [BrFu1]. Proof of Theorem 3.51. Using Borcherds products, Theorem 3.51 can be proved as follows (see [Bo6]). In view of Corollary 3.50 it suﬃces to show that ˜ c(0) c1(M−1/2) + n<0 ˜ c(n)π∗(T−n) = 0 ∈CH1( X)Q for every f = n c(n)qn in W + 0 (p, χp) with integral Fourier coeﬃcients. But this is an immediate consequence of Theorem 3.44: Up to torsion, the Borcherds lift of f is a rational section of Mc(0) with divisor n<0 ˜ c(n) · π∗(T−n). □ Notice that we have only used (i) and (ii) of Theorem 3.44. Using the product expansion (iii) in addition, one can prove an arithmetic version of Theorem 3.51, saying that certain arithmetic Hirzebruch–Zagier divisors are the coeﬃcients of a modular form in M + 2 (p, χp) with values in an arithmetic Chow group, see [BBK], [Br3]. Finally, we mention that this argument gener-alizes to Heegner divisors on quotients of O(2, n). Remark 3.52. With some further work it can be proved that the dimension of the subspace of CH( X)Q generated by the Hirzebruch–Zagier divisors is equal to dim M + 2 (p, χp), see Corollary 3.62. Examples Recall that p is a prime congruent to 1 modulo 4. By a result due to Hecke [He] the dimension of S+ 2 (p, χp) is equal to [ p−5 24 ]. In particular there exist three such primes for which S+ 2 (p, χp) is trivial, namely p = 5, 13, 17. In these cases W + 0 (p, χp) is a free module of rank p+1 2 over the ring C[j(pτ)]. Therefore it is not hard to compute explicit bases. For any m ∈Z>0 with χp(m) ̸= −1 there is a unique fm = n≥−m cm(n)qn ∈W + 0 (p, χp) whose Fourier expansion starts with fm = q−m + cm(0) + O(q), if p ∤m , 1 2q−m + cm(0) + O(q), if p \| m . The fm (m ∈Z>0) form a base of the space W + 0 (p, χp). The Borcherds lift Ψm of fm is a Hilbert modular form for ΓF of weight cm(0) = −B+ 2 (m)/2 with divisor Tm. Here B+ 2 (m) denotes the m-th coeﬃcient of the Eisenstein series E+ 2 (τ) as before. Hilbert Modular Forms 161 The case p = 5. We consider the real quadratic ﬁeld F = Q( √ 5). The fundamental unit is given by ε0 = 1 2(1 + √ 5). Here the ﬁrst few fm were computed in [BB]. One obtains: f1 = q−1 + 5 + 11 q −54 q4 + 55 q5 + 44 q6 −395 q9 + 340 q10 + . . . , f4 = q−4 + 15 −216 q + 4959 q4 + 22040 q5 −90984 q6 + 409944 q9 + . . . , f5 = 1 2 q−5 + 15 + 275 q + 27550 q4 + 43893 q5 + 255300 q6 + . . . , f6 = q−6 + 10 + 264 q −136476 q4 + 306360 q5 + 616220 q6 + . . . , f9 = q−9 + 35 −3555 q + 922374 q4 + 7512885 q5 −53113164 q6 + . . . , f10 = 1 2 q−10 + 10 + 3400 q + 3471300 q4 + 9614200 q5 + 91620925 q6 + . . . . The Eisenstein series E+ 2 (τ) ∈M + 2 (5, χ5) has the Fourier expansion E+ 2 (τ, 0) = 1 −10q −30q4 −30q5 −20q6 −70q9 −20q10 −120q11 −60q14 −. . . . One easily shows that the weight of any Borcherds product is divisible by 5. By a little estimate one concludes that there is just one holomorphic Borcherds product of weight 5, namely Ψ1. There exist precisely 3 holomorphic Borcherds products in weight 10, namely Ψ 2 1 , Ψ6, and Ψ10. In weight 15 there are the holomorphic Borcherds products Ψ4, Ψ5, Ψ 3 1 , Ψ1Ψ6, and Ψ1Ψ10. It follows from Lemma 3.37 that Tm does not go through the cusp ∞when m is not the norm of some λ ∈OF . In particular, T6 and T10 do not meet ∞. This also implies that S(6) = S(10) = ∅. There is just one Weyl chamber of index 6 and 10 (namely (R>0)2) and the corresponding Weyl vector is 0. The divisor T1 does meet ∞. As in Example 3.41, let W be the unique Weyl chamber of index 1 containing (ε−1 0 , ε0). The corresponding Weyl vector is ρ1 = ε0 √ D 1 tr(ε0). We obtain the Borcherds product expansions Ψ1 = qρ1 1 qρ′ 1 2 ν∈d−1 F ε0ν′−ε′ 0ν>0 1 −qν 1qν′ 2 ˜ c1(5νν′) , Ψ6 = ν∈d−1 F ν≫0 1 −qν 1qν′ 2 ˜ c6(5νν′) , Ψ10 = ν∈d−1 F ν≫0 1 −qν 1qν′ 2 ˜ c10(5νν′) . Gundlach [Gu] constructed a Hilbert modular form s5 for ΓF with divisor T1 as a product of 10 theta functions of weight 1/2, see Section 1.5. We have s5 = Ψ1. Moreover, s15, the symmetric cusp form of weight 15, is equal to Ψ5. For further examples we refer to [Ma]. 162 J. H. Bruinier 3.3 Automorphic Green Functions By Theorem 3.49 of the previous section we know precisely which linear com-binations of Hirzebruch–Zagier divisors occur as divisors of Borcherds prod-ucts on Y (ΓF ). It is natural to ask, whether every Hilbert modular form on Y (ΓF ) whose divisor is a linear combination of Hirzebruch–Zagier divisors is a Borcherds product, i.e., in the image of the lift of Theorem 3.44. In this section we discuss this question in some detail. To answer it, we ﬁrst simplify the problem. We extend the Borcherds lift to a larger space of “input modular forms” and answer the question for this extended lift. In that way we are led to automorphic Green functions associated with Hirzebruch–Zagier divisors. Let k be an integer, let Γ be a subgroup of SL2(Q) which is commensur-able with SL2(Z), and χ a character of Γ. A twice continuously diﬀerentiable function f : H →C is called a weak Maass form (of weight k and eigenvalue λ with respect to Γ and χ), if (i) f aτ+b cτ+d = χ(γ)(cτ + d)kf(τ) for all a b c d ∈Γ; (ii) there is a C > 0 such that for any cusp s ∈Q ∪{∞} of Γ and δ ∈SL2(Z) with δ∞= s the function fs(τ) = j(δ, τ)−kf(δτ) satisﬁes fs(τ) = O(eCv) as v →∞; (iii) Δkf = λΔ for some λ ∈C. Here Δk = −v2 ∂2 ∂u2 + ∂2 ∂v2 + ikv ∂ ∂u + i ∂ ∂v (3.21) denotes the usual hyperbolic Laplace operator in weight k and τ = u + iv. In the special case where the eigenvalue λ is zero, f is called a harmonic weak Maass form. This is the case we are interested in here. If we compare the deﬁnition of a harmonic weak Maass form with the deﬁ-nition of a weakly holomorphic modular form, we see that we simply replaced the condition that f be holomorphic on H by the weaker condition that f be annihilated by Δk, and the meromorphicity at the cusps by the correspond-ing growth condition. In particular, any weakly holomorphic modular form is a harmonic weak Maass form. The third condition implies that f is actu-ally real analytic. Because of the transformation behavior, it has a Fourier expansion, which involves besides the exponential function a second type of Whittaker function. (See [BrFu1] Section 3 for more details.) There are two fundamental diﬀerential operators on modular forms for Γ, the Maass raising and lowering operators Rk = 2i ∂ ∂τ + kv−1 and Lk = −2iv2 ∂ ∂¯ τ . If f is a diﬀerentiable function on H satisfying the transformation law (i) in weight k, then Lkf transforms in weight k −2, and Rkf in weight k + 2. It can be shown that the assignment Hilbert Modular Forms 163 f(τ) →ξk(f)(τ) := vk−2Lkf(τ) = R−kvkf(τ) deﬁnes an antilinear map ξk from harmonic weak Maass forms of weight k to weakly holomorphic modular forms of weight 2 −k. Its kernel is precisely the space of weakly holomorphic modular forms in weight k. We write Nk(p, χp) for the space of harmonic weak Maass forms of weight k with respect to Γ0(p) and χp. Let us have a closer look at the map ξk : Nk(p, χp) →W2−k(p, χp). We denote by Nk(p, χp) the inverse image of S2−k(p, χp) under ξk, and its plus subspace by N + k (p, χp). (Note that our notation diﬀers from the notation of [BrFu1].) Theorem 3.53. We have the following exact sequence: 0 W + k (p, χp) N + k (p, χp) ξk S+ 2−k(p, χp) 0 . Proof. This can be proved using Serre duality for the Dolbeault resolution of the structure sheaf on a modular curve (see [BrFu1] Theorem 3.7) or by means of Hejhal–Poincaré series (see [Br2] Chapter 1). □ Let k ≤0. For every harmonic weak Maass form f ∈N + k (p, χp) there is a unique Fourier polynomial P(f) = n<0 c(n)qn ∈C[q−1] (with c(n) = 0 if χp(n) = −1) such that f(τ) −P(f)(τ) is bounded as v →∞. It is called the principal part of f. This generalizes the notion of the principal part of a weakly holomorphic modular form. Proposition 3.54. Let Q = n<0 c(n)qn ∈C[q−1] be a Fourier polynomial satisfying c(n) = 0 if χp(n) = −1. There exists a unique f ∈N + k (p, χp) whose principal part is equal to Q. Proof. See [BrFu1] Proposition 3.11. □ This Proposition is a key fact, which suggests to study the regularized theta lift of harmonic weak Maass forms. If f ∈N + 0 (p, χp), then we deﬁne its regularized theta lift Φ(z, f) by (3.18), in the same way as for weakly holomorphic modular forms. Theorem 3.55. Let f ∈N + 0 (p, χp) be a harmonic weak Maass form with principal part P(f) = n<0 c(n)qn and constant term c(0). (i) The regularized theta integral Φ(z, f) deﬁnes a ΓF -invariant function on H2 with a logarithmic singularity along −4Z(f), where Z(f) = n<0 ˜ c(n)T−n . (ii) It is a Green function for the divisor 2Z(f) on Y (ΓF ) in the sense of [SABK], that is, it satisﬁes the identity of currents ddc[Φ(z, f)] + δ2Z(f) = [ω(z, f)] on Y (ΓF ). Here δD denotes the Dirac current associated with a divisor D on Y (ΓF ) and ω(z, f) is a smooth (1, 1)-form. 164 J. H. Bruinier (iii) If Δ(j) = −y2 j ∂2 ∂x2 j + ∂2 ∂y2 j denotes the SL2(R)-invariant Laplace opera-tor on H2 in the variable zj, then Δ(j)Φ(z, f) = −2c(0) . Proof. See [Br1] and [BBK]. □ In view of Proposition 3.54, for every positive integer m with χp(m) ̸= −1, there exists a unique harmonic weak Maass form fm ∈N + 0 (p, χp), whose principal part is given by P(fm) = q−m, if p ∤m , 1 2q−m, if p \| m . Its theta lift Φm(z) = 1 2Φ(z, fm) can be regarded as an automorphic Green function for Tm. Let π : X →X(ΓF ) be a desingularization. The Fourier expansion of Φ(z, f) can be computed explicitly. It can be used to determine the growth behavior at the boundary of Y (ΓF ) in X. It turns out that the boundary singularities are of log and log-log type. More precisely, one can view π∗Φ(z, f) as a pre-log-log Green function for the divisor 2π∗(Z(f)) on X in the sense of [BKK] (see [BBK] Proposition 2.16). So the current equation in (ii) does not only hold for test forms with compact support on Y (ΓF ), but also for test forms which are smooth on X. Moreover, one ﬁnds that Φ(z, f) can be split into a sum Φ(z, f) = −2 log \|Ψ(z, f)\|2 + ξ(z, f) , (3.22) where ξ(z, f) is real analytic on the whole domain H2 and Ψ(z, f) is a mero-morphic function on H2 whose divisor equals Z(f). If f is weakly holomorphic, the function ξ(z, f) is simply equal to −2c(0) (log(y1y2) + log(2π) + Γ ′(1)), and we are back in the case of Borcherds’ original lift. However, if f is an honest harmonic weak Maass form, then ξ is a complicated function and Ψ far from being modular. The splitting (3.22) implies that the smooth form ω(z, f) in Theorem 3.55 is given by ω(z, f) = ddcξ(z, f) . By the usual Poincaré–Lelong argument, 1 2ω(z, f) represents the Chern class of the divisor Z(f) in the second cohomology H2(Y (ΓF )). One can further show that it is a square integrable harmonic representative. Moreover, 1 2π∗ω(z, f) is a pre-log-log form on X, representing the class of π∗Z(f) in H2( X, C). We now discuss the relationship between the Borcherds lift (Theorem 3.44) and its generalization in the present section. For simplicity, we write Nk, Hilbert Modular Forms 165 Wk, Mk, Sk for the spaces N + k (p, χp), W + k (p, χp), M + k (p, χp), S+ k (p, χp), res-pectively. We denote by Wk0 the subspace of elements of Wk with vanishing constant term. Moreover, we denote by M ∨ k the dual of the vector space Mk. Theorem 3.56. We have the following commutative diagram with exact rows: 0 W00 N0 M ∨ 2 0 0 Rat( X)C Div( X)C CH1( X)C 0 . Here the map N0 →M ∨ 2 is given by fm →am, where am denotes the func-tional taking a modular form in M2 to its m-th Fourier coeﬃcient. The map M ∨ 2 →CH1( X)C is deﬁned by am →π∗Tm for m > 0 and a0 →c1(M−1/2). The map N0 →Div( X)C is deﬁned by f →π∗Z(f). Proof. The exactness of the ﬁrst row is an immediate consequence of The-orem 3.53. Moreover, by Theorem 3.44, if f ∈W00, then Z(f) ∈Rat( X)C. □ Remark 3.57. The map N0 →Div( X)C does not really depend on the analytic properties of the harmonic weak Maass forms. In particular the Green func-tion Φ(z, f) associated to f ∈N0 does not play a role. However, there is an analogue of the above diagram in Arakelov geometry. If X is a regular model of X over an arithmetic ring and Tm denotes the Zariski closure of π∗Tm in X , then the pair ) Tm = (Tm, π∗Φm) deﬁnes an arithmetic divisor in the sense of [BKK]. The map N0 → Div( X), deﬁned by fm →) Tm, gives rise to a diagram as above for the ﬁrst arithmetic Chow group of X. So the generalized Borcherds lift can be viewed as a map to the group of arithmetic divisors on X (see [BBK], [Br3]). Theorem 3.58. Let h be a meromorphic Hilbert modular form of weight r for ΓF , whose divisor div(h) = n<0 ˜ c(n)T−n is a linear combination of Hirzebruch–Zagier divisors. Then −2 log \|h(z)2(y1y2)r\| = Φ(z, f) + constant , where f is the unique harmonic weak Maass form in N0 with principal part n<0 c(n)qn. Proof. (See [Br2] Chapter 5.) Let f be the unique harmonic weak Maass form in N0 with principal part n<0 c(n)qn. Then Φ(z, f) is real analytic on H2 \ Z(f) and has a logarithmic singularity along −4Z(f). Hence d(z) := Φ(z, f) + 2 log \|h(z)2(y1y2)r\| is a smooth function on Y (ΓF ). By Theorem 3.55 (iii), it is subharmonic. 166 J. H. Bruinier One can show that d(z) is in L1+ε(Y (ΓF )) for some ε > 0 (with re-spect to the invariant measure coming from the Haar measure). By results of Andreotti–Vesentini and Yau (see e.g. [Yau]) on sub-harmonic functions on complete Riemann manifolds that satisfy such integrability conditions it follows that d(z) is constant. □ The question regarding the surjectivity of the Borcherds lift raised at the beginning of this section is therefore reduced to the question whether the harmonic weak Maass form f in the Theorem is actually weakly holomorphic. It is answered aﬃrmatively in Theorem 3.61 below. Corollary 3.59. The assignment π∗Tm →1 2ddcξ(z, fm) deﬁnes a linear map CH1 HZ( X)C − →H1,1(Y (ΓF )) from the subspace of CH1( X)C generated by the Hirzebruch–Zagier divisors to the space of square integrable harmonic (1, 1)-forms on Y (ΓF ). □ Composing the map M ∨ 2 →CH1( X)C with the map CH1 HZ( X)C → H1,1(Y (ΓF )) from Corollary 3.59, we obtain a linear map M ∨ 2 − →H1,1(Y (ΓF )) . On the other hand, we have the Doi–Naganuma lift S2 →S2(ΓF ), and there is a natural map from Hilbert cusp forms of weight 2 to harmonic (1, 1)-forms on Y (ΓF ) (see e.g. [Br1] Section 5). Summing up, we get the following diagram: M ∨ 2 CH1 HZ( X)C H1,1(Y (ΓF )) S2 f→(·,f) S2(ΓF ) . (3.23) Theorem 3.60. The above diagram (3.23) commutes. Proof. See [Br1] Theorem 8. □ So the above construction can be viewed as a diﬀerent approach to the Doi–Naganuma lift, making its geometric properties quite transparent. Using, for instance, the description of the Doi–Naganuma lifting in terms of Fourier expansions, it can be proved that S2 →S2(ΓF ) is injective. As a consequence, we obtain the following converse theorem for the Borcherds lift (see [Br1], [Br2] Chapter 5). Theorem 3.61. Let h be a meromorphic Hilbert modular form for ΓF , whose divisor div(F) = n<0 ˜ c(n)T−n is given by Hirzebruch–Zagier divisors. Then there is a weakly holomorphic modular form f ∈W0 with principal part n<0 c(n)qn, and, up to a constant multiple, h is equal to the Borcherds lift of f in the sense of Theorem 3.44. □ Hilbert Modular Forms 167 Corollary 3.62. The dimension of CH1 HZ( X)C is equal to dim(M2). □ Notice that the analogue of Theorem 3.58 holds for arbitrary congruence subgroups of ΓF (more generally also for O(2, n)), whereas the analogue of Theorem 3.61 is related to the injectivity of a theta lift and therefore more complicated. So far it is only known for particular arithmetic subgroups of O(2, n), see [Br2], [Br3]. For example, if we go to congruence subgroups of the Hilbert modular group ΓF , it is not clear whether the analogue of Theorem 3.61 holds or not. See also [BrFu2] for this question. A Second Approach The regularized theta lift Φm(z) = 1 2Φ(z, fm) of the weak Maass form fm ∈N0 is real analytic on H2 \ Tm and has a logarithmic singularity along −2Tm. Here we present a diﬀerent, more naive, construction of Φm(z). For details see [Br1]. The idea is to construct Φm(z) directly as a Poincaré series by summing over the logarithms of the deﬁning equations of Tm. We consider the sum (a,b,λ)∈Z2⊕d−1 F ab−λλ′=m/D log az1¯ z2 + λz1 + λ′¯ z2 + b az1z2 + λz1 + λ′z2 + b . (3.24) The denominators of the summands ensure that this function has a logarith-mic singularity along −2Tm in the same way as Φm(z). The enumerators are smooth on the whole H2. They are included to make the sum formally ΓF -invariant. Unfortunately, the sum diverges. However, it can be regularized in the following way. If we put Q0(z) = 1 2 log z+1 z−1 , we may rewrite the sum-mands as log az1¯ z2 + λz1 + λ′¯ z2 + b az1z2 + λz1 + λ′z2 + b = Q0 1 + \|az1z2 + λz1 + λ′z2 + b\|2 2y1y2m/D . Now we replace Q0 by the 1-parameter family Qs−1 of Legendre functions of the second kind (cf. [AbSt] §8), deﬁned by Qs−1(z) = ∞ 0 (z + z2 −1 cosh u)−sdu . (3.25) Here z > 1 and s ∈C with ℜ(s) > 0. If we insert s = 1, we get back the above Q0. Hence we consider φm(z, s) = a,b∈Z λ∈d−1 F ab−λλ′=m/D Qs−1 1 + \|az1z2 + λz1 + λ′z2 + b\|2 2y1y2m/D . (3.26) 168 J. H. Bruinier It is easily seen that this series converges normally for z ∈H2 \ Tm and ℜ(s) > 1 and therefore deﬁnes a ΓF -invariant function, which has logarith-mic growth along −2Tm. It is an eigenfunction of the hyperbolic Laplacians Δ(j) with eigenvalue s(s −1), because of the diﬀerential equation satisﬁed by Qs−1. Notice that for D = m = 1 the function Φm(z, s) is simply the classical resolvent kernel for SL2(Z) (cf. [Hej], [Ni]). One can compute the Fourier expansion of φm(z, s) explicitly and use it to obtain a meromor-phic continuation to s ∈C. At s = 1 there is a simple pole, reﬂecting the divergence of the formal sum (3.24). We deﬁne the regularization φm(z) of (3.24) to be the constant term in the Laurent expansion of φm(z, s) at s = 1. It turns out that φm is up to an additive constant equal to Φm. The Green functions φm can be used to give diﬀerent proofs of the results of the previous section and of Theorem 3.44. Similar Green functions on O(2, n) are investi-gated in the context of the theory of spherical functions on real Lie groups in [OT]. 3.4 CM Values of Hilbert Modular Functions In this section we consider the values of Borcherds products on Hilbert modu-lar surfaces at certain CM cycles. We report on some results obtained in joint work with T. Yang, see [BY]. This generalizes work of Gross and Zagier on CM values of the j-function [GZ]. Singular Moduli We review some of the results of Gross and Zagier on the j-function. We begin by recalling some background material (see also pp. 77–79). Let k be a ﬁeld and E/k an elliptic curve, that is, a non-singular projective curve over k of genus 1 together with a k-rational point. If char k ̸= 2, 3, then by the Riemann–Roch theorem one ﬁnds that E has a Weierstrass equation of the form y2 = 4x3 −g2x −g3 , with g2, g3 ∈k and g3 2 −27g2 3 ̸= 0. The j-invariant of E is deﬁned by j(E) = 1728 g3 2 g3 2 −27g2 3 . A basic result of the theory of elliptic curves says that if k is algebraically closed then two elliptic curves over k are isomorphic if and only if they have the same j-invariant. Moreover, for every given a ∈k there is an elliptic curve with j-invariant a. So the assignment E →j(E) deﬁnes a bijection {elliptic curves over k}/ ∼− →k . Hilbert Modular Forms 169 Over C, the theory of the elliptic functions implies that any elliptic curve is complex analytically isomorphic to a complex torus C/L, where L ⊂C is a lattice. (Here g2 = 60G4(L) and g3 = 140G6(L) where G4, G6 are the usual Eisenstein series of weight 4 and 6.) Two elliptic curves E, E′ over C are isomorphic if and only if the corresponding lattices L, L′ satisfy L = a L′ for some a ∈C∗. On the other hand it is easily seen that we have a bijection SL2(Z)\H − →{lattices in C}/C∗, [τ] →[Zτ + Z] . Summing up, we obtain a bijection SL2(Z)\H − →{elliptic curves over C}/ ∼, [τ] →[C/(Zτ + Z)] . (3.27) Hence, the j-invariant induces a function on Y (1) := SL2(Z)\H. A more detailed examination of the map in (3.27) shows that j is a holomorphic function on Y (1) with the Fourier expansion j(τ) = q−1 +744+196884q +. . . at the cusp ∞. So we may view the j-function as a function on the coarse moduli space of isomorphism classes of elliptic curves over C. There are special points on Y (1) which correspond to special elliptic curves, namely to elliptic curves with complex multiplication. Let K/Q be an imaginary quadratic ﬁeld with ring of integers OK. A point τ ∈H is called a CM point of type OK if the corresponding elliptic curve Eτ = C/(Zτ + Z) has complex multiplication OK # →End(Eτ), or equiv-alently if Zτ + Z ⊂K is a fractional ideal. We may consider the 0-cycle CM(K) ⊂Y (1) given by the points τ for which Eτ has complex multiplica-tion by OK. The values of the j-function at CM points are classically known as singular moduli. If τ0 is a CM point of type OK, then, by the theory of complex multiplication, j(τ0) is an algebraic integer generating the Hilbert class ﬁeld of K. Moreover, the Galois group Gal(H/K) acts transitively on CM(K) ⊂ Y (1). This implies that j(CM(K)) = [τ]∈CM(K) j(τ) is an integer. It is a natural question to ask for the shape of this number. At the beginning of the 20-th century, Berwick made extensive computations of these numbers and conjectured various congruences [Be]. We listed some values in Table 2. In [GZ], Gross and Zagier found an explicit formula for the prime factor-ization of j(CM(K)) and proved Berwick’s conjectures. More precisely, they considered the function j(z1) −j(z2) on Y (1) × Y (1). 170 J. H. Bruinier Table 2. Some CM values of the j-function \| disc(K)\| h(K) (j(CM(K)))1/3 3 1 0 4 1 22 · 3 7 1 3 · 5 8 1 22 · 5 11 1 25 19 1 25 · 3 23 3 53 · 11 · 17 31 3 33 · 11 · 17 · 23 43 1 26 · 3 · 5 47 5 55 · 112 · 23 · 29 59 3 216 · 11 67 1 25 · 3 · 5 · 11 71 7 113 · 172 · 23 · 41 · 47 · 53 Let K1 and K2 be two imaginary quadratic ﬁelds of discriminants d1 and d2, respectively. Assume (d1, d2) = 1, and put D = d1d2. We consider the CM cycle CM(K1) × CM(K2) on Y (1) × Y (1) and put J(d1, d2) = [τ1]∈CM(K1) [τ2]∈CM(K2) (j(τ1) −j(τ2)) 4 w1w2 , where wi is the number of units in Ki. Theorem 3.63 (Gross, Zagier). We have J(d1, d2)2 = ± x,n,n′∈Z, n,n′>0 x2+4nn′=D nϵ(n′) . (3.28) Here ϵ is the genus character deﬁned as follows: ϵ(n) = ϵ(li)ai if n has the prime factorization n = lai i , and ϵ(l) = ⎧ ⎨ ⎩ ( d1 l ) if l ∤d1, ( d2 l ) if l ∤d2, for primes l with ( D l ) ̸= −1. In particular, this result implies that the prime factors of J(d1, d2) are bounded by D/4. Since j(CM(Q(√−3))) = j(e2πi/3) = 0, we obtain an ex-plicit formula for the CM values of j as a special case. It leads to the values in Table 2. Hilbert Modular Forms 171 The surface Y (1) × Y (1) can be viewed as the Hilbert modular surface corresponding to the real quadratic “ﬁeld” Q ⊕Q of discriminant 1. Moreover, j(z1) −j(z2) is a Borcherds product on this surface given by j(z1) −j(z2) = q−1 1 m>0 n∈Z (1 −qm 1 qn 2 )c(mn) . (3.29) Here qj = e2πizj, and c(n) is the n-th Fourier coeﬃcient of j(τ)−744. In fact, this is the celebrated denominator identity of the monster Lie algebra, which is crucial in Borcherds’ proof of the moonshine conjecture. From this viewpoint it is natural to ask if the formula of Gross and Zagier has a generalization to Hilbert modular surfaces. In the rest of this section we report on joint work with T. Yang on this problem [BY]. See also [Ya] for further motivation and background information. CM Extensions As before, let F ⊂R be a real quadratic ﬁeld. Let K be a CM extension of F, that is, K = F( √ Δ), where Δ ∈F is totally negative. We view both K and F( √ Δ′) as subﬁelds of C with √ Δ, √ Δ′ ∈H. The ﬁeld M = F( √ Δ, √ Δ′) is Galois over Q. There are three possibilities for the Galois group Gal(M/Q) of M over Q: Gal(M/Q) = ⎧ ⎪ ⎨ ⎪ ⎩ Z/2Z × Z/2Z, if K/Q is biquadratic , Z/4Z, if K/Q is cyclic , D4, if K/Q is non Galois . Lemma 3.64. Let the notation be as above, and let ˜ F = Q( √ ΔΔ′). (i) K/Q is biquadratic if and only if ˜ F = Q. (ii) K/Q is cyclic if and only if ˜ F = F. (iii) K/Q is non-Galois if and only if ˜ F ̸= F is a real quadratic ﬁeld. □ Gross and Zagier considered a biquadratic case. Here we assume that K is non-biquadratic, i.e., ˜ F is a real quadratic ﬁeld. Then M/Q has an auto-morphism σ of order 4 such that σ( √ Δ) = √ Δ′, σ( √ Δ′) = − √ Δ . (3.30) Notice that K has four CM types, i.e., pairs of non complex conjugate com-plex embeddings: Φ = {1, σ}, σΦ = {σ, σ2}, σ2Φ, and σ3Φ. Since K is not biquadratic, these CM types are primitive. We write ( ˜ K, ˜ Φ) for the reﬂex of (K, Φ). Then ˜ K = Q( √ Δ + √ Δ′) and ˜ F is the real quadratic subﬁeld of ˜ K. We refer to [Sh2] for details about CM types and reﬂex ﬁelds. For the rest of this section we assume that the discriminant of F is a prime p ≡1 (mod 4). Moreover, we suppose that the discriminant dK of K is given 172 J. H. Bruinier Table 3. CM extensions of Q( √ 5) q K = F( √ Δ) hK Cl(K) 5 Δ = −5+ √ 5 2 1 OK = OF + √ ΔOF 41 Δ = −13+ √ 5 2 1 OK = OF 1 2 “√ Δ + 3+ √ 5 2 ” OF 61 Δ = −(9 + 2 √ 5) 1 OK = OF 1 2 “√ Δ + 1 ” OF 109 Δ = −21+ √ 5 2 1 OK = OF 1 2 “√ Δ + 3+ √ 5 2 ” OF 241 Δ = −33+5 √ 5 2 3 OK = OF 1 2 “√ Δ + 3+ √ 5 2 ” OF , A = 2OF 1 2 “√ Δ + 9+3 √ 5 2 ” OF , B = 4OF 1 2 “√ Δ + 9+3 √ 5 2 ” OF 281 Δ = −37+7 √ 5 2 3 OK = OF 1 2 “√ Δ + 1+ √ 5 2 ” OF , A = 2OF 1 2 “√ Δ + 1+ √ 5 2 ” OF , B = 4OF 1 2 “√ Δ + 9+ √ 5 2 ” OF 409 Δ = −41+3 √ 5 2 3 OK = OF 1 2 “√ Δ + 1+ √ 5 2 ” OF , A = 2OF 1 2 “√ Δ + 7+3 √ 5 2 ” OF , B = 4OF 1 2 “√ Δ + −1+3 √ 5 2 ” OF by dK = p2q for a prime q ≡1 (mod 4). This assumption guarantees that the class number of K is odd, which is crucial in the argument of [BY]. It implies that ˜ F = Q(√q) and d ˜ K = q2p. In Table 3 we listed a few CM extensions of F = Q( √ 5) satisfying the assumption, including the class number hK, and a system of representatives for the ideal class group of K. CM Cycles We now deﬁne CM points on Hilbert modular surfaces analogously to the CM points on the modular curve Y (1) above. Recall that the Hilbert modular surface Y (ΓF ) corresponding to ΓF = SL2(OF ) parameterizes isomorphism classes of triples (A, ı, m), where (i) A is an abelian surface over C, (ii) ı : OF →End(A) is a real multiplication by OF , (iii) and m : (PA, P + A ) → d−1 F , d−1,+ F is an OF -linear isomorphism between the polarization module PA = Homsym OF (A, A∨) of A and d−1 F , taking the subset of polarizations to totally positive elements of d−1 F . Hilbert Modular Forms 173 (See e.g. [Go], Theorem 2.17 and [BY] Section 3.) The moduli interpretation can be used to construct a model of the Hilbert modular surface Y (ΓF ) over Q, see [Rap], [DePa], [Ch]. Let Φ = (σ1, σ2) be a CM type of K. A point z = (A, ı, m) ∈Y (ΓF ) is said to be a CM point of type (K, Φ) if one of the following equivalent conditions holds (see [BY] Section 3 for details): (i) As a point z ∈H2, there is τ ∈K such that Φ(τ) = (σ1(τ), σ2(τ)) = z and such that Λτ = OF τ + OF is a fractional ideal of K. (ii) (A, ı) is a CM abelian variety of type (K, Φ) with complex multiplication ı′ : OK # →End(A) such that ı = ı′\|OF . We consider the CM type Φ = {1, σ} of K, where σ is deﬁned by (3.30). Let CM(K, Φ, OF ) be the CM 0-cycle in Y (ΓF ) of CM abelian surfaces of type (K, Φ). By the theory of complex multiplication [Sh2], the ﬁeld of moduli for CM(K, Φ, OF ) is the reﬂex ﬁeld ˜ K of (K, Φ). In fact, one can show that the ﬁeld of moduli for CM(K) = CM(K, Φ, OF ) + CM(K, σ3Φ, OF ) is Q (see [BY], Remark 3.5). Therefore, if Ψ is a rational function on Y (ΓF ), i.e., a Hilbert modular function for ΓF over Q, then Ψ(CM(K)) is a rational number. The purpose of the following section is to ﬁnd a formula for this number, when Ψ is given by a Borcherds product. CM Values of Borcherds Products We keep the above assumptions on F and K. We denote by WK the number of roots of unity in K. For an ideal a of ˜ F we consider the representation number ρ(a) = #{A ⊂O ˜ K; N ˜ K/ ˜ F A = a} of a by integral ideals of ˜ K. We brieﬂy write \|a\| for the norm of a. For a non-zero element t ∈d−1 ˜ K/ ˜ F and a prime ideal l of ˜ F, we put Bt(l) = (ordl t + 1)ρ(td ˜ K/ ˜ F l−1) log \|l\| if l is non-split in ˜ K, 0 if l is split in ˜ K , and Bt = l Bt(l) . We remark that ρ(a) = 0 for a non-integral ideal a, and that for every t ̸= 0, there are at most ﬁnitely many prime ideals l such that Bt(l) ̸= 0. In fact, when t > 0 > t′, then Bt = 0 unless there is exactly one prime ideal l such that χl(t) = −1, in which case Bt = Bt(l) (see [BY], Remark 7.3). Here χ = l χl is the quadratic Hecke character of ˜ F associated to ˜ K/ ˜ F. The following formula for the CM values of Borcherds products is proved in [BY]. 174 J. H. Bruinier Theorem 3.65. Let f = n≫−∞c(n)qn ∈W + 0 (p, χp), and assume that ˜ c(n) ∈Z for all n < 0, and c(0) = 0. Then the Borcherds lift Ψ = Ψ(z, f) (see Theorem 3.44) is a rational function on Y (ΓF ), whose value at the CM cycle CM(K) satisﬁes log \|Ψ(CM(K))\| = W ˜ K 4 m>0 ˜ c(−m)bm, where bm = t= n+m√q 2p ∈d−1 ˜ K/ ˜ F \|n\|0 ˜ c(−m)bm(l) , and bm(l) log l = l\|l t= n+m√q 2p ∈d−1 ˜ K/ ˜ F \|n\|0; ˜ c(−m) ̸= 0} and some integer \|n\| < m√q. Corollary 3.68. Let the notation and assumption be as in Corollary 3.66. Every prime factor of Ψ(CM(K)) is less than or equal to N 2q 4p , where N = max(M). We now indicate the idea of the proof of Theorem 3.65. It roughly follows the analytic proof of Theorem 3.63 given in [GZ], although each step requires Hilbert Modular Forms 175 some new ideas. By the construction of the Borcherds lift and by the results of Section 3.3, we have −4 log \|Ψ(z, f)\| = Φ(z, f) = m>0 ˜ c(−m)φm(z) , where φm(z) denotes the automorphic Green function for Tm. Consequently, it suﬃces to compute φm(CM(K)). Using a CM point, the lattice Z2 ⊕d−1 F deﬁning the automorphic Green function can be related to some ideal of the reﬂex ﬁeld ˜ K of (K, Φ). In that way, one derives an expression for φm(CM(K)) as an inﬁnite sum involving arithmetic data of ˜ K/ ˜ F. To come up with a ﬁnite sum for the CM value φm(CM(K)), we consider an auxiliary function. It is constructed using an incoherent Eisenstein series (see e.g. [Ku1]) of weight 1 on ˜ F associated to ˜ K/ ˜ F. We consider the central derivative of this Eisenstein series, take its restriction to Q, and compute its holomorphic projection. In that way we obtain a holomorphic cusp form h ∈S+ 2 (p, χp) of weight 2. Its m-th Fourier coeﬃcient is the sum of two parts. One part is the inﬁnite sum for φm(CM(K)), the other part is a linear combination of the quantity bm (what we want) and the logarithmic derivative of the Hecke L-series of ˜ K/ ˜ F. Finally, the duality between W + 0 (p, χp) and S+ 2 (p, χp) of Theorem 3.49, applied to f and h, implies a relation for the Fourier coeﬃcients of h, which leads to the claimed formula. Notice that the assumption in Theorem 3.65 that the constant term of f vanishes can be dropped. Then the Borcherds lift of f is a meromor-phic modular form of non-zero weight, and one can prove a formula for log ∥Ψ(CM(K), f)∥Pet, where ∥· ∥Pet denotes the Petersson metric on the line bundle of modular forms (see [BY] Theorem 1.4). In a recent preprint [Scho], Schofer obtained a formula for the evalua-tion of Borcherds products on O(2, n) at CM 0-cycles associated with bi-quadratic CM ﬁelds by means of a diﬀerent method. It would be interest-ing to use his results to derive explicit formulas as in Theorem 3.65 for the values of Hilbert modular functions at CM cycles associated to biquadratic CM ﬁelds. Finally, notice that Goren and Lauter have recently proved re-sults on the CM values of Igusa genus two invariants using arithmetic meth-ods [GL]. Examples We ﬁrst consider the real quadratic ﬁeld F = Q( √ 5) and the cyclic CM extension K = Q(ζ5), where ζ5 = e2πi/5. So p = q = 5. If σ denotes the complex embedding of K taking ζ5 to ζ2 5 then Φ = {1, σ} is a CM type of K. We have OK = OF + OF ζ5, and the corresponding CM cycle CM(K, Φ) is represented by the point (ζ5, ζ2 5) ∈H2. 176 J. H. Bruinier In Section 3.2 we constructed some Borcherds products for ΓF . Using the basis (fm) of W + 0 (p, χp) we see that the Borcherds products R1(z) = Ψ(z, f6 −2f1) = Ψ6 Ψ 2 1 , R2(z) = Ψ(z, f10 −2f1) = Ψ10 Ψ 2 1 are rational functions on Y (ΓF ) with divisors T6 −2T1 and T10 −2T1, re-spectively. Let us see what the above results say about R1(CM(K)). We have M = {1, 6} and N = 6. According to Corollary 3.68, the prime divisors of R1(CM(K)) are bounded by 9. Consequently, only the primes 2, 3, 5, 7 can occur in the factorization. The divisibility criterion given in Corollary 3.67 actually shows that only 2, 3, 5 can occur. The exact value is given by Corol-lary 3.66. It is equal to R1(CM(K)) = 220 · 310. In Table 4 we listed some further CM values of R1 and R2. Table 4. The case F = Q( √ 5) q R1(CM(K)) R2(CM(K)) 5 (cyclic) 220 · 310 220 · 510 41 214 · 310 · 61 · 73 214 · 59 · 37 · 41 61 220 · 36 · 13 · 97 · 109 220 · 59 · 61 109 220 · 38 · 61 · 157 · 193 220 · 512 · 73 149 220 · 310 · 312 · 37 · 229 220 · 512 · 17 · 113 269 220 · 310 · 13−2 · 372 · 61 · 97 · 349 · 433 220 · 514 · 13−1 · 53 · 73 · 233 References [AbSt] M. Abramowitz and I. Stegun, Pocketbook of Mathematical Functions, Ver-lag Harri Deutsch, Thun (1984). [Be] W. H. Berwick, Modular Invariants, Proc. Lond. Math. Soc. 28 (1927), 53–69. [Bo1] R. E. Borcherds, Automorphic forms on Os+2,2(R) and inﬁnite products, Invent. Math. 120 (1995), 161–213. [Bo2] R. E. Borcherds, Automorphic forms on Os+2,2(R)+ and generalized Kac-Moody algebras, Proceedings of the ICM 1994, Birkhäuser, Basel (1995), 744-752. [Bo3] R. E. Borcherds, Automorphic forms and Lie algebras, Current Develop-ments in mathematics 1996, International Press (1998). [Bo4] R. E. Borcherds, Automorphic forms with singularities on Grassmannians, Invent. Math. 132 (1998), 491–562. Hilbert Modular Forms 177 [Bo5] R. E. Borcherds, What is moonshine?, Proceedings of the International congress of Mathematicians, Doc. Math., Extra Vol. I (1998), 607-615. [Bo6] R. E. Borcherds, The Gross-Kohnen-Zagier theorem in higher dimensions, Duke Math. J. 97 (1999), 219–233. [Br1] J. H. Bruinier, Borcherds products and Chern classes of Hirzebruch–Zagier divisors, Invent. math. 138 (1999), 51–83. [Br2] J. H. Bruinier, Borcherds products on O(2, l) and Chern classes of Heegner divisors, Lect. Notes Math. 1780, Springer-Verlag, Berlin (2002). [Br3] J. H. Bruinier, Inﬁnite propducts in number theory and geometry, Jahres-ber. Dtsch. Math. Ver. 106 (2004), Heft 4, 151–184. [BB] J. H. Bruinier and M. Bundschuh, On Borcherds products associated with lattices of prime discriminant, Ramanujan J. 7 (2003), 49–61. [BrFr] J. H. Bruinier and E. Freitag, Local Borcherds products, Annales de l’Institut Fourier 51.1 (2001), 1–26. [BrFu1] J. H. Bruinier and J. Funke, On two geometric theta lifts, Duke Math. J. 125 (2004), 45–90. [BrFu2] J. H. Bruinier and J. Funke, On the injectivity of the Kudla– Millson lift and surjectivity of the Borcherds lift, preprint (2006). [BY] J. H. Bruinier and T. Yang, CM values of Hilbert modular functions, Invent. Math. 163 (2006), 229–288. [BBK] J. H. Bruinier, J. Burgos, and U. Kühn, Borcherds products and arith-metic intersection theory on Hilbert modular surfaces, Duke Math. J. 139 (2007), 1–88. [BKK] J. Burgos, J. Kramer, and U. Kühn, Cohomological Arithmetic Chow groups, J. Inst. Math. Jussieu. 6, 1–178 (2007). [Ch] C.-L. Chai, Arithmetic minimal compactiﬁcation of the Hilbert-Blumen-thal moduli spaces, Ann. Math. 131 (1990), 541–554. [DePa] P. Deligne and G. Pappas, Singularités des espaces de modules de Hilbert, en les caractéristiques divisant le discriminant, Compos. Math. 90 (1994), 59–79. [DI] F. Diamond and J. Im, Modular forms and modular curves, Canadian Mathematical Society Conference Proceedings 17 (1995), 39–133. [DN] K. Doi and H. Naganuma, On the functional equation of certain Dirichlet series, Invent. Math. 9 (1969), 1–14. [EZ] M. Eichler and D. Zagier, The Theory of Jacobi Forms, Progress in Math. 55 (1985), Birkhäuser. [Fra] H.-G. Franke, Kurven in Hilbertschen Modulﬂächen und Humbertsche Flächen im Siegelraum, Bonner Math. Schriften 104 (1978). [Fr] E. Freitag, Hilbert Modular Forms, Springer-Verlag, Berlin (1990). [Fu] J. Funke, Heegner divisors and nonholomorphic modular forms, Compos. Math. 133 (2002), 289-321. [Ga] P. Garrett, Holomorphic Hilbert modular forms. Wadsworth and Brooks Advanced Books, Paciﬁc Grove, (1990). [Ge1] G. van der Geer, Hilbert Modular Surfaces, Springer-Verlag, Berlin (1988). [Ge2] G. van der Geer, On the geometry of a Siegel modular threefold, Math. Ann. 260 (1982), 317–350. [Go] E. Z. Goren, Lectures on Hilbert Modular Varieties and Modular Forms, CRM Monograph Series 14, American Mathematical Society, Providence (2002). 178 J. H. Bruinier [GL] E. Z. Goren and K. E. Lauter, Class invariants for quartic CM ﬁelds, Ann. Inst. Fourier 57 (2007), 457–480. [GZ] B. Gross and D. Zagier, On singular moduli, J. Reine Angew. Math. 355 (1985), 191–220. [Gu] K.-B. Gundlach, Die Bestimmung der Funktionen zur Hilbertschen Modul-gruppe des Zahlkörpers Q( √ 5), Math. Annalen 152 (1963), 226–256. [HM] J. Harvey and G. Moore, Algebras, BPS states, and strings, Nuclear Phys. B 463 (1996), no. 2-3, 315–368. [Ha] W. Hausmann, Kurven auf Hilbertschen Modulﬂächen, Bonner Math. Schriften 123 (1980). [He] E. Hecke, Analytische Arithmetik der positiv deﬁniten quadratischen For-men, Kgl. Danske Vid. Selskab. Math. fys. Med. XIII 12 (1940). Werke, 789–918. [Hej] D. A. Hejhal, The Selberg Trace Formula for PSL(2, R), Lect. Notes Math. 1001, Springer-Verlag, Berlin (1983). [Hi] H. Hironaka, Resolution of singularities of an algebraic variety over a ﬁeld of characteristic zero I, II, Ann. Math. 79 (1964), 109–203 , 205–326. [Hz] F. Hirzebruch, The Hilbert modular group, resolution of the singularities at the cusps and related problems. Sem. Bourbaki 1970/71, Lect. Notes Math. 244 (1971), 275–288. [HZ] F. Hirzebruch and D. Zagier, Intersection Numbers of Curves on Hilbert Modular Surfaces and Modular Forms of Nebentypus, Invent. Math. 36 (1976), 57–113. [Ho] R. Howe, θ-series and invariant theory, Proceedings of Symposia in Pure Mathematics 33, part 1, American Mathematical Society (1979), 275–285. [Ki] Y. Kitaoka, Arithmetic of quadratic forms, Cambridge Tracts in Mathem-atics 106, Cambridge University Press, Cambridge (1993). [Kon] M. Kontsevich, Product formulas for modular forms on O(2, n), Séminaire Bourbaki 821 (1996). [Ku1] S. Kudla, Central derivatives of Eisenstein series and height pairings, Ann. Math. 146 (1997), 545–646. [Ku2] S. Kudla, Algebraic cycles on Shimura varieties of orthogonal type, Duke Math. J. 86 (1997), 39–78. [Ku3] S. Kudla, Integrals of Borcherds forms, Compos. Math. 137 (2003), 293– 349. [KM1] S. Kudla and J. Millson, The theta correspondence and harmonic forms I, Math. Ann. 274, (1986), 353–378. [KM2] S. Kudla and J. Millson, The theta correspondence and harmonic forms II, Math. Ann. 277, (1987), 267–314. [KM3] S. Kudla and J. Millson, Intersection numbers of cycles on locally symmet-ric spaces and Fourier coeﬃcients of holomorphic modular forms in several complex variables, IHES Publi. Math. 71 (1990), 121–172. [KR] S. Kudla and M. Rapoport, Arithmetic Hirzebruch–Zagier divisors, J. Reine Angew. Math. 515 (1999), 155–244. [Lo] E. Looijenga, Compactiﬁcations deﬁned by arrangements. II. Locally sym-metric varieties of type IV. Duke Math. J. 119 (2003), 527–588. [Ma] S. Mayer, Hilbert Modular Forms for the Fields Q( √ 5), Q( √ 13), and Q( √ 17), Thesis, University of Aachen (2007). [Mü] R. Müller, Hilbertsche Modulfunktionen zu Q( √ 5), Arch. Math. 45, 239– 251 (1985). Hilbert Modular Forms 179 [Na] H. Naganuma, On the coincidence of two Dirichlet series associated with cusp forms of Hecke’s “Neben”-type and Hilbert modular forms over a real quadratic ﬁeld, J. Math. Soc. Japan 25 (1973), 547–555. [Ne] J. Neukirch, Algebraic number theory, Grundlehren der Mathematischen Wissenschaften 322, Springer-Verlag, Berlin (1999). [Ni] D. Niebur, A class of nonanalytic automorphic functions, Nagoya Math. J. 52 (1973), 133–145. [Od1] T. Oda, On Modular Forms Associated with Indeﬁnite Quadratic Forms of Signature (2, n −2), Math. Ann. 231 (1977), 97–144. [Od2] T. Oda, A note on a geometric version of the Siegel formula for quadratic forms of signature (2, 2k), Sci. Rep. Niigata Univ. Ser. A No. 20 (1984), 13–24. [OT] T. Oda and M. Tsuzuki, Automorphic Green functions associated with the secondary spherical functions. Publ. Res. Inst. Math. Sci. 39 (2003), 451–533. [Ral] S. Rallis, Injectivity properties of liftings associated to Weil representa-tions, Compositio Math. 52 (1984), 139–169. [RS] S. Rallis and G. Schiﬀmann, On a relation between SL2 cusp forms and cusp forms on tube domains associated to orthogonal groups, Trans. AMS 263 (1981), 1–58. [Rap] M. Rapoport, Compactiﬁcations de l’espace de modules de Hilbert-Blumen-thal, Compos. Math. 36 (1978), 255–335. [Scha] W. Scharlau, Quadratic and Hermitian forms, Grundlehren der Mathema-tischen Wissenschaften 270, Springer-Verlag, Berlin (1985). [Scho] J. Schofer, Borcherds Forms and Generalizations of Singular Moduli, preprint (2006). [Sh1] G. Shimura, Introduction to the Arithmetic Theory of Automorphic Func-tions, Princeton University Press, Princeton (1971). [Sh2] G. Shimura, Abelian varieties with complex multiplication and modular functions, Princeton Univ. Press (1998). [SABK] C. Soulé, D. Abramovich, J.-F. Burnol, and J. Kramer, Lectures on Arakelov Geometry, Cambridge Studies in Advanced Mathematics 33, Cambridge University Press, Cambridge (1992). [Ya] T. Yang, Hilbert modular functions and their CM values, preprint (2005). [Yau] S.-T. Yau, Some function-theoretic properties of complete Riemannian Manifolds and their applications to geometry, Indiana Univ. Math. J. 25 (1976), 659–670. [Za1] D. Zagier, Modular Forms Associated to Real Quadratic Fields, Invent. Math. 30 (1975), 1–46. [Za2] D. Zagier, Modular forms whose Fourier coeﬃcients involve zeta-functions of quadratic ﬁelds. In: Modular Functions of One Variable VI, Lecture Notes in Math. 627, Springer-Verlag (1977), 105–169. Siegel Modular Forms and Their Applications Gerard van der Geer Korteweg-de Vries Instituut, Universiteit van Amsterdam, Plantage Muidergracht 24, 1018 TV Amsterdam, The Netherlands E-mail: geer@science.uva.nl Summary. These are the lecture notes of the lectures on Siegel modular forms at the Nordfjordeid Summer School on Modular Forms and their Applications. We give a survey of Siegel modular forms and explain the joint work with Carel Faber on vector-valued Siegel modular forms of genus 2 and present evidence for a conjecture of Harder on congruences between Siegel modular forms of genus 1 and 2. 1 Introduction Siegel modular forms generalize the usual modular forms on SL(2, Z) in that the group SL(2, Z) is replaced by the automorphism group Sp(2g, Z) of a uni-modular symplectic form on Z2g and the upper half plane is replaced by the Siegel upper half plane Hg. The integer g ≥1 is called the degree or genus. Siegel pioneered the generalization of the theory of elliptic modular forms to the modular forms in more variables now named after him. He was moti-vated by his work on the Minkowski–Hasse principle for quadratic forms over the rationals, cf., . He investigated the geometry of the Siegel upper half plane, determined a fundamental domain and its volume and proved a central result equating an Eisenstein series with a weighted sum of theta functions. No doubt, Siegel modular forms are of fundamental importance in number theory and algebraic geometry, but unfortunately, their reputation does not match their importance. And although vector-valued rather than scalar-valued Siegel modular forms are the natural generalization of elliptic modular forms, their reputation amounts to even less. A tradition of ill-chosen notations may have contributed to this, but the lack of attractive examples that can be han-dled decently seems to be the main responsible. Part of the beauty of elliptic modular forms is derived from the ubiquity of easily accessible examples. The accessible examples that we have of Siegel modular forms are scalar-valued Siegel modular forms given by Fourier series and for g > 1 it is diﬃcult to extract the arithmetic information (e.g., eigenvalues of Hecke operators) from the Fourier coeﬃcients. 182 G. van der Geer The general theory of automorphic representations provides a generaliza-tion of the theory of elliptic modular forms. But despite the obvious merits of this approach some of the attractive explicit features of the g = 1 theory are lost in the generalization. The elementary theory of elliptic modular forms (g = 1) requires little more than basic function theory, while a good grasp of the elementary the-ory of Siegel modular forms requires a better understanding of the geom-etry involved, in particular of the compactiﬁcations of the quotient space Sp(2g, Z)\Hg. A singular compactiﬁcation was provided by Satake and Baily-Borel and a smooth compactiﬁcation by Igusa in special cases and by Mumford c.s. by an intricate machinery in the general case. The fact that Sp(2g, Z)\Hg is the moduli space of principally polar-ized abelian varieties plays an important role in the arithmetic theory of modular forms. Even for g = 1 one needs the understanding of the ge-ometry of moduli space as a scheme (stack) over the integers and its co-homology as Deligne’s proof of the estimate \|a(p)\| ≤2p(k−1)/2 (the Ra-manujan conjecture) for the Fourier coeﬃcients of a Hecke eigenform of weight k showed. For quite some time the lack of a well-developed the-ory of moduli spaces of principally polarized abelian varieties over the in-tegers formed a serious hurdle for the development of the arithmetic the-ory. Fortunately, Faltings’ work on the moduli spaces of abelian varieties has provided us with the ﬁrst necessary ingredients of the arithmetic the-ory, both the smooth compactiﬁcation over Z as well as the Satake compact-iﬁcation over Z. It also gives the analogue of the Eichler–Shimura theorem which expresses Siegel modular forms in terms of the cohomology of local systems on Sp(2g, Z)\Hg. The fact that the vector-valued Siegel modular forms are the natural generalization of the classical elliptic modular forms becomes apparent if one studies the cohomology of the universal abelian variety. Examples of modular forms for SL(2, Z) are easily constructed using Eisen-stein series or theta series. These methods are much less eﬀective when deal-ing with the case g ≥2, especially if one is interested in vector-valued Siegel modular forms. Some examples can be constructed using theta series, but it is not always easy to calculate the Fourier coeﬃcients and more diﬃcult to extract the eigenvalues of the Hecke operators. We show that there is an alternative approach that uses the analogue of the classical Eichler–Shimura theorem. Since cohomology of a variety over a ﬁnite ﬁeld can be calculated by determining the number of rational points over extension ﬁelds one can count curves over ﬁnite ﬁelds to calculate traces of Hecke operators on spaces of vector-valued cusp forms for g = 2. This is joint work with Carel Faber. It has the pleasant additional feature that our forms all live in level 1, i.e. on the full Siegel modular group. We illustrate this by providing convincing evidence for a conjecture of Harder on congruences between the eigenvalues of Siegel modular forms of genus 2 and elliptic modular forms. Siegel Modular Forms and Their Applications 183 In these lectures we concentrate on modular forms for the full Siegel modu-lar group Sp(2g, Z) and leave modular forms on congruence subgroups aside. We start with the elementary theory and try to give an overview of the vari-ous interesting aspects of Siegel modular forms. An obvious omission are the Galois representations associated to Siegel modular forms. A good introduction to the Siegel modular group and Siegel modular forms is Freitag’s book . The reader may also consult the introductory book by Klingen . Two other references to the literature are the two books [94,95] by Shimura. Vector-valued Siegel modular forms are also discussed in a paper by Harris, . Acknowledgements. I would like to thank Carel Faber, Alex Ghitza, Chris-tian Grundh, Robin de Jong, Winfried Kohnen, Sam Grushevsky, Martin Weissman and Don Zagier for reading the manuscript and/or providing help-ful comments. Finally I would like to thank Kristian Ranestad for inviting me to lecture in Nordfjordeid in 2004. 2 The Siegel Modular Group The ingredients of the deﬁnition of ‘elliptic modular form’ are the group SL(2, Z), the upper half plane H, the action of SL(2, Z) on H, the concept of a holomorphic function and the factor of automorphy (cz + d)k. So if we want to generalize the concept ‘modular form’ we need to generalize these notions. But the upper half plane can be expressed in terms of the group as SL(2, Z)/SO(2), where SO(2) = U(1), a maximal compact subgroup, is the stabilizer of the point i = √−1. Therefore, the group is the central object and we start by generalizing the group. The group SL(2, Z) is the automorphism group of the lattice Z2 with the standard alternating form ⟨, ⟩with ⟨(a, b), (c, d)⟩= ad −bc . This admits an obvious generalization by taking for g ∈Z≥1 the lattice Z2g of rank 2g with basis e1, . . . , eg, f1, . . . , fg provided with the symplectic form ⟨, ⟩with ⟨ei, ej⟩= 0, ⟨fi, fj⟩= 0 and ⟨ei, fj⟩= δij , where δij is Kronecker’s delta. The symplectic group Sp(2g, Z) is by deﬁnition the automorphism group of this symplectic lattice Sp(2g, Z) := Aut(Z2g, ⟨, ⟩) . By using the basis of the e’s and the f’s we can write the elements of this group as matrices A B C D , where A, B, C and D are g × g integral matrices satisfying ABt = BAt, CDt = DCt and ADt −BCt = 1g. Here we write 1g for the g × g identity 184 G. van der Geer matrix. For g = 1 we get back the group SL(2, Z). The group Sp(2g, Z) is called the Siegel modular group (of degree g) and often denoted Γg. Exercise 1. Show that the conditions on A, B, C and D are equivalent to Ct · A −At · C = 0, Dt · B −Bt · D = 0 and Dt · A −Bt · C = 1g. The upper half plane H can be given as a coset space SL(2, R)/K with K = U(1) a maximal compact subgroup, and this admits a general-ization, but the desired generalization also admits a description as a half plane and with this we start: the Siegel upper half plane Hg is deﬁned as Hg = {τ ∈Mat(g × g, C): τ t = τ, Im(τ) > 0} , consisting of g × g complex symmetric matrices which have positive deﬁnite imaginary part (obtained by taking the imaginary part of every matrix entry). Clearly, we have H1 = H. An element γ = A B C D of the group Sp(2g, Z), sometimes denoted by (A, B; C, D), acts on the Siegel upper half plane by τ →γ(τ) = (Aτ + B)(Cτ + D)−1 . (1) Of course, we must check that this is well-deﬁned, in particular that Cτ + D is invertible. For this we use the identity (C¯ τ + D)t(Aτ + B) −(A¯ τ + B)t(Cτ + D) = τ −¯ τ = 2iy , (2) where we write τ = x + iy with x and y symmetric real g × g matrices. We claim that det(Cτ + D) ̸= 0. Indeed, if the equation (Cτ + D)ξ = 0 has a solution ξ ∈Cg then equation (2) implies ¯ ξtyξ = 0 and by the assumed positive deﬁniteness of y that ξ = 0. One can also check directly the identity (Cτ + D)t(γ(τ) −γ(τ)t)(Cτ + D) = (Cτ + D)t(Aτ + B) −(Aτ + B)t(Cτ + D) = τ −τ t = 0 that shows that γ(τ) is symmetric. Moreover, again by (2) and this last iden-tity we ﬁnd the relation between y′ = Im(γ(τ)) and y (C¯ τ + D)ty′(Cτ + D) = 1 2i(C¯ τ + D)t(γ(τ) −(γ(τ))t)(Cτ + D) = y and this shows that y′ = Im(γ(τ)) is positive deﬁnite. Using these details one easily checks that (1) deﬁnes indeed an action of Sp(2g, Z), and even of Sp(2g, R) on Hg. Siegel Modular Forms and Their Applications 185 The group Sp(2g, R)/{±1} acts eﬀectively on Hg and it is the biholomor-phic automorphism group of Hg. The action is transitive and the stabilizer of i 1g is U(g) := $ A B −B A ∈Sp(2g, R): A · At + B · Bt = 1g % , the unitary group. We may thus view Hg as the coset space Sp(2g, R)/U(g) of a simple Lie group by a maximal compact subgroup (which is unique up to conjugation). The disguise of H1 as the unit disc {z ∈C: \|z\| < 1} also has an ana-logue for Hg. The space Hg is analytically equivalent to a bounded symmetric domain Dg := {Z ∈Mat(g × g, C): Zt = Z, Zt · Z < 1g} and the generalized Cayley transform τ →z = (τ −i1g)(τ + i1g)−1, z →τ = i · (1g + z)(1g −z)−1 makes the correspondence explicit. The ‘symmetric’ in the name refers to the existence of an involution on Hg (or Dg) τ →−τ −1 (z →−z) having exactly one isolated ﬁxed point. Note that we can write Hg also as Sg +iS+ g with Sg (resp. S+ g ) the R-vector space (resp. cone) of real symmetric (resp. real positive deﬁnite symmetric) matrices of size g × g. The group Sp(2g, Z) is a discrete subgroup of Sp(2g, R) and acts properly discontinuously on Hg, i.e., for every τ ∈Hg there is an open neighborhood U of τ such that {γ ∈Sp(2g, Z): γ(U) ∩U ̸= ∅} is ﬁnite. In fact, this follows immediately from the properness of the map Sp(2g, R) →Sp(2g, R)/U(g). For g = 1 usually one proceeds after these introductory remarks on the action to the construction of a fundamental domain for the action of SL(2, Z) and all the texts display the following archetypical ﬁgure. Siegel (see ) constructed also a fundamental domain for g ≥2, namely the set of τ = x + iy ∈Hg satisfying the following three conditions: 186 G. van der Geer 1. We have \| det(Cτ + D)\| ≥1 for all (A, B; C, D) ∈Γg; 2. the matrix y is reduced in the sense of Minkowski; 3. the entries xij of x satisfy \|xij\| ≤1/2. Here Minkowski reduced means that y satisﬁes the two properties 1) htyh ≥ ykk (k = 1, . . . , g) for all primitive vectors h in Zg and 2) yk,k+1 ≥0 for 0 ≤k ≤g −1. Already for g = 2 the boundary of this fundamental domain is complicated; Gottschling found that it posesses 28 boundary pieces, cf., , and the whole thing does not help much to understand the nature of the quotient space Sp(2g, Z)\Hg. The group Sp(2g, Z) does not act freely on Hg, but the subgroup Γg(n) := {γ ∈Sp(2g, Z): γ ≡12g (mod n)} acts freely if n ≥3 as is easy to check, cf. . The quotient space (orbit space) Yg(n) := Γg(n)\Hg is then for n ≥3 a complex manifold of dimension g(g + 1)/2. Note that the ﬁnite group Sp(2g, Z/nZ) acts on Yg(n) as a group of biholomorphic auto-morphisms and we can thus view Sp(2g, Z)\Hg as an orbifold (quotient of a manifold by a ﬁnite group). The Poincaré metric on the upper half plane also generalizes to the Siegel upper half plane. The corresponding volume form is given by (det y)−(g+1) i≤j dxij dyij which is ∂∂log det Im(τ)g. The volume of the fundamental domain was calcu-lated by Siegel, . If we normalize the volume such that it gives the orbifold Euler characteristic the result is (cf. Harder ) vol(Sp(2g, Z)\Hg) = ζ(−1)ζ(−3) · · · ζ(1 −2g) with ζ(s) the Riemann zeta function. In particular, for n ≥3 the Euler number of the manifold Γg(n)\Hg equals [Γg(1) : Γg(n)]ζ(−1) · · · ζ(1 −2g). We ﬁrst present two exercises for the solution of which we refer to . Exercise 2. Show that the Siegel modular group Γg is generated by the elem-ents 1g s 0 1g with s = st symmetric and the element 0 1g −1g 0 . Exercise 3. Show that Sp(2g, Z) is contained in SL(2g, Z). Siegel Modular Forms and Their Applications 187 We close with another model of the domain Hg that can be obtained as follows. Extend scalars of our symplectic lattice (Z2g, ⟨, ⟩) to C and let Yg be the Lagrangian Grassmann variety parametrizing totally isotropic subspaces of dimension g: Yg := {L ⊂C2g : dim(L) = g, ⟨x, y⟩= 0 for all x, y ∈L} . Since the group Sp(2g, C) acts transitively on the set of totally isotropic sub-spaces we may identify Yg with the compact manifold Sp(2g, C)/Q, where Q is the parabolic subgroup that ﬁxes the ﬁrst summand Cg. Consider now in Yg the open set Y + g of Lagrangian subspaces L such that −i⟨x, ¯ x⟩> 0 for all non-zero x in L. Then Y + g is stable under the action of Sp(2g, R) and the stabilizer of a point is isomorphic to the unitary group U(g). A basis of such an L is given by the columns of a unique 2g × g matrix −1g τ with τ ∈Hg and this embeds Hg in Yg as the open subset Y + g ; for g = 1 we get the upper half plane in P1. The manifold Yg is called the compact dual of Hg. Remark 1. Just as for g = 1 we could consider congruence subgroups of Sp(2g, Z), like for example Γg(n), the kernel of the natural homomorphism Sp(2g, Z) →Sp(2g, Z/nZ) for natural numbers n. We shall stick to the full symplectic group Sp(2g, Z) here. 3 Modular Forms To generalize the notion of modular form as we know it for g = 1 we still have to generalize the ‘automorphy factor’ (cz + d)k. To do this we consider a representation ρ: GL(g, C) →GL(V ) with V a ﬁnite-dimensional C-vector space. For reasons that become clear later, it is useful to provide V with a her-mitian metric ( , ) such that (ρ(g)v1, v2) = (v1, ρ(gt)v2) and we shall put ∥v∥= (v, v)1/2. Such a hermitian metric can always be found and is unique up to a scalar for irreducible representations. Deﬁnition 1. A holomorphic map f : Hg →V is called a Siegel modular form of weight ρ if f(γ(τ)) = ρ(Cτ + D)f(τ) for all γ = A B C D ∈Sp(2g, Z) and all τ ∈Hg, plus for g = 1 the requirement that f is holomorphic at ∞. Before we proceed, a word about notations. The subject has been plagued with unfortunate choices of notations, and the tradition of using capital letters for the matrix blocks of elements of the symplectic group is one of them. 188 G. van der Geer I propose to use lower case letters, so I will write f(γ(τ)) = ρ(cτ + d)f(τ) for all γ = (a, b; c, d) ∈Γg for our condition. The modular forms we consider here are vector-valued modular forms. As it turns out, the holomorphicity condition is not necessary for g > 1, see the Koecher principle hereafter. Modular forms of weight ρ form a C-vector space Mρ = Mρ(Γg) and we shall see later (in Section 13) that all the Mρ are ﬁnite-dimensional. If ρ is a direct sum of two representations ρ = ρ1 ⊕ρ2 then Mρ is isomorphic to the direct sum Mρ1 ⊕Mρ2 and this allows us to restrict ourselves to studying Mρ for the irreducible representations of GL(g, C). As is well-known (see , but see also the later Section 12), the irreducible ﬁnite-dimensional representations of GL(g, C) correspond bijectively to the g-tuples (λ1, . . . , λg) of integers with λ1 ≥λ2 ≥· · · ≥λg, the highest weight of the representation ρ. That is, for each irreducible V there exists a unique 1-dimensional subspace ⟨vρ⟩of V such that ρ(diag(a1, . . . , ag)) acts on vρ by multiplication by g i=1 aλi i . For example, the g-tuple (1, 0, . . . , 0) corresponds to the tautological representation ρ(x) = x for x ∈GL(g, C), while the deter-minant representation corresponds to λ1 = . . . = λg = 1. Tensoring a given irreducible representation with the k-th power of the determinant changes the λi to λi + k. We thus can arrange that λg = 0 or that λg ≥0 (i.e. that the representation is ‘polynomial’). Let R be the set of isomorphism classes of representations of GL(g, C). This set forms a ring with ⊕as addition and ⊗ as multiplication. It is called the representation ring of GL(g, C). For g = 1 one usually forms a graded ring of modular forms by taking M∗(Γ1) = ⊕Mk(Γ1). We can try do something similar for g > 1 and try to make the direct sum ⊕ρ∈RMρ(Γg) into a graded ring. But of course, this is a huge ring, even for g = 1 much larger than M∗(Γ1) since it involves also the reducible representations and it is not really what we want. The classes of the irreducible representations of GL(g, C) form a subset of all classes of representations. For g = 1 and g = 2 the fact is that the tensor product of two irreducible representations is a direct sum of irreducible repre-sentations with multiplicity 1. In fact, for g = 1 the tensor product of the irre-ducible representations ρk1 and ρk2 of degree k1+1 and k2+1 is the irreducible representation ρk1+k2. For g = 2, a case that will play a prominent role in these lecture notes, we let ρj,k denote the irreducible representation of GL(2, C) that is Symj(W) ⊗det(W)k with W the standard 2-dimensional representation; it corresponds to highest weight (λ1, λ2) = (j + k, k). Then there is the formula ρj1,k1 ⊗ρj2,k2 ∼ = min(j1,j2) r=0 ρj1+j2−2r,k1+k2+r . So we can decompose Mρj1,k1 as a direct sum min(j1,j2) r=0 Mρj1+j2−2r,k1+k2+r, but this is not canonical as it depends upon a choice of isomorphism in the above formula. Nevertheless, this decomposition is useful to construct modular forms in new weights by multiplying modular forms. Siegel Modular Forms and Their Applications 189 To make ⊕ρ∈IrrMρ(Γ2) into a ring requires a consistent choice for all these identiﬁcations. We can avoid this by viewing the symmetric power Symj(W) as a space of polynomials of degree j in two variables and then by remarking that multiplication of polynomials deﬁnes a canonical map Symj1(W) ⊗Symj2(W) →Symj1+j2(W). Using this and the obvious map det(W)k1 ⊗det(W)k2 →det(W)k1+k2 the direct sum ⊕ρ∈IrrMρ(Γ2) be-comes a ring; we just ‘forgot’ the terms in the above sum with r > 0. For g ≥3 the tensor products come in general with multiplicities, given by Littlewood–Richardson numbers. Nevertheless, one can deﬁne a ring struc-ture on ⊕ρ∈IrrMρ(Γg) that extends the multiplication of modular forms for g = 1 and the one given here for g = 2 as Weissman shows. We refer to his interesting paper, . For every g one obtains a subring of the representation ring by taking the powers of the determinant det : GL(g, C) →C∗. This leads to a ring of ‘classical’ modular forms. Deﬁnition 2. A classical Siegel modular form of weight k (and degree g) is a holomorphic function f : Hg →C such that f(γ(τ)) = det(cτ + d)kf(τ) for all γ = (a, b; c, d) ∈Sp(2g, Z) (with for g = 1 the usual holomorphicity requirement at ∞). Classical Siegel modular forms are also known as scalar-valued Siegel modular forms. Let Mk = Mk(Γg) be the vector space of classical Siegel modular forms of weight k. Together these spaces form a graded ring M cl := ⊕Mk of M of classical Siegel modular forms. Of course, for g = 1 the notion of classical modular form reduces to the usual notion of modular form on SL(2, Z). 4 The Fourier Expansion of a Modular Form The classical Fourier expansion of a modular form on SL(2, Z) has an analogue. To deﬁne it we need the following deﬁnition. Deﬁnition 3. A symmetric g × g-matrix n ∈GL(g, Q) is called half-integral if 2n is an integral matrix the diagonal entries of which are even. Every half-integral g × g-matrix n deﬁnes a linear form with integral co-eﬃcients in the coordinates τij with 1 ≤i ≤j ≤g of Hg, namely Tr(nτ) = g i=1 niiτii + 2 1≤i c · 1g} with c > 0. Proof. For g = 1 the boundedness comes from the requirement in the def-inition that the Fourier expansion f = n a(n)qn has no negative terms. So suppose that g ≥2 and let f = n a(n)e2πiTrnτ ∈Mρ(Γg). Since f Siegel Modular Forms and Their Applications 191 converges absolutely on Hg we see by substitution of τ = i · 1g that there exists a constant c > 0 such that for all half-integral matrices we have \|a(n)\| ≤ce2πTrnτ. We ﬁrst will show that a(n) vanishes for n that are not positive semi-deﬁnite. Suppose that n is not positive semi-deﬁnite. Then there exists a primitive integral (column) vector ξ such that ξtnξ < 0. We can complete ξ to a uni-modular matrix u. Using the relation a(utnu) = ρ(ut)a(n) and replacing n by utnu we may assume that entry n11 of n is negative. Consider now for m ∈Z the matrix v = ⎛ ⎝ 1 m 0 1 1g−2 ⎞ ⎠∈GL(g, Z) , where the omitted entries are zero. We have \|a(n)\| = \|ρ(vt)−1\| \|a(vtnv)\| ≤ce2πTrvtnv . But Tr(vtnv) = Tr(v) + n11m2 + 2n12m and if m →∞then this expression goes to −∞, so \|a(n)\| = 0. We conclude that f = n≥0 a(n)e2πiTrnτ. We can now majorize by the value at c i · 1g of f, viz. n≥0 \|a(n)\|e−2πTrnc, uniformly in τ on {τ ∈ Hg : Im(τ) > c · 1g}. The proof of this theorem shows the validity of the so-called Koecher principle announced above. Theorem 2. (Koecher Principle) Let f = n a(n)qn ∈Mρ(Γg) with qn = e2πiTr(nτ) be a modular form of weight ρ. Then a(n) = 0 if the half-integral matrix n is not positive semi-deﬁnite. The Koecher principle was ﬁrst observed in 1928 by Götzky for Hilbert modu-lar forms and in general by Koecher in 1954, see and Bruinier’s lectures. Corollary 2. A classical Siegel modular form of negative weight vanishes. Proof. Let f ∈Mk(Γg) with k < 0. Then the function h = det(y)k/2\|f(τ)\| is invariant under Γg since Im(γ(τ)) = (cτ + d)−t(Im(τ))(cτ + d) −1. It is not diﬃcult to see that a fundamental domain is contained in {τ ∈Hg : Tr(x2) < 1/c, y > c · 1g} for some suitable c. This implies that for negative k the expression det(y)k/2 is bounded on a fundamental domain, and by the Koecher principle f is bounded on {τ ∈Hg : det y ≥c}. It follows that h is bounded on Hg, say h ≤c′ and with a(n)e−2πTrny = x mod 1 f(τ)e−2πTrnxdx we get \|a(n)\|e−2πTrny ≤ sup x mod 1 \|f(x + iy)\| ≤c′ det y−k/2 . If we let y →0 then for k < 0 we see \|a(n)\| = 0 for all n ≥0. 192 G. van der Geer This corollary admits a generalization for vector-valued Siegel modular forms, cf., : Proposition 1. Let ρ be a non-trivial irreducible representation of GL(g, C) with highest weight λ1 ≥. . . ≥λg. If Mρ ̸= {0} then we have λg ≥1. One proves this by taking a totally real ﬁeld K of degree g over Q and by identifying the symplectic space OK ⊕O∨ K (with O∨ K the dual of OK with respect to the trace) with our standard symplectic space (Z2g, ⟨, ⟩). This induces an embedding SL(2, OK) →Sp(2g, Z) and a map SL(2, OK)\Hg 1 → Sp(2g, Z)\Hg. Pulling back Siegel modular forms yields Hilbert modular forms on SL(2, OK). Now use that a Hilbert modular form of weight (k1, . . . , kg) vanishes if one of the ki ≤0, cf., . By varying K one sees that if λg ≤0 then a non-constant f vanishes on a dense subset of Hg. 5 The Siegel Operator and Eisenstein Series Since modular forms f ∈Mρ(Γg) are bounded in the sets of the form {τ ∈ Hg : Im(τ) > c · 1g} we can take the limit. Deﬁnition 4. We deﬁne an operator Φ on Mρ(Γg) by Φf = lim t→∞f τ ′ 0 0 it with τ ′ ∈Hg−1, t ∈R. In view of the convergence we can also apply this limit to all terms in the Fourier series and get (Φf)(τ′) = n′≥0 a n′ 0 0 0 e2πiTr(n′τ ′) . The values of Φf generate a subspace V ′ ⊆V that is invariant under the action of the subgroup of matrices {(a, 0; 0, 1): a ∈GL(g −1, C)} and that deﬁnes a representation ρ′ of GL(g −1, C). The operator Φ deﬁned on Siegel modular forms of degree g is called the Siegel operator and deﬁnes a linear map Mρ(Γg) →Mρ′(Γg−1). If ρ is the irreducible representation with highest weight (λ1, . . . , λg) then Φ maps Mρ(Γg) to Mρ′(Γg−1) with ρ′ the irreducible representation of GL(g −1, C) with highest weight (λ1, . . . , λg−1). Deﬁnition 5. A modular form f ∈Mρ is called a cusp form if Φf = 0. The subspace of Mρ of cusp forms is denoted by Sρ = Sρ(Γg). Exercise 4. Show that a modular f = a(n)e2πiTr(nτ) ∈Mρ is a cusp form if and only if a(n) = 0 for all semi-deﬁnite n that are not deﬁnite. Siegel Modular Forms and Their Applications 193 We can apply the Siegel operator repeatedly (say r ≤g times) to a Siegel modular form on Γg and one thus obtains a Siegel modular form on Γg−r. If ρ is irreducible with highest weight (λ1, . . . , λg) and ΦF = f ̸= 0 for some F ∈Mρ(Γg) then necessarily λg ≡0(mod2) because with γ also −γ lies in Γg. Let now f1 and f2 be modular forms of weight ρ, one of them a cusp form. Then we deﬁne the Petersson product of f1 and f2 by ⟨f1, f2⟩= F (ρ(Im(τ))f1(τ), f2(τ))dτ , where dτ = det(y)−(g+1) i≤j dxijdyij is an invariant measure on Hg, F is a fundamental domain for the action of Γg on Hg and the brackets ( , ) refer to the Hermitian product deﬁned in Section 3. One checks that it converges exactly because at least one of the two forms is a cusp form. Furthermore, we deﬁne Nρ = S⊥ ρ , for the orthogonal complement of Sρ and then have an orthogonal decompo-sition Mρ = Sρ ⊕Nρ. Just as in the case g = 1 one can construct modular forms explicitly using Eisenstein series. We ﬁrst deal with the case of classical Siegel modular forms. Let g ≥1 be the degree and let r be a natural number with 0 ≤r ≤g. Suppose that f ∈Sk(Γr) is a (classical Siegel modular) cusp form of even weight k. For a matrix τ1 z z τ2 with τ1 ∈Hr and τ2 ∈Hg−r we write τ∗= τ1 ∈Hr. (For r = 0 we let τ∗be the unique point of H0.) If k is positive and even we deﬁne the Klingen Eisenstein series, a formal series, Eg,r,k(f) := A=(a,b;c,d)∈Pr\Γg f((aτ + b)(cτ + d)−1)∗) det(cτ + d)−k , where Pr is the subgroup Pr := ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ ⎛ ⎜ ⎜ ⎝ a′ 0 b′ ∗ ∗u ∗ ∗ c′ 0 d′ ∗ 0 0 0 u−t ⎞ ⎟ ⎟ ⎠∈Γg : a′ b′ c′ d′ ∈Γr, u ∈GL(g −r, Z) ⎫ ⎪ ⎪ ⎬ ⎪ ⎪ ⎭ . For an interpretation of this subgroup we refer to Section 11. In case r = 0, f constant, say f = 1, we get the old Eisenstein series Eg,0,k = (a,b;c,d) det(cτ + d)−k , where the summation is over a full set of representatives for the cosets GL(g, Z)\Γg. 194 G. van der Geer Theorem 3. Let g ≥1 and 0 ≤r ≤g and k > g + r + 1 be integers with k even. Then for every cusp form f ∈Sk(Γr) the series Eg,r,k(f) converges to a classical Siegel modular form of weight k in Mk(Γg) and Φg−rEg,r,k(f) = f. This theorem was proved by Hel Braun1 in 1938 for r = 0 and k > g + 1. The Fourier coeﬃcients of these Eisenstein series were determined by Maass, see . Often we shall restrict the summation over co-prime (c, d) in order to avoid an unnecessary factor. Corollary 3. The Siegel operator Φ : Mk(Γg) →Mk(Γg−1) is surjective for even k > 2g. Weissauer improved the above result and proved that Φr is surjective if k > (g + r + 3)/2, see . He also treated the case of vector-valued modular forms and showed that the image Φ(Mρ(Γg)) contains the space of cusp forms Sρ′(Γg−1) if k = λg ≥g + 2, see loc. cit. p. 87. If k is odd we have no good Eisenstein series; for example look at the Siegel operator Mk(Γg) →Mk(Γg−1) for k ≡g ≡1 (mod2). Then Mk(Γg) = (0) while the target space Mk(Γg−1) is non-zero for suﬃciently large k (e.g. M35(Γ2) ̸= (0) as we shall see later). Just as for g = 1 one can construct Poincaré series and use these to generate the spaces of cusp forms if the weight is suﬃciently high. These Poincaré series behave well with respect to the Petersson product. We refer to , Ch. 6, or for the general setting. 6 Singular Forms A particularity of g > 1 are the so-called singular modular forms. Deﬁnition 6. A modular form f = n a(n)e2πiTrnτ ∈Mk(Γg) is called sin-gular if a(n) ̸= 0 implies that n is a singular matrix (det(n) = 0). Modular forms of small weight are singular as the following theorem shows, see . Theorem 4. (Freitag, Saldaña, Weissauer) Let ρ be irreducible with highest weight (λ1, . . . , λg). A non-zero modular form f ∈Mρ(Γg) is singular if and only if 2λg < g. In particular, there are no cusp forms of weight 2λg < g. One deﬁnes the co-rank of an irreducible representation as #{1 ≤i ≤g: λi = λg}. For a modular form f = n a(n) exp(2πiTrnτ) ∈Mρ(Γg), Weissauer introduced the rank and co-rank of f by 1 Hel Braun was a student of Carl Ludwig Siegel (1896–1981), the mathematician after whom our modular forms are named. She sketches an interesting portrait of Siegel in Siegel Modular Forms and Their Applications 195 rank(f) = max{rank(n): a(n) ̸= 0} and co-rank(f) = g −min{rank(n): a(n) ̸= 0} . In particular, modular forms of rank <g are singular while cusp forms have co-rank 0 and Siegel–Eisenstein forms Eg,0,k have co-rank g; Φ applied k + 1 times to forms of co-rank k should be zero. Weissauer proved (see ) for irreducible ρ that co-rank(f) ≤co-rank(ρ) and also that Mρ(Γg) = (0) if λg ≤g/2 −co-rank(ρ). More precisely, he proved Theorem 5. Let ρ = (λ1, . . . , λg) be an irreducible representation of co-rank < g −λg. If #{i: 1 ≤i ≤g, λi = λg + 1} < 2(g −λg −co-rank (ρ)) then Mρ = (0). Finally, Duke and Imamoˇ glu prove in that there are no cusp forms of small weights; for example, S6(Γg) = (0) for all g. 7 Theta Series Besides Eisenstein series one can construct Siegel modular forms using theta series. We begin with the so-called theta-constants. Let ϵ = ϵ′ ϵ′′ with ϵ′, ϵ′′ ∈ {0, 1}g and consider the rapidly converging series θ[ϵ] = m∈Zg exp 2πi m + 1 2ϵ′ t τ m + 1 2ϵ′ + 1 2 m + 1 2ϵ′ t (ϵ′′) 0 . This vanishes identically if ϵ is odd, that is, if ϵ′(ϵ′′)t is odd. The other 2g−1(2g + 1) cases (the ‘even’ ones) yield the so-called even theta characteris-tics. These are modular forms on a level 2 congruence subgroup of Sp(2g, Z) of weight 1/2, cf. . These can be used to construct classical Siegel modular forms on Sp(2g, Z). For example, for g = 1 one has θ[0 0]θ[0 1]θ[1 0] 8 = 28Δ ∈S12(Γ1) . For g = 2 the product −2−14 θ[ϵ]2 of the squares of the ten even theta characteristics gives a cusp form χ10 of weight 10 on Sp(4, Z), cf. [52–54]. Similarly, an expression θ[ϵ]) ±(θ[ϵ1]θ[ϵ2]θ[ϵ3] 20 , where the product is over the even theta characteristics and the sum is over so-called azygous triples of theta characteristics (i.e., triples such that ϵ1+ϵ2+ϵ3 is odd) deﬁnes (up to a normalization −2−395−3i) a cusp form χ35 of weight 35 on Sp(4, Z). Similarly, for g = 3 the product of the 36 even theta characteristics 196 G. van der Geer deﬁnes a cusp form of weight 18 on Sp(6, Z). The reason why one needs such a complicated expression is that the theta characteristics are modular forms on a subgroup Γg(4, 8) of Sp(2g, Z) and the quotient group Sp(2g, Z)/Γg(4, 8) permutes them and creates signs in addition so that we need a sort of sym-metrization to get something invariant. Another source of Siegel modular forms are theta series associated to even unimodular lattices. Let B be a positive deﬁnite symmetric even unimodular matrix of size r ≡0(mod 8). We denote by Hk(r, g) the space of harmonic polynomials P : Cr×g →C satisfying for M ∈GL(g, C) the identity P(zM) = det(M)kP(z). Recall that harmonic means that i,j ∂2/∂z2 ij P(z) = 0 if zij are the coordinates on Cr×g. For a pair (B, P) with P ∈Hk(r, g) we set θB,P (τ) = A∈Zr×g P( √ BA)eπiTr(AtBAτ) , where √ B is a positive matrix with square B. Then θB,P is a classical Siegel modular form in Mk+r/2(Γg), see . Such theta series for P ∈Hk−r/2,g and B as above span a subspace of Mk(Γg) that is invariant under the Hecke-operators that will be introduced later, cf. Section 16. There are analogues of these that give vector-valued Siegel modular forms if we require that P is a vector-valued polynomial satisfying the relation P(zM) = ρ(M)P(z). See also Section 25 and [48,49] for an example. Finally, we would like to make a reference to Siegel’s Hauptsatz (or , p. 285) on representations of quadratic forms by quadratic forms which can be viewed as an identity between an Eisenstein series and a weighted sum of theta series, and to its far-reaching generalizations, cf. . 8 The Fourier–Jacobi Development of a Siegel Modular Form As we saw above, just as for g = 1 we have a Fourier expansion of a Siegel modular form f = n≥0 a(n)e2πiTr(nτ). But for g > 1 there are other de-velopments that provide more information, like the so-called Fourier–Jacobi development, a concept due to Piatetski–Shapiro. We consider classical Siegel modular forms of weight k on Γg. We write τ ∈Hg as τ = τ′ z zt τ ′′ with τ ′ ∈H1 , z ∈Cg−1and τ ′′ ∈Hg−1 . (4) From the deﬁnition of modular form it is clear that f is invariant under τ′ →τ ′ + b for b ∈Z (given by an element of Sp(2g, Z)), hence we have a Fourier series f = ∞ m=0 φm(τ ′′, z)e2πimτ ′ . Siegel Modular Forms and Their Applications 197 Here the function φm is a holomorphic function on Hg−1 × Cg−1 satisfying certain transformation rules. More generally, if we split τ as in (4) but with τ′ ∈Hr, z ∈Cr(g−r) and τ ′′ ∈Hg−r we ﬁnd a development m φm(τ ′′, z)e2πiTr(mτ ′) , where the sum is over positive semi-deﬁnite half-integral matrices r ×r matri-ces m and the functions φm are holomorphic on Hr × Cr(g−r). For r = g we get back the Fourier expansion and for r = 1 we get what is called the Fourier–Jacobi development. For ease of explanation and to simplify matters we start with g = 2. Then the function φm(τ ′, z) turns out to be a Jacobi form of weight k and index m, i.e., φm ∈Jk,m which amounts to saying that it satisﬁes 1. φm((aτ′ + b)/(τ ′ + d), z/(cτ′ + d)) = (cτ′ + d)ke2πimcz2/(cτ ′+d)φm(τ ′, z), 2. φm(τ ′, z + λτ ′ + μ) = e−2πim(λ2τ ′+2λz)φm(τ ′, z), 3. φm has a Fourier expansion of the form φm = ∞ n=0 r∈Z, r2≤4mn c(n, r)e2π(nτ ′+rz) . This gives a relation between Siegel modular forms for genus 2 and Jacobi forms (see ) that we shall exploit later. In the general case, if we split τ as τ = τ′ z zt τ ′′ with τ ′ ∈Hr, z ∈Cr(g−r) and τ ′′ ∈Hg−r and a symmetric matrix n as n′ ν νt n′′ and if we use the fact that Tr(nτ) = Tr(n′τ ′) + 2Tr(νz) + Tr(n′′τ ′′) then we can decompose the Fourier series of f ∈Mρ(Γg) as n′′≥0 φn′′(τ ′, z)e2πiTr(n′′τ ′′) with V -valued holomorphic functions φn′′(τ ′, z) that satisfy the rules 1. For λ, μ ∈Zg we have φn′′(τ ′, z + τ ′λ + μ) = ρ 1r −λ 0 1g−r e−2πiTr(2λtz+λtτ ′λ)φn′′(τ ′, z) . 2. For γ′ = (a′, b; c′, d′) ∈Γg−1 we have φn′′(γ′(τ ′), (c′τ ′ + d′)−tz) = e2πiTr(n′′zt(c′τ ′+d′)−1c′z)ρ c′τ ′ + d′ c′z 0 1g−r φn′′(τ ′, z) . 3. φn′′(τ ′, z) is regular at inﬁnity. 198 G. van der Geer The last condition means that φn′′(τ ′, z) has a Fourier expansion φn′′(τ ′, z) = c(m, r) exp(2πiTr(mτ ′ + 2rtz)) for which c(m, r) ̸= 0 implies that m r rt n′′ is positive semi-deﬁnite. A holomorphic V -valued function φ(τ′, z) satisfying 1), 2) and 3) is called a Jacobi form of weight (ρ′, n′′). The sceptical reader may frown upon this unattractive set of transformation formulas, but there is a natural geometric explanation for this transformation behavior that we shall see in Section 11. 9 The Ring of Classical Siegel Modular Forms for Genus Two So far we have not met any striking examples of Siegel modular forms. To convince the reader that the subject is worthy of his attention we turn to the ﬁrst non-trivial case: classical Siegel modular forms of genus 2. For g = 1 we know the structure of the graded ring M∗(Γ1) = ⊕kMk(Γ1). It is a polynomial ring generated by the Eisenstein series e4 = E(1) 4 and e6 = E(1) 6 and the ideal of cusp forms is generated by the famous cusp form Δ = (e3 4 −e2 6)/1728 of weight 12. In comparison to this our knowledge of the graded ring ⊕ρ∈IrrMρ of Siegel modular forms for g = 2 is rather restricted and most of what we know con-cerns classical Siegel modular forms. A ﬁrst basic result was the determination by Igusa of the ring of classical Siegel modular forms for g = 2. We now know also the structure of the ring of classical Siegel modular forms for g = 3, a result of Tsuyumine, . Recall that we have the Eisenstein Series E(g) k ∈Mk(Γg) for k > g + 1. In particular, for g = 2 we have E4 = E(2) 4 ∈M4(Γ2) and E6 = E(2) 6 ∈M6(Γ2). Let us normalize them here so that Ek = (c,d) det(cτ + d)−k , where the sum is over non-associated pairs of co-prime symmetric integral matrices (non-associated w.r.t. to the multiplication on the left by GL(g, Z)). The Fourier expansion of these modular forms is known. If we write τ = τ1 z z τ2 then Ek = N a(N)e2πiTr(Nτ) , with constant term 1 and for non-zero N = n r/2 r/2 m the coeﬃcient a(N) given as a(N) = d\|(n,r,m) dk−1H k −1, 4mn −r2 d2 Siegel Modular Forms and Their Applications 199 with H(k −1, D) Cohen’s function, i.e., H(k −1, D) = L−D(2 −k), where LD(s) = L(s, D ) is the Dirichlet L-series associated to D if D is 1 or a discriminant of a real quadratic ﬁeld, cf., , p. 21. (This H(k −1, D) is essentially a class number.) Explicitly we have with qj = e2πiτj and ζ = e2πiz the developments (cf., ) E4 = 1 + 240(q1 + q2)+2160 q2 1 + q2 2 (240 ζ−2+13440 ζ−1 + 30240 + 13440 ζ + 240 ζ2)q1q2 + . . . and E6 = 1 −504(q1 + q2) −16632 q2 1 + q2 2 + +(−504 ζ−2 + 44352 ζ−1 + 166320 + 44352 ζ −504 ζ2)q1q2 + . . . . Under Siegel’s operator Φ: Mk(Γ2) →Mk(Γ1) the Eisenstein series Ek on Γ2 maps to the Eisenstein series ek on Γ1 for k ≥4. In particular, the modular form E10 −E4E6 maps to e10 −e4e6, and this is zero since dim M10(Γ1) = 1 and the ek are normalized so that their Fourier expansions have constant term 1. We thus ﬁnd a cusp form. Similarly, E12 −E2 6 deﬁnes a cusp form of weight 12 on Γ2. To see that these are not zero we restrict to the ‘diagonal’ locus as follows. Consider the map δ: H1 × H1 →H2 given by (τ1, τ2) → τ1 0 0 τ2 . There is a corresponding map SL(2, Z) × SL(2, Z) →Sp(4, Z) by sending a b c d , a′ b′ c′ d′ to (A, B; C, D) (diﬃcult to avoid capital letters here) with A = a 0 0 a′ , etc. that induces δ (on (SL(2, R)/U(1))2 →Sp(4, R)/U(2)). If we use the coordinates τ = τ1 z z τ2 ∈H2 then the image of the map δ is given by z = 0 and it is the ﬁxed point locus of the involution on H2 given by (τ1, z, τ2) →(τ1, −z, τ2) induced by the element (A, B; C, D) with A = (1, 0; 0, −1) = D and B = C = 0. An element F ∈Mk(Γ2) can be developed around this locus z = 0 F = f(τ1, τ2)zn + O(zn+1) for some n ∈Z≥0 . (5) It is now easy to check that 1. f(τ1, τ2) ∈Mk+n(Γ1) ⊗Mk+n(Γ1); 2. f(τ2, τ1) = (−1)kf(τ1, τ2); 3. f(τ1, −z, τ2) = (−1)kf(τ1, z, τ2). the ﬁrst by looking at the action of SL(2, Z) × SL(2, Z) and the second by applying the involution (A, B; C, D) with A = D = (0, 1; 1, 0) and B = C = 0 200 G. van der Geer which interchanges τ1 and τ2 and the last by using the involution z →−z. The idea of developing along the diagonal locus was ﬁrst used by Witt, . Developing E10 −E4E6 along z = 0 and writing qj = e2πiτj one ﬁnds cq1q2z2 + O(z3), with c ̸= 0, so we normalize to get a cusp form χ10 = E(1) 10 (τ1) ⊗E(1) 10 (τ2)z2 + O(z3). Similarly, the form E(2) 12 −(E(2) 6 )2 gives after normalization a non-zero cusp form χ12 = Δ(τ1) ⊗Δ(τ2)z2 + O(z3). As we saw above in Section 7 we also know the existence of a cusp form χ35 of odd weight 35. We now describe the structure of the ring of classical Siegel modular forms for g = 2. The theorem is due to Igusa and various proofs have been recorded in the literature, cf. [5,33,43,52–54]. Here is another variant. Theorem 6. (Igusa) The graded ring M = ⊕kMk(Γ2) of classical Siegel modular forms of genus 2 is generated by E4, E6, χ10, χ12 and χ35 and M ∼ = C[E4, E6, χ10, χ12, χ35]/(χ2 35 = R) , where R is an explicit (isobaric) polynomial in E4, E6, χ10 and χ12 (given on , p. 849). Proof. (Isobaric means that every monomial has the same weight (here 70) if E4, E6, χ10 and χ12 are given weights 4, 6, 10 and 12.) We start by introducing the vector spaces of modular forms: M ≥n k (Γ1) = {f ∈Mk(Γ1) : f = O(qn) at ∞} = Δn Mk−12n(Γ1) and M ≥n k (Γ2) = {F ∈Mk(Γ2) : F = O(zn) near δ(H1 × H1) } We distinguish two cases depending on the parity of k. k even. As we saw above (use properties (1), (2), (5)) any element F ∈M ≥2n k (Γ2) can be written as F(τ1, z, τ2) = f(τ1, τ2)z2n + O(z2n+2) with f ∈Mk+2n(Γ1) ⊗Mk+2n(Γ1) symmetric (i.e. f(τ1, τ2) = f(τ2, τ1)) and f = O(qn 1 , qn 2 ). This last fact follows from the observation that each Fourier– Jacobi coeﬃcient φm(τ1, z) of F is also O(z2n), so is zero if 2n > 2m. We ﬁnd an exact sequence 0 →M ≥2n+2 k (Γ2) →M ≥2n k (Γ2) r − →Sym2 M ≥n k+2n(Γ1) →0 , where the surjectivity of r is a consequence of the fact that Sym2 M ≥n k+2n(Γ1) = C[e4 ⊗e4, e6 ⊗e6, Δ ⊗Δ] and χ10 = Δ(τ1)Δ(τ2)z2 + O(z4) so that a modular form χn 10P(E4, E6, χ12) with P an isobaric polynomial maps to P(e4 ⊗e4, e6 ⊗e6, Δ ⊗Δ). It follows that Siegel Modular Forms and Their Applications 201 dim Mk(Γ2) = ∞ n=0 dim Sym2(M ≥n k+2n(Γ1)) = 0≤n≤k/10 dim Sym2(Mk−10n(Γ1)) , i.e., we get k even dim Mk(Γ2)tk = 1 1 −t10 k≥0 dim Sym2(Mk(Γ1))tk = 1 1 −t10 Hilbert series of C[e4 ⊗e4, e6 ⊗e6, Δ ⊗Δ] = 1 (1 −t4)(1 −t6)(1 −t10)(1 −t12) . k odd. For F ∈M ≥2n+1 k (Γ2) we ﬁnd f = O(qn+2 1 , qn+2 2 ). Since our Fourier–Jacobi coeﬃcients φm(τ1, z) have a zero of order 2n + 1 at z = 0 and another three at the 2-torsion points we see 2m ≥(2n + 1) + 3 for non-zero φm. Also we know that f is anti-symmetric now, so dim Mk(Γ2) ≤ n≥0 dim ∧2(M ≥n+2 k+2n+1(Γ1)) and this shows that for odd k < 35 dim Mk(Γ2) = 0. Since we have a non-trivial form of weight 35 we see that k odd dim Mk(Γ2)tk = t35 (1 −t4)(1 −t6)(1 −t10)(1 −t12) . The square χ2 35 is a modular form of even weight, hence can be expressed as an polynomial R in E4, E6, χ10 and χ12. This was done by Igusa in . This completes the proof. 10 Moduli of Principally Polarized Complex Abelian Varieties For g = 1 the quotient space Γ1\H1 has an interpretation as the moduli space of elliptic curves over the complex numbers (complex tori of dimen-sion 1). To a point τ ∈H1 we associate the complex torus C/Z + Zτ. Then to a point (aτ + b)/(cτ + d) in the Γ1-orbit of τ we associate the torus C/Z + Z(aτ + b)/(cτ + d), and the homothety z →(cτ + d)z deﬁnes an isomorphism of this torus with C/Z(cτ + d) + Z(aτ + b) = C/Z + Zτ since (cτ + d, aτ + b) is a basis of Z + Zτ as well. Conversely, every 1-dimensional complex torus can be represented as C/Z + Zτ. This can be generalized to g > 1 as follows. A point τ ∈Hg determines a complex torus Cg/Zg + Zgτ, but we do not get all complex g-dimensional tori. The fol-lowing lemma, usually ascribed to Lefschetz, tells us what conditions this imposes. 202 G. van der Geer Lemma 1. The following conditions on a complex torus X = V/Λ are equiva-lent: 1. X admits an embedding into a complex projective space; 2. X is the complex manifold associated to an algebraic variety; 3. There is a positive deﬁnite Hermitian form H on V such that Im(H) takes integral values on Λ × Λ. A complex torus satisfying these requirements is called a complex abelian variety. For g = 1 we could take H(z, w) = z ¯ w/Im(τ) on Λ = Z + Zτ and indeed, the map C/Λ →P2 given by z →(℘(z) : ℘′(z) : 1) for z / ∈Λ with ℘ the Weierstrass ℘-function deﬁnes the embedding. For g > 1 we can take H(z, w) = zt(Im(τ))−1w. An H as in the lemma is called a polarization. It is called a principal polarization if the map Im(H) : Λ × Λ →Z is unimodular. We shall write E = Im(H) for the alternating form that is the imaginary part of H. Given a complex torus X = V/Λ and a principal polarization on Λ we can normalize things as follows. We choose an isomorphism V ∼ = Cg and choose a symplectic basis e1, . . . , e2g of the lattice Λ such that E takes the standard form J = 0g 1g −1g 0g with respect to this basis. These two bases yield us a period matrix Ω ∈ Mat(g × 2g, C) expressing the ei in terms of the chosen C-basis of V . A nat-ural question is which period matrices occur. For this we note that E is the imaginary part of a Hermitian form H(x, y) = E(ix, y) + √−1E(x, y) if and only if E satisﬁes the condition E(iz, iw) = E(z, w) for all z, w ∈V and this translates into (Exercise!) Ω J−1Ωt = 0 while the positive deﬁniteness of H translates into the condition 2i( ¯ ΩJ−1Ωt)−1is positive deﬁnite . These conditions were found by Riemann in his brilliant 1857 paper . If we now associate to Ω = (Ω1 Ω2) with Ωi complex g × g matrices we see that the two conditions just found say that if we put τ = Ω−1 2 Ω1 we have τ = τt, Im(τ) > 0 i.e., τ lies in Hg. A change of basis of Λ changes (τ 1g) into (τa + c, τb + d) with (a, b; c, d) ∈Sp(2g, Z), but the corresponding torus is isomorphic to Cg/Zg(τb + d)−1(τa + c) + Zg. In this way we see that the isomorphism classes of complex tori with a principal polarization are in 1–1 correspondence with the points of the orbit space Hg/Sp(2g, Z). If we transpose we can identify this orbit space with the orbit space Sp(2g, Z)\Hg for the usual action τ →(aτ + b)(cτ + d)−1. Proposition 2. There is a canonical bijection between the set of isomorphism classes of principally polarized abelian varieties of dimension g and the orbit space Γg\Hg. Siegel Modular Forms and Their Applications 203 If we try to construct the whole family of abelian varieties we encounter a diﬃculty. The action of the semi-direct product Γg⋉Z2g on Hg×Cg given by the usual action of Γg on Hg and the action of (λ, μ) ∈Z2g on a ﬁbre {τ}×C2g by z →z + τλ + μ forces −12g ∈Γg to act by −1 on a ﬁbre, so instead of ﬁnding the complex torus Cg/Zg + τZg we get its quotient by the action z →−z. However, if we replace Γg by the congruence subgroup Γg(n) with n ≥3 (see ) then we get an honest family Xg(n) = Γg(n) ⋉Z2g\Hg × Cg of abelian varieties over Γg(n)\Hg. If we insist on using Γg then we have to work with orbifolds or stacks to have a universal family available; the orbifold in question is the quotient of Xg(n) under the action of the ﬁnite group Sp(2g, Z/nZ). The cotangent bundle of the family of abelian varieties over Ag(n) = Γg(n)\Hg along the zero section deﬁnes a vector bundle of rank g on Ag(n). It can be constructed explicitly as a quotient Γg(n)\Hg × Cg un-der the action of γ ∈Γg(n) by (τ, z) →(γ(τ), (cτ + d)−tz). The bun-dle is called the Hodge bundle and denoted by E = Eg. The ﬁnite group Sp(2g, Z/nZ) acts on the bundle E on Ag(n). A section of det(E)⊗k that is Sp(2g, Z/nZ)-invariant comes from a holomorphic function on Hg that is a classical Siegel modular form of weight k. Classical modular forms thus get a geometric interpretation. In particular, the determinant of the cotan-gent bundle of Ag(n), i.e., the canonical bundle, is isomorphic to det(E)⊗g+1; so to a modular form f of weight g + 1 we can associate a top diﬀeren-tial form on Hg that is Γg-invariant via f →f(τ) i≤j dτij. In a sim-ilar way one can construct for each ρ a vectorbundle over Ag(n) whose Sp(2g, Z/nZ)-invariant sections are the Siegel modular forms of weight ρ by taking the quotient Hg × V by Γg under (τ, z) →(γ(τ), ρ(cτ + d)z), see Sec-tion 13. The Hermitian form H on the lattice Λ ⊂Cg can be viewed as the ﬁrst Chern class (in H2(X, Z) ∼ = ∧2(H1(X, Z)∨) ∼ = (∧2Λ)∨) of a line bundle L on X = Cg/Λ with dimC H0(X, L) = 1. A non-zero section determines an eﬀective divisor Θ on X. The line bundle L and the corresponding divisor Θ are determined by H up to translation over X. If we require that Θ be invariant under z →−z then Θ is unique up to translation over a point of order 2 on X and then 2Θ is unique. If we pull a non-zero section s of L back to the universal cover Cg then we obtain a holomorphic function with a certain transformation behavior under translations by elements of Λ. An example of such a function is provided by Riemann’s theta function θ(τ, z) = n∈Zg eπi(ntτn+2ntz), (τ ∈Hg, z ∈Cg) a series that converges very rapidly and deﬁnes a holomorphic function that satisﬁes for all λ, μ ∈Zg θ(τ, z + τλ + μ) = e−πi(λtτλ+2λtz)θ(τ, z) . 204 G. van der Geer Conversely, a holomorphic function f on Cg that satisﬁes for all λ, μ ∈Zg f(z + τλ + μ) = e−πi(λtτλ+2λtz)f(z) is up to a multiplicative constant precisely θ(τ, z) as one sees by developing f in a Fourier series f = n c(n) exp (2πintz) and observing that addition of a column τk of τ to z produces f(z + τk) = n c(n) exp (2πint(z + τk)) = n c(n) exp (2πintτk) exp (2πintz) from which one obtains c(n + ek) = c(n) exp (2πintτk + πiτkk) and gets that f is completely determined by c(0). If S is a compact Riemann surface of genus g it determines a Jacobian variety Jac(S) which is a principally polarized complex abelian variety of dimension g. Sending S to Jac(S) provides us with a map Mg(C) →Γg\Hg from the moduli space of compact Riemann surfaces of genus g to the moduli of complex principally polarized abelian varieties of dimension g which is injective by a theorem of Torelli. The geometric interpretation given for Siegel modular forms thus pulls back to the moduli of compact Riemann surfaces. 11 Compactiﬁcations It is well known that Γ1\H1 is not compact, but can be compactiﬁed by adding the cusp, that is, the orbit of Γ1 acting on Q ⊂¯ H1. Or if we use the equivalence of H1 with the unit disc D1 given by τ →(τ −i)/(τ + i) then we add to D1 the rational points of the boundary of the unit disc and take the orbit space of this enlarged space. We can do something similar for g > 1 by considering the bounded symmetric domain Dg = {z ∈Mat(g × g, C) : zt = z, zt · ¯ z < 1g} which is analytically equivalent to Hg. We now enlarge this space by adding not the whole boundary but only part of it as follows. Let Dr = $ z′ 0 0 1g−r : z′ ∈Dr % ⊂¯ Dg and deﬁne now D∗ g to be the union of all Γg-orbits of these Dr for 0 ≤r ≤g. Note that Γg acts on Dg and on its closure ¯ Dg. Then Γg acts in a natural way on D∗ g and the orbit space decomposes naturally as a disjoint union Γg\D∗ g = ⊔g i=0Γi\Di . Going back to the upper half plane model this means that we consider Γg\H∗ g = ⊔g i=0Γi\Hi Siegel Modular Forms and Their Applications 205 Satake has shown how to make this space into a normal analytic space, the Satake compactiﬁcation. One ﬁrst deﬁnes a topology on H∗ g and then a sheaf of holomorphic functions. The quotient Γg\H∗ g then becomes a normal analytic space. By using explicitly constructed modular forms one then shows that classical modular forms of a suitably high weight separate points and tangent vectors and thus deﬁne an embedding of Γg\H∗ g into projective space. By Chow’s lemma it is then a projective variety. The following theorem is a special case of a general theorem due to Baily and Borel, . Theorem 7. Scalar Siegel modular forms of an appropriately high weight de-ﬁne an embedding of Γg\H∗ g into projective space and the image of Γg\Hg (resp. Γg\H∗ g) is a quasi-projective (resp. a projective) variety. The resulting Satake or Baily–Borel compactiﬁcation is for g > 1 very singular. As a ﬁrst attempt at constructing a smooth compactiﬁcation we reconsider the case g = 1. In H1,c = {τ ∈H1 : Im(τ) ≥c} with c > 1 the action of Γ1 reduces to the action of Z by translations τ →τ + b. So consider the map H1,c →C∗, τ →q = exp 2πiτ. It is clear how to compactify H1,c/Z: just add the origin q = 0 to the image in C∗⊂C. In other words, glue Γ1\H1 with Z\H∗ 1,c over Z\H1,c. To do something similar for g > 1 we consider the subset (for a suitable real symmetric g × g-matrix c ≫0 which is suﬃciently positive deﬁnite) Hg,c = $ τ = τ1 z zt τ2 ∈Hg : Im(τ2) −Im(zt)Im(τ1)−1Im(z) ≥c % The action of Γg in Hg,c reduces to the action of the subgroup P ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ ⎛ ⎜ ⎜ ⎝ a 0 b ∗ ∗±1 ∗∗ c 0 d ∗ 0 0 0 ±1 ⎞ ⎟ ⎟ ⎠∈Γg, a b c d ∈Γg−1 ⎫ ⎪ ⎪ ⎬ ⎪ ⎪ ⎭ , the normalizer of the ‘boundary component’ Hg−1. We now make a map Hg,c →Hg−1 × Cg−1 × C∗, τ →(τ1, z, q2 = exp(2πiτ2)) . The associated parabolic subgroup P acts on Hg−1 × Cg−1 × C∗and this action can be extended to an action on Hg−1 × Cg−1 × C, where ⎛ ⎜ ⎜ ⎝ 1g−1 0 0 0 0 1 0 b 0 0 1g−1 0 0 0 0 1 ⎞ ⎟ ⎟ ⎠ acts now by (τ1, z, q2) →(τ1, z, e2πibq2) while the matrix 206 G. van der Geer ⎛ ⎜ ⎜ ⎝ 1g−1 0 0 l m 1 l 0 0 0 1g−1 −m 0 0 0 1 ⎞ ⎟ ⎟ ⎠ acts by (τ1, z, q2) →(τ1, z + τ1m + l, e2πi(mτ1m+2mz+lm)q2), and the diagonal matrix with entries (1, . . . , −1, 1, . . ., 1, −1) acts by (τ1, z, ζ) →(τ1, −z, ζ) and ﬁnally (a, b; c, d) ∈Γg−1 acts on Hg−1 × Cg−1 × C by (τ1, z, q2) →(γ(τ1), (a −(γ(τ1)c) z, τ2 −zt(cτ1 + d)−1cz) and this action can be extended similarly. We now have an embedding Γg\Hg,c − →P\Hg−1 × Cg−1 × C and by taking the closure of the image we obtain a ‘partial compactiﬁcation’. The quotient of Hg−1 × Cg−1 × {0} by this action is the ‘dual universal abelian variety ˆ Xg−1 = Γg−1 ⋉Z2g−2\Hg−1 × Cg−1 over Γg−1\Hg−1 (in the orbifold sense). Note that a principally polarized abelian variety is isomorphic to its dual, so we can enlarge our orbifold Γg\Hg by adding this orbifold quotient Xg−1 = Γg−1 × Z2g−2\Hg−1 × C2g−2. The result is a partial compactiﬁcation A(1) g = Ag ⊔X ′ g−1 , where the prime refers to the fact that we are dealing with orbifolds and have to divide by (at least) an extra involution since a semi-abelian variety generically has Z/2 × Z/2 as its automorphism group, while a generic abelian variety has only Z/2. This space parametrizes principally polarized complex abelian varieties of dimension g or degenerations of such (so-called semi-abelian varieties of torus rank 1) that are extensions 1 →Gm →˜ X →X →0 of a g−1-dimensional principally polarized complex abelian variety by a rank 1 torus Gm = C∗. Such extension classes are classiﬁed by the dual abelian varie-ty ˆ X ∼ = X (associate to a line bundle on X the Gm-bundle obtained by deleting the zero section) which explains why we ﬁnd the universal abelian variety of dimension g −1 in the ‘boundary’ of Ag. (There is the subtlety whether one allows isomorphisms to be −1 on Gm or not.) This partial compactiﬁcation is canonical. If we wish to construct a full smooth compactiﬁcation one can use Mumford’s theory of toroidal compactiﬁcations, but unfortunately there is (for g ≥4) no unique such compactiﬁcation. We refer e.g. to . This partial compactiﬁcation enables us to reinterpret the Fourier–Jacobi series of a Siegel modular form. In particular, the formulas in Section 8 tell us that the pull back of f to a ﬁbre of Xg−1 →Ag−1 is an abelian function and that f restricted to the zero-section of Xg−1 →Ag−1 is a Siegel modular form of weight k −1 on Γg−1. Siegel Modular Forms and Their Applications 207 We can be more precise. We work with a group Γg(n) with n ≥3 or interpret everything in the orbifold sense. The normal bundle of Xg−1 is the line bundle O(−2Θ), as one can deduce from the action given above. We can also extend the Hodge bundle E = Eg to a vector bundle on A(1) g . On the boundary divisor X ′ g−1 it is the extension of the pull back π∗Eg−1 from Ag−1 to Xg−1 by a line bundle. So if we are given a classical Siegel modular form of weight k we can interpret it as a section of det(E)⊗k and develop (the pull back of) f along the boundary Xg−1 where the m-th term in the development is a section of (det(E)\|Xg−1)⊗k ⊗O(−2mΘ) on Xg−1. This gives us a geometric interpretation of the Fourier–Jacobi de-velopment. Of course, it is useful to have not only a partial compactiﬁcation, but a smooth compactiﬁcation. The theory of toroidal compactiﬁcations devel-oped by Mumford and his co-workers Ash, Rapoport and Tai provides such compactiﬁcations ˜ Ag. They depend on the choice of a certain cone decompo-sition of the cone of positive deﬁnite bilinear forms in g variables, cf. . The ‘boundary’ ˜ Ag −Ag is a divisor with normal crossings and one has a universal semi-abelian variety over ˜ Ag in the orbifold sense. 12 Intermezzo: Roots and Representations Here we record a few concepts and notations that we shall need in the later sections. The reader may want to skip this on a ﬁrst reading. Recall that we started out in Section 2 with a symplectic lattice (Z2g, ⟨, ⟩) with a basis e1, . . . , eg, f1, . . . , fg with ⟨ei, fj⟩= δij and ⟨e1, . . . , eg⟩and ⟨f1, . . . , fg⟩isotropic subspaces. We let G := GSp(2g, Q) be the group of rational symplectic similitudes (transformations that preserve the form up to a scalar), viz., G := GSp(2g, Q) = {γ ∈GL(Q2g) : γtJγ = η(γ)J} and G+ = {γ ∈G: η(γ) > 0}. Note that det(γ) = η(γ)g for γ ∈G and that G0 = Sp(2g, Z) is the kernel of the map that sends γ to η(γ) on G+(Z). For γ ∈G the element η(γ) is called the multiplier. Note that we view elements of Z2g as column vectors and G acts from the left. There are several important subgroups that play a role in the sequel. Given our choice of basis there is a natural Borel subgroup B respecting the symplec-tic ﬂag ⟨e1⟩⊂⟨e1, e2⟩⊂. . . ⊂⟨e1, e2⟩⊥⊂⟨e1⟩⊥. It consists of the matrices (a, b; 0, d) with a upper triangular and d lower triangular. Other natural subgroups are: the subgroup M of elements respecting the decomposition Zg ⊕Zg of our symplectic space. It is isomorphic to GL(g)×Gm 208 G. van der Geer and consists of the matrices γ = (a, 0; 0, d) with adt = η(γ)1g. Furthermore, we have the Siegel (maximal) parabolic subgroup Q of elements that stabilize the ﬁrst summand Zg = ⟨e1, . . . , eg⟩; it consists of the matrices (a, b; 0, d). It con-tains the subgroup U (unipotent radical) of matrices of the form (1g, b; 0, 1g) with b symmetric that act as the identity of the ﬁrst summand Zg. Another important subgroup of G is the diagonal torus T isomorphic to Gg+1 m of matrices γ = diag(a1, . . . , ag, d1, . . . , dg) with aidi = η(γ). Let X be the character group of T; it is generated by the characters ϵi : γ →ai for i = 1, . . . , g and ϵ0(γ) = η(γ). Let Y be the co-character group of Tm, i.e., Y = Hom(Gm, T). This group is isomorphic to the group Zg+1 of g + 1-tuples with (α1, . . . , αg, c) corresponding to the co-character t → diag(tα1, . . . , tαg, tc−α1, . . . , tc−αg). We ﬁx a basis of Y by letting χi for i = 1, . . . , g correspond to αj = δij and c = 0 and χ0 to αj = 0 and c = 1. Then the characters and co-characters pair via ⟨ϵi, χj⟩= δij. The adjoint action of T on the Lie algebras of M and G deﬁnes root systems ΦM and ΦG in X. Concretely, we may take as simple roots αi = ϵi −ϵi+1 for i = 1, . . . , g−1 and αg = 2ϵg−ϵ0 and coroots α∨ i = χi−χi+1 for i = 1, . . . , g−1 and α∨ g = χg. The set Φ+ G of positive roots (those occuring in the Lie algebra of the nilpotent radical of B) consists of the so-called compact roots Φ+ M = {ϵi − ϵj : 1 ≤i < j ≤g} and the non-compact roots Φ+ nc = {ϵi+ϵj−ϵ0: 1 ≤i, j ≤g}. We let 2ϱ = 2ϱG (resp. 2ϱM) be the sum of the positive roots in Φ+ G (res. Φ+ M). When viewed as characters 2ϱM corresponds to γ →g i=1 ag+1−2i i and 2ϱG to γ →η(γ)−g(g+1)/2 g i=1 a2g+2−2i i . There is a symmetry group acting on our situation, the Weyl group WG = N(T)/T, with N(T) the normalizer of T in G. This group WG is iso-morphic to Sg ⋉(Z/2Z)g, where the generator of the i-th factor Z/2Z acts on a matrix of the form diag(a1, . . . , ag, d1, . . . , dg) by interchanging ai and di and the symmetric group Sg acts by permuting the a’s and d’s. The Weyl group of M (normalizer this time in M) is isomorphic to the symmetric group Sg. We have positive Weyl chambers P + G = {χ ∈Y : ⟨χ, α⟩≥0 for all α ∈Φ+ G } and similarly for M: P + M = {χ ∈Y : ⟨χ, α⟩≥0 for all α ∈Φ+ M } giving the dominant weights. Lemma 2. The irreducible complex representations of G (resp. M) corres-pond to integral weights in the chamber P + G (res. P + M) that come from char-acters of T. Sometimes we just work with G0 and M 0 = M ∩G0. This means that we forget about the action of the multiplier η. We can give a set W0 of 2g canonical coset representatives of WM\WG, the Kostant representatives, which are characterized by the conditions W0 = w ∈WG : Φ+ M ⊂w Φ+ G = w ∈WG : w(ϱ) −ϱ ∈P + M . With our normalizations we have ϱ = (g, g −1, . . . , 1, 0) and 2ϱM = (g + 1, . . . , g + 1, −g(g + 1)/2). If we restrict to G0 and M 0 then dominant weights Siegel Modular Forms and Their Applications 209 for M 0 ∼ = GL(g) are given by g-tuples (λ1, . . . , λg) with λi ≥λi+1 for i = 1, . . . , g −1. A coset in WM\WG is given by a vector s (in {±1}g) of g signs. The Kostant representative of s is the element σ s such that (sσ(1)λσ(1), . . . , sσ(g)λσ(g)) is in P + M, i.e., sσ(i)λσ(i) ≥sσ(i+1)λσ(i+1) for i = 1, . . . , g −1 for all (λ1 ≥. . . ≥λg). 13 Vector Bundles Deﬁned by Representations Let π: Xg →Ag be the universal family of abelian varieties over Ag. The Hodge bundle E = π∗ΩXg/Ag, a holomorphic bundle of rank g, and the de Rham bundle R1π∗C on Ag, a locally constant sheaf of rank 2g, are examples of vector bundles associated to representations of GL(g) and GSp(2g). Their ﬁbres at a point [X] ∈Ag are H0(X, Ω1 X) and H1(X, C). The ﬁrst is a holo-morphic vector bundle, the second a local system. Both are important for understanding Siegel modular forms. To deﬁne these bundles recall the description of Hg as an open part Y + g of the symplectic Grassmann variety Yg given in Section 2. We can identify Yg with G(C)/Q(C) with Q the subgroup ﬁxing the totally isotropic ﬁrst summand Cg of our complexiﬁed symplectic lattice (Zg, ⟨, ⟩) ⊗C. If ρ: Q0 → End(V ) is a complex representation (with Q0 = Q ∩G0) then we can deﬁne a G0(C)-equivariant vector bundle Vρ on Yg by Vρ = G0(C) ×Q0(C) V as the quotient of G0(C) × V under the equivalence relation (g, v) ∼(g q, ρ(q)−1v) for all g ∈G0(C) and q ∈Q0(C). Then Γg (or any ﬁnite index subgroup Γ ′) acts on Vρ and the quotient is a vector bundle Vρ on Ag in the orbifold sense (or a true one if Γ ′ acts freely on Hg). Recall that M is the subgroup of GSp(2g, Q) respecting the decomposition Qg ⊕Qg of our symplectic space and M 0 = M ∩Sp(2g, Q). If we are given a complex representation of M 0(C) ∼ = GL(g) (or of M ∼ = GL(g)× Gm) we can obtain a vector bundle by extending the representation to a representation on Q0(C) by letting it be trivial on the unipotent radical U of Q. (Note that Q = M · U.) If we do this with the tautological representation of M 0 we get the Hodge bundle E. But there is a subtle point here. If we work with M instead of M 0 then the Hodge bundle is given by the representation of M that acts by η(γ)−1a on Cg for γ = (a, 0; 0, d). In any case we thus get a holomorphic vector bundle W(λ) associated to each dominant weight (λ1 ≥. . . ≥λg) of GL(g). Another way of getting these vector bundles thus associated to the irreducible representations of M 0 (or M) is by starting from the Hodge bundle and applying Schur operators (idempotents) to the symmetric powers of E analogously to the way one gets the corresponding representations from the standard one. Since the Hodge bundle E extends over a toroidal compactiﬁcation ˜ Ag this makes it clear that these vector bundles W(λ) can be extended over any toroidal compactiﬁcation as constructed by Mumford (or Faltings–Chai). The space of sections can be identiﬁed with a space of modular forms Mρ and it thus follows from general 210 G. van der Geer theorems in algebraic geometry that these spaces of Siegel modular forms Mρ are ﬁnite dimensional. Another important vector bundle is the bundle associated to the ﬁrst co-homology of the universal abelian variety Xg with ﬁbre H1(X, C); more pre-cisely, it is given by V := R1π∗C with π: Xg →Ag the universal abelian variety. It can be gotten from the construction just given by taking the dual or contragredient of the standard or tautological representation of Sp(2g, C) and restricting it to Q0(C). (Again, if one takes the multiplier into account – as one should – then R1π∗C corresponds to η−1 times the standard representa-tion.) In this case we ﬁnd a ﬂat bundle: all the bundles Vρ on Yg come with a trivialization given by [(g, v)] →ρ(g)v. So the quotient bundle carries a nat-ural integrable connection. The resulting V is a local system (locally constant sheaf). We thus ﬁnd for each dominant weight λ = (λ1 ≥. . . ≥λg, c) of G a local system Vλ(c) on Ag. The multiplier representation deﬁnes a local sys-tem of rank 1 denoted by C(1) and we can twist Vλ(c) by the nth power of C(1) to change c, cf. Section 12. 14 Holomorphic Diﬀerential Forms Let Γ ′ ⊂Γg be a subgroup of ﬁnite index which acts freely on Hg, e.g., Γ ′ = ker{Sp(2g, Z) →Sp(2g, Z/nZ} for n ≥3. Let Ωi be the sheaf of holomorphic i-forms on Hg. A section of Ω1 can be written as ω = Tr(f(τ)dτ) , where dτ = (dτij) and f is a symmetric matrix of holomorphic functions on Hg. Then ω is invariant under the action of Γ ′ if and only if f(γ(τ)) = (cτ + d)f(τ)(cτ + d)t for all γ = (a, b; c, d) ∈Γ ′. Note that if r is the standard representation of GL(g, C) on V = Cg then the action on symmetric bilinear forms Sym2(V ) is given by b →r(g) b r(g)t. So the space of holomorphic 1-forms on Γ ′\Hg can be identiﬁed with Mρ(Γ ′), with ρ the second symmetric power of the standard representation and the space of holomorphic i-forms with Mρ′(Γ ′) with ρ′ equal to the ith exterior power of Sym2V . So we ﬁnd an isomorphism Ω1 Γ ′\Hg ∼ = Sym2E and this can be extended over a toroidal compactiﬁcation ˜ A to an isomorphism Ω1 ˜ A(log D) ∼ = Sym2(E) with D the divisor at inﬁnity. (But again, one should be aware of the action of the multiplier: if one looks at the action of GSp(2g, R)+ one has d((aτ + b)(cτ + d)−1) = η(γ)(cτ + d)−1 dτ (cτ + d)−1.) The question arises which representations occur in ∧iSym2(V ). The answer is given in terms of roots. A theorem of Kostant tells that the irreducible representations ρ of GL(g, C) that occur in the exterior Siegel Modular Forms and Their Applications 211 algebra ∧∗Sym2(V ) with V the standard representation of GL(g, C) are those ρ for which the dual ˆ ρ is of the form wδ −δ with δ = (g, g −1, . . . , 1) the half-sum of the positive roots and w in the set W0 of Kostant representatives. Now if ˆ ρ = (λ1 ≥λ2 . . . ≥λg) occurs in this exterior algebra then wδ is of the form (g −λg, g −1 −λg−1, . . . , 1 −λ1). If α is the largest integer that occurs among the entries of wδ then either α = −1 or 1 ≤α ≤g. In the latter case wδ is of the form (α, ∗, . . . , ∗, −α −1, −α −2, . . . , −g) and it follows that λg−α = g + 1. This implies that the number of λj with λj = λg (the co-rank of ˆ ρ, cf., Section 5) plus the number of those with λj = λg + 1 is at most α. The vanishing theorem of Weissauer (Thm. 5) now implies that non-zero diﬀerentials can only come from representations that are of the form ρ = (g + 1, g + 1, . . . , g + 1) which corresponds to top diﬀerentials (∧g(g+1)/2Ω1) and classical Siegel modu-lar forms of weight g + 1, or of the form ρ = (g + 1, g + 1, . . . , g + 1, g −α, . . . , g −α) , with 1 ≤α ≤g and these occur in ∧pΩ1 with p = g(g + 1)/2 −α(α + 1)/2. For the following theorem of Weissauer we refer to . Theorem 8. Let ˜ Ag be a smooth compactiﬁcation of Ag. If p is an integer 0 ≤p < g(g + 1)/2 then the space of holomorphic p-forms on ˜ Ag is zero unless p is of the form g(g + 1)/2 −α(α + 1)/2 with 1 ≤α ≤g and then H0( ˜ Ag, Ωp ˜ Ag) ∼ = Mρ(Γg) with ρ = (g + 1, . . . , g + 1, g −α, . . . , g −α) with g −α occuring α times. If f is a classical Siegel modular form of weight k = g + 1 on the group Γg then f(τ) i≤j dτij is a top diﬀerential on the smooth part of quotient space Γg\Hg = Ag. It can be extended over the smooth part of the rank-1 com-pactiﬁcation A(1) g if and only if f is a cusp form. It is not diﬃcult to see that this form can be extended as a holomorphic form to the whole smooth compactiﬁcation ˜ Ag. Proposition 3. The map that associates to a classical cusp form f ∈Sg+1(Γg) of weight g+1 the top diﬀerential ω = f(τ) i≤j dτij gives an isomorphism be-tween Sg+1(Γg) and the space of holomorphic top diﬀerentials H0( ˜ Ag, Ω g(g+1) 2 ) on any smooth compactiﬁcation ˜ Ag. For this and an analysis of when the other forms extend over the singular-ities in these cases we refer to [30,106]. Finally we refer to two papers of Salvati–Manni where he proves the exis-tence of diﬀerential forms of some weights, [83,84] and a paper of Igusa, , where Igusa discusses the question whether certain Nullwerte of jacobians of odd thetafunctions can be expressed as polynomials or rational functions in theta Nullwerte. 212 G. van der Geer 15 Cusp Forms and Geometry The very ﬁrst cusp forms that one encounters often have a beautiful geometric interpretation. We give some examples. For g = 1 the ﬁrst cusp form is Δ = τ(n)qn ∈S12(Γ1). It is up to a normalization the discriminant g2(τ)3 −27g2 3 of the equation y2 = 4x3 − g2x −g3 for the Riemann surface C/Zτ + Z and does not vanish on H1. Here g2 = (4π4/3)E4(τ) and g3 = (8π6/27)E6(τ) are the suitably normalized Eisenstein series. For g = 2 there is a similar cusp form χ10 of weight 10 with development χ10 τ1 z z τ2 = (exp(2πiτ1) exp(2πiτ2) + . . .)(πz)2 + . . . which vanishes (with multiplicity 2) along the ‘diagonal’ z = 0. So its zero divisor in A2 is the divisor of abelian surfaces that are products of elliptic curves with multiplicity 2. There is the Torelli map M2 →A2 that associates to a hyperelliptic complex curve of genus 2 given by y2 = f(x) its Jacobian. Then the pull back of χ10 to M2 is related to the discriminant of f, cf. Igusa’s paper or , Prop. 2.2. For g = 3 the ring of classical modular forms is generated by 34 elements, cf. . As we saw above, there is a cusp form of weight 18, namely the product of the 36 even theta constants θ[ϵ] and its zero divisor is the closure of the hyperelliptic locus. This expresses the fact that a genus 3 Riemann surface with a vanishing theta characteristic is hyperelliptic. For g = 4 there is the following beautiful example. There is up to isometry only one isomorphism class of even unimodular positive deﬁnite quadratic forms in 8 variables, namely E8. In 16 variables there are exactly two such classes, E8 ⊕E8 and E16. To each of these quadratic forms in 16 variables we can associate a Siegel modular form on Γ4 by means of a theta series: θE8⊕E8 and θE16. The diﬀerence θE8⊕E8 −θE16 is a cusp form of weight 8. Its zero divisor is the closure of the locus of Jacobians of Riemann surfaces of genus 4 in A4 as shown by Igusa, cf., . Here we also refer to for a proof. We shall encounter this form again in Section 21. Similarly, the theta series associated to the 24 diﬀerent Niemeier lat-tices (even, positive deﬁnite) of rank 24 produce in genus 12 a linear sub-space of M12(Γ12) of dimension 12. It intersects the space of cusp forms in a 1-dimensional subspace, as was proved in . We thus ﬁnd a cusp form of weight 12. As we shall see later, it is an Ikeda lift of the cusp form Δ for g = 1 (proven in ). Question 1. What is the geometric meaning of this cusp form? The paper contains explicit results on Siegel modular forms of weight 12 obtained from lattices in dimension 24. For example, it gives a non-zero cusp form of weight 12 on Γ11, hence one has a top diﬀerential on ˜ A11, cf., Prop. 3, implying that this modular variety is not rational or unirational. It is a well-known result of Mumford () that ˜ Ag is of general type for g > 6. Siegel Modular Forms and Their Applications 213 16 The Classical Hecke Algebra In the arithmetic theory of elliptic modular forms Hecke operators play a piv-otal role. They enable one to extract arithmetic information from the Fourier coeﬃcients of a modular form: if f = n a(n)qn is a common eigenform of the Hecke operators which is normalized (a(1) = 1) then the eigenvalue λ(p) of f under the Hecke operator T (p) equals the Fourier coeﬃcient a(p). The classical theory of Hecke operators as for example exposed in Shimura’s book ( ) can be generalized to the setting of g > 1 as Shimura showed in , though the larger size of the matrices involved is a discouraging as-pect of it. It is worked out in the books [1,4,30], of which the last, by Freitag, is certainly the most accessible. In this section we sketch this approach, in the next section we give another approach. We refer to loc. cit. for details. Recall the group G := GSp(2g, Q) = {γ ∈GL(Q2g) : γtJγ = η(γ)J, η(γ) ∈ Q∗} of symplectic similitudes of the symplectic vector space (Q2g, ⟨, ⟩) and G+ = {γ ∈G : η(γ) > 0}. We start by deﬁning the abstract Hecke algebra H(Γ, G) for the pair (Γ, G) with Γ = Γg and G = GSp(2g, Q). Its elements are ﬁnite formal sums (with Q-coeﬃcients) of double cosets ΓγΓ with γ ∈G+. Each such double coset ΓγΓ can be written as a ﬁnite disjoint union of right cosets Li = Γγi by virtue of the following lemma. Lemma 3. Let m be a natural number. The set Og(m) = {γ ∈Mat(2g × 2g, Z) : γtJγ = mJ} can be written as a ﬁnite disjoint union of right cosets. Every right coset has a representative of the form (a, b; 0, d) with at d = m 1g and such that a has zeros below the diagonal So to each double coset ΓγΓ we can associate a ﬁnite formal sum of right cosets. Let L be the Q-vector space of ﬁnite formal expressions i ciLi with Li = Γγi a right coset and ci ∈Q. The map H(Γ, G) →L is injective and induces an isomorphism H(Γ, G) ∼ = LΓ , where the action of Γ on L is Γγ1 →Γγ1γ. We now make this into an algebra by specifying the product of ΓγΓ = i Γγi and ΓδΓ = j Γδj by (ΓγΓ) · (ΓδΓ) = i,j Γγiδj . To deal with these double cosets the following proposition is very helpful. Proposition 4. (Elementary divisors) Let γ ∈GSp+(2g, Q) be an element with integral entries. Then double coset ΓγΓ has a unique representative of the form α = diag(a1, . . . , ag, d1, . . . , dg) with integers aj, dj satisfying aj > 0, ajdj = η(γ) for all j, and furthermore ag\|dg, aj\|aj+1 for j = 1, . . . , g −1. 214 G. van der Geer On G we have the anti-involution γ = a b c d →γ∨= dt −bt −ct at = η(γ) γ−1 . (Note that η(γ∨) = η(γ).) Another involution is given by γ →J γ J−1 = d −c −b a = η(γ)γ−t . Because of the proposition we have ΓγΓ = Γγ∨Γ since we may choose γ di-agonal and then γ∨= JγJ−1 and J ∈Γ. This implies that for a sum of right cosets ΓγΓ = Γγi we have ΓγΓ = γ∨ i Γ. And it is easy to see that γ →γ∨deﬁnes an anti-involution of H(Γ, G) which acts trivially so that the Hecke algebra is commutative. We can decompose these diagonal matrices as a product of matrices so that in each of the factors only powers of one prime occur as non-zero entries. This leads to a decomposition H(Γ, G) = ⊗pHp as a product of local Hecke algebras Hp = H(Γ, G ∩GL(2g, Z[1/p])) , where we allow in the denominators only powers of p. Now Hp has a subring H0 p generated by integral matrices. We have Hp = H0 p[1/T ] with T the element deﬁned by T = Γg(p 12g)Γg. By induction one proves the following theorem, cf., [4,30]. Theorem 9. The local Hecke algebra H0 p is generated by the element T (p) given by Γg 1g 0g 0g p 1g Γg and the elements Ti(p2) for i = 1, . . . , g given by Γg ⎛ ⎜ ⎜ ⎝ 1g−i p1i p21g−i p1i ⎞ ⎟ ⎟ ⎠Γg For completeness sake we also introduce the element T0(p2) given by the double coset Γp 1g 0 0g p21g Γg. Note that Tg(p2) equals the T = Γg(p 12g)Γg given above. Deﬁnition 7. Let T (m) be the element of H(Γ, G) deﬁned by the set M = Og(m) which is a ﬁnite disjoint union of double cosets. Siegel Modular Forms and Their Applications 215 If m = p is prime then M = Og(m) is one double coset and T (m) coincides with T (p), introduced above. For m = p2 the set Og(p2) is a union of g + 1 double cosets and the element T (p2) is a sum g i=0 Ti(p2). The Hecke algebra can be made to act on the space of Siegel modular forms Mρ(Γg). We ﬁrst deﬁne the ‘slash operator’. Deﬁnition 8. Let ρ : GL(g, C) →End(V ) be a ﬁnite-dimensional irreducible complex representation corresponding to (λ1 ≥. . . ≥λg). For a function f : Hg →V and an element γ ∈GSp+(2g, Q) we set f\|γ,ρ(τ) = ρ(cτ + d)−1f(γ(τ)) γ = ∗∗ c d . (For a good action on integral cohomology one might wish to add a factor η(γ) P λi−g(g+1)/2.) Note that f\|γ1,ρ\|γ2,ρ = f\|γ1γ2,ρ. So if g > 1 then f is a modular form of weight ρ if and only if f is holomorphic and f\|γ,ρ = f for every γ ∈Γg. Let now M ⊂GSp(2g, Q) be a subset satisfying the two properties 1. M = ⊔h i=1Γg γi is a ﬁnite disjoint union of right cosets Γgγi; 2. M Γg ⊂M. The ﬁrst condition implies that if for a modular form f ∈Mρ we set TMf := h i=1 f\|γi,ρ then this is independent of the choice of the representatives γi, while the second condition implies that (TMf)\|γ = TMf for all γ ∈Γg. Together these conditions imply that TM is a linear operator on the space Mρ. Double cosets ΓγΓ satisfy condition 2) if Γ = Γg and γ ∈Sp(2g, Q) and also condition 1) by what was said above. An important observation is that ⟨T f, g⟩= ⟨f, T ∨g⟩, where ⟨, ⟩gives the Petersson product and thus the Hecke operators deﬁne Hermitian operators on the space of cusp forms Sρ. Just as in the classical case g = 1 we can associate correspondences (i.e. divisors on Ag × Ag) to Hecke operators. The correspondence associated to T (p) sends a principally polarized abelian variety X to the sum X′ of principally polarized X′ which admit an isogeny X →X′ with kernel an isotropic (for the Weil pairing) subgroup H ⊂X[p] of order pg. Similarly, the correspondence associated to Ti(p2) sends X to the sum X′ with the X′ quotients X/H, where H ⊂X[p2] is an isotropic subgroup of order p2g with H ∩X[p] of order pg+i. 17 The Satake Isomorphism We can identify the local Hecke algebra Hp with the Q-algebra of Q-valued locally constant functions on GSp(2g, Qp) with compact support and which 216 G. van der Geer are invariant under the (so-called hyperspecial maximal compact) subgroup K = GSp(2g, Zp) acting both from the left and right. The multiplication in this algebra is convolution f1 · f2 = GSp(2g,Qp)) f1(g)f2(g−1h)dg, where dg denotes the unique Haar measure normalized such that the volume of K is 1. The correspondence is obtained by sending the double coset KγK to the characteristic function of KγK. A compactly supported function in Hp is constant on double cosets and its support is a ﬁnite linear combination of characteristic functions of double cosets. Note that Theorem 9 tells us that Hp is generated by the double cosets of diagonal matrices. In order to describe this algebra conveniently we compare it with the p-adic Hecke algebras of two subgroups, the diagonal torus and the Levi subgroup of the standard parabolic subgroup. To be precise, recall the diagonal torus T of GSp(2g, Q) isomorphic to Gg+1 m and the Levi subgroup M = $ a 0 0 d ∈GSp(2g, Q) % of the standard parabolic Q = {(a, b; 0, d) ∈GSp(2g, Z)} that stabilizes the ﬁrst summand Zg of Zg ⊕Zg. In particular for an element (a, 0; 0, d) ∈M we have adt = η and the group M is isomorphic to GL(g) × Gm. Let Y ∼ = Zg+1 be the co-character group of Tm, i.e., Y = Hom(Gm, T), cf. Section 12. We can construct a local Hecke algebra Hp(T) = Hp(T, TQ) for the group T too as the Q-algebra of Q-valued, bi-T(Zp)-invariant, locally constant func-tions with compact support on T(Qp). This local Hecke-algebra is easy to de-scribe: Hp(T) ∼ = Q[Y ], the group algebra over Q of Y where λ ∈Y corresponds to the characteristic function of the double coset Dλ = Kλ(p)K. Concretely, Hp(T) is isomorphic to the ring Q[(u1/v1)±, . . . , (ug/vg)±, (v1 · · · vg)±] under a map that sends (a1, . . . , ag, c) to the element (u1/v1)a1 · · · (ug/vg)ag(v1 · · · vg)c . Similarly, we have a p-adic Hecke algebra Hp(M) = Hp(M, MQ) for M. Recall that the Weyl group WG = N(T)/T, with G = GSp(2g, Q) and N(T) the normalizer of T in G, acts. This group WG is isomorphic to Sg ⋉ (Z/2Z)g, where the generator of the i-th factor Z/2Z acts on a matrix of the form diag(α1, . . . , αg, δ1, . . . , δg) by interchanging αi and δi and the symmetric group Sg acts by permuting the α’s and δ’s. The Weyl group of M (normalizer this time in M) is isomorphic to the symmetric group Sg. The algebra of invariants Hp(T)WG is of the form Q[y± 0 , y1, . . . , yg], cf. . We now give Satake’s so-called spherical map of the Hecke algebra Hp(Γ, G) to the Hecke algebras Hp(M) and Hp(T), cf., [17,29,40,85]. The images will land in the WM-invariant (resp. the WG-invariant) part. We ﬁrst need the following characters. The Borel subgroup B of matrices (a, b; 0, d) with a upper triangular and d lower triangular determines a set Φ+ Siegel Modular Forms and Their Applications 217 of positive roots in the set of all roots Φ (= characters that occur in the adjoint representation of G on Lie(B)). We let 2ρ = Φ+ α. Deﬁne e2ρn : M →Gm by γ = (a, 0; 0, d) →det(a)g+1η(γ)−g(g+1)/2, where the multiplier η(γ) is deﬁned by a · dt = η(γ)1g. (This corresponds to the adjoint action of T on the Lie algebra of the unipotent radical of P.) Secondly, we have the character e2ρM : T →Gm given by diag(α1, . . . , αg, δ1, . . . , δg) → g i=1 αg+1−2i i = g i=1 δ2i−(g+1) . and 2ρM is the sum of the positive roots in Φ+ M = {ai/aj : 1 ≤i < j ≤g}. Together they give a character e2ρ : T →Gm given by e2ρ(t) = e2ρn(t)e2ρM (t) for t ∈T; explicitly, diag(α1, . . . , αg, δ1, . . . , δg) →η−g(g+1)/2 g i=1 α2g+2−2i i . Satake’s spherical map SG,M : Hp(Γ, G) →Hp(M) is deﬁned by integrating SG,M(φ)(m) = \|eρn(m)\| U(Qp) φ(mu)du , where \|p\| = 1/p. Similarly, we have a map SM,T : Hp(M) →Hp(T) given by ST (φ)(t) = \|eρM(t)\| M∩N φ(tn)dn . In the authors deﬁne a ‘twisted’ version of these spherical maps where they put \|e2ρn(m)\| and \|e2ρM(t)\| instead of the multipliers above. In this way one avoids square roots of p. If one uses this twisted version one should also twist the action of the Weyl group on the co-character group Y of T by eρ too: in the usual action Sg permutes the ai and di and the i-th generator τi of (Z/2Z)g interchanges ai and di. Under the twisted action τi sends (ui, vi) to (pg+1−ivi, pi−g−1ui), while the permutation (i i + 1) ∈Sg sends (ui/vi) to pui+1/vi+1. The formula is w · φ(t) = \|eρ(w−1t)−ρ(t)\|φ(w−1t) for w ∈W and t ∈T, cf., . The basic result is the following theorem. Theorem 10. Satake’s spherical maps SG,M and SM,T deﬁne isomorphisms of Q-algebras Hp(G) ∼ − →Hp(T)WG and Hp(M) ∼ − →Hp(T)WM . For the untwisted version there is a similar result but one needs to tensor with Q(√p). One can calculate these maps explicitly. A right coset Kλ(p) with 218 G. van der Geer λ ∈Y is mapped under SGT to p⟨λ,ρ⟩λ. Concretely, if γ = diag(pα1, . . . , pc−αg) then SG,T(Kγ) equals pcg(g+1)/4(v1 · · · vg)c g i=1 (ui/pivi)αi . If we write a double coset Kλ(p)K as a ﬁnite sum of right cosets Kγ then we may take γ = λ(p) as one of these coset representatives. Then the image of the double coset Kλ(p)K is a sum p⟨λ,ρ⟩λ + μ nλ,μμ where the μ satisfy μ < λ (i.e. λ −μ is positive on Φ+) and the nλ,μ are non-negative integers, cf., [17,40]. 18 Relations in the Hecke Algebra We derive some relations in the Hecke algebras. We ﬁrst deﬁne elements φi in the Hecke algebra Hp(M) by pi(i+1)/2φi = M(Zp) ⎛ ⎝ 1g−i p1g 1i ⎞ ⎠M(Zp) i = 0, . . . , g From , p. 142–145 one can derive the following result. Proposition 5. We have SG,M(T (p)) = g i=0 φi and for i = 1, . . . , g SG,M(Ti(p2)) = g j,k≥0,j+i≤k mk−j(i)p−( k−j+1 2 )φjφk , where mh(i) = #{A ∈Mat(h × h, Fp): At = A, corank (A) = i}. Moreover, for i = 0, . . . , g we have SM,T(φi) = (v1 · · · vg)σi(u1/v1, . . . , ug/vg) , where σi denotes the elementary symmetric function of degree i. Example 1. g = 1. We have T (p) →φ0+φ1, T0(p2) →φ2 0+((p−1)/p)φ0φ1+φ2 1 and T1(p2) →φ0φ1/p. We derive that T (p2) = T0(p2) + T1(p2) satisﬁes the well-known relation T (p2) = T (p)2 −pT1(p2). g = 2. We ﬁnd T (p) →φ0 + φ1 + φ2 and T1(p2) →1 pφ0φ1 + p2−1 p3 φ0φ2 + 1 pφ1φ2 and similarly T2(p2) → 1 p3 φ0φ2. We denote the element φ0 corresponding to (1g, 0; 0, p1g) by Frob. This elem-ent of Hp(M) generates the fraction ﬁeld of Hp(M) over the fraction ﬁeld of Hp(Γ, G) as we can see from the calculation above. Indeed, we have that Siegel Modular Forms and Their Applications 219 ST (φ0) = v1 · · · vg and this element of Hp(T) is ﬁxed by Sg, but not by any other element of WG. In particular, it is a root of the polynomial w∈(Z/2Z)g (X −w(φ0)) = I⊂{1,...,g} 1 X − i∈I ui i/ ∈I vi 2 . For example, for g = 1 we ﬁnd by elimination that φ0 is a root of X2 −T (p)X + pT1(p2), while for g = 2 we have that φ0 is a root of X4 −T (p)X3 + (p T1(p2) + (p3 + p)T2(p2))X2 −p3 T (p)T2(p2)X + p6 T2(p2)2 . Using the relation T (p)2 = T0(p2) + (p + 1)T1(p2) + (p3 + p2 + p + 1)T2(p2) this can be rewritten as a polynomial F(X) given by X4 −T (p)X3 +(T (p)2 −T (p2)−p2T2(p2)) X2 −p3 T (p)T2(p2) X +p6 T2(p2)2 . Moreover, in the power series ring over the Hecke ring of Sp(4, Q) one has the formal relation (cf., , , p. 152) ∞ i=0 T (pi) zi = 1 −p2 T2(p2) z2 z4F(1/z) . For a slightly diﬀerent approach we refer to a paper by Krieg and a preprint by Ryan with an algorithm to calculate the images, cf., . 19 Satake Parameters The usual argument that uses the Petersson product shows that the spaces Sρ possess a basis of common eigenforms for the action of the Hecke algebra. If F is a Siegel modular form in Mρ(Γg) for an irreducible representation ρ = (λ1, . . . , λg) of GL(g, C) which is an eigenform of the Hecke algebra H then we get for each Hecke operator T an eigenvalue λF (T ) ∈C, a real algebraic number. Now the determination of the local Hecke algebra Hp ⊗C ∼ = C[Y ]WG says that HomC(Hp, C) ∼ = (C∗)g+1/WG . In particular, for a ﬁxed eigenform F the map Hp →C given by T →λF (T ) is determined by (the WG-orbit of) a (g + 1)-tuple (α0, α1, . . . , αg) of non-zero complex numbers, the p-Satake parameters of F. So for i = 1, . . . , g the parameter αi is the image of ui/vi and α0 that of v1 · · · vg and τi ∈WG acts 220 G. van der Geer by τi(α0) = α0αi, τi(αi) = 1/αi and τi(αj) = αj if j ̸= 0, i. These Satake parameters satisfy the relation α2 0α1 · · · αg = p Pg i=1 λi−(g+1)g/2 . This follows from the fact that Tg(p2), which corresponds to the double coset of p · 12g, is mapped to p−g(g+1)/2(v1 · · · vg)2 g i=1(ui/vi) as we saw above. For example, if f = n a(n)qn ∈Sk(Γ1) is a normalized eigenform and if we write a(p) = β + ¯ β with β ¯ β = pk−1 then (α0, α1) = (β, ¯ β/β) or (α0, α1) = (¯ β, β/¯ β). Or if f ∈Mk(Γg) is the Siegel Eisenstein series of weight k then the Satake parameters at p are: α0 = 1, αi = pk−i for i = 1, . . . , g. The formulas from Proposition 5 give now formulas for the eigenvalues of the Hecke operators T (p) and Ti(p2) in terms of these Satake parameters: λ(p) = α0(1 + σ1 + . . . + σg) and similarly λi(p2) = g j,k≥0,j+i≤k mk−j(i)p−(k−j+1 2 )α2 0σiσj , where σj is the jth elementary symmetric function in the αi with i = 1, . . . , αg and the mh(i) are deﬁned as in Proposition 5. 20 L-functions It is customary to associate to an eigenform f = a(n)qn ∈Mk(Γ1) of the Hecke algebra a Dirichlet series n≥1 a(n)n−s with s a complex pa-rameter whose real part is > k/2 + 1. It is well-known that for a cusp form this L-function admits a holomorphic continuation to the whole s-plane and satisﬁes a functional equation. The multiplicativity properties of the coeﬃcients a(n) ensure that we can write it formally as an Euler product n>0 a(n)n−s = p (1 −a(p)p−s + pk−1−2s)−1 . In deﬁning L-series for Siegel modular forms one uses Euler products. Suppose now that f ∈Mρ(Γg) is an eigenform of the Hecke algebra with eigenvalues λf(T ) for T ∈H0 p. Then the assignment T →λf(T ) deﬁnes an element of HomC(H0 p, C). We called the corresponding (g + 1)-tuple of α’s the p-Satake parameters of f. The fact that Z[Y ]WG is also the representation ring of the complex dual group ˆ G of G = GSP(2g, Q) (determined by the dual ‘root datum’) is responsible for a connection with L-functions. In our case we can use the Satake parameters to deﬁne the following formal L-functions. Siegel Modular Forms and Their Applications 221 Firstly, there is the spinor zeta function Zf(s) with as Euler factor at p the expression Zf,p(p−s)−1 with Zf,p(t) given by (1 −α0t) g r=1 1≤i1<···0 λf(m)m−s. We set Δ(f, s) = (2π)−gsπ−s/2Γ s + ϵ 2 g j=1 Γ(s + k −j)D(f, s) , where ϵ = 0 for g even and ϵ = 1 for g odd. Then the function Δ(f, s) can be extended meromorphically to the whole s-plane and satisﬁes a functional equa-tion Δ(f, s) = Δ(f, 1−s), cf. papers by Böcherer , Andrianov–Kalinin , Piatetski–Shapiro and Rallis . If f ∈Sk(Γg) is a cusp form and k ≥g then Δ(f, s) is holomorphic except for simple poles at s = 0 and s = 1. It is even holomorphic if the eigenform does not lie in the space generated by theta series coming from unimodular lattices of rank 2g. Also for k < g we have informa-tion about the poles, cf., . Andrianov proved that for g = 2 the function Φf(s) = Γ(s)Γ(s−k+2)(2π)−2sZf(s) is meromorphic with only ﬁnitely many poles and satisﬁes a functional equation Φf(2k −2 −s) = (−1)kΦf(s). One instance where spinor zeta functions associated to Siegel classi-cal modular forms of weight 2 occur is as L-functions associated to the 1-dimensional cohomology of simple abelian surfaces. We end by giving two additional references: the lectures notes by Courtieu and Panchishkin and a paper by Yoshida on motives associated to Siegel modular forms. 21 Liftings It is well-known that for a normalized cusp form which is an eigenform f = n≥1 a(n)qn of weight k on Γ1 we have the inequality \|a(p)\| ≤2p(k−1)/2 for 222 G. van der Geer every prime p, or equivalently, the roots of the Euler factor 1−a(p)X+pk−1X2 at p have absolute value p−(k−1)/2. This was shown by Eichler for cusp forms of weight k = 2 on the congruence subgroups Γ0(N) ⊂SL(2, Z) and by Deligne for general k in two steps, by ﬁrst reducing it to the Weil conjectures in 1968 () and then by proving the Weil conjectures in 1974. For g = 2 the analogous Euler factor at p for an eigenform F of the Hecke algebra is the expression Fp = 1 −λ(p)X + (λ(p)2 −λ(p2) −p2k−4)X2 −λ(p)p2k−3X3 + p4k−6X4 , with λ(p) the eigenvalue of the cusp form F ∈Sk(Γ2); cf., the polynomial at the end of Section 18. The tacit assumption of many mathematicians in the 1970’s was that the absolute values of the roots of Fp were equal to p−(2k−3)/2. For example, for k = 3 a classical cusp form F of weight 3 on a congruence subgroup Γ2(n) with n ≥3 determines a holomorphic 3-form F(τ) i≤j dτij on the complex 3-dimensional manifold Γ2(n)\H2 that can be extended to a compactiﬁcation and we thus ﬁnd an element of the cohomology group H3, so we expect to ﬁnd absolute value p−3/2. But then in 1978 Kurokawa and independently H. Saito ( ) found examples of Siegel modular forms of genus 2 contradicting this expectation. Their examples are the very ﬁrst ex-amples that one encounters, like the cusp form χ10 ∈S10(Γ2). On the basis of explicit calculations Kurokawa guessed that L(χ10, s) = ζ(s −9)ζ(s −8)L(f18, s) , with f18 = Δ e6 ∈S18(Γ1) the normalized cusp form of weight 18 on SL(2, Z) and L(χ10, s) = p F(p−s)−1 the spinor L-function. For example, he found for p = 2 F2 = (1 −28X)(1 −29X)(1 + 528 X + 217X2) giving the absolute values p8, p9 and p17/2 for the inverse roots. The examples he worked out suggested that in these cases L(Fk, s) = ζ(s −k + 1)ζ(s − k + 2)L(f2k−2, s) with f2k−2 ∈S2k−2(Γ1) a normalized cusp form and Fk a corresponding Siegel modular form of weight k which is an eigenform of the Hecke algebra. On the basis of this he conjectured the existence of a ‘lift’ S2k−2(Γ1) − →Sk(Γ2), f →F with L(F, s) = ζ(s −k + 1)ζ(s −k + 2)L(f, s). A little later, Maass identiﬁed in Mk(Γ2) a subspace (‘Spezialschar’, nowadays called the Maass subspace, cf., ) consisting of modular forms F with a Fourier development F = N≥0 a(N)e2πiTrNτ satisfying the property that a(N) depends only on the discriminant d(N) and the content e(N), i.e., if we write N = n r/2 r/2 m Siegel Modular Forms and Their Applications 223 then N corresponds to the positive deﬁnite quadratic form [n, r, m] := nx2 + rxy+my2 with discriminant d = 4mn−r2 and content e = g.c.d.(n, r, m). We shall write a([n, r, m]) for a(N). The condition that F belongs to the Maass space can be formulated alternatively as a([n, r, m]) = d>0, d\|(n,r,m) dk−1a([1, r/d, mn/d2]) We shall write M ∗ k(Γ2) or S∗ k(Γ2) for the Maass subspace of Mk(Γ2) or Sk(Γ2). It was then conjectured (‘Saito–Kurokawa Conjecture’) that there is a 1-1 correspondence between eigenforms in S2k−2(Γ1) and eigenforms in the Maass space S∗ k(Γ2) given by an identity between their L-functions. More precisely, we now have the following theorem. Theorem 11. The Maass subspace S∗ k(Γ2) is invariant under the action of the Hecke algebra and there is a 1-1 correspondence between eigenspaces in S2k−2(Γ1) and Hecke eigenspaces in S∗ k(Γ2) given by f ↔F ⇐ ⇒ L(F, s) = ζ(s −k + 1)ζ(s −k + 2)L(f, s) with L(F, s) the spinor L-function of F. The lion’s share of the theorem is due to Maass, but it was completed by Andrianov and Zagier, see [2,71,109]. We can make an extended picture as follows. The map F →φk,1 that sends a Siegel modular form to its ﬁrst Fourier–Jacobi coeﬃcient induces an isomorphism M ∗ k(Γ2) ∼ = Jk,1, the space of Jacobi forms, and the map h = c(n)qn → n≡−r2 (mod 4) c(n)q(n+r2)/4ζr gives an isomorphism of the Kohnen plus space M + k−1/2 with Jk,1 ﬁtting in a diagram M ∗ k(Γ2) ∼ − →Jk,1 ∼ ← − M + k−1/2 ↓∼ = M2k−2(Γ1) where the vertical map is the Kohnen isomorphism. Note that the vertical map is quite diﬀerent from the horizontal two maps. The vertical isomorphism is not canonical at all, but depends on the choice of a discriminant D. We now sketch a proof of Theorem 11. A classical Siegel modular form F ∈Mk(Γ2) has a Fourier–Jacobi series F(τ, z, τ′) = φm(τ, z)e2πimτ ′ with φm(τ, z) ∈Jk,m, the space of Jacobi forms of weight k and index m. The reader may check this by himself. We have on the Jacobi forms a sort of Hecke operators Vm : Jk,m →Jk,ml with φ\|k,mVl(τ, z) given explicitly by lk−1 Γ1\O(l) (cτ + d)−ke2πiml(−cz2/(cτ+d))φ((aτ + b)/(cτ + d), lz/(cτ + d)) . 224 G. van der Geer On coeﬃcients, if φ = n,r c(n, r)qnζr then φ\|k,mVl = n,r a\|(n,r,l) ak−1c(nl/a2, r/a)qnζr . One now checks using generators of Γ2 that for φ ∈Jk,1 the expression v(φ) := m≥0 (φ\|Vm)(τ, z)e2πimτ ′ is a Siegel modular form in Mk(Γ2). We then have a map Mk(Γ2) →⊕∞ m=0Jk,m by associating to a modular form its Fourier–Jacobi coeﬃcients; we also have a map in the other direction Jk,1 →Mk(Γ2) given by φ →v(φ) and the composition Jk,1 →Mk(Γ2) →⊕mJk,m pr − →Jk,1 is the identity. So v : Jk,1 →Mk(Γ2) is injective and the image consists of those modular forms F with the property that πm = φ1\|Vm. This implies the following relation for the Fourier coeﬃcients for [n, r, m] ̸= [0, 0, 0] a([n, r, m]) = d\|(n,r,m) dk−1c((4mn −r2)/d2) , where c(N) is given by c(N) = ⎧ ⎨ ⎩ a([n, 0, 1]) N = 4n a([n, 1, 1]) N = 4n −1 . In particular, we see that the image is the Maass subspace because a([n, r, m]) = d\|(n,r,m) dk−1a([nm/d2, r/d, 1]) . On the other hand, it is known that Jk,1 ∼ = M + k−1/2. Combination of the two isomorphisms yields what we want. Duke and Imamoˇ glu conjectured in a generalization of this and some evidence was given by Breulmann and Kuss . Then Ikeda generalized the Saito–Kurokawa lift of modular forms from one variable to Siegel modular forms of degree 2 in in 1999 under the condition that g ≡k (mod 2) to a lifting from an eigenform f ∈S2k(Γ1) to an eigenform F ∈Sg+k(Γ2g) such that the standard zeta function of F is given in terms of the usual L-function of f by ζ(s) 2g j=1 L(f, s + k + g −j) . Siegel Modular Forms and Their Applications 225 The Satake parameters of F are β0, β1, . . . , β2g with β0 = pgk−g(g+1)/2, βi = α pi−1/2, βg+i = α−1pi−1/2 for i = 1, . . . , g with f = a(n)qn and (1 −αpk−1/2X)(1 −α−1pk−1/2X) = 1 −a(p)X + p2k−1X2 , cf., . (In particular, such lifts do not satisfy the Ramanujan inequal-ity.) Kohnen ( ) has interpreted it as an explicit linear map S+ k+1/2 − → Sk+g(Γ2g) given by f = (−1)kn≡0,1( mod 4) c(n)qnF → N a(N)e2πTriNτ , with a(N) given by an expression a\|fN ak−1φ(a, N)c(\|DN\|/a2) and φ(a, N) an explicitly given integer-valued numbertheoretic function. One deﬁnes also a Maass space with M ∗ k(Γg) consisting of F such that a(N) = a(N ′) if the discriminants of N and N ′ are the same and in addit-ion φ(a, N) = φ(a, N ′) for all divisors a of fN = fN ′. Under the additional assumption that g ≡0, 1 (mod 4) Kohnen and Kojima prove in that the image of the lifting is the Maass space. Example 2. Let k = 6 and g = 2. Then the Ikeda lift is a map from S12(Γ1) → S8(Γ4) and the image of Δ is a cusp form that vanishes on the closure of the Jacobian locus (i.e., the abelian 4-folds that are Jacobians of curves of genus 4), . Or take k = g = 6 and get a lift S12(Γ1) →S12(Γ12). This lifted form occurs in the paper . Miyawaki observed in that the standard L-function of a non-zero cusp form F of weight 12 on Γ3 is a product DΔ(F, s)L(φ20, s + 10)L(φ20, s + 9), with Δ ∈S12(Γ1) and φ20 ∈S20(Γ1) the normalized Hecke eigenforms of weight 12 and 20. He conjectured a lifting and his idea was reﬁned by Ikeda to the following conjecture. Conjecture 1. (Miyawaki–Ikeda) Let k and n be natural numbers with k −n even. Furthermore, let f ∈S2k(Γ1) be a normalized Hecke eigenform. Then there exists for every eigenform g ∈Sk+n+r(Γr) with n, r ≥1 a Siegel modular eigenform Ff,g ∈Sk+n+r(Γ2n+r) such that DFf,g(s) = Zg(s) 2n j=1 Lf(s + k + n −j) , with Lf = Zf the usual L-function. In Ikeda constructs a lifting from Siegel modular cusp forms of degree r to Siegel cusp forms of degree r + 2n. This is a partial conﬁrmation of this conjecture. 226 G. van der Geer Finally, I would like to mention a conjectured lifting from vector-valued Siegel modular forms of half-integral weight to vector-valued Siegel modular forms of integral weight due to Ibukiyama. He predicts in the case of genus g = 2 for even j ≥0 and k ≥3 an isomorphism S+ j,k−1/2(Γ0(4), ψ) ∼ − →S2k−6,j+3(Γ2) which should generalize the Shimura–Kohnen lifting S+ k−1/2(Γ0(4)) ∼ = S2k−2(Γ1), see . Here ψ(γ) = −4 det(d) . 22 The Moduli Space of Principally Polarized Abelian Varieties It is a fundamental fact, due to Mumford, that the moduli space of principally polarized abelian varieties exists as an algebraic stack Ag over the integers. The orbifold Γg\Hg is the complex ﬁbre Ag(C) of this algebraic stack. This fact has very deep consequences for the arithmetic theory of Siegel modular forms, but an exposition of this exceeds the framework of these lectures. Also the various compactiﬁcations, the Baily–Borel or Satake compactiﬁcation and the toroidal compactiﬁcations constructed by Igusa and Mumford et. al. exist over Z as was shown by Faltings. We refer to an extensive, but very con-densed survey of this theory in . In particular, Faltings constructed the Satake compactiﬁcation over Z as the image of a toroidal compactiﬁcation ˜ Ag by the sections of a suﬃciently big power of det(E), the determinant of the Hodge bundle. A corollary of Faltings’ results is that the ring of classical Siegel modular forms with integral Fourier coeﬃcients is ﬁnitely generated over Z. In the following sections we shall sketch how one can use some of these facts to extract information on the Hecke eigenvalues of Siegel modular forms. The action of the Galois group of Q on the points of Ag(¯ Q) that correspond to abelian varieties with complex multiplication is described in Shimura’s theory of canonical models. This theory can also explain the integrality of the eigenvalues of Hecke operators. For this we refer to two papers by Deligne, see [21,22]. 23 Elliptic Curves over Finite Fields Suppose we did not have the elementary approach to g = 1 modular forms using holomorphic functions on the upper half plane like the Eisenstein series and Δ. How would we get the arithmetic information hidden in the Fourier coeﬃcients of Hecke eigenforms? Would we encounter Δ? Siegel Modular Forms and Their Applications 227 We claim that one would by playing with elliptic curves over ﬁnite ﬁelds. Let Fq with q = pm be a ﬁnite ﬁeld of characteristic p and cardinality q. An elliptic curve E deﬁned over Fq can be given as an aﬃne curve by an equation y2 + a1xy + a3y = x3 + a2x2 + a4x + a6 , with ai ∈Fq and with non-zero discriminant (a polynomial in the coeﬃcients). We can then count the number #E(Fq) of Fq-rational points of E. A result of Hasse tells us that #E(Fq) is of the form q + 1 −α −¯ α for some algebraic in-teger α with \|α\| = √q. We can do this for all elliptic curves E deﬁned over Fq up to Fq-isomorphism and we could ask (as Birch did in ) for the average of #E(Fq), or better for E q + 1 −#E(Fq) #AutFq(E) , where AutFq(E) is the group of Fq-automorphisms of E, or more generally we could ask for the average of the expression h(k, E) := αk + αk−1 ¯ α + . . . + α¯ αk−1 + ¯ αk , i.e. we sum σk(q) = − E h(k, E) #AutFq(E) where the sum is over all elliptic curves E deﬁned over Fq up to Fq-isomorph-ism. (As a rule of thumb, whenever one counts mathematical objects one should count them with weight 1/#Aut with Aut the group of automorphisms of the object.) If we do this for F3 we get the following table, where we also give the j-invariant of the curve y2 = f f #E(k) 1/#Autk(E) j x3 + x2 + 1 6 1/2 −1 x3 + x2 −1 3 1/2 1 x3 −x2 + 1 5 1/2 1 x3 −x2 −1 2 1/2 −1 x3 + x 4 1/2 0 x3 −x 4 1/6 0 x3 −x + 1 7 1/6 0 x3 −x −1 1 1/6 0 and obtain the following frequencies for the number of F3-rational points: n 1 2 3 4 5 6 7 freq 1/6 1/2 1/2 2/3 1/2 1/2 1/6 228 G. van der Geer Note that 1/AutFq(E) = q and E : j(E)=j 1/AutFq(E) = 1 (see for a proof); so a ‘physical point’ of the moduli space contributes 1. If we work this out not only for p = 3, but for several primes (p = 2, 3, 5, 7 and 11) we get the following values: p 2 3 5 7 11 σ10 −23 253 4831 −16743 534613 Anyone who remembers the cusp form of weight 12 Δ = n>0 τ(n)qn = q −24 q2 + 252 q3 −3520 q4 + 4830 q5 + . . . will not fail to notice that σ10(p) = τ(p) + 1 for the primes listed in this example. And in fact, the relation σ10(p) = τ(p)+1 holds for all primes p. The reason behind this is that the cohomology of the nth power of the universal elliptic curve E →A1 is expressed in terms of cusp forms on SL(2, Z). To describe this we recall the local system W on A1 associated to η−1 times the standard representation of GSp(2, Q) in Section 12. The ﬁbre of this local system over a point [E] given by the elliptic curve E can be identiﬁed with the cohomology group H1(E, Q). Or consider the universal elliptic curve (in the orbifold sense) π : E →A1 obtained as the quotient SL(2, Z)×Z2\H1 ×C, where the action of (a, b; c, d) ∈SL(2, C) on (τ, z) ∈H1 × C is ((aτ + b)/ (cτ + d), (cτ + d)−1z). Associating to an elliptic curve its homology H1(E, Q) deﬁnes a local system that can be obtained as a quotient SL(2, Z)\H1 × Q2. Then the dual of this local system is W := R1π∗Q. We now put Wk := Symk(W) , a local system with a k + 1-dimensional ﬁbre for k ≥0. We now have the following cohomological interpretation of cusp forms on SL(2, Z), cf. . Theorem 12. (Eichler–Shimura) For even k ∈Z≥2 we have an isomorphism of the compactly supported cohomology of Wk H1 c (A1, Wk ⊗C) ∼ = Sk+2 ⊕¯ Sk+2 ⊕C with Sk+2 the space of cusp forms of weight k + 2 on SL(2, Z) and ¯ Sk+2 the complex conjugate of this space. Replacing W by WR we have the exact sequence 0 →E →W ⊗R O →E∨→0 with O the structure sheaf and an induced map E⊗k →Wk ⊗R O. Now the de Rham resolution 0 →Wk ⊗R C →Wk ⊗R O d − →Wk ⊗Ω1 →0 Siegel Modular Forms and Their Applications 229 deﬁnes a connecting homomorphism H0(A1, Ω1(Wk)) →H1(A1, Wk ⊗C) . The right hand space has a natural complex conjugation and we thus ﬁnd also a complex conjugate map H0(A1, Ω1(Wk)) →H1(A1, Wk ⊗C) . A cusp form f ∈Sk+2 deﬁnes a section of H0(A1, Ω1(Wk)) by putting f(τ) → f(τ)dτdzk. We thus have a cohomological interpretation of the space of cusp forms. As observed above the moduli space A1 is deﬁned over the integers Z. This means that we also have the moduli space A1 ⊗Fp of elliptic curves in characteristic p > 0. It is well-known that one can obtain a lot of information about cohomology by counting points over ﬁnite ﬁelds. (Here we work with ℓ-adic étale cohomology for ℓ̸= p.) And, indeed, there exists an analogue of the Eichler–Shimura isomorphism in characteristic p and the relation σ10(p) = τ(p) + 1 is a manifestation of this. In fact a good notation for writing this relation is H1 c (A1, W10) = S + 1 , where the formula H1 c (A1, W2k) ∼ = S[2k + 2] + 1 for k ≥1 may be interpreted complex-analytically as the Eichler–Shimura isomorphism and in characteristic p as the relation σ2k(p) = 1 + Trace of T (p) on S2k+2. (A better interpretation is as a relation in a suitable K-group and with S[2k + 2] as the motive associated to S2k+2. This motive can be constructed in the kth power of E as done by Scholl or using moduli space of n-pointed elliptic curves as done by Consani and Faber, .) This 1 in the formula H1 c (A1, W2k) ∼ = S[2k + 2] + 1 is really a nuisance. To get rid of it in a conceptual way we consider the natural map H1 c (A1, Wk) →H1(A1, Wk) the image of which is called the interior cohomology and denoted by H1 !( A1, Wk). We thus have an elegant and sophisticated form of the Eichler–Shimura iso-morphism H1 c (A1, Wk) = S[k + 2] + 1, H1 ! (A1, Wk) = S[k + 2] . The 1 is the 1 in 1+pk+1, the eigenvalue of the action of T (p) on the Eisenstein series Ek+2 of weight k + 2 on SL(2, Z). The moral of this is that we can obtain information on the traces of Hecke operators on the space Sk+2 by calculating σk(p), i.e., by counting points on elliptic curves over Fp. Even from a purely computational point of view this is not a bad approach to calculating the traces of Hecke operators. 230 G. van der Geer 24 Counting Points on Curves of Genus 2 With the example of g = 1 in mind it is natural to ask whether also for g = 2 we could obtain information on modular forms using curves of genus 2 over ﬁnite ﬁelds. In joint work with Carel Faber () we showed that we can. For g = 2 the quotient space Γ2\H2 is the analytic space of the moduli space A2 of principally polarized abelian surfaces. A principally polarized abelian surface is the Jacobian of a smooth projective irreducible algebraic curve or it is a product of two elliptic curves. If the characteristic is not 2 a curve of genus 2 can be given as an aﬃne curve with equation y2 = f(x) with f a polynomial of degree 5 or 6 without multiple zeros. The moduli space A2 exists over Z and provides us with a moduli space A2 ⊗Fp for every characteristic p > 0. Also here we have a local system which is the analogue of the local system W that we saw for g = 1: V := GSp(4, Z)\H2 × Q4 , where the action of γ = (a, b; c, d) ∈GSp(4, Z) is given by η−1 times the stan-dard representation. Or in more functorial terms, we consider the universal family π : X2 →A2 and then V is the direct image R1π∗(Q). The ﬁbre of this local system over the point [X] corresponding to the polarized abelian surface X is H1(X, Q). The local system V comes equipped with a symplectic pairing V × V →Q(−1). Just as for g = 1 where we made the local systems Wk out of the basic one W we can construct more local systems out of V but now parametrized by two indices l and m with l ≥m ≥0. Namely, the irreducible representations of Sp(4, Q) are parametrized by such pairs (l, m) and we thus have local systems Vl,m with l ≥m ≥0 such that Vl,0 = Syml(V) and V1,1 is the ‘primitive part’ of ∧2V. A local system Vl,m is called regular if l > m > 0. Just as in the case g = 1 we are now interested in the cohomology of the local systems Vl,m. We put ec(A2, Vl,m) = i (−1)i[Hi c(A2, Vl,m)] . Here we consider the alternating sum of the cohomology groups with compact support in the Grothendieck group of mixed Hodge structures. We also have an ℓ-adic analogue of this that can be used in positive char-acteristic. It is obtained from R1π∗(Qℓ) and lives over A2⊗Z[1/ℓ]; we consider the étale cohomology of this sheaf. We simply use the same name Vl,m and assume that ℓis diﬀerent from the characteristic p. Using a theorem of Getzler (on M2) tells us what the Euler charac-teristic i(−1)i dim Hi c(A2, Vl,m) over C is. This Euler characteristic equals the Euler characteristic of the ℓ-adic variant over a ﬁnite ﬁeld, cf., . The ﬁrst observation is that because of the action of the hyperelliptic involution these cohomology groups are zero for l + m odd. Siegel Modular Forms and Their Applications 231 Our strategy is now to make a list of all Fq-isomorphism classes of curves of genus 2 over Fq and to determine for each of them #AutFq(C) and the characteristic polynomial of Frobenius. So for each curve C we determine algebraic integers α1, ¯ α1, α2, ¯ α2 of absolute value √q such that #C(Fqi) = qi + 1 −αi 1 −¯ αi 1 −αi 2 −¯ αi 2 for all i ≥1. These α’s can be calculated using this identity for i = 1 and i = 2. We also must calculate the contribution from the degenerate curves of genus 2, i.e., the contribution from the principally polarized abelian surfaces that are products of elliptic curves. Having done that we are able to calculate the trace of Frobenius on the alternating sum of Hi c(A2 ⊗Fq, Vl,m), where by Vl,m we mean the ℓ-adic variant, a smooth ℓ-adic sheaf on A2 ⊗Fq. In practice, it means that we sum a certain symmetric expression in the α’s divided by #AutFq(C), analoguous to the σk(q) for genus 1. What does this tell us about Siegel modular forms of degree g = 2? To get the connection with modular forms we have to replace the com-pactly supported cohomology by the interior cohomology, i.e., by the image of Hi c(A2, Vl,m) →Hi(A2, Vl,m) which is denoted by Hi ! (A2, Vl,m). So let us deﬁne eEis(A2, Vl,m) = ec(A2, Vl,m) −e!(A2, Vl,m) . If we do the same thing for g = 1 we ﬁnd eEis(A1, Wk) = −1 for even k > 0. Let L be the 1-dimensional Tate Hodge structure of weight 2 . It corres-ponds to the second cohomology of P1. In terms of counting points one reads q for L. Our ﬁrst result is (cf., ) Theorem 13. Let (l, m) be regular. Then eEis(A2, Vl,m) is given by −S[l + 3] −sl+m+4Lm+1 + S[m + 2] + sl−m+2 · 1 + 1 l even 0 l odd , where sn = dim Sn(Γ1). Faltings has shown (see ) that H3 ! (A2, Vl,m) possesses a Hodge ﬁltra-tion 0 ⊂F l+m+3 ⊂F l+2 ⊂F m+1 ⊂F 0 = H3 ! (A2, Vl,m) . Moreover, if (l, m) is regular then Hi ! (A2, Vl,m) = (0) for i ̸= 3. Furthermore, Faltings shows that F l+m+3 ∼ = Sl−m,m+3(Γ2) . Here Sj,k(Γ2) is the space of Siegel modular forms for the representation Symj ⊗detk of GL(2, C). This is the sought-for connection with vector valued Siegel modular forms and the analogue of H1 ! (A1, Wk) = F 0 ⊃F k+1 ∼ = Sk+2(Γ1) for g = 1. Faltings gives an interpretation of all the steps in the Hodge ﬁltration in terms of the cohomology of the bundles W(λ). 232 G. van der Geer However, although for g = 1 the Eichler–Shimura isomorphism tells us that we know H1 ! (A1, Wk) once we know Sk+2(Γ1), for g = 2 there might be pieces of cohomology hiding in F l+2 ⊂F m+1 that are not detectable in F l+m+3 or in F 0/F m+1 and indeed there is such cohomology. The contribution to this part of the cohomology is called the contribution from endoscopic lifting from N = GL(2) × GL(2)/Gm. We conjecture on the basis of our numerical calculations that this endo-scopic contribution is as follows. Conjecture 2. Let (l, m) be regular. Then the endoscopic contribution is given by eendo(A2, Vl,m) = −sl+m+4S[l −m + 2] Lm+1 . There is a very extensive literature on endoscopic lifting (cf. ), but a pre-cise result on the image in our case seems to be absent. Experts on endoscopic lifting should be able to prove this conjecture. Actually, since we know the Euler characteristics of the interior cohomology and have Tsushima’s dimen-sion formula it suﬃces to construct a subspace of dimension 2sl+m+4sl−m+2 in the endoscopic part via endoscopic lifting for regular (l, m). In terms of Galois representations a Siegel modular form (with rational Fourier coeﬃcients) should correspond to a rank 4 part of the cohomology or a 4-dimensional irreducible Galois representation. A modular form in the endoscopic part corresponds to a rank 2 part and a 2-dimensional Galois representation. Modular forms coming from the Saito–Kurokawa lift give 4-dimensional representations that split oﬀtwo 1-dimensional pieces. In analogy with the case of g = 1 we now set S[l −m, m + 3] := H3 ! (A2, Vl,m) −H3 endo(A2, Vl,m) . This should be a motive analogous to the motive S[k] we encountered for g = 1 and lives in a power of the universal abelian surface over A2. The trace of Frobenius on étale ℓ-adic H3 ! (A2, Vl,m) −H3 endo(A2, Vl,m) should be the trace of the Hecke operator T (p) on the space of modular forms Sl−m,m+3. 25 The Ring of Vector-Valued Siegel Modular Forms for Genus 2 The quest for vector-valued Siegel modular forms starts with genus 2. We can consider the direct sum M = ⊕ρMρ(Γ2) (see Section 3), where ρ runs through the set of irreducible polynomial representations of GL(2, C). Each such ρ is given by a pair (j, k) such that ρ = Symj(W) ⊗det(W)k, with W the standard representation of GL(2, C). (Note that in the earlier notation we have (λ1 −λ2, λ2) = (j, k).) So we may write M = ⊕j,k≥0Mj,k(Γ2) and we know that Mj,k(Γ2) = (0) if j is odd. If F and F ′ are Siegel modular forms of weights (j, k) and (j′, k′) then the product is a modular forms of weight of Siegel Modular Forms and Their Applications 233 weight (j + j′, k + k′). The multiplication is obtained from the canonical map Symj1(W)⊗det(W)k1⊗Symj2(W)⊗det(W)k2 →Symj1+j2(W)⊗det(W)k1+k2 obtained from multiplying polynomials in two variables. There is the Siegel operator that goes from Mj,k(Γ2) to Mj+k(Γ1). For j > 0 the Siegel operator gives a map to Sj+k(Γ1) and for j > 0, k > 4 the map Φ : Mj,k(Γ2) →Sj+k(Γ1) is surjective. For these facts on the Siegel operator we refer to Arakawa’s paper . The Siegel operator is multiplicative: Φ(F · F ′) = Φ(F) Φ(F ′). There is a dimension formula for dim Mj,k(Γ2), due to Tsushima, . But apart from this not much is known about vector-valued Siegel modular forms. The direct sum ⊕kMj,k(Γ2) for ﬁxed j is a module over the ring M cl = ⊕M0,k(Γ2) of classical Siegel modular forms and we know generators of this module for j = 2 and j = 4 and even j = 6 due to Satoh and Ibukiyama, cf. [48,49,86]. One way to construct vector-valued Siegel modular forms from classical Siegel modular forms is diﬀerentiation, the simplest example being given by a pair f ∈Ma(Γ2), g ∈Mb(Γ2) for which one sets [f, g] := 1 b f∇g −1 ag∇f with ∇f deﬁned by 2πi∇f = a (2iy)−1f + ∂/∂τ11 ∂/∂τ12 ∂/∂τ12 ∂/∂τ22 f . The point is that [f, g] is then a modular form in M2,a+b(Γ2). Using this operation (an instance of Cohen–Rankin operators) Satoh showed in that ⊕k≡0(2)M2,k is generated over the ring ⊕kMk(Γ2) of classical Siegel modular forms by such [f, g] with f and g classical Siegel modular forms. We give a little table with dimensions for dim Sj,k(Γ2) for 4 ≤k ≤20, 0 ≤j ≤18 with j even: ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ j\k 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 0 0 0 0 0 0 0 1 0 1 0 1 0 2 0 2 0 3 2 0 0 0 0 0 0 0 0 0 0 1 0 2 0 2 0 3 4 0 0 0 0 0 0 1 0 1 0 2 1 3 1 4 2 6 6 0 0 0 0 1 0 1 1 2 1 3 2 5 3 7 4 9 8 0 0 0 0 1 1 2 1 3 2 5 4 7 5 9 7 13 10 0 0 0 0 0 1 2 1 3 2 5 5 8 6 11 9 15 12 0 0 1 1 2 2 4 4 6 5 9 8 13 11 17 15 22 14 0 0 0 1 2 2 4 4 6 6 10 10 15 13 19 18 26 16 0 0 1 1 3 3 6 5 9 8 13 13 19 17 25 23 33 18 0 1 1 2 4 5 7 8 11 11 17 17 23 23 31 30 40 ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ 234 G. van der Geer The ring ⊕j,kMj,k(Γ2) is not ﬁnitely generated as was explained to me by Christian Grundh. Here is his argument. Lemma 4. The ring ⊕j,kMj,k(Γ2) is not ﬁnitely generated. Proof. Suppose that gn for n = 1, . . . , r are the generators with weights (jn, kn). If we have a modular form g of weight (j, k) with j > max(jn, n = 1, . . . , r) then g is a sum of products of gn, two of which at least have jn > 0, hence by the properties of Φ we see that then Φ(g) is a sum of products of cusp forms, hence lies in the ideal generated by Δ2 of the ring of elliptic modular forms. But for j > 0, k > 4 the map Φ : Mj,k(Γ2) →Sj+k(Γ1) is surjective, so we have forms g in Mj,k(Γ2) that land in the ideal generated by Δ, but not in the ideal generated by Δ2. Thus the ring cannot be generated by gn for n = 1, . . . , r. Just as Δ is the ﬁrst cusp form for g = 1 that one encounters the ﬁrst vector-valued cusp form that one encounters for g = 2 is the generator of S6,8(Γ2). The adjective ‘ﬁrst’ refers to the fact that the weight of the local system Vj+k−3,k−3 is j + 2k −6. Our calculations (modulo the endoscopic conjecture given in Section 24) allow the determination of the eigenvalues λ(p) and λ(p2) for p = 2, 3, 5, 7. We then can calculate the characteristic polynomial of Frobenius and even the slopes of it on S6,8(Γ2). p λ(p) λ(p2) slopes 2 0 −57344 13/2, 25/2 3 −27000 143765361 3, 7, 12, 16 5 2843100 −7734928874375 2, 7, 12, 17 7 −107822000 4057621173384801 0, 6, 13, 19 At our request Ibukiyama () has constructed a vector-valued Siegel modular form 0 ̸= F ∈S6,8, using a theta series for the lattice Γ = {x ∈Q16 : 2xi ∈Z, xi −xj ∈Z, 16 i=1 xi ∈2Z} . One puts a = (2, i, i, i, i, 0, . . ., 0) ∈C16 and one denotes by ( , ) the usual scalar product. If F = (F0, . . . , F6) is the vector of functions on H2 deﬁned by Fν = x,y∈Γ (x, a)6−ν(y, a)νeπi((x,x)τ11+2(x,y)τ12+(y,y)τ22) (ν = 0, . . . , 6) with τ = (τ11, τ12; τ12, τ22) ∈H2, then Ibukiyama’s result is that F ̸= 0 and F ∈S6,8. The vanishing of λ(2) in the table above agrees with this. Here are two more examples of 1-dimensional spaces, the space S18,5 and the last one, S28,4. In these examples and the other ones we assume the validity of our conjecture on the endoscopic contribution. The eigenvalues λ(p) grow approximately like p(j+2k−3)/2, i.e. p25/2 and p33/2. Siegel Modular Forms and Their Applications 235 p λ(p) on S18,5 λ(p) on S28,4 2 −2880 35040 3 −538920 30776760 5 118939500 522308049900 7 1043249200 18814963644400 11 −9077287359096 132158356344353064 13 −133873858788740 −1710588414695522180 17 667196591802660 −17044541241181641180 19 2075242468196920 888213094972004807320 23 −8558834216776560 −43342643806617018857520 29 64653981488634780 −172663192093972503614820 31 −5977672283905752896 1826186223285615270299584 37 56922208975445092780 −29747516862655204839491540 In principle our database allows for the calculation of the traces of the Hecke operators T (p) with p ≤37 on the spaces Sj,k for all values j, k. In the cases at hand these numbers tend to be ‘smooth’, i.e., they are highly composite numbers as we illustrate with the two 1-dimensional spaces Sj,k for (j, k) = (8, 8) and (12, 6) (where the trace equals the eigenvalue of T (p)). p λ(p) on S8,8 λ(p) on S12,6 2 26 · 3 · 7 −24 · 3 · 5 3 −23 · 32 · 89 23 · 35 · 5 · 7 5 −22 · 3 · 52 · 132 · 607 22 · 3 · 52 · 7 · 79 · 89 7 24 · 7 · 109 · 36973 −24 · 52 · 7 · 119633 11 23 · 3 · 4759 · 114089 23 · 3 · 23 · 2267 · 2861 13 −22 · 13 · 17 · 109 · 3404113 22 · 5 · 7 · 13 · 50083049 17 22 · 32 · 17 · 41 · 1307 · 168331 −22 · 32 · 5 · 7 · 13 · 47 · 14320807 19 −23 · 5 · 74707 · 9443867 −23 · 5 · 73 · 19 · 2377 · 35603 Satoh had calculated a few eigenvalues of Hecke operators T (m) acting on S14,2(Γ2), (for m = 2, 3, 4, 5, 9 and 25) cf. , and our values agree with his. 26 Harder’s Conjecture In his study of the contribution of the boundary of the moduli space to the cohomology of local systems on the symplectic group, more precisely of the Eisenstein cohomology, Harder arrived at a conjectural congruence between modular forms for g = 1 and Siegel modular forms for g = 2, cf., [44,45]. The second reference is his colloquium talk in Bonn (February 2003) which can be found in this volume and where this conjectural relationship was formulated in precise terms. One can view his conjectured congruences as a generalization 236 G. van der Geer of the famous congruence for the Fourier coeﬃcients of the g = 1 cusp form Δ = τ(n)qn of weight 12 τ(p) ≡p11 + 1 (mod 691) . To formulate it we start with a g = 1 cusp form f ∈Sr(Γ1) of weight r that is a normalized eigenform of the Hecke operators. We write f = n≥1 a(n)qn with a(n) = 1. To f we can associate the L-series L(f, s) deﬁned by L(f, s) = n≥1 a(n)/ns for complex s with real part > k/2 + 1. If we deﬁne Λ(f, s) by Λ(f, s) = Γ(s) (2π)s L(f, s) = ∞ 0 f(iy)ys−1dy then Λ(f, s) admits a holomorphic continuation to the whole s-plane and satisﬁes a functional equation Λ(f, s) = ikΛ(f, k −s). It is customary to call the values Λ(f, t) for t = k −1, k −2, . . . , 0 the critical values. In view of the functional equation we may restrict to the values t = k −1, . . . , k/2. A basic result due to Manin and Vishik is the following. Theorem 14. There exist two real numbers (‘periods’) ω+, ω−such that the ratios Λ(f, k −1)/ω−, Λ(f, k −2)/ω+, . . . , Λ(f, k/2)/ω(−1)k/2 are in the ﬁeld of Fourier coeﬃents Qf = Q(a(n) : n ∈Z≥1). If the Fourier coeﬃcients are rational integers we may normalize these ratios so that we get integers in a minimal way. In practice one observes that one usually ﬁnds many small primes dividing these coordinates. By small we mean here less than k (or something close to this). Occasionally, there is a larger prime dividing these critical values of Λ(f, s). Instead of calculating the integrals one may use a slightly diﬀerent ap-proach by employing the so-called period polynomials, , which are deﬁned for f ∈Sk(Γ1) by r = i r+ + r−with r+(f) = 0≤n≤k−2,n even (−1)n/2 k −2 n rn(f)Xk−2−n and r−(f) = 0 r dividing the critical L-value then we checked the con-gruence λ(p) −a(p) −pj+k−1 −pk−2 ≡0(mod ℓ) for all primes p ≤37. In case dim Sr(Γ1) = 2 and dim Sj,k(Γ2) = 1 I checked that in the quadratic ﬁeld Q(a(p)) the expression λ(p) −a(p) −pj+k−1 −pk−2 has a norm divisible by ℓfor all primes p ≤37. With a bit of additional eﬀort one can check the congruence in the real quadratic ﬁeld. For example, take r = 24 and let f = a(n)qn = q −(54 −12 √ 144169) q2 + . . . be a normalized eigenform in S24(Γ1). In the quadratic ﬁeld Q( √ 144169) the prime 73 splits as π · π′ with π = (73, 53 + 36 √ 144169). Let λ(p) be the eigenvalue under T (p) of the generator of S12,7(Γ2). Then we can check the congruence λ(p) ≡p5 + a(p) + p18 (mod π) for all p ≤37. In case dim Sj,k(Γ2) = 2 I can calculate the characteristic polynomial g of T (2). In general this is an irreducible polynomial g of degree 8 over Q. The corresponding number ﬁeld L possesses just one subﬁeld L of degree 2 over Q and g decomposes in two polynomials of degree 4 that are irreducible over K. I then checked that the expression λ(p) −a(p) −pj+k−1 −pk−2 has a norm in the composite ﬁeld (Q(a(2)), K) which is divisible by our congruence prime ℓ. Siegel Modular Forms and Their Applications 239 For example, we treat the case of the local system V18,6 with (ℓ, m) = (18, 6). The characteristic polynomial g of Frobenius at the prime 2 is: 1 + t1 X + t2 X2 + t3 X3 + t4 X4 + 227 t3X5 + 254t2 X6 + 281t1X7 + 2108 X8 . with the coeﬃcients t1 = 12432, t2 = 193574912, t3 = 3043199287296 and t4 = 31380514975776768. The corresponding degree 8 ﬁeld extension K of Q has one quadratic subﬁeld Q( √ 7 · 3607). Our polynomial g splits into the product of a quartic polynomial h 18014398509481984 X4 + (834297397248 −9663676416 √ 25249) X3 + (142913536 −110592 √ 25249) X2 + (6216 −72 √ 25249) X + 1 and its conjugate over this quadratic subﬁeld Q( √ 25249) and we get λ(2) = −6216 ± 72 √ 25249. The normalized eigenform in S28 has Fourier coeﬃcient a(2) = −4140 ± 108 √ 18209) and one checks that the norm of 6216 + 72 √ 25249 + 27 + 220 −(4140 + 108 √ 18209) in the ﬁeld Q( √ 25249, √ 18209) is divisible by 4057 as predicted by Harder. But there are cases where the characteristic polynomial g decomposes. These are the cases (j, k) = (18, 7) where we have two factors of degree 4 and (j, k) = (8, 13) where g is a product of four quadratic factors. In the cases (j, k) = (18, 7) there is a congruence modulo 3779. In fact, g decomposes as the product of 288230376151711744X4 −4252017623040X3 + 45752320X2 −7920X + 1 and 288230376151711744X4 + 17575006175232X3 + 857571328X2 + 32736X + 1 and one calculates Norm(4320 + 96 √ 51349 + 224 + 25 + 32736) = 282720345772032 and this is divisible by 3779. In the cases (j, k, r) = (32, 4, 38) there are two congruence primes and one ﬁnds indeed a congruence for both of them. The following table lists the congruence primes in question. All of these are checked in the sense explained above. Let me ﬁnish by expressing the hope that these explicit examples will convince the reader that Siegel modular forms are not less fascinating than elliptic modular forms and moreover that in this corner of nature there are many exciting secrets that await discovery. 240 G. van der Geer r dim(Sr) (j, k) dim(Sj,k) L-value primes 20 1 (6, 8) 1 22 · 3 · 112 22 1 (4, 10) 1 −2 · 3 · 17 · 41 41 22 1 (8, 8) 1 3 · 7 · 13 · 17 22 1 (12, 6) 1 −2 · 7 · 132 24 2 (12, 7) 1 24 · 5 · 72 · 11 · 73 73 24 2 (6, 10) 1 3 · 112 · 132 · 17 24 2 (8, 9) 1 3 · 72 · 11 · 19 · 179 179 26 1 (4, 12) 1 2 · 11 · 17 · 19 26 1 (6, 11) 1 3 · 5 · 11 · 19 26 1 (10, 9) 1 −2 · 7 · 11 · 29 29 26 1 (14, 7) 1 5 · 7 · 97 97 26 1 (16, 6) 1 −2 · 11 · 17 · 19 26 1 (8, 10) 2 −32 · 7 · 11 · 19 26 1 (12, 8) 2 3 · 52 · 11 · 17 28 2 (2, 14) 1 23 · 52 · 132 · 172 · 19 · 23 28 2 (16, 7) 1 25 · 34 · 5 · 7 · 13 · 367 367 28 1 (14, 8) 2 24 · 11 · 132 · 17 · 19 · 23 · 647 647 28 2 (12, 9) 2 23 · 7 · 11 · 13 · 23 · 4057 4057 28 2 (8, 11) 1 5 · 112 · 13 · 23 · 2027 2027 28 2 (18, 6) 1 24 · 32 · 52 · 11 · 132 · 172 · 19 28 1 (10, 10) 2 22 · 52 · 112 · 132 · 17 · 23 · 157 157 28 2 (6, 12) 2 5 · 112 · 132 · 19 · 23 · 823 823 28 2 (20, 5) 1 29 · 34 · 5 · 193 193 30 2 (14, 9) 2 28 · 3 · 5 · 13 · 1039 1039 30 2 (6, 13) 1 24 · 5 · 11 · 13 · 19 · 23 30 2 (10, 11) 1 34 · 11 · 13 · 23 · 97 97 30 2 (24, 4) 1 210 · 34 · 55 · 7 · 97 97 30 2 (20, 6) 2 26 · 33 · 7 · 11 · 13 · 17 · 19 · 23 · 593 593 30 2 (4, 14) 2 32 · 5 · 72 · 13 · 192 · 23 · 4289 4289 30 2 (18, 7) 2 24 · 32 · 5 · 11 · 3779 3779 32 2 (4, 15) 1 22 · 5 · 72 · 13 · 19 · 23 · 61 61 32 2 (2, 16) 2 33 · 52 · 72 · 192 · 23 · 211 211 32 2 (22, 6) 2 23 · 33 · 5 · 7 · 13 · 17 · 19 · 23 · 7687 7687 32 2 (24, 5) 2 29 · 35 · 5 · 3119 3119 32 2 (8, 13) 2 2 · 73 · 113 · 132 · 23 34 2 (10, 13) 2 23 · 32 · 5 · 7 · 132 · 232 · 292 34 2 (28, 4) 1 210 · 38 · 55 · 7 · 103 103 34 2 (26, 5) 2 211 · 33 · 53 · 15511 15511 34 2 (6, 15) 2 2 · 52 · 7 · 13 · 232 · 29 · 233 233 38 2 (32, 4) 2 28 · 38 · 54 · 72 · 67 · 83 67, 83 Siegel Modular Forms and Their Applications 241 References 1. A.N. Andrianov: Quadratic forms and Hecke operators. Grundlehren der Mathematik 289, Springer Verlag, 1987. 2. A.N. Andrianov: Modular descent and the Saito–Kurokawa conjecture. Invent. Math. 53 (1979), p. 267–280. 3. A.N. Andrianov, V.L. Kalinin: On the analytic properties of standard zeta functions of Siegel modular forms. Math. USSR Sb. 35 (1979), p. 1–17. 4. A.N. Andrianov, V.G. Zhuravlev: Modular forms and Hecke operators. Trans-lated from the 1990 Russian original by Neal Koblitz. Translations of Mathem-atical Monographs, 145. AMS, Providence, RI, 1995. 5. H. Aoki: Estimating Siegel modular forms of genus 2 using Jacobi forms. J. Math. Kyoto Univ. 40 (2000), p. 581–588. 6. T. Arakawa: Vector valued Siegel’s modular forms of degree 2 and the associ-ated Andrianov L-functions, Manuscr. Math. 44 (1983) p. 155–185. 7. A. Ash, D. Mumford, M. Rapoport, Y. Tai: Smooth compactiﬁcation of locally symmetric varieties. Lie Groups: History, Frontiers and Applications, Vol. IV. Math. Sci. Press, Brookline, Mass., 1975. 8. W. Baily, A. Borel: Compactiﬁcation of arithmetic quotients of bounded sym-metric domains. Ann. of Math, 84 (1966), p. 442–528. 9. J. Bergström, G. van der Geer: The Euler characteristic of local systems on the moduli of curves and abelian varieties of genus 3. arXiv:0705.0293 10. B. Birch: How the number of points of an elliptic curve over a ﬁxed prime ﬁeld varies. J. London Math. Soc. 43 (1968), p. 57–60. 11. D. Blasius, J.D. Rogawski: Zeta functions of Shimura varieties. In: Motives (2), U. Jannsen, S. Kleiman, J.-P. Serre, Eds., Proc. Symp. Pure Math. 55 (1994), p. 447–524. 12. S. Böcherer: Siegel modular forms and theta series. Proc. Symp. Pure Math. 49, Part 2, (1989), p. 3–17. 13. S. Böcherer: Über die Funktionalgleichung automorpher L-Funktionen zur Siegelschen Modulgruppe. J. Reine Angew. Math. 362 (1985), p. 146–168. 14. R.E. Borcherds, E. Freitag, R. Weissauer: A Siegel cusp form of degree 12 and weight 12. J. Reine Angew. Math. 494 (1998), p. 141–153. 15. S. Breulmann, M. Kuss: On a conjecture of Duke-Imamoˇ glu. Proc. A.M.S. 128 (2000), p. 1595–1604. 16. H. Braun: Eine Frau und die Mathematik 1933–1940. Der Beginn einer wissenschaftlichen Laufbahn. Herausgegeben von Max Koecher. Berlin etc.: Springer-Verlag, 1990. 17. P. Cartier: Representations of p-adic groups: A survey. Proc. Symp. Pure Math. 33,1, p. 111–155. 18. C. Consani, C. Faber: On the cusp form motives in genus 1 and level 1. In: Moduli and arithmetic geometry (Kyoto, 2004). Advanced Studies in Pure Mathematics 45, Math. Soc. of Japan, 2006, p. 297–314. 19. M. Courtieu, A. Panchishkin: Non-Archimedean L-functions and arithmetical Siegel modular forms. Second edition. Lecture Notes in Mathematics, 1471. Springer-Verlag, Berlin, 2004. 20. P. Deligne: Formes modulaires et représentations ℓ-adiques. Sém. Bourbaki 1968/9, no. 355. Lecture Notes in Math. 179 (1971), p. 139–172. 242 G. van der Geer 21. P. Deligne: Travaux de Shimura. Séminaire Bourbaki 1971. 23ème année (1970/71), Exp. No. 389, pp. 123–165. Lecture Notes in Math., Vol. 244, Springer, Berlin, 1971. 22. P. Deligne: Variétés de Shimura: Interpretation modulaire et techniques de construction de modèles canoniques. Automorphic forms, representations and L-functions, Part 2, pp. 247–289, Proc. Sympos. Pure Math., XXXIII, Amer. Math. Soc., Providence, R.I., 1979. 23. W. Duke, Ö. Imamoˇ glu: A converse theorem and the Saito–Kurokawa lift. Int. Math. Res. Notices 7 (1996), p. 347–355. 24. W. Duke, Ö. Imamoˇ glu: Siegel modular forms of small weight. Math. Annalen 310 (1998), p. 73–82. 25. N. Dummigan: Period ratios of modular forms. Math. Ann. 318 (2000), p. 621–636. 26. M. Eichler, D. Zagier: The theory of Jacobi forms. Progress in Mathematics, 55. Birkhäuser Boston, Inc., Boston, MA, 1985. 27. C. Faber, G. van der Geer: Sur la cohomologie des systèmes locaux sur les espaces de modules des courbes de genre 2 et des surfaces abéliennes. I, II C. R. Math. Acad. Sci. Paris 338, (2004) No.5, p. 381–384 and No.6, 467–470. 28. G. Faltings: On the cohomology of locally symmetric Hermitian spaces. Paul Dubreil and Marie-Paule Malliavin algebra seminar, Paris, 1982, 55–98, Lecture Notes in Math., 1029, Springer, Berlin, 1983. 29. G. Faltings, C-L. Chai: Degeneration of abelian varieties. Ergebnisse der Math. 22. Springer Verlag 1990. 30. E. Freitag: Siegelsche Modulfunktionen. Grundlehren der Mathematischen Wissenschaften 254. Springer-Verlag, Berlin 31. E. Freitag: Singular modular forms and theta relations. Lecture Notes in Math. 1487. Springer Verlag 32. E. Freitag: Eine Verschwindungssatz für automorphe Formen zur Siegelschen Modulgruppe. Math. Zeitschrift 165, (1979), p. 11–18. 33. E. Freitag: Zur theorie der Modulformen zweiten Grades. Nachr. Akad. Wiss. Göttingen Math.-Phys. Kl. II (1965) p. 151–157. 34. W. Fulton, J. Harris: Representation theory. A ﬁrst course. Graduate Texts in Mathematics, 129. Readings in Mathematics. Springer-Verlag, New York, 1991. 35. G. van der Geer: Hilbert Modular Surfaces. Springer Verlag 1987. 36. G. van der Geer, M. van der Vlugt: Supersingular curves of genus 2 over ﬁnite ﬁelds of characteristic 2. Math. Nachr. 159 (1992), p. 73–81. 37. E. Getzler: Euler characteristics of local systems on M2. Compositio Math. 132 (2002), 121–135. 38. R. Godement: Fonctions automorphes, vol. 1. Seminaire H. Cartan, 1957/8 Paris. 39. E. Gottschling: Explizite Bestimmung der Randﬂächen des Fundamentalbe-reiches der Modulgruppe zweiten Grades. Math. Ann. 138, 1959, p. 103–124. 40. B.H. Gross: On the Satake isomorphism. Galois representations in arithmetic algebraic geometry (Durham, 1996), p. 223–237, London Math. Soc. Lecture Note Ser., 254, Cambridge Univ. Press, Cambridge, 1998. 41. C. Grundh: Master Thesis. Stockholm. 42. S. Grushevsky: Geometry of Ag and its compactiﬁcations. To appear in Proc. Symp. Pure Math. Siegel Modular Forms and Their Applications 243 43. W.F. Hammond: On the graded ring of Siegel modular forms of genus two. Amer. J. Math. 87 (1965), p. 502–506. 44. G. Harder: Eisensteinkohomologie und die Konstruktion gemischter Motive. Lecture Notes in Mathematics, 1562. Springer-Verlag, Berlin, 1993. 45. G. Harder: A congruence between a Siegel and an elliptic modular form. Manuscript, February 2003. In this volume. 46. M. Harris: Arithmetic vector bundles on Shimura varieties. Automorphic forms of several variables (Katata, 1983), p. 138–159, Progr. Math., 46, Birkhäuser Boston, Boston, MA, 1984. 47. K. Hulek, G.K. Sankaran: The geometry of Siegel modular varieties. Higher dimensional birational geometry (Kyoto, 1997), p. 89–156, Adv. Stud. Pure Math., 35, Math. Soc. Japan, Tokyo, 2002. 48. T. Ibukiyama: Vector valued Siegel modular forms of symmetric tensor repre-sentations of degree 2. Unpublished preprint. 49. T. Ibukiyama: Vector valued Siegel modular forms of detkSym(4) and detkSym(6). Unpublished preprint. 50. T. Ibukiyama: Letter to G. van der Geer, July 2001. 51. T. Ibukiyama: Construction of vector valued Siegel modular forms and conjec-ture on Shimura correspondence. Preprint 2004. 52. J. Igusa: On Siegel modular forms of genus 2. Am. J. Math. 84 (1962), p. 612–649. 53. J. Igusa: Modular forms and projective invariants. Am. J. Math. 89 (1967), p. 817–855. 54. J. Igusa: On the ring of modular forms of degree two over Z. Am. J. Math. 101 (1979), p. 149–183. 55. J. Igusa: Schottky’s invariant and quadratic forms. E. B. Christoﬀel (Aachen/Monschau, 1979), p. 352–362, Birkhäuser, Basel-Boston, Mass., 1981. 56. J. Igusa: Theta Functions. Springer Verlag. Die Grundlehren der mathemati-schen Wissenschaften, Band 194. Springer-Verlag, New York-Heidelberg, 1972. 57. J. Igusa: On Jacobi’s derivative formula and its generalizations. Amer. J. Math. 102 (1980), no. 2, p. 409–446. 58. T. Ikeda: On the lifting of elliptic cusp forms to Siegel cusp forms of degree 2n. Ann. of Math. (2) 154 (2001), p. 641–681. 59. T. Ikeda: Pullback of the lifting of elliptic cusp forms and Miyawaki’s conjec-ture. Duke Math. Journal 131, (2006), p. 469–497. 60. R. de Jong: Falting’s Delta invariant of a hyperelliptic Riemann surface. In: Number Fields and Function Fields, two parallel worlds. (Eds. G. van der Geer, B. Moonen, R. Schoof). Progress in Math. 239, Birkhäuser 2005. 61. H. Klingen: Introductory lectures on Siegel modular forms. Cambridge Studies in advanced mathematics 20. Cambridge University Press 1990. 62. W. Kohnen: Lifting modular forms of half-integral weight to Siegel modular forms of even genus. Math. Ann. 322, (2003), p. 787–809. 63. M. Koecher: Zur Theorie der Modulformen n-ten Grades. I. Math. Zeitschrift 59 (1954), p. 399–416. 64. W. Kohnen, H. Kojima: A Maass space in higher genus. Compositio Math. 141 (2005), 313–322. 65. W. Kohnen, D. Zagier: Modular forms with rational periods. In: Modular forms (Durham, 1983), p. 197–249. Ellis Horwood Ser. Math. Appl.: Statist. Oper. Res., Horwood, Chichester, 1984. 244 G. van der Geer 66. B. Kostant: Lie algebra cohomology and the generalized Borel-Weil theorem. Ann. of Math. 74 (1961), p. 329–387. 67. A. Krieg: Das Vertauschungsgesetz zwischen Hecke-Operatoren und dem Siegelschen φ-Operator. Arch. Math. 46 (1986), p. 323–329. 68. S.S. Kudla, S. Rallis: A regularized Siegel-Weil formula: the ﬁrst term identity. Annals of Math. 140 (1994), 1–80. 69. N. Kurokawa: Examples of eigenvalues of Hecke operators on Siegel cusp forms of degree two. Invent. Math. 49 (1978), p. 149–165. 70. H. Maass: Siegel’s modular forms and Dirichlet series. Lecture Notes in Math. 216, Springer Verlag, 1971. 71. H. Maass: Über eine Spezialschar von Modulformen zweiten Grades. Invent. Math. 52 (1979), p. 95–104. II Invent. Math. 53 (1979), p. 249–253. III Invent. Math. 53 (1979), p. 255–265. 72. I. Miyawaki: Numerical examples of Siegel cusp forms of degree 3 and their zeta-functions. Mem. Fac. Sci. Kyushu Univ. Ser. A 46 (1992), p. 307–339. 73. S. Mizumoto: Poles and residues of standard L-functions attached to Siegel modular forms. Math. Annalen 289 (1991), p. 589–612. 74. K. Murokawa: Relations between symmetric power L-functions and spinor L-functions attached to Ikeda lifts. Kodai Math. J. 25 (2002), p. 61–71. 75. D. Mumford: On the Kodaira dimension of the Siegel modular variety. Alge-braic geometry - open problems (Ravello, 1982), p. 348–375, Lecture Notes in Math., 997, Springer, Berlin, 1983. 76. D. Mumford: Abelian varieties. Tata Institute of Fundamental Research Stud-ies in Mathematics, No. 5 Published for the Tata Institute of Fundamental Research, Bombay; Oxford University Press, London 1970. 77. Y. Namikawa: Toroidal compactiﬁcation of Siegel spaces. Lecture Notes in Mathematics, 812. Springer, Berlin, 1980. 78. G. Nebe, B. Venkov: On Siegel modular forms of weight 12. J. reine angew. Math. 351 (2001), p. 49–60. 79. I. Piatetski–Shapiro, S. Rallis: L-functions of automorphic forms on simple classical groups. In: Modular forms (Durham, 1983), p. 251–261, Ellis Horwood, Horwood, Chichester, 1984. 80. C. Poor: Schottky’s form and the hyperelliptic locus. Proc. Amer. Math. Soc. 124 (1996), 1987–1991. 81. B. Riemann: Theorie der Abel’schen Funktionen. J. für die reine und angew. Math. 54 (1857), p. 101–155. 82. N.C. Ryan: Computing the Satake p-parameters of Siegel modular forms. math.NT/0411393. 83. R. Salvati Manni: On the holomorphic diﬀerential forms of the Siegel modular variety. Arch. Math. (Basel) 53 (1989), no. 4, 363–372. 84. R. Salvati–Manni: On the holomorphic diﬀerential forms of the Siegel modular variety. II. Math. Z. 204 (1990), no. 4, 475–484. 85. I. Satake: On the compactiﬁcation of the Siegel space. J. Indian Math. Soc. 20 (1956), p. 259–281. 86. T. Satoh: On certain vector valued Siegel modular forms of degree 2. Math. Ann. 274 (1986) p. 335–352. 87. A.J. Scholl: Motives for modular forms. Invent. Math. 100 (1990), p. 419–430. 88. J. Schwermer: On Euler products and residual Eisenstein cohomology classes for Siegel modular varieties. Forum Math. 7 (1995), p. 1–28. Siegel Modular Forms and Their Applications 245 89. J.-P. Serre: Rigidité du foncteur de Jacobi d’échelon n ≥3. Appendice d’exposé 17, Séminaire Henri Cartan 13e année, 1960/61. 90. C.L. Siegel: Einführung in die Theorie der Modulfunktionen n-ten Grades. Math. Annalen 116 (1939), p. 617–657 (= Gesammelte Abhandlungen, II, p. 97–137). 91. C.L. Siegel: Symplectic geometry. Am. J. of Math. 65 (1943), p. 1–86 (= Gesammelte Abhandlungen, II, p. 274–359. Springer Verlag.) 92. G. Shimura: Introduction to the arithmetic theory of automorphic functions. Reprint of the 1971 original. Publications of the Math. So. of Japan, 11. Kanô Memorial Lectures, 1. Princeton University Press, Princeton, NJ, 1994. 93. G. Shimura: On modular correspondences for Sp(n, Z) and their congruence relations. Proc. Ac. Sci. USA 49, (1963), p. 824–828. 94. G. Shimura: Arithmeticity in the theory of automorphic forms. Mathematical Surveys and Monographs, 82. American Mathematical Society, Providence, RI, 2000. 95. G. Shimura: Arithmetic and analytic theories of quadratic forms and Cliﬀord groups. Mathematical Surveys and Monographs, 109. American Mathematical Society, Providence, RI, 2004. 96. C.L. Siegel: Über die analytische Theorie der quadratischen Formen. Annals of Math. 36 (1935), 527–606. 97. C.L. Siegel: Symplectic geometry. Amer. J. Math. 65 (1943), p. 1–86. 98. C.L. Siegel: Zur Theorie der Modulfunktionen n-ten Grades. Comm. Pure Appl. Math. 8 (1955), p. 677–681. 99. R. Taylor: On the ℓ-adic cohomology of Siegel threefolds. Invent. Math. 114 (1993), 289–310. 100. R. Tsushima: A formula for the dimension of spaces of Siegel cusp forms of degree three. Am. J. Math. 102 (1980), p. 937–977. 101. R. Tsushima: An explicit dimension formula for the spaces of generalized auto-morphic forms with respect to Sp(2, Z). Proc. Jap. Acad. 59A (1983), 139–142. 102. S. Tsuyumine: On Siegel modular forms of degree three. Amer. J. Math. 108 (1986), p. 755–862. Addendum. Amer. J. Math. 108 (1986), p. 1001–1003. 103. T. Veenstra: Siegel modular forms, L-functions and Satake parameters. J. Num-ber Theory 87 (2001), p. 15–30. 104. H. Yoshida: Motives and Siegel modular forms. American Journal of Math. 123 (2001), p. 1171–1197. 105. M. Weissman: Multiplying modular forms. Preprint, 2007. 106. R. Weissauer: Vektorwertige Modulformen kleinen Gewichts. Journal für die reine und angewandte Math. 343 (1983), p. 184–202. 107. R. Weissauer: Stabile Modulformen und Eisensteinreihen. Lecture Notes in Mathematics, 1219. Springer-Verlag, Berlin, 1986. 108. E. Witt: Eine Identität zwischen Modulformen zweiten Grades. Math. Sem. Hamburg 14 (1941), p. 323–337. 109. D. Zagier: Sur la conjecture de Saito–Kurokawa (d’après H. Maass). Sem-inaire Delange-Pisot-Poitou, Paris 1979–80, pp. 371–394, Progr. Math., 12, Birkhäuser, Boston, Mass., 1981. A Congruence Between a Siegel and an Elliptic Modular Form Günter Harder Mathematisches Institut, Universität Bonn, Beringstraße 1, 53115 Bonn, Germany E-mail: harder@math.uni-bonn.de Preface The winter semester 2002/2003 was the last semester before my retirement from the university. It also happened that I was the chairman of the Collo-quium and the speaker foreseen for February 7 had to cancel his visit. At about the same time I found some numerical support for a very ge-neral conjecture relating divisibilities of certain special values of L-functions to congruences between modular forms. I have been thinking about this kind of relationship for many years, but I never had any idea how one could ﬁnd experimental evidence. But in the early 2003 C. Faber and G. van der Geer had written a program that produced lists of eigenvalues of Hecke operators on some special Siegel modular forms. After a few days of suspense we could compare their list with my list of eigenvalues of elliptic modular forms and verify the congruence in our examples. I was very exited about this and spontaneously invited myself to give the Colloquium lecture, which is documented in the text below. (Bonn Spring 2007) 1 Elliptic and Siegel Modular Forms I have to recall some well known facts from the classical theory of modular forms. We have the upper half plane H = {z \| x + iy with y > 0} . On this upper half plane we have an action of Sl2(R), which is given by Sl2(R) × H − →H a b c d , z →az + b cz + d . 248 G. Harder The stabilizer of i ∈H is the maximal compact subgroup SO(2) and we can identify H = Sl2(R)/SO(2). Let k be a positive (even) integer. A holomorphic modular form of weight k with respect to Sl2(Z) is a holomorphic function f : H →C, which satisﬁes f az + b cz + d = (cz + d)kf(z) for all matrices a b c d ∈Sl2(Z) , and which satisﬁes a growth condition. To formulate this growth condition we restrict f to a “neighborhood of inﬁnity” H(c) = {z\|ℑ(z) > c}. On this neighborhood the group Γ∞= 1 n 0 1 with n ∈Z acts and the map z →e2πiz identiﬁes Γ∞\H(c) to a punctured disk. Since f satisﬁes f(z) = f(z + 1) we can view its restriction to Γ∞\H(c) as a function in the variable q. The growth condition requires that f has a (Fourier or Laurent) expansion f(q) = a0 + a1q + a2q2 . . . , i.e. it extends to a holomorphic function on the disk. If a0 = 0, then f is called a cusp form. Remark: The quotient Sl2(Z)\H has the structure of a Riemann surface, which can be compactiﬁed to a compact Riemann surface Sl2(Z)\H by adding one point at ∞. We write the maximal compact subgroup SO(2) = U(1) = K = $ e(φ) \| e(φ) = cos(φ) sin(φ) −sin(φ) cos(φ) % . Since H = Sl2(R)/SO(2), the representation ρk : SO(2) →C×, which is given by e(φ) →e(φ)k deﬁnes a Sl2(R)-invariant holomorphic line bundle Lk on H, this gives us a line bundle, also called Lk, on Sl2(Z)\H. This line bundle can be extended in a speciﬁc way to a line bundle on the compactiﬁcation. Then the space of modular forms of weight k can be canonically identiﬁed with the space of sections H0(Sl2(Z)\H, Lk). We have the two modular forms of weight 4 and 6 E4(z) = 1 2 (c,d)=1 1 (cz + d)4 , E6(z) = 1 2 (c,d)=1 1 (cz + d)6 , and then we have the q-expansions E4(q) = 1 + 240q + 2160q2 + 6720q3 + 17520q4 + 30240q5 . . . E6(q) = 1 −504q −16632q2 −122976q3 −532728q4 −1575504q5 + . . . . A Congruence Between a Siegel and an Elliptic Modular Form 249 The space of cusp forms has dimension 1 for the values k = 12, 16, 18, 20, 22, 26. The modular form Δ(z) = E4(q)3 −E6(q)2 123 = q −24q2 + 252q3 −1472q4 + 4830q5 + . . . is the generator of the space of cusp forms of weight 12. The space of cusp forms of weight 22 is generated by f(q) =E6(q)E4(q)4 −E6(q)3 · E4(q) 123 = q −288q2 −128844q3 −2014208q4 + 21640950q5 + 37107072q6 −768078808q7 + 1184071680q8 + 6140423133q9 −6232593600q10 −94724929188q11 ± . . . . Now I have to say a few words on Siegel modular forms. We start from a lattice L = Z4 = Ze1 ⊕Ze2 ⊕Zf2 ⊕Zf1 on which we have an alternating pairing which on the basis vectors is given by ⟨e1, f1⟩= ⟨e2, f2⟩= −⟨f1, e1⟩= −⟨f2, e2⟩= 1 , and all other values of the pairing are zero. The group of automorphisms of this symplectic form is a semi-simple group scheme Sp2/ Spec(Z). This is the symplectic group of genus 2. Its group of real points Sp2(R) = {g ∈GL4(R) \| ⟨gx, gy⟩= ⟨x, y⟩} contains U(2) as a maximal compact subgroup and we can form the quotient space H2 = Sp2(R)/U(2) . This is the space of symmetric 2 × 2 matrices Z = X + iY with complex entries whose imaginary part Y is positive deﬁnite. Hence we have a complex structure on this space. This complex structure can also be seen in the following way: let P(C) in Sp2(C) be the stabilizer of the isotropic plane {e1 −if1, e2 −if2} ⊂C4, then we have an open embed-ding H2 = Sp2(R)/U(2)# − →Sp2(C)/P(C) , the group SU(2) is the group of real points of P(C) intersected with its complex conjugate ¯ P(C). The object on the right is the Grassmann vari-ety of isotropic complex planes in (C4, ⟨, ⟩). It is projective and of di-mension 3. The group Γ = Sp2(Z) acts upon H2 and the quotient Γ\H2 is a quasiprojective algebraic variety over C. We have a homomorphism 250 G. Harder P(C) →GL2(C). For any pair of integers i ≥0, j the holomorphic repre-sentation ρ : GL2(C) − →Symi(C2) ⊗detj deﬁnes a holomorphic vector bundle Eij on the ﬂag variety Sp2(C)/P(C) which is Sp2(C)-equivariant. Hence its restriction – also called Eij – to H2 is a Sp2(R) equivariant holomorphic bundle on H2 and hence descends to a holomorphic bundle on Γ\H2. We can consider the space of holomorphic sections H0(Γ\H2, Eij) , and deﬁne the subspace of modular forms Mij (which satisfy some growth condition) and the subspace Sij of cusp forms; these are rapidly decreasing at inﬁnity. These spaces are called the spaces of modular forms (cusp forms) of weight i, j. (See remark above.) There are formulas by R. Tsushima for the dimensions of these spaces Sij (Riemann–Roch–Hirzebruch or the trace formula), and for small values i, j the dimensions are zero. We say that i, j is a regular pair if i > 0, j > 3. We have 29 cases of regular pairs i, j where Sij is of dimension one. 2 The Hecke Algebra and a Congruence Whenever we have such a space of modular forms we have an action of the algebra of Hecke operators on it. This is an algebra generated by operators Tp (for Sl2(Z)) and T (ν) p , ν = 1, 2 (for Sp2(Z)), which are attached to a prime p and which induce endomorphisms T (ν) p : Sij →Sij, and which commute with each other. If we pick a prime, then we can consider the matrix ⎛ ⎜ ⎜ ⎝ p 0 p 1 0 1 ⎞ ⎟ ⎟ ⎠ which is in GSp2(Q), and if f(Z) ∈H0(Γ\H2, Eij), then f ⎛ ⎜ ⎜ ⎝ ⎛ ⎜ ⎜ ⎝ p 0 p 1 0 1 ⎞ ⎟ ⎟ ⎠Z ⎞ ⎟ ⎟ ⎠ is not invariant under Γ; it is only a section in H0(Γ0(p)\H2, Eij) , where Γ0(p) ⊂Γ is a subgroup of ﬁnite index. We can form a trace by summing over Γ0(p)\Γ and up to a normalizing factor this will be our operator T (1) p : Sij − →Sij . A Congruence Between a Siegel and an Elliptic Modular Form 251 If now dim Sij = 1, then the operator T (1) p : Sij →Sij induces the multi-plication by a number λ(p) on Sij and if j ≥3, then we get a sequence of integers {λ(p)}p∈Primes which, of course, depends on i, j. We also have the Hecke operators for classical modular forms, and our cusp form f of weight 22 is also an eigenform for the operators Tp. In this case the situation is simple. Because f is normalized, i.e. a1 = 1, we have Tpf = apf where ap is the p-th Fourier coeﬃcient. We have dim S4,10 = 1 and formulate the conjecture Conjecture: For S4,10 we have a congruence λ(p) ≡p8 + ap + p13 mod 41 for all primes p . Prop. The conjecture holds for 2 ≤p ≤11. One might say that this is really not so much evidence for the conjec-ture. But here are certain numbers, namely, 4, 10, 22, 8, 13 and 41, which seem to be somewhat arbitrary. I did not play with these numbers un-til I found a congruence. I picked all these numbers in advance and only then I checked the congruence, which I expected to be true for this speciﬁc choice. The congruence is a generalization of a classical congruence. If we write the Δ-function Δ(z) = q −24q2 + 252q3 −1472q4 + 4830q5 ± . . . = ∞ n=1 τ(n)qn , then we have the famous Ramanujan congruence τ(p) ≡p11 + 1 mod 691 for all primes p . But there is a diﬀerence: Usually people interpret this last congruence as a congruence between the q-expansions of two modular forms, namely the Δ-function and the Eisenstein series E12(z). Since the Fourier coeﬃcients are the same as the eigenvalues of the Hecke operators we also get the congruence between the eigenvalues. For the congruence between the Siegel modular form and the elliptic modular form we only have a congruence be-tween Hecke eigenvalues. I do not see a congruence between Fourier coeﬃ-cients. I want to say something about the numbers, how I get them and I want to say a few words about the meaning of this congruence. 252 G. Harder 3 The Special Values of the L-function We start from our modular cusp form of weight 22 f(q) = q −288q2 −128844q3 −2014208q4 + 21640950q5 + . . . = ∞ n=1 anqn , we have its associated L-function L(f, s) = ∞ n=1 an ns , and because f is an eigenform for the Hecke algebra this L-function has an Euler product expan-sion L(f, s) = ∞ n=1 an ns = p 1 1 −app−s + p21−2s . Actually it is better to consider the Mellin transform ∞ 0 f(iy)ys dy y = Γ(s) (2π)s · L(f, s) = Λ(f, s) . From this integral representation we easily get the functional equation Λ(f, 22 −s) = −Λ(f, s) . Now we consider the “special” values Λ(f, 21), Λ(f, 20), . . ., Λ(f, 11). It follows from the theory of modular symbols (Manin–Vishik) that there exist two real numbers Ω−, Ω+ ̸= 0 (the periods) such that Λ(f, 21) Ω− , Λ(f, 20) Ω+ , Λ(f, 19) Ω− , . . . ∈Q These periods are only deﬁned up to elements in Q×, but a closer look allows us to pin them down up to a factor in Z× = {±1}. In this case we can simply try to normalize them such that $Λ(f, 21) Ω− , Λ(f, 19) Ω− , . . . , Λ(f, 11) Ω− % and $Λ(f, 20) Ω+ , . . . , Λ(f, 14) Ω+ , Λ(f, 12) Ω+ % are sets of co-prime integers. Of course, it is not so diﬃcult to produce these lists of integers. (From this list we conclude that the normalization of Ω−was not the right one. This is related to the fact that 131 · 593 \| ζ(−21) , and this produces a congruence between f and an Eisenstein series ap ≡p21 + 1 mod 131 · 593 . A Congruence Between a Siegel and an Elliptic Modular Form 253 This forces us to replace Ω−by 131 ·593 ·Ω−.) With this modiﬁcation the list for the odd case is {25·33·56·7·13·17·19/(131·593), 25·3·52·13·17 , 2·3·53·7·13, 2·52·13·17, 537, 0} and for the even case {25 · 33 · 5 · 19, 23 · 7 · 132, 3 · 5 · 7 · 13, 2 · 3 · 41, 2 · 3 · 7} . We have exactly one “large” prime dividing a value. This is 41 \| Λ(f, 14) Ω+ , and this divisibility is the source for the congruence above. 4 Cohomology with Coeﬃcients To explain this connection I have to recall some other facts from the theory of Siegel modular varieties. The space Γ\H2 can be interpreted as the parameter space of principally polarized abelian surface over C. Roughly, we can attach to a point in H2 a triple ⟨L, ⟨, ⟩, I⟩= AI where I is a complex structure on L ⊗R, which is an isometry for the pairing and s.t. the associated hermitian form is positive deﬁnite. (I personally prefer to view H2 as the space of such complex structures on L ⊗R.) This AI is an abelian surface and AI ∼ = AI′ if there is a γ ∈Γ such that γI = I′, and this γ provides an isomorphism γ∗: AI ∼ = AI′. Here we encounter a minor diﬃculty, because γ is not unique, and γ∗depends on the choice of γ. Therefore we can not attach an abelian variety to a point ˜ I ∈Γ\H2. But if we pass to a suitably small normal congruence subgroup Γ ′ ⊂Γ then it is clear that we have a family π : A →Γ ′\H2 of principally polarized abelian varieties over Γ ′\H2. Then the family of cohomology groups H1(A˜ I, Z) deﬁnes a local system of free Z-modules of rank 4 over Γ ′\H2. This local system descends to a sheaf on Γ\H2. This sheaf is also obtained from the standard representation ρ10 : Γ − →Gl(L) = Gl(M1,0) . We deﬁne a representation ρ01 : Γ →Gl(M0,1) where the module M0,1 is deﬁned by Λ2M1,0 = M0,1 ⊕Z . 254 G. Harder We can form the modules Symm(M1,0) ⊗Symn(M0,1) and these modules have a unique submodule (or quotient) ρm,n : Γ →Gl(Mm,n) , which is deﬁned be the requirement that it has the largest dominant weight amoung all highest weights of submodules. (The representations ρ10 (resp. ρ01) have highest weight γβ (resp. γα), which are the two fundamental dominant weights. Then Mm,n is the unique irreducible submodule with highest weight λ = mγβ + nγα. At this point is an ambiguity: Instead of taking the submodule we could take the unique irreducible quotient having this fundamental weight (actually this ambiguity already occurs when we form the symmetric products). Then the submodule will map injectively into the quotient and the image is a submodule of ﬁnite index. This index will be only divisible by “small” primes ≤m, n, they do not play a role in our considerations, in other words it does not matter whether we take the submodule or the quotient.) These representations yield sheaves ˜ Mm,n of Z-modules. For an open set U ⊂Γ\H2 and its inverse image ˜ U ⊂H2 we have ˜ Mm,n(U) = {f : ˜ U →Mm,n\|f locally constant and f(γu) = ρn,m(γ)f(u) for all γ ∈Γ} . For m even, these modules give us sheaves ˜ Mm,n on the space Γ\H2 which are almost local systems. We can consider the cohomology groups Hi c(Γ\H2, ˜ Mm,n), Hi(Γ\H2, ˜ Mm,n) where H• c denotes the cohomology with compact support. These cohomology groups sit in an exact sequence →Hi−1(∂(Γ\H2), ˜ Mm,n) →Hi c(Γ\H2, ˜ Mm,n) →Hi(Γ\H2, ˜ Mm,n) →Hi(∂(Γ\H2), ˜ Mm,n) →, where ∂(Γ\H2) is the boundary of the Borel–Serre compactiﬁcation. We denote by Hi ! (Γ\H2, ˜ Mm,n) the image of the cohomology with compact supports in the cohomology. The coeﬃcient system ˜ Mm,n is called regular if n, m > 0. In this case all cohomology groups Hi ! (Γ\H2, ˜ Mm,n ⊗Q) vanish for i ̸= 3. We have a Hodge ﬁltration on H3 ! (Γ\H2, ˜ Mm,n ⊗C), and the lowest step of this ﬁltration is given by (Faltings) Sm,n+3 # − →H3 ! (Γ\H2, ˜ Mm,n ⊗C) . To get the connection to the conjecture we choose m = 4, n = 7. Now I formulate a second conjecture. We invert some small primes (say ≤22), and we denote the resulting ring by R = Z[ 1 2, . . . , 1 19]. A Congruence Between a Siegel and an Elliptic Modular Form 255 Then we get an exact sequence (assumption) 0 →H3(Γ\H2, ˜ M4,7 ⊗R) →H3(Γ\H2, ˜ M4,7 ⊗R) →H3 ∂(Γ\H2), ˜ M4,7 ⊗R →0 , I know that this is true if I replace R by Q. Now we can show that we have an action of the Hecke operators on these modules and we have H3(∂(Γ\H2), ˜ M4,7 ⊗R) = R where T (1) (p) acts on R with the eigenvalue p8 + ap + p13 . (For this assertion I refer to my lecture notes volume or [Modsym].) Now we formulate another assumption H3 ! (Γ\H2, ˜ M4,7 ⊗R) ∼ = R4 . Then we know that T (1) p acts as a scalar by multiplication by λ(p) on this cohomology group. We have λ(2) ̸= 28 +a2+213 and hence we can decompose H3(Γ\H2, ˜ M4,7 ⊗Q) = H3 ! (Γ\H2, ˜ M4,7 ⊗Q) ⊕H3 Eis(Γ\H2, ˜ M4,7 ⊗Q) . Now the main assertion of the second conjecture is: If we intersect this decomposition with the integral cohomology, then H3(Γ\H2, ˜ M4,7 ⊗R) ⊃H3 ! (Γ\H2, ˜ M4,7 ⊗R) ⊕H3 Eis(Γ\H2, ˜ M4,7 ⊗R) . and the index of the direct sum in H3(Γ\H2, ˜ M4,7 ⊗R) is divisible by 41. (The denominator of the Eisenstein class is divisible by 41.) The point with this second conjecture is that it implies the ﬁrst conjecture and it can be veriﬁed on a computer. To do this we have to ﬁnd a way to compute the cohomology groups. This can be done by using a suitable acyclic covering of Γ\H2, and then the cohomology is computed from the Čech com-plex of this covering. We could also try to use a cell decomposition. This method will allow us to check the two assumptions. I think the problem will be that the number of cells will not be so big, but we have nontrivial coef-ﬁcient systems, its dimension in a general point is 1820. It will be still more diﬃcult to implement the action of the Hecke operator, because one has to pass to a ﬁner cell decomposition, which also computes the cohomology and where the Hecke operators can be implemented as a homomorphism between the two complexes. Of course, we could also compute mod 41, then we ﬁnd λ(2) ≡28+a2+213 mod 41, and our conjecture would say that T (1) 2 mod 41 is not diagonalizable. 256 G. Harder Why didn’t I do this earlier? In my lecture notes volume (Chap III, 3.1) I discuss the above conjecture in greater generality and I raise the question whether computer experiments should be made. There I say that these com-putations would “. . . einen beträchtlichen Aufwand erfordern, aber die Frage entscheiden, ob es sich lohnt, das Problem zu behandeln”. Recently I got some kind of unexpected help from C. Faber und G. van der Geer. They produced some tables of eigenvalues λ(p) for certain local systems Mm,n with some small values n, m. They make use of the fact that Γ\H2 is actually the set of the complex points of a quasiprojective scheme A2/ Spec(Z), and that our local systems ˜ Mm,n have an algebraic-geometric meaning. They are “motivic” sheaves, and it is not quite clear what that means. But in any case we can pick a prime ℓand then ˜ Mm,n⊗Zℓwill be an ℓ-adic sheaf on A2. Then we have the Grothendieck ﬁxed point formula tr (Φp \| H• c (A2 ×Z ¯ Fp, ˜ Mm,n ⊗Zℓ)) = x∈A2(Fp) tr (Φp \| ˜ Mm,n,x) , where Φp is the Frobenius at p. The right hand side can be computed because we have the modular interpretation. The left hand side consists of several pieces (Eisenstein cohomology, endo-scopic contributions, if m = 0 there may be some Saito–Kurokawa lifts), and the trace of Φp on these pieces can be computed explicitly (for small n, m) and can be expressed in terms of modular forms for Sl2(Z) and in terms of algebraic Hecke characters. This can be brought to the right hand side, and the resulting expression can be computed explicitly for small values n, m. Then we are left with the “genuine” part in H3 c (A2 ×Z ¯ Fp, ˜ Mm,n ⊗Zℓ), and this part will be of rank 4 · dim Sm,n+3. If now dim Sm,n+3 = 1, then this “genuine” part will be of rank 4 and we have tr (Φp \| H3 genuine(A2 ×Z ¯ Fp, ˜ Mm,n ⊗Zℓ)) = λ(p) . But now the λ(p) can be computed from the right hand side, if we take the eﬀort to compute the sum over A2(Fp) and the non “genuine” traces. After I saw the preprint by C. Faber and G. van der Geer I realized that I might be able to check the ﬁrst conjecture in a special case. I had to go through the values L(f,k) Ωε(k) for the modular cusp form f of weight ≤22. (For higher weights except 26 the dimension of these space are ≥2. The eigenvalues of the Hecke operators are algebraic integers and also the normalized L-values will be algebraic integers, and the computations will be much more complicated.) I had to ﬁnd a “large” prime dividing one of the values, and I found for our form of weight 22 41 \| L(f, 14) Ω+ . I computed the numbers 4, 7 and 7 + 3 = 10 from these data and wrote an e-mail to G. van der Geer inquiring the dimension of S4,10. Several answers A Congruence Between a Siegel and an Elliptic Modular Form 257 were possible. The dimension could be zero. This would be devastating. The dimension could be >1, this would mean a horrible additional computational eﬀort. But the answer was Re: Kohomologie lokaler Systemen Lieber Guenter, die Dimension ist dann 1. Die ersten Eigenwerte sind wie folgt: −24 · 3 · 5 · 7 23 · 34 · 5 · 17 −22 · 3 · 52 · 17 · 1439 24 · 52 · 72 · 17 · 31 · 59 23 · 3 · 11 · 17 · 5650223 d.h. fuer die Primzahlen 2, 3, 5, 7, 11. Mit bestem Gruss, Gerard I read this message in my oﬃce in the Beringstrasse and I had the values of the ap at home on my laptop. After two oral examinations of computer science students I went home and checked the numbers. I was extremely pleased when I found that the congruences hold. (Actually van der Geer was also pleased because he considered it as con-ﬁrmation of his computations with Faber. (I have multiplied the values in his table by −1, probably this has to be done because the trace occurs in odd degree)) 5 Why the Denominator? We stick to the case M4,7, and f is still our modular cusp form of weight 22. If we had a splitting under the Hecke-algebra H3(Γ\H2, ˜ M4,7 ⊗R) = H3 ! (Γ\H2, ˜ M4,7 ⊗R) ⊕H3(∂(Γ\H2), ˜ M4,7 ⊗R) , then we could construct a mixed Tate motive X(f) which sits in an exact sequence 0 →R(−8) →X(f) →R(−13) →0 and hence deﬁnes an element in the extension group [X(f)] ∈Ext1 MM(R(−13), R(−8)) = Ext1 MM(R(−5), R(0)) . (For more details see [Mixmot].) This Ext1 group is some kind of undeﬁned object, but we can attach to our object X(f) elements in two other extension groups, namely: 258 G. Harder (i) an extension class in the category of mixed Hodge structures [X(f)]BdRh ∈Ext1 BdRh(R(−13), R(−8)) = Ext1 BdRh(R(−5), R(0)) = R (See MixMot 1.5.2). It is some kind of general belief that those elements in the extension group of mixed Hodge structures, which come from mixed motives X over Z, are in fact elements of the form [XBdRh] = a(X)ζ′(−4) with a(X) ∈Q . This last conjecture can be veriﬁed in our particular case we have the formula [XBdRh(f)] = c · Λ(f,13) Ω− Λ(f,14) Ω+ ζ′(−4) where c is a rational number containing only small primes. (ii) For any prime ℓwe can attach an ℓ-adic extension class [X(f)]ℓ∈Ext1 Gal(Rℓ(−13), Rℓ(−8)) = H1(Gal(¯ Q/Q, )Rℓ(5)) and this cohomology group contains certain speciﬁc elements cℓ(5), these are the Soulé elements. These elements should also be generators of the image of Ext1 MM(Rℓ(−13), Rℓ(−8)) →H1(Gal(¯ Q/Q), Rℓ(5)) if we tensor by Q. We write the Galois cohomology group multiplicatively and now it is general belief that we must have [X(f)]ℓ= cℓ(5) c· Λ(f,13) Ω− Λ(f,14) Ω+ . From now on we choose ℓ= 41 (of course we could replace ℓby 41 in the following considerations but this causes some confusion), then the value of the ℓ-adic ζ-function ζℓ(5) ̸≡0 mod ℓand this implies that cℓ(5) is a primitive element in the Galois cohomology group. But the rational exponent has ℓin its denominator, this contradicts the existence of our mixed motive and this motive has been constructed under the assumption that ℓdoes not divide the denominator of the Eisenstein class. 6 Arithmetic Implications We get a diagram (still ℓ= 41) of ℓ-adic Galois modules 0 →H3 ! (A2 ×Z ¯ Q, ˜ M4,7 ⊗Rℓ) →H3(A2 ×Z ¯ Q, ˜ M4,7 ⊗Rℓ) →Rℓ(−13) →0 ∪ r ↗ H3 ! (A2 ×Z ¯ Q, ˜ M4,7 ⊗Rℓ) ⊕Rℓ(−13) A Congruence Between a Siegel and an Elliptic Modular Form 259 where the image of the homomorphism r is contained in ℓRℓ(−13). This gives us an injective homomorphism ψ : Z/(ℓ)(−13) # − →H3 ! (A2 ×Z ¯ Q, ˜ M4,7 ⊗Z/(ℓ)) . The module H3 ! (A2 ×Z ¯ Q, ˜ M4,7 ⊗Z/(ℓ)) is of dimension 4 over Fℓand the cup product provides a non -degenerated pairing of this module with itself into Z/(ℓ)(−21). The orthogonal complement Y of the image ψ(Z/(ℓ)(−13)) is of dimension 3 over Fℓand we get two exact sequences 0 →Z/(ℓ)(−13) →Y →X →0 and 0 →X →H3 ! (A2 ×Z ¯ Q, ˜ M4,7 ⊗Z/(ℓ))/ψ(Z/(ℓ)(−13)) →Z/(ℓ)(−8) →0 . The module X is actually the reduction of the ℓ-adic representation attached to f mod ℓ. It also has a non degenerate pairing with itself with values in Z/(ℓ)(−21) and the two sequences are dual to each other. The sequences give us two extension classes, the ﬁrst one a class [Y ] ∈Ext1 Gal(X, Z/(ℓ)(−13)) =H1(Gal(¯ Q/Q), Hom(X, Z/(ℓ)(−13))) ∼ − → H1(Gal(¯ Q/Q), X ⊗Z/(ℓ)(8)) and under the isomorphism [Y ] is mapped to the extension class of the second sequence. Now we can hope that this extension class is actually an element in the Selmer group of the Scholl-Deligne motive M(f) attached to f, and that it is in fact an element of order ℓ. If this turns out to be the case, then we have produced an element in the Selmer group whose existence is predicted by the general philosophy of the Bloch–Kato–Birch–Swinnerton Dyer conjecture. References Fa-vdG Faber, Carel; van der Geer, Gerard: Sur la cohomologie des systèmes locaux sur les espaces de modules des courbes de genre 2 et des surfaces abéliennes.I, C. R. Math. Acad. Sci. Paris 338 (2004), no. 5, 381–384. II, no. 6, 467–470 Ha-Eis G. Harder, Eisensteinkohomologie und die Konstruktion gemischter Motive, SLN 1562 The following two manuscripts on mixed motives and modular sym-bols can be found on my home page www.math.uni-bonn.de/people/harder/ in my ftp-directory folder Eisenstein. Mixmot Modular construction of mixed motives II Modsym Modular Symbols and Special Values of Automorphic L-Functions 260 G. Harder Appendix In the meanwhile C. Faber, G. van der Geer and I did some further compu-tations. We have still another one dimensional space of modular cusp forms, this is spanned by the modular form g(q) = Δ(q)E6(q)E4(q)2 = q −48q2 −195804q3 −33552128q4 −741989850q5 + 9398592q6 + . . . of weight 26. We have the following divisibilities by “large” primes 43\|L(g, 23) Ω− , 97\|L(g, 21) Ω− , 29\|L(g, 19) Ω− . The corresponding spaces of modular forms S18,5, S14,7, S10,9 have dimen-sion 1. Now let ℓbe one of the primes 41, 43, 97, 29. Let f(q) = q + a2q2 + a3q3 . . . be the corresponding modular form of weight 22 or 26. Let Si,j be the cor-responding one dimensional space of Siegel modular forms and let ˜ Mi,j−3 = ˜ Mm,n be the corresponding local system. The Hecke algebra acts on the co-homology H3 ! (A2 ×Z ¯ Q, ˜ Mm,n ⊗Rℓ) and we should ﬁnd an isotypical submodule of rank 4 on which the Hecke operators act by the scalar by which they acts on Si,j. The local Hecke algebra at a prime p is generated by two Hecke operators Tp,α, Tp,β which correspond to the double classes Sp2(Zp) ⎛ ⎜ ⎜ ⎝ p 0 p 1 0 1 ⎞ ⎟ ⎟ ⎠Sp2(Zp) and Sp2(Zp) ⎛ ⎜ ⎜ ⎝ p2 0 p p 0 1 ⎞ ⎟ ⎟ ⎠Sp2(Zp) . (See [Ha-Eis] 3.1.2.1) So we get sequences of eigenvalues {λα(p), λβ(p)}p∈Primes Now we state the conjecture that in all four cases we have congruences λα(p) ≡pn+1 + ap + pn+m+2 mod ℓ and λβ(p) ≡ap(1 + pm+1) + (p2 −1)pnα+nβ mod ℓ for all primes p. A Congruence Between a Siegel and an Elliptic Modular Form 261 For our four primes ℓabove and the corresponding modular forms the conjecture for λα(p) has been checked for all p ≤37. The general rule is: If k is even and f an eigenform of weight k. Let K = Q(f) be its ﬁeld of deﬁnition. Let us assume that a large prime l divides L(f, ν)/Ωϵ(ν). Then we solve the equations k = 2n + m + 4, ν = n + m + 3 . Then we can construct an Eisenstein class in H3 ! (Γ\H2, ˜ Mm,n ⊗K) whose denominator is divisible by l. Added on April 3, 2003 (the day when the ﬁrst Abel-Prize was given to J.-P. Serre): I also checked congruences for the modular cusp forms of weight 24. In this case we have two eigenforms f(q) = ∞ n anqn = q −(540 −12 √ 144169)q2 + (169740 + 576 √ 144169)q3 . . . where we take the positive root-, and we have the conjugate eigenform f ′(q) = ∞ n a′ nqn . We put ω = 1+ √ 144169 2 . In this case we ﬁnd periods Ω±, Ω′ ± such that L(f, k) Ωϵ(k) ∈Z[ω], L(f ′, k) Ω′ ϵ(k) ∈Z[ω] . We normalize the periods such that these numbers for a ﬁxed choice of the sign ϵ(k) are coprime and such that L(f,k) Ωϵ(k) and L(f ′,k) Ω′ ϵ(k) are conjugate. The primes 73 and 179 split in Q( √ 144169) and for 73 the decomposition is l = (73, 53 + 36 √ 144169), l′ = (73, 53 −36 √ 144169) (73) = ll′ . We ﬁnd L(f,19) Ω− ∈l. The corresponding space S12,7 has dimension 1, if λ(p) is the sequence of eigenvalues the congruence λ(p) ≡p5 + ap + p18 mod l has been checked for all primes p ≤19. Of course we get a second congruence if we conjugate it. For 179 we have a splitting (179) = ll′ , 262 G. Harder with l = (179, 54+61 √ 144169), l′ = (179, 54−61 √ 144169). We ﬁnd L(f,17) Ω− ∈l, again the corresponding space S8,9 has dimension 1. If λ(p) is the sequence of eigenvalues the congruences λ(p) ≡p7 + ap + p16 mod l and of course its conjugates have been checked for the same set of primes p. (There is a slight risk that I mixed up the two primes l, l′.) Added on March 25, 2005: When lecturing on this subject, I had sometimes diﬃculties to get the numbers right. Therefore I formulate the rules: We start from an elliptic modular form f for Sl2(Z) which is of (even) weight k, it should be an eigenform for the Hecke-algebra. Then its eigenvalues generate a ﬁeld Q(f). Then we look at the values $Λ(f, k −1) Ω+ , Λ(f, k −3) Ω+ , . . . , Λ(f, k −ν) Ω+ , . . . , Λ(f, k −μ(k)) Ω+ % and $Λ(f, k −2) Ω+ , Λ(f, k −4) Ω+ , . . . , Λ(f, k −ν) Ω+ , . . . , Λ(f, k −μ′(k)) Ω+ % , where in the ﬁrst row the ν are odd and in the second row they are even. The last value is the one which is nearest to the central point k 2 from above. Then we look for large primes ℓ\|Λ(f, k −ν) Ωϵ(ν) . Now we choose the highest weight λ = mγβ + nγα. The numbers m, n must satisfy 2n + m + 3 = n + 1 + n + m + 2 = k −1 and n + m + 3 = k −ν hence we get n = ν −1 and m = k −2ν −2 Index abelian variety 202 anisotropic vector 128 automorphic Green function 164, 168 Baily–Borel compactiﬁcation 109, 205, 226 Baily–Borel topology 109 Betti number 117 Birch–Swinnerton-Dyer conjecture 80, 95, 97 Borcherds product 83, 103, 156 CM values 173 local 153 boundary component 139 rational 139 boundary point 109 Braun 194 canonical automorphism 131 canonical involution 131 canonical model 226 Cayley transform 185 Chow group 159 Chowla–Selberg formula 84, 85, 103 class ﬁeld theory 75 class numbers 106 Hurwitz 73–75, 81 of binary quadratic forms 8, 41, 74, 81, 100 of imaginary quadratic ﬁelds 8, 41 class polynomial 72, 73, 83 classical Siegel modular form 189 Cliﬀord algebra 130 center 132 even 131 Cliﬀord group 133 Cliﬀord norm 131 CM cycle 169, 173 CM extension 171 co-rank of a modular form 195 Cohen–Kuznetsov series 53–55, 102 compact dual 187 compactiﬁcations 204 complex multiplication (CM) 67–99, 102, 103, 169 CM modular forms 93–99, 103 elliptic curves with CM 67, 90–92 converse theorem strong 166 weak 165 critical value 236 cusp 107 cusp form 23, 115, 192 Dedekind zeta function 120 special value 123 desingularization 110 diagonal 122 diﬀerent 106 Dirichlet character 17, 32, 78, 84 discriminant of a quadratic ﬁeld 106 of a quadratic space 129, 132 discriminant function Δ(z) 11, 20–22, 36, 212, 249 divisor sum 121 Doi–Naganuma lift 149 264 Index dual lattice 138 Eichler element 128 Eichler–Shimura 228 Eisenstein series 13, 120, 193 for congruence subgroups 17, 96, 147 for the Hilbert modular group 119–123 for the Siegel modular group 193, 198, 199 non-holomorphic 19, 55, 58, 78, 89 of weight 2 18–20, 28, 49, 55, 86, 87 on the full modular group 12–20, 48, 49, 56 elliptic ﬁxed point 107 endoscopic contribution 232 eta function η(z) 29, 30, 43, 45, 64, 65, 94, 98, 103 Euler characteristic 117, 186 Fourier expansion 5, 114, 190 Fourier–Jacobi development 197, 207 Fourier–Jacobi series 206 Frobenius 218 fundamental domain 3, 6–9, 11, 14, 23, 72, 73, 112, 185, 186 fundamental set 111 Götzky–Koecher principle 114 Gram matrix 129 Grassmann algebra 131 Grassmannian 136 grossencharacter 86, 89, 90, 93–95 Gundlach theta function 124, 161 half-integral 189 Hamilton quaternion algebra 130 Harder’s conjecture 237 harmonic weak Maass form 162 Hauptmodul 61, 77, 81 Hecke L-function 126 Hecke algebra 213 Hecke estimate 23, 125 Hecke operators 15, 22, 37, 38, 45, 74, 213–215, 250 eigenforms for 37, 39, 40, 45, 94, 220–225, 236, 237 Heegner divisor 140 Heegner points 76, 77, 79–81, 97, 102, 103 height 80 Hilbert modular group 106, 143 Hirzebruch–Zagier divisor 146 local 150 Hirzebruch–Zagier theorem 159 Hodge bundle 203 holomorphic diﬀerential form 210 Hurwitz numbers 85, 86, 89 Hurwitz–Kronecker class number relations 74, 82 Ibukiyama lift 226 ideal 106 ideal class group 106 Ikeda lift 224 interior cohomology 229 isometry 128 isotropic subspace 139 isotropic vector 128 Jacobi forms 28, 34, 100, 197 Kac–Wakimoto conjecture 31 Klingen–Eisenstein series 193 Koecher principle 141, 191 Kohnen isomorphism 223 Kostant representatives 208 Kronecker limit formula 85 L-series (or L-function) 67, 95, 220 of a grossencharacter 86, 90, 93, 95, 96 of a modular form 39–41, 43–47, 93, 95 of an elliptic curve 44–46, 93, 95 Lagrangian Grassmann variety 187 lattices 32–36, 100, 138 even 32–36 extremal 35, 36 in C 4–6, 14, 67 unimodular 33–36 lifting 221, 222 local Hecke algebra 214 local system 230 Maass subspace 222 Manin–Vishik theorem 236 Index 265 Minkowski–Siegel mass formula 35 Miyawaki–Ikeda conjecture 225 modular form 4, 5, 113, 141 almost holomorphic 20, 58–60 Fourier expansion of 52 number of zeros of 9, 11, 12, 31 of half-integral weight 8, 26, 29, 30, 81, 82, 88, 96 on the full modular group 8–11 Taylor expansion of 52, 87–90, 96 modular function 3, 4, 11, 46, 61, 63, 68 modular group SL(2, Z) 3, 33 fundamental domain for 6–8 generators 6, 7 modular invariant j(z) 22, 36, 63, 67 motive 229 multiplier 207, 208 obstructions 158 orthogonal group 128 partial compactiﬁcation 206 period matrix 202 periods 83, 84, 86 Petersson product 118, 193 plus space 147 Poincaré series 13, 194 polarization 202 principal part 155, 163 projective model 137 properly discontionuous 112 quadratic ﬁeld 106 quadratic forms 31–33, 128 binary 7, 8, 41–43, 73 even unimodular 33 quadratic module 128 quasi-recursion 88, 89, 96 quasimodular forms 20, 48, 49, 58–60 quaternion algebra 132 rank of a modular form 195 Rankin–Cohen brackets 53, 54, 82 reﬂection 128 regular local system 230 Riemann 202 Riemann’s theta function 203 ring of classical modular forms 198 ring of integers 106 ring of modular forms 189, 232 Saito–Kurokawa Conjecture 223 Satake compactiﬁcation 205, 226 Satake isomorphism 217 Satake parameter 219 Satake’s spherical map 217 semi-abelian variety 206 series 55 Serre derivative 48, 54, 55, 62 Serre duality 158 Shimura–Kohnen lifting 226 Siegel 181, 194 Siegel domain 111 Siegel modular form 3, 28, 187 Siegel modular group 184 Siegel operator 192 Siegel theta function 142 Siegel upper half plane 184 Siegel’s Hauptsatz 196 signature 129 singular modular form 194 singular moduli 66–71, 75, 168–170 norms of 77–79, 103 traces of 81–83, 103 singularity 110 slash operator 113, 215 slope 234 small weights 195 special orthogonal group 128 Spezialschar 222 Spin group 134 spinor norm 134 spinor zeta function 221 standard zeta function 221 Sylvester’s problem 77, 97, 98, 102, 103 symplectic group 183 Tate Hodge structure 231 theta constant 195 theta lift 142, 148 regularized 156 theta series 24, 25, 31, 195, 196 for binary quadratic forms 26, 27, 41, 42, 93, 94 for unary quadratic forms 25, 30 Jacobi’s theta functions 25–29, 63, 82, 96, 97 Siegel theta function 142 266 Index with spherical coeﬃcients 37, 67, 93, 94 Torelli’s theorem 204 tube domain model 137 type of a quadratic form 129 unit group 106 vector representation 133 weak Maass form 162 weakly holomorphic modular form 154 weight 113, 187 Weyl chamber 151, 155 Weyl group 208 Weyl vector 153, 155
15923	https://www.lenovo.com/us/en/glossary/octal/?srsltid=AfmBOoqv9OkQ_P4DQ6ZcTZAgkim7_D5OjbtxTIT9dpN6lT3XLgAfu6pE	Octal Explained: The Programmer's Base-8 Code \| Lenovo US Please note: This website includes an accessibility system. Press Control-F11 to adjust the website to people with visual disabilities who are using a screen reader; Press Control-F10 to open an accessibility menu. Accessibility Popup heading Press enter for Accessibility for blind people who use screen readers Press enter for Keyboard Navigation Press enter for Accessibility menu What is an octal? 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Sign in or Create an Account to Join Rewards View Cart Remove Your cart is empty! Don’t miss out on the latest products and savings — find your next favorite laptop, PC, or accessory today. item(s) in cart Some items in your cart are no longer available. Please visit cart for more details. has been deleted Please review your cart as items have changed. of Contains Add-ons Subtotal Proceed to Checkout Yes No Popular Searches What are you looking for today ? Trending Recent Searches Items All Cancel Top Suggestions View All > Starting at Fall Tech Fest!Power up your season with these unbeatable deals on PCs and tech Buy online, pick up select products at Best Buy.Shop Pick Up > My Lenovo Rewards!Earn 3%-9% in rewards and get free expedited delivery on select products.Join for Free > Lease-to-own today with Katapult. Get started with an initial lease payment as low as $1! Learn More > Shopping for a business?New Lenovo Pro members get $100 off first order of $1,000+, exclusive savings & 1:1 tech support.Learn More > Home>Glossary> What is an octal? Learn More Annual Sale Laptop Deals Desktop Deals Workstation Deals Gaming PC Deals PC Accessories Deals Monitor Deals Tablets & Phones Deals Server & Storage Deals Clearance Sale Knowledgebase AI-Glossary SMB-Glossary ISG Glossary Monitors Glossary What is an octal? Octal is a numeral system crucial in computing, offering a base-8 representation that simplifies binary code expression. In the realm of computers understanding octal is valuable. It aids programmers in efficiently translating binary machine code and finds applications in areas like file permissions, memory addressing, and American Standard Code for Information Interchange (ASCII) encoding. Why is octal used in computing? Octal is utilized in computing as a compact numeral system, representing binary code more efficiently. In the realm of computer architecture, it serves as a concise way to express memory addresses and machine-level instructions. Octal simplifies the translation between binary and human-readable forms, aiding programmers in writing and interpreting low-level code for computers, laptops, and desktops. Its historical significance in early computing and continued relevance in specific applications make octal a valuable tool in navigating the intricacies of computing systems. How might octal be relevant in data serialization and deserialization? In data serialization, where data is converted into a format suitable for storage or transmission, octal could be used for compactly representing certain types of information. However, its use in this context is less common compared to other numeral systems like binary or hexadecimal. How is octal related to binary? Octal is closely linked to binary in computing, serving as a convenient representation. In binary, each digit is a power of 2, making it intricate for human interpretation. Octal simplifies this by grouping binary digits into sets of three, aiding programmers. For instance, the octal number 23 corresponds to 10011 in binary. This relationship facilitates a smoother transition between human-readable code and the machine-level binary language, streamlining programming and system design tasks. Does octal have any advantages over hexadecimal in computing? Octal offers a user-friendly advantage over hexadecimal in computing, notably in code readability. While both systems represent binary code compactly, octal's simplicity makes it more human-readable, aiding programmers in deciphering machine-level code. For tasks like memory addressing and bitwise operations, octal's straightforward representation can enhance comprehension. In contrast, hexadecimal, though widely used, may be less intuitive due to its larger digit set. When prioritizing ease of interpretation in computing scenarios, octal's streamlined approach can be advantageous. How is octal used in file permissions? In Unix-like operating systems, file permissions are often represented using octal numbers. Each digit corresponds to the permission settings for the owner, group, and others. For example, 755 means read, write, and execute permissions for the owner, and read and execute permissions for others. How is octal applied in the ASCII system? ASCII, the character encoding standard, assigns a unique number to each character. These numbers are often represented in octal. For instance, the octal code for the letter 'A' is 101. What role does octal play in the addressing of memory locations? Octal plays a crucial role in addressing memory locations within computers, laptops, and desktops. In computer architecture, memory addresses are often expressed in octal, simplifying communication between software and hardware. This base-8 numeral system streamlines the representation of binary-coded locations, aiding programmers in efficiently managing and accessing memory. Understanding octal is essential for optimizing memory usage and enhancing overall system performance in computing devices. How does octal help simplify subnetting in networking? Octal simplifies subnetting in networking by offering a concise representation of subnet masks. In this system, each digit corresponds to three binary bits, streamlining the expression of complex configurations. This aids network administrators in efficiently managing internet protocol (IP) address ranges and optimizing communication between devices. Octal's clear structure enhances readability, facilitating precise subnetting decisions crucial for effective network design and maintenance. Why might octal be used in error codes? Octal is employed in error codes within computer and laptop systems due to its efficiency in representing binary information. With each octal digit corresponding to three binary bits, it offers a concise and human-readable format. This makes it practical for conveying error states, aiding programmers in quickly identifying and addressing issues. The use of octal in error codes facilitates streamlined communication between hardware and software, enhancing the diagnostic process and contributing to efficient troubleshooting in computer, laptop, and desktop environments. How does octal facilitate bitwise operations in programming? In programming, bitwise operations are common for manipulating individual bits in binary numbers. Octal representation is sometimes used as a shorthand in bitwise calculations, making it easier to perform complex operations with fewer characters. In what scenarios would octal be less practical than other numeral systems? Octal is less practical when dealing with large binary numbers, as it requires more digits than hexadecimal to represent the same value. Hexadecimal is often preferred in these cases due to its more compact representation. Does octal play a role in modern encryption algorithms? Octal doesn't play a prominent role in modern encryption algorithms. Encryption relies on hexadecimal for its compatibility with binary operations, making it the preferred choice. Octal's sparse usage in cryptographic contexts is limited, as hexadecimal provides a more seamless mapping to binary, the core language of computing. In computer security, hexadecimal encoding remains the prevalent method, ensuring robust encryption practices for safeguarding sensitive data on computers, laptops, and desktops. How might octal be relevant in embedded systems programming? In embedded systems, where memory and resources are often limited, octal may be used in specific contexts to optimize space. However, the choice between octal and other numeral systems depends on the specific requirements of the embedded application. In what ways does octal contribute to code optimization? In certain scenarios, using octal representations can lead to more concise code, especially when dealing with bitwise operations or memory addresses. This can contribute to code optimization by reducing the overall size of the program. Does octal have any implications for data compression in computing? Octal's impact on data compression in computing is minimal. While it can be used for compact representation, modern compression techniques, predominantly based on binary or hexadecimal, offer more efficient algorithms. Octal lacks the widespread support and versatility needed for advanced compression methods, making it less relevant in the dynamic landscape of data optimization. When considering data compression strategies for computer, laptop, or desktop applications, focus on widely accepted numeral systems to ensure compatibility and optimal performance. Looking for a Great Deal? 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15924	https://artofproblemsolving.com/wiki/index.php/P-adic_valuation?srsltid=AfmBOoo9Rdus_2AeUBuw5h7PwLDvJZIPiCaAA4aJquwpZ0yrQxDwzmwp	Art of Problem Solving p-adic valuation - 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 Wikip-adic valuation Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search p-adic valuation The title of this article has been capitalized due to technical restrictions. The correct title should be -adic valuation. For some integer and prime, the -adic valuation of n, denoted , represents the largest power of which divides. In other words, it is the value of the exponent of in the prime factorization of . Contents 1 Basic Examples 2 Properties 3 Extension to Rational Numbers 4 See Also Basic Examples . . . Properties For positive integers and , This property follows from the fact that . Furthermore, This follows because we can factor out copies of from the sum . Note that equality holds if , because, in this case, after factoring out copies of from the sum , the remaining factor cannot be congruent to modulo, because one of the terms will be congruent to , while the other will not (because all common factors of have already been factored out). If is a positive integer, because , we deduce that because logarithms are monotone increasing for all bases greater than , which includes all primes. Lifting the Exponent: A series of identities, among which the most prominent is: for odd primes if . Legendre's Formula: . Extension to Rational Numbers is defined to be infinite. Furthermore, as seen in the properties above, From this inspiration, we can define fractional inputs as follows: Note that it does not matter if is simplified or not, because ν p(k x k y)=ν p(k x)−ν p(k y)=(ν p(k)+ν p(x))−(ν p(k)+ν p(y))=ν p(x)−ν p(y)=ν p(x y). See Also Lifting the Exponent Legendre's Formula p-adic number This article is a stub. Help us out by expanding it. Retrieved from " Categories: Titles incorrect due to Limitations Number theory Definition Stubs 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.
15925	https://www.math.utah.edu/~zwick/Classes/Fall2012_2270/Lectures/Lecture37_with_Examples.pdf	Math 2270 - Lecture 37 : Linear Transformations, Change of Bases, and Why Matrix Multiplication Is The Way It Is Dylan Zwick Fall 2012 This lecture covers section 7.2 of the textbook. Today we’re going to delve deeper into linear transformations, and their connections with matrices. We’re also ﬁnally going to see why matrix multiplication works the way it does! The assigned problems for this section are: Section 7.2 - 5, 14, 15, 17, 26 Linear Transformation and Vector Spaces We know that matrix multiplication represents a linear transformation, but can any linear transformation be represented by a matrix? The answer, it turns out, is yes. That’s where matrices originated. Thus far in our class we’ve restricted ourselves to dealing with ﬁnite-dimensional vector spaces over the real numbers. These restrictions aren’t necessary for a vector space. We’ll still maintain the assumption that our vector space is ﬁnite-dimensional, but besides that we’ll deal with things in greater generality. 1 Any ﬁnite-dimensional vector space V will have a basis. Suppose this basis is v1, . . . , vn. Then any vector v in our vector space can be written as: v = c1v1 + c2v2 + · · · + cnvn. Suppose we have another ﬁnite-dimensional vector space w with basis w1, w2, . . . , wm, and a linear transformation, T, that takes vectors in V to vectors in W. If we know what the transformation T does to the basis vectors of V, then we know what it does to every other vector in V. Read that last sentence again, because it’s that important. Suppose we know: T(v1) = a11w1 + a21w2 + · · · + am1wm T(v2) = a12w1 + a22w2 + · · · + am2wm . . . T(vn) = a1nw1 + a2nw2 + · · · + amnwm. Then if v is any vector in V, we can write it in terms of the basis v1, . . ., vn: v = c1v1 + c2v2 + · · · + cnvn. The action of T on v will then be given by T(v) = c1T(v1) + c2T(v2) + · · · + cnT(vn). Note how important the idea of linearity is here! In fact, we can repre-sent v compactly as v =      c1 c2 . . . cn      2 And the action of T on v will just be: a 11 a 12 ai a 21 a 22 a 2 c 2 T(v)= ami am2 amn That’s right, the linear transformation has an associated matrix! Any linear transformation from a finite dimension vector space V with dimension n to another finite dimensional vector space W with dimension m can be represented by a matrix. This is why we study matrices. Example - Suppose we have a linear transformation T taking V to W, where both V and W are 2-dimensional vector spaces. If the action of this transformation on the basis vectors of V is: T(v 1 ) 2w 1 + 3w 2 , T(v 2 ) = 3w 1 + 1w 2 , what is the matrix representing this transformation? Also, what is T(v), written in terms of the basis vector of W, if v = 1v 1 + 4v 2 ? A= (3) (Ls)(I) ( I ( ± I 3 + lIJ 3 Matrix Multiplication Suppose we have a linear transformation S from a 2-dimensional vector space U, to another 2-dimension vector space V, and then another linear transformation T from V to another 2-dimensional vector space W. Sup-pose we have a vector u ∈U: u = c1u1 + c2u2. Suppose S maps the basis vectors of U as follows: S(u1) = a11v1 + a21v2, S(u2) = a12v1 + a22v2. Then the action of S on u will be: S(u) = c1(a11v1 + a21v2) + c2(a12v1 + a22v2) = (a11c1 + a12c2)v1 + (a21c1 + a22c2)v2. Now, suppose T maps the basis vectors of V as follows: T(v1) = b11w1 + b21w2, T(v2) = b12w1 + b22w2. Then the action of T on S(u) will be: T(S(u)) = (a11c1 + a12c2)(b11w1 + b21w2) + (a21c1 + a22c2)(b12w1 + b22w2) = ((a11b11 + a21b12)c1 + (a12b11 + a22b12)c2)w1 + ((a11b21 + a21b22)c1 + (a12b21 + a22b22)c2)w2. Or, in matrix form: (a11b11 + a21b12)c1 + (a12b11 + a22b12)c2 (a11b21 + a21b22)c1 + (a12b21 + a22b22)c2 4 Clear as mud? Well, we can rewrite this as: b11a11 + b12a21 b11a12 + b12a22 b21a11 + b22a21 b21a12 + b22a22 c1 c2 . Looking more familiar? How about now: b11 b12 b21 b22 a11 a12 a21 a22 c1 c2 . That’s right, it’s just matrix multiplication! We ﬁnally see the reason we multiply matrices in that weird way. Matrix multiplication corresponds to composition of linear transformations. At last it makes some sense!1 Linear Transformations in Different Bases Suppose we have two bases for R2: the standard basis, and another basis given by 2 1 , 1 1 . To transform a vector written in terms of the second basis into a vector written in terms of the standard basis, we multiply it by the basis change matrix: M = 2 1 1 1 To go the other way, taking a vector written in terms of the standard basis and writing it in terms of the second basis, we would multiply by the inverse of the basis change matrix: 1Note this is true for composition of linear transformations for any size vector spaces, I just chose 2 × 2 because it’s relatively easy. 5 M1=(1 -1) Suppose we have a linear transformation T. A is the matrix that rep resents this transformation in the standard basis, while B is the matrix representing this transformation in the second basis. How are these ma trices related? Well, Suppose we’re given A, and a vector v represented in terms of the second basis. To find the action of T on v we’d first multiply v by M to get it in terms of the standard basis, and then multiply Mv by A to get the transformation. This would give us the transformed vector, but in terms of the standard basis. To get this transformed vector in terms of the second basis we’d multiply AMy by M’. So, the matrix for our transformation is B = M’AM. Look familiar? It should. If two matrices represent the same transfor mation, just in different bases, then the matrices are similar. That’s what’s so important about similar matrices, and why mathematicians are inter ested in studying them. If you want to understand linear transformations and don’t want to have to tie them to certain bases, you want to under stand the equivalence class of similar matrices. Example - If the linear transformation T is represented in the second basis above as: (2 3 i’\1 4 what is its representation in the standard basis? / / I -I L) H [L (-5 /5 I R-3 71 L/ .j 6
15926	https://brainly.com/question/28412711	[FREE] Is the length of a horizontal segment the absolute difference of the x-coordinates of the endpoints? Yes - brainly.com 2 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 +23,7k Smart guidance, rooted in what you’re studying Get Guidance Test Prep +32,7k Ace exams faster, with practice that adapts to you Practice Worksheets +8,6k Guided help for every grade, topic or textbook Complete See more / Mathematics Textbook & Expert-Verified Textbook & Expert-Verified Is the length of a horizontal segment the absolute difference of the x-coordinates of the endpoints? Yes or no? 1 See answer Explain with Learning Companion NEW Asked by beiles0501a • 08/31/2022 0:00 / -- Read More Community by Students Brainly by Experts ChatGPT by OpenAI Gemini Google AI Community Answer This answer helped 17397844 people 17M 0.0 1 Upload your school material for a more relevant answer Yes, the length of a horizontal segment is the absolute difference of the x-coordinates of the endpoints. Yes, the length of a horizontal segment is the absolute difference of the x-coordinates of the endpoints. For example, if we have two points A(x1, y1) and B(x2, y2) on a horizontal line, then the length of the segment AB is given by \|x2 - x1\|. So, the answer to your question is yes. Learn more about length of a horizontal segment here: brainly.com/question/32035080 SPJ11 Answered by CarrLola •32.7K answers•17.4M people helped Thanks 1 0.0 (0 votes) Textbook &Expert-Verified⬈(opens in a new tab) This answer helped 17397844 people 17M 0.0 1 Simple nature - Benjamin Crowell Fields and Circuits - Benjamin Crowell Big Ideas in Cosmology - Kim Coble, Kevin McLin, Lynn Cominsky Upload your school material for a more relevant answer Yes, the length of a horizontal segment is the absolute difference of the x-coordinates of its endpoints. This is because the distance between two points on a horizontal line is determined only by their x-coordinates. The formula used is \|x₂ - x₁\|. Explanation Yes, the length of a horizontal segment is indeed the absolute difference of the x-coordinates of its endpoints. To understand this, let’s consider two points A and B that are defined as follows: Point A has coordinates (x₁, y₁) Point B has coordinates (x₂, y₂) When the segment connecting these points is horizontal, the y-coordinates remain the same, meaning y₁ = y₂. Therefore, the length of the horizontal segment, AB, can be calculated by looking only at the x-coordinates. The formula to find the length of the segment is: L e n g t h=∣x 2−x 1∣ Here, the absolute value is used because distance cannot be negative. For example, if point A is at (2, 3) and point B is at (5, 3), the calculation would be: L e n g t h=∣5−2∣=∣3∣=3 Thus, the answer to the question is yes, as the length of a horizontal segment is indeed the absolute difference of the x-coordinates of its endpoints. Examples & Evidence For example, if point A is at (1, 4) and point B at (4, 4), the length of the segment AB is calculated as \|4 - 1\| = 3. If point A is at (-5, 2) and point B at (2, 2), the length is \|2 - (-5)\| = \|2 + 5\| = 7. This concept is based on the definition of distance in a coordinate plane, where the horizontal distance is simply calculated using the x-coordinates. The absolute difference ensures that we always have a non-negative measure of length. Thanks 1 0.0 (0 votes) Advertisement beiles0501a has a question! Can you help? Add your answer See Expert-Verified Answer ### Free Mathematics solutions and answers Community Answer 4.6 12 Jonathan and his sister Jennifer have a combined age of 48. If Jonathan is twice as old as his sister, how old is Jennifer Community Answer 11 What is the present value of a cash inflow of 1250 four years from now if the required rate of return is 8% (Rounded to 2 decimal places)? Community Answer 13 Where can you find your state-specific Lottery information to sell Lottery tickets and redeem winning Lottery tickets? (Select all that apply.) 1. Barcode and Quick Reference Guide 2. Lottery Terminal Handbook 3. Lottery vending machine 4. OneWalmart using Handheld/BYOD Community Answer 4.1 17 How many positive integers between 100 and 999 inclusive are divisible by three or four? Community Answer 4.0 9 N a bike race: julie came in ahead of roger. julie finished after james. david beat james but finished after sarah. in what place did david finish? Community Answer 4.1 8 Carly, sandi, cyrus and pedro have multiple pets. carly and sandi have dogs, while the other two have cats. sandi and pedro have chickens. everyone except carly has a rabbit. who only has a cat and a rabbit? Community Answer 4.1 14 richard bought 3 slices of cheese pizza and 2 sodas for $8.75. Jordan bought 2 slices of cheese pizza and 4 sodas for $8.50. How much would an order of 1 slice of cheese pizza and 3 sodas cost? A. $3.25 B. $5.25 C. $7.75 D. $7.25 Community Answer 4.3 192 Which statements are true regarding undefinable terms in geometry? Select two options. A point's location on the coordinate plane is indicated by an ordered pair, (x, y). A point has one dimension, length. A line has length and width. A distance along a line must have no beginning or end. A plane consists of an infinite set of points. Community Answer 4 Click an Item in the list or group of pictures at the bottom of the problem and, holding the button down, drag it into the correct position in the answer box. Release your mouse button when the item is place. If you change your mind, drag the item to the trashcan. Click the trashcan to clear all your answers. Express In simplified exponential notation. 18a^3b^2/ 2ab New questions in Mathematics Find the area bounded by the lines y=2 3x−4 and y=−2 x+7 and the x axis. Find the vertex for the parabola. Then determine a reasonable viewing rectangle on your graphing utility and use it to graph the quadratic function: y=0.05 x 2+0.5 x−130. Consider the function f(x)=−2 x 2+20 x−1. a. Determine, without graphing, whether the function has a minimum value or a maximum value. b. Find the minimum or maximum value and determine where it occurs. c. Identify the function's domain and its range. Consider the function f(x)=−2 x 2+20 x−1. a. Determine, without graphing, whether the function has a minimum value or a maximum value. b. Find the minimum or maximum value and determine where it occurs. c. Identify the function's domain and its range. Consider the function f(x)=−2 x 2+20 x−1. a. Determine, without graphing, whether the function has a minimum value or a maximum value. b. Find the minimum or maximum value and determine where it occurs. c. Identify the function's domain and its range. 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15927	https://byjus.com/greater-than-calculator/	Check which is Greater Than Enter two numbers to find whether Number 1 is greater than 2 or not? Number 1 Number 2 Greater Than Calculator is a free online tool that displays the greatest of two numbers. BYJU’S online greater than calculator tool makes the calculation faster, and it displays a greater number in a fraction of seconds. How to Use the Greater Than Calculator? The procedure to use the greater than calculator is as follows: Step 1: Enter two numbers (Integer/Decimal Number) in the respective input field Step 2: Now click the button “Solve” to get the result Step 3: Finally, the result “Yes / No” will be displayed in the output field Greater Than Symbol in Maths In Mathematics, to compare two numbers the equality and inequality symbols are used. Equality symbol is used when two numbers are equal. Inequalities are used when the first number is greater than or less than the second number. If the first number is greater than the second number, greater than symbol “>” is used. If the first number is less than the second number, less than the symbol “<” is used. Example: 7 > 5 (Greater than) 3 < 5 (Less than) Disclaimer: This calculator development is in progress some of the inputs might not work, Sorry for the inconvenience. Comments notanumber November 17, 2020 at 12:08 am is 21 greater than 10 Madhukrishna kabade December 2, 2020 at 2:26 pm Yes 21 is greater than 10 julissa brown November 24, 2020 at 5:58 am thank you for the answer Elizabeth Holland January 26, 2021 at 3:26 am is 1 . 1 x 15 greater than 15 or less than 15? chandraprabha February 24, 2021 at 2:59 pm 1 . 1 x 15 = 16.5, which is greater than 15. (16.5 > 15) brooklynn tanner February 2, 2021 at 2:56 am This was helpful thankyou! Chloejohnson February 11, 2021 at 10:57 pm Is 3/12 greater than 8/9 Register with BYJU'S & Download Free PDFs
15928	https://math.stackexchange.com/questions/3527915/prove-union-of-equivalence-classes-is-the-whole-set	Prove union of equivalence classes is the whole set - 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 Prove union of equivalence classes is the whole set Ask Question Asked 5 years, 8 months ago Modified5 years, 8 months ago Viewed 2k times This question shows research effort; it is useful and clear 1 Save this question. Show activity on this post. Prove union of equivalence classes is the whole set: Given a set X X and let ∀x∈X∀x∈X , [x][x] be the equivalence class of x x , then we want to show that ⋃x∈X[x]=X⋃x∈X[x]=X or equivalently ⋃[x]∈X/∼[x]=X⋃[x]∈X/∼[x]=X proofwiki proves this theorem but it says ∃x∈X:x∉[x]∃x∈X:x∉[x] is equivalent to ∃x∈X:x∉⋃[x]∃x∈X:x∉⋃[x] which is not right because it is not what union of sets states. I've tried myself like this: From the definition of equivalence relation and using the symmetric property of ∼∼ we know ∀x∈X:x∈[x]∀x∈X:x∈[x] if and only if ¬(∃x∈X:x∉[x])¬(∃x∈X:x∉[x]) holds, then from the definition of intersection it follows : ¬(x∉⋂x∈X[x])¬(x∉⋂x∈X[x]) This is true if and only if: x∈⋂x∈X[x]x∈⋂x∈X[x] But this is not what I wanted, so how can I prove that? equivalence-relations Share Share a link to this question Copy linkCC BY-SA 4.0 Cite Follow Follow this question to receive notifications edited Jun 12, 2020 at 10:38 CommunityBot 1 asked Jan 30, 2020 at 8:12 user715522 user715522 Add a comment\| 2 Answers 2 Sorted by: Reset to default This answer is useful 1 Save this answer. Show activity on this post. The proof that ⋃x∈X[x]=X⋃x∈X[x]=X is simpler than what proofwiki does and simpler than what you're trying to do. By definition, [x]⊆X[x]⊆X for every x∈X x∈X, and therefore ⋃x∈X[x]⊆X⋃x∈X[x]⊆X. Let y∈X y∈X be arbitrary. It is a theorem (and easy to show) that y∈[y]y∈[y], and therefore y∈⋃x∈X[x]y∈⋃x∈X[x] (since y∈X y∈X is one of the indices in the union). Since y∈X y∈X was arbitrary, this proves that X⊆⋃x∈X[x]X⊆⋃x∈X[x]. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications answered Jan 30, 2020 at 8:18 Greg MartinGreg Martin 93.6k 6 6 gold badges 107 107 silver badges 157 157 bronze badges 1 so in the first one you use the definition of equivalence relation and for the second one you use the definition of union, I got it, but can you tell me proofwiki is right or not, in my opinion it s not user715522 –user715522 2020-01-30 08:31:30 +00:00 Commented Jan 30, 2020 at 8:31 Add a comment\| This answer is useful 0 Save this answer. Show activity on this post. You are correct that there is an error in proofwiki. The def'n of "big union" ⋃⋃ is ∀w∀y(y∈⋃w⟺∃z(y∈z∈w)).∀w∀y(y∈⋃w⟺∃z(y∈z∈w)). In particular, for any set x x we have ⋃{x}=x.⋃{x}=x. For example if x∉x x∉x and [x]={x}[x]={x} then x∉x=⋃{x}=⋃[x].x∉x=⋃{x}=⋃[x]. An easier way than proofwiki's (attempted) approach is that X=∪x∈X{x}⊂∪x∈X[x]⊂∪x∈X X=X X=∪x∈X{x}⊂∪x∈X[x]⊂∪x∈X X=X because for all x∈X x∈X we have x∈{y:x∼y∈X}⊂X x∈{y:x∼y∈X}⊂X and [x]={y:x∼y}={y:x∼y∈X}⊂X.[x]={y:x∼y}={y:x∼y∈X}⊂X. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications edited Jan 30, 2020 at 9:53 answered Jan 30, 2020 at 9:47 DanielWainfleetDanielWainfleet 59.7k 4 4 gold badges 39 39 silver badges 76 76 bronze badges Add a comment\| You must log in to answer this question. 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 Related 5Working with Equivalence Classes and Quotient Sets 6Uncountability of the equivalence classes of R/Q R/Q 0suppose R R is an equivalence relation on A A such that there are only finitely many distinct equivalence classes A 1,A 2,…,A k A 1,A 2,…,A k w.r.t R R 1Find equivalence classes (Solution with questions) 3Show that X X can be represented as a union of disjoint equivalence classes 0Equivalence relation on R R: x∼y⟺x−y∈Q x∼y⟺x−y∈Q - disjoint equivalence classes 1Set of equivalence classes as the image of a map 0Theorem about equivalence classes, how to prove it Hot Network Questions Riffle a list of binary functions into list of arguments to produce a result An odd question What is the meaning and import of this highlighted phrase in Selichos? 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15929	https://neuropathology-web.org/chapter6/chapter6aMs.html	Multiple sclerosis NEUROPATHOLOGY An illustrated interactive course for medical students and residents By Dimitri P. Agamanolis, M.D. NEUROPATHOLOGY Home Neuropathology CourseNeurocytologyHypoxia and StrokePerinatal DisordersTraumaCNS InfectionsDemyelinating DiseasesBrain TumorsNutritional DisordersDegenerative DiseasesMetabolic DisordersCongenital MalformationsPeripheral NeuropathyMyopathologyCerebrospinal Fluid Neuropathology Tests~~Neurocytology~~ (No Test)Hypoxia and StrokePerinatal DisordersTraumaCNS InfectionsDemyelinating DiseasesBrain TumorsNutritional DisordersDegenerative DiseasesMetabolic DisordersCongenital MalformationsPeripheral NeuropathyMyopathology~~Cerebrospinal Fluid~~ (No Test) M2 Course & LabsIntroductionLearning ObjectivesNEUROPATH-LABLecturesQuiz 1Quiz 2Quiz 3Quiz 4Quiz 5Quiz 6Quiz 7Quiz 8Quiz 9Quiz 10Quiz 11Quiz 12Quiz 13Quiz 14Quiz 15Test Answers About Donate × Custom Search Sort by: Relevance Relevance Date Chapter 6 DEMYELINATIVE DISEASES Go to page... Page 1 Page 2 Page 3 Page 4 Page 5 Test Learning Objectives Go to topic... MS Pathology CSF Findings in MS Pathogenesis of MS Pathophysiology of MS Experimental Allergic Encephalomyelitis Acute Disseminated Encephalomyelitis Neuromyelitis Optica Spectrun Disorders Progressive Multifocal Leukoencephalopathy Central Pontine Myelinolysis DEMYELINATIVE DISEASES Demyelinative diseases of the central nervous system are characterized by loss of myelin with variable loss of axons. In contrast, infarcts, contusions, encephalitis, and other conditions destroy myelin and axons equally. The main demyelinative disease of the CNS is multiple sclerosis (MS) and its variants. Its counterpart in the peripheral nervous system is inflammatory demyelinative polyradiculoneuropathy (Guillain-Barré syndrome-GBS) and its chronic variants. MS and GBS are autoimmune inflammatory diseases. There are also virus-induced demyelinative diseases, such as progressive multifocal leukoencephalopathy. Demyelinative diseases should be distinguished from leukodystrophies, which are inherited metabolic disorders of myelin lipids and proteins. MS affects one in every 500 persons, women three times as frequently as men. It is more common in young adults and causes a variety of neurological deficits (visual loss, paralysis, sensory loss, ataxia, brainstem signs, psychiatric disorders, dementia). Many MS cases evolve over a long period (20-30 years) with remissions and exacerbations. Some cases have an acute, fulminant, even fatal course, and others go into a relentlessly progressive phase after a period of remissions and exacerbations. PATHOLOGY OF MS The pathology is characterized by multifocal lesions, the MSplaques. The usual evolution of the MS plaqueis as follows: in the acute phase (active plaque), activated mononuclear cells, including lymphocytes, microglia, and macrophages destroy myelin and, to a variable degree, oligodendrocytes. Myelin debris are picked up by macrophages and degraded. At an early stage, macrophages contain myelin fragments; later, they contain proteins and lipids from chemical degradation of myelin. This process takes a few weeks. With time, gliosis develops, and plaques reach a burned-out stage consisting of demyelinated axons traversing glial scar tissue (inactive plaque). Remaining oligodendrocytes attempt to make new myelin. If the inflammatory process is arrested at an early phase, plaques are partially remyelinated (shadow plaque). In more advanced lesions, remyelination is ineffective because gliosis creates a barrier between the myelin producing cells and their axonal targets. The pathological process may be arrested at any time, sometimes after partial demyelination. The pattern described above is variable. In most cases, the inflammatory reaction subsides only to appear at another location or at another time. Some lesions expand at their periphery while activity in their center dies down (smoldering plaque). In fulminant MS cases, large lesions with diffuse activity develop and expand inexorably. Although myelin is preferentially affected, axonal loss is significant, and necrosis and cavitation may develop, especially in severe, acute lesions. MS-perivascular lymphocytes MS plaque-myelin stain In H&E stains, plaques appear pale compared to normal white matter. Active lesions are cellular because they contain inflammatory cells and reactive astrocytes. Diagnosis of acute MS, especially with stereotactic needle biopsies, may be tricky because cellularity and reactive astrocytes in the lesions may be misinterpreted as a neoplasm. Activity is often confined to the borders of plaques. Myelin stains show complete loss of myelin or pallor of myelin staining. The "normal appearing white matter" around MS plaques is not entirely normal but shows milder pathology. Immunopathology. The inflamatory cells in MS include primarily CD8 T-lymphocytes, microglia, and macrohages. In addition, components of humoral immunity, including B-lymphocytes, plasma cells, immunoglobulins, and complement have been identified in plaques. Grossly, MS plaques appear as irregular, sharply demarcated, gray areas in the white matter. MS-periventricular plaques Spinal cord plaques Spinal cord plaques. They are usually multiple. Long-standing plaques are firm (sclerosis) because of gliosis. Plaques are randomly distributed. They have a predilection for the periventricular white matter,optic nerves, and spinal cord but spare no part of the CNS. They may involve gray matter such as cerebral cortex, deep nuclei, and brainstem. In these locations, they involve myelinated axons while sparing the neuronal bodies. Because of the predilection of plaques for the optic nerves, most MS patients present with visual loss (optic neuritis). This may lead to loss of retinal ganglion cells. Spinal lesions cause paralysis and sensory loss (transverse myelitis). Usually, these patients have plaques elsewhere in the brain or develop them later. These other plaques may be clinically silent, whereas the optic and spinal lesions always cause symptoms. Concentric sclerosis of Balo Clinical and radiological variants of MS include tumefactive MS (characterized by a large, acute, tumor like lesion with cerebral edema and mass effect), Marburg type MS ( a fulminant form of MS similar to tumefactive MS), and concentric sclerosis of Balo (another severe form of MS, with an unusual pattern of concentric rings of demyelination and partial preservation of myelin, which can be detected by MRI). The pathogenesis of this pattern is unclear. Schilder's disease is an acute relentlessly progressive form of MS seen most commonly in children and young adults. It causes extensive confluent demyelination instead of multiple focal lesions. This MS variant has been confused in the past with X-linked adrenoleukodystrophy. IMAGING STUDIES The best test for diagnosis of MS is MRI. Old plaques are hyperintense on T2-weighted and FLAIR studies. Active plaques show gadolinium enhancement. The latter correlates with inflammation and increased vascular permeability, and disappears after treatment (or with time), when the integrity of the blood brain barrier is restored. Mass effect, mimicking a malignant brain tumor, may also be present in acute MS (tumefactive MS). Schilder's disease tends to cause bilateral lesions that join across the corpus callosum, a feature seen also in some glioblastomas. MRI reveals inactive plaques, usually around the lateral ventricles, n most cases. Advanced MS causes brain atrophy. Extensive periventricular plaques cause dilatation of the lateral ventricles. CSF FINDINGS IN MS CSF protein is moderately elevated, and there is mild mononuclear pleocytosis. The latter is a measure of the activity of the disease. Total protein exceeding 110 mg/dl and cell counts higher than 50/cubic mm make the diagnosis of MS unlikely. The IgG fraction is elevated above 11% of total CSF protein, especially in chronic MS. The IgG/albumin index in CSF is elevated in 90 percent of MS patients, including some who have normal total protein. Elevation of IgG/albumin index in CSF but not in serum means that IgG is produced inside the blood-brain barrier. Oligoclonal IgG bands (OCBs) are detected on agarose electrophoresis in 90% of patients. This pattern may be present even when the total amount of IgG is normal. OCBs indicate that IgG represents antibodies to specific antigens. About 70% of MS patients and only 5% of controls have antibodies to measles. A smaller number have antibodies to rubella, mumps, and herpes simplex. Similar CSF changes are seen in some chronic CNS infections such as chronic measles encephalitis and syphilis. Myelin proteins such as myelin basic protein leak from plaques into the CSF and can be detected by radioimmunoassay. ETIOLOGY-PATHOGENESIS OF MS MS is thought to be an autoimmune disorder which is probably triggered by a viral infection. Genetic susceptibility and environmental factors play important roles in its pathogenesis. Genetic factors:First-degree relatives of MS patients have a 2 to 4% risk of MS. The risk in the population at large is approximately 0.1%. Monozygotic twins are are 30 to 50% concordant for MS; dizygotic twins are only 2.3% concordant. Genetic susceptibility is probably conferred by MHC molecules that modulate the immune response (particularly autoimmunity) and cell-cell interactions. MS patients express with high frequency certain class I and II HLA antigens, particularly DW2 and DR. Environmental factors:The incidence of MS is higher in high latitude zones. Prevalence in the northern US is 4-6 times higher than in the South. Individuals who grow up in high prevalence areas retain the high risk even if they subsequently migrate to low-risk regions. These findings suggest that an unknown predisposing factor is acquired by prolonged exposure to some environments. Viruses, particularly measles and HTLV-1, have been suspected (but not proven) to be involved in the pathogenesis of MS. Mononucleosis increases the risk of MS. There are several immunological abnormalities apparent in MS, including perivascular T- and B-lymphocytes, activation of T-lymphocytes, intrathecal immunoglobulin production, and the presence of immunoglobulins, complement, and cytokines in the plaques. Pregnancy, which causes a diffuse immunosuppression, suppresses MS activity. The disease flares up postpartum. Interferon (INF) gamma, which enhances the immune response, provokes MS attacks. Infections such as URIs stimulate secretion of INF gamma by immune cells and exacerbate MS. On the other hand, INF beta, which suppresses the immune response, decreases the frequency of attacks. The first event in the pathogenesis of MS may be a viral infection in childhood. Activated T- lymphocytes generated during such an infection cross the blood- brain barrier and become sensitized to myelin antigens. Alternatively, lymphocytes are sensitized to viral proteins that may have some similarity to myelin proteins. How, after remaining latent for years, these lymphocytes re-enter the CNS and initiate an immune reaction against myelin is a subject of speculation. B-lymphocytes entering acute plaques become sensitized and produce antibodies to myelin antigens. Myelin, oligodendroglial cells, and axons are damaged by inflammatory cytokines, glutamate, NO, and other toxic substances produced by microglia/macrophages. While the agents of CNS damage are the same, the immune reactions that initiate and propagate the process may vary among different forms of MS. PATHOPHYSIOLOGY OF MS The neurologic deficits, in MS, are due to loss of myelin and axons. Demyelination causes loss ofsaltatory conduction. Linear conduction along demyelinated axons is slow because the internodal axon membrane has few ion channels. In addition, lack of insulation of axons allows impulses to disperse laterally to adjacent demyelinated axons. The abnormal physiology of demyelinated axons results in inefficient conduction or conduction block. This is reflected by abnormal evoked response potentials, an electrodiagnostic test that measures conduction velocity in the CNS. Loss of axons, which occurs during the acute inflammatory phase of the disease, explains the permanent disability. While loss of function is easy to explain, clinical recovery is not. Remyelination is limited and does not fully explain the remissions. The neurological deficit from an acute MS plaque is caused not only by the loss of myelin and axons, but also by inflammation and edema that involve a wide area around the lesion. Even without remyelination, neurological function returns to some extent when the inflammatory reaction subsides and homeostasis is restored. In tracts that are partially involved by MS lesions, remaining axons may carry out the function. New ion channels may develop on axonal membrane, helping demyelinated axons conduct more efficiently. Conductivity in demyelinated areas is also influenced by electrolyte concentration and other changes in the extracellular fluid and by physical factors such as body temperature. These factors explain why the severity of neurological deficits fluctuates. Recovery probably depends on structural and functional reserves and on a potential for CNS repair and regeneration that we do not fully appreciate. Anatomical observations alone do not explain adequately the recovery seen in some MS patients. The autopsy is like a snapshot and is not ideally suited to follow the changes of an evolving disease. MRI imaging is better for this purpose. Further Reading Lucchinetti C. Pathological heterogeneity of idiopathic central nervous system demyelinating disorders. Curr Top Microbiol Immunol 2008;318:19-43. PubMed Ratelade J, Verkman AS. Neuromyelitis optica: Aquaporin-4 based pathogenesis mechanisms and new therapies. Int J Biochem Cell Biol. 2012;44(9):1519-30. PubMed Frischer JM, Weigand SD, Guo Y et al. Clinical and Pathological Insights into the Dynamic Nature of the White Matter Multiple Sclerosis Plaque. Ann Neurol 2015;Aug 3. doi: 10.1002/ana.24497. [Epub ahead of print]. PubMed Wingerchuk D, Banwell B, Bennett JL, et al. International consensus diagnostic criteria for neuromyelitis optica spectrum disorders. Neurology 2015;85:177-89. PubMed Reich DS, Lucchinetti CF, Calabresi PA. Multiple Sclerosis. N Engl J Med 2018;378:169-80. PubMed Hardy TA, Reddel SW, Barnett MH et al. Atypical inflammatory demyelinating syndromes of the CNS. Lancet Neurol 2016; 15: 967-81. PubMed Weber MS, Derfuss T, Metz I and Br�ck W. Defining distinct features of anti-MOG antibody associated central nervous system demyelination. Ther Adv Neurol Disord 2018; 11: 1-15.PubMed Updated: Aug, 2022 Back to top of page Neuropathology Copyright © 2023 Dimitri Agamanolis. All rights reserved. Unauthorized duplication is prohibited. Home \| About \| Donate \| Feedback \| Privacy 177 Shares Share Tweet Share Share Email Print
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15931	https://planbee.com/pages/prime-numbers?srsltid=AfmBOoqIodJbZYV3LYZ7LrbfWvtzaYLKGlI7av11Gy2pAGRoBGFDSZiM	Prime Numbers What is a prime number? A prime number is a whole number greater than 1 with only two factors - itself and 1. This means that it can only be divided by itself and 1, without remainders. For example, 11 is a prime number because it can only be divided by 1 and 11. However, 12 is not a prime number because it can be divided by 1, 2, 3, 4, 6 and 12. Why is 1 not a prime number? 1 can only be divided by one number, itself. This means it has only one factor. By definition, prime numbers have exactly two factors. Are there any even prime numbers? There is only one even prime number - 2. This is the only even number that has only two factors. Prime numbers to 100: There are 25 prime numbers below 100. They are: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89 and 97. Download our FREE Prime Numbers to 100 Grid to use as a teaching aid or part of a display. How to find prime numbers from 1 - 100: This is a fun, visual activity which allows children to find all of the prime numbers up to 100 easily. They will need a blank hundred square grid. Give the following instructions one at a time: Cross off number 1. 1 is not a prime number as it only has one factor - itself. 2 is a prime number, because it has exactly two factors - 1 and itself. However, you can cross off all of the multiples of 2 (other than 2, no even number is a prime number). 3 is a prime number, because it has exactly two factors - 1 and itself. However, you can cross off all of the multiples of 3. 5 is a prime number, because it has exactly two factors - 1 and itself. However, you can cross off all of the multiples of 5. 7 is a prime number, because it has exactly two factors - 1 and itself. However, you can cross off all of the multiples of 7. The numbers that are left on the grid are prime numbers. Prime numbers Year 5: The concept of prime numbers is introduced to children in Year 5. In line with the national curriculum, they are expected to: Prime numbers Year 6: Children develop their knowledge and understanding of prime numbers in Year 6. In line with the national curriculum, they are expected to: LESSON PACK Primes, Squares and Cubes \| Prime Numbers LESSON PACK Factors, Multiples and Primes \| Prime Factor Trees FREE Prime Numbers to 100 Grid PlanBee Resources Ltd Unit B Bayhorne Lane, Horley, Surrey RH6 9ES, United Kingdom info@planbee.com About PlanBee Helpful Links Support Follow us Copyright © 2025 PlanBee. VAT Number: 127 1439 27 Company Number: 08012907 Added to your cart: What's Your Email? \| Order # \| Date \| Resources \| --- Liquid error (snippets/flits_custom_snippet line 28): Array 'customer.orders' is not paginateable. Let customers speak for us Very detailed. Colourful and explanatory. Great for display purposes. These were so helpful when creating our schedule. My students are able to see the time shown on the clock and match it to the time on our schedule. Files wouldn’t download This is a well structured unit. Starting with reading and vocabulary leading into a structured piece of work. Thank you for taking the time to leave us a review - we're pleased to hear that you found this scheme useful. We are on the 3rd lesson and I love it! Easy to follow fantastic slides and motivating lessons for the kids! We're so pleased to hear that! Thank you for taking the time to leave us a review :-)
15932	https://www.dpbh.nv.gov/globalassets/dpbh/programs/std/dta/publications/HPV.pdf	CS280191E Genital HPV Infection - CDC Fact Sheet Human papillomavirus (HPV) is the most common sexually transmitted infection in the United States. Some health effects caused by HPV can be prevented with vaccines. What is HPV? HPV is the most common sexually transmitted infection (STI). HPV is a different virus than HIV and HSV (herpes). HPV is so common that nearly all sexually active people get it at some point in their lives. There are many different types of HPV. Some types can cause health problems including genital warts and cancers. But there are vaccines that can stop these health problems from happening. How is HPV spread? You can get HPV by having vaginal, anal, or oral sex with someone who has the virus. It is most commonly spread during vaginal or anal sex. HPV can be passed even when an infected person has no signs or symptoms. Anyone who is sexually active can get HPV, even if you have had sex with only one person. You also can develop symptoms years after you have sex with someone who is infected. This makes it hard to know when you first became infected. Does HPV cause health problems? In most cases, HPV goes away on its own and does not cause any health problems. But when HPV does not go away, it can cause health problems like genital warts and cancer. Genital warts usually appear as a small bump or group of bumps in the genital area. They can be small or large, raised or flat, or shaped like a cauliflower. A healthcare provider can usually diagnose warts by looking at the genital area. Does HPV cause cancer? HPV can cause cervical and other cancers including cancer of the vulva, vagina, penis, or anus. It can also cause cancer in the back of the throat, including the base of the tongue and tonsils (called oropharyngeal cancer). Cancer often takes years, even decades, to develop after a person gets HPV. The types of HPV that can cause genital warts are not the same as the types of HPV that can cause cancers. There is no way to know which people who have HPV will develop cancer or other health problems. People with weak immune systems (including those with HIV/AIDS) may be less able to fight off HPV. They may also be more likely to develop health problems from HPV. National Center for HIV/AIDS, Viral Hepatitis, STD, and TB Prevention Division of STD Prevention How can I avoid HPV and the health problems it can cause? You can do several things to lower your chances of getting HPV. Get vaccinated. The HPV vaccine is safe and effective. It can protect against diseases (including cancers) caused by HPV when given in the recommended age groups. (See “Who should get vaccinated?” below) CDC recommends 11 to 12 year olds get two doses of HPV vaccine to protect against cancers caused by HPV. For more information on the recommendations, please see: Get screened for cervical cancer. Routine screening for women aged 21 to 65 years old can prevent cervical cancer. If you are sexually active • • Use latex condoms the right way every time you have sex. This can lower your chances of getting HPV. But HPV can infect areas not covered by a condom - so condoms may not fully protect against getting HPV; • • Be in a mutually monogamous relationship – or have sex only with someone who only has sex with you. Who should get vaccinated? All boys and girls ages 11 or 12 years should get vaccinated. Catch-up vaccines are recommended for males through age 21 and for females through age 26, if they did not get vaccinated when they were younger. The vaccine is also recommended for gay and bisexual men (or any man who has sex with a man) through age 26. It is also recommended for men and women with compromised immune systems (including those living with HIV/AIDS) through age 26, if they did not get fully vaccinated when they were younger. How do I know if I have HPV? There is no test to find out a person’s “HPV status.” Also, there is no approved HPV test to find HPV in the mouth or throat. There are HPV tests that can be used to screen for cervical cancer. These tests are only recommended for screening in women aged 30 years and older. HPV tests are not recommended to screen men, adolescents, or women under the age of 30 years. Most people with HPV do not know they are infected and never develop symptoms or health problems from it. Some people find out they have HPV when they get genital warts. Women may find out they have HPV when they get an abnormal Pap test result (during cervical cancer screening). Others may only find out once they’ve developed more serious problems from HPV, such as cancers. How common is HPV and the health problems caused by HPV? HPV (the virus): About 79 million Americans are currently infected with HPV. About 14 million people become newly infected each year. HPV is so common that most sexually-active men and women will get at least one type of HPV at some point in their lives. Health problems related to HPV include genital warts and cervical cancer. Genital warts: Before HPV vaccines were introduced, roughly 340,000 to 360,000 women and men were affected by genital warts caused by HPV every year. Also, about one in 100 sexually active adults in the U.S. has genital warts at any given time. Cervical cancer: More than 11,000 women in the United States get cervical cancer each year. There are other conditions and cancers caused by HPV that occur in people living in the United States. Every year, approximately 17,600 women and 9,300 men are affected by cancers caused by HPV. These figures only look at the number of people who sought care for genital warts. This could be an underestimate of the actual number of people who get genital warts. I’m pregnant. Will having HPV affect my pregnancy? If you are pregnant and have HPV, you can get genital warts or develop abnormal cell changes on your cervix. Abnormal cell changes can be found with routine cervical cancer screening. You should get routine cervical cancer screening even when you are pregnant. Can I be treated for HPV or health problems caused by HPV? There is no treatment for the virus itself. However, there are treatments for the health problems that HPV can cause: 1. Genital warts can be treated by your healthcare provider or with prescription medication. If left untreated, genital warts may go away, stay the same, or grow in size or number. 2. Cervical precancer can be treated. Women who get routine Pap tests and follow up as needed can identify problems before cancer develops. Prevention is always better than treatment. For more information visit www.cancer.org. 3. Other HPV-related cancers are also more treatable when diagnosed and treated early. For more information visit www.cancer.org Where can I get more information? HPV Topic Page www.cdc.gov/hpv/index.html HPV Vaccination www.cdc.gov/vaccines/vpd/ hpv/index.html Cancer Prevention and Control www.cdc.gov/cancer/ Cervical Cancer – What Should I Know About Screening? www.cdc.gov/cancer/cervical/ basic_info/screening.htm CDC’s National Breast and Cervical Cancer Early Detection Program www.cdc.gov/cancer/nbccedp/ Division of STD Prevention (DSTDP) Centers for Disease Control and Prevention www.cdc.gov/std CDC-INFO Contact Center 1-800-CDC-INFO (1-800-232-4636) wwwn.cdc.gov/dcs/ContactUs/ Form CDC National Prevention Information Network (NPIN) npin.cdc.gov/disease/stds P.O. Box 6003 Rockville, MD 20849-6003 E-mail: npin-info@cdc.gov American Sexual Health Association (ASHA) P. O. Box 13827 Research Triangle Park, NC 27709-3827 1-800-783-9877 July 2017
15933	https://bestpractice.bmj.com/topics/en-us/350	Skip to main content Skip to search  Menu  Close Overview  Theory  Diagnosis  Management  Follow up  Resources  Log in or subscribe to access all of BMJ Best Practice Last reviewed: 8 Aug 2025 Last updated: 25 Jun 2024 Summary Acute paronychia is an acute infection of the nail folds and periungual tissues, usually caused by Staphylococcus aureus. Treatment of acute paronychia includes incision and drainage of any purulent fluid, soaks, and topical and/or oral antibacterials. Chronic paronychia is a chronic irritant dermatitis of the periungual tissues resulting from barrier damage to the protective nail tissues, including the cuticle and the proximal and lateral nail folds. Water and irritant avoidance is the hallmark of treatment of chronic paronychia. Definition Paronychia is the inflammation of the nail apparatus. Acute paronychias are infections of the periungual tissues, usually presenting with an acutely painful, purulent infection. Chronic paronychia represents barrier damage to the protective nail tissues, including the cuticle and the proximal and lateral nail folds. The altered nail barrier predisposes the nail to irritant dermatitis, most importantly from water, soap, chemicals, and microbes. Avoidance of such irritants is the hallmark of treatment.[Figure caption and citation for the preceding image starts]: Acute paronychiaFrom the collection of Dr N.J. Jellinek and Professor C.R. Daniel III [Citation ends].[Figure caption and citation for the preceding image starts]: Chronic paronychiaFrom the collection of Dr N.J. Jellinek and Professor C.R. Daniel III [Citation ends].[Figure caption and citation for the preceding image starts]: Chronic paronychiaFrom the collection of Dr N.J. Jellinek and Professor C.R. Daniel III [Citation ends]. History and exam Key diagnostic factors presence of risk factors pain, swelling, drainage (acute) swollen, purulent nail fold (acute) nail plate irregularities (chronic) swelling/redness of nail folds (chronic) pink, swollen nail folds (chronic) missing cuticle (chronic) underlying nail plate abnormalities (chronic) Full details Risk factors microscopic or macroscopic injury to the nail folds (acute) occupational risks (acute and chronic) barrier damage to the nail folds, cuticle (chronic) ingrown nail medications (chronic) toddler and adult female Full details Log in or subscribe to access all of BMJ Best Practice Diagnostic tests 1st tests to order swab for Gram stain, culture, and sensitivity (acute or acute-on-chronic) swab for Tzanck smear (acute, herpetic) Full details Tests to consider potassium hydroxide or fungal culture (chronic) x-ray MRI biopsy of skin/bone Full details Log in or subscribe to access all of BMJ Best Practice Treatment algorithm ACUTE acute paronychia ONGOING chronic paronychia retronychia Log in or subscribe to access all of BMJ Best Practice Contributors Authors Shirin Zaheri, MBBS, BSc, MRCP Dermatology Consultant Charing Cross Hospital Imperial College NHS Trust London UK Disclosures SZ declares that she has no competing interests. Khawar Hussain, BSc, MRCP, FHEA Specialty Registrar in Dermatology Charing Cross Hospital Imperial College NHS Trust London UK Disclosures KH has been reimbursed by Sanofi for conference travel expenses. Acknowledgements Dr Shirin Zaheri and Dr Khawar Hussain would like to gratefully acknowledge Dr Catherine Hardman, Dr Nathaniel J. Jellinek, Professor C. Ralph Daniel III, and Dr Shaimaa Nassar, previous contributors to this topic. Disclosures NJJ, CH, and SN declare that they have no competing interests. CRD is an author of a number of references cited in this topic. Peer reviewers Shehla Admani, MD Clinical Assistant Professor of Dermatology Stanford University School of Medicine San Jose CA Disclosures SA declares that he has no competing interests. Peer reviewer acknowledgements BMJ Best Practice topics are updated on a rolling basis in line with developments in evidence and guidance. The peer reviewers listed here have reviewed the content at least once during the history of the topic. Disclosures Peer reviewer affiliations and disclosures pertain to the time of the review. References Our in-house evidence and editorial teams collaborate with international expert contributors and peer reviewers to ensure that we provide access to the most clinically relevant information possible. Key articles Daniel CR 3rd. Paronychia. Dermatol Clin. 1985 Jul;3(3):461-4. Abstract Leggit JC. Acute and chronic paronychia. Am Fam Physician. 2017 Jul 1;96(1):44-51. Full text Abstract Relhan V, Goel K, Bansal S, et al. Management of chronic paronychia. Indian J Dermatol. 2014 Jan;59(1):15-20. Full text Abstract Shafritz AB, Coppage JM. Acute and chronic paronychia of the hand. J Am Acad Orthop Surg. 2014 Mar;22(3):165-74. Abstract Reference articles A full list of sources referenced in this topic is available to users with access to all of BMJ Best Practice. Differentials Herpetic whitlow Arthropod bite or sting Traumatic injury More Differentials #### Patient information Dermatitis Fungal nail infections More Patient information Log in or subscribe to access all of BMJ Best Practice Use of this content is subject to our disclaimer Log in or subscribe to access all of BMJ Best Practice Log in or subscribe to access all of BMJ Best Practice Log in to access all of BMJ Best Practice person personal subscription or user profile Access through your institution OR SUBSCRIPTION OPTIONS Cookies and privacy We and our 225 partners store and access personal data, like browsing data or unique identifiers, on your device. Selecting I Accept enables tracking technologies to support the purposes shown under we and our partners process data to provide. Selecting Reject All or withdrawing your consent will disable them. If trackers are disabled, some content and ads you see may not be as relevant to you. 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15934	https://www.youtube.com/watch?v=TrpM-YEYH3A	Excel square and ratio pixel grids made easy Learn Microsoft Office with Helen Bradley 2170 subscribers 17 likes Description 5853 views Posted: 7 Jun 2024 Excel square and ratio pixel grids made easy Learn to make printable grids in an Excel worksheet. I'll show you how to make the cells in the worksheet perfectly square and also how to create ratio grids such as a 4:5 grid such as might be used to create knitting patterns. I'll show you how to build a mini calculator to determine how a width to height ratio translates to actual cell sizes too. VIEW MORE VIDEOS JUST LIKE THIS ON MY YOUTUBE CHANNEL Microsoft Office ► TIP ME TO SAY THANK YOU? ► paypal.me/projectwoman VISIT PROJECTWOMAN.COM FOR MORE ON MICROSOFT OFFICE APPLICATIONS: Website ► Blog ► LET'S CONNECT! 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Students - Essential Skills Made Easy ► Org Chart Basics: Data Prep and Creation in Excel ► Place Text in Shapes in Excel & Word - Understand the Differences ► helenbradley #microsoftexcel #microsoftoffice #projectwoman #Excelforstudents #excelforbeginners #excelformulaforjobinterview 4 comments Transcript: hello and welcome to this video tutorial today we're looking at creating Square grids in Excel and also ratio grids so let's start with our Square grids I've got a worksheet already set up for this we want to make a square grid so we're going to select over all of the columns that we want in our grid you're going to position your cursor between two of those columns doesn't matter which two and you're going to click and drag you'll see that you get a width value that's changing while that's useless in this sit situation because you can't use that same value for a row height because they're measured in different quantities Excel is not telling you what those quantities are but the width and the height values are not the same unit but the pixels are so let's say we want a 50 pixel grid so I'm just going to wind back until the pixels Read 50 and I'm going to let go of the mouse button and now what I'll do is just go and grab on all the rows I want to be affected and I'm going to do the same thing I'm going to position my cursor between the rows click and drag and here you'll see that the height value is 15 whereas before the width was a much smaller value and yet the width of the column was bigger and the height you can see the values are totally different units but we've got pixels here and that's going to be the secret to this we're just going to read off the pixels and as soon as we get it to 50 we're just going to let go and now all of these cells are square and so we could go ahead and print that but what if we want a ratio GD so let's look at the scenario say of a knitting pattern in knitting when you knit up a little sample you may discover that for every four stitches on a row you have to do five rows to make a square so if you mark out a square on your knitting it might be four stitches across but five rows tall and that's a 4 to five ratio so we need four stitches across and then five rows to make that square so that's our width to height ratio and you may find that you've got another scenario where you need cells that the width to height ratio is 3:2 or 4 to 6 whatever it is you're just going to put those numbers in here and then we're going to work out in this case what our cell height is I want my cell height to be 100 so I'm just going to type in 100 here and so now we're going to work out what the width needs to be well the Height's going to have to be 100 because that's what we've just specified so I'm just typing a plus sign or an equal sign and just putting the contents of that cell in here just want to be able to read data off this area and I am using a formula here because if I change this value I want this one to change and now we need to build our formula so we're going to type an equal or a plus sign we're going to be multiplying the anticipated or desired value by this ratio so the width is going to be bigger than 100 it has to be bigger than 100 cuz we need four of those to five of the height so we're going to divide by four and we're going to multiply by five so I'm going to put in my division sign going to click on the cell that has four in it I'm going to hit my multiplication sign the asterisk I'm going to click on the cell it has five in it I'm going to press enter now if you get these two round the wrong way you're just going to go rebuild your formula but use a sort of guess here is that if we've got four stitches across or four somethings across and we have to do five high to make that same distance then the acrosses are going to be bigger than the Downs so this is what we've got we've got a width of 125 a height of 100 so let's go and make that grid on a new worksheet so we're just going to select over some cells that we want to include in our knitting pattern we're going to go up to 125 so there's our width and now we're going to select our rows and these were going to have to be 100 so again just clicking and dragging between and this is going to be a ratio grid where in these cells here if we select four across and five down this is a square so this is a knitting pattern grid now a word of warning here if you go and say you want your cell height to be 50 for example you're going to get a width that's a fractional amount and you can't do that so when we go and change these values you can see that there is no fractional amount they're all whole numbers so if you want it to be accurate you probably going to just need to change these values down a little bit so if I go to 48 then I can get a width and a height value that are whole numbers and so that's going to give me my grid a couple of pixels difference in height but this is going to be a true 4 to5 ratio grid so there's how to achieve Square pixel grids in Excel but also how to create your own ratio grids given a width to height ratio if you like carefully research content like this clearly present in a step-by-step format so that you can get great results every time then you'll love my other YouTube videos so give this video a thumbs up and click to subscribe to the channel and on the screen now you'll see a video that I've handpicked for you to watch next
15935	https://www.cantorsparadise.com/two-palindromes-f94c39677e26	Sitemap Open in app Sign in Sign in ## Cantor’s Paradise · Medium’s #1 Math Publication Two Palindromes 13 min readOct 10, 2020 Last week, in a blog post titled Finding Patterns, I introduced a puzzle. I asked my teenager to solve it using only a pen/pencil and paper. The Puzzle Two integers differ by 22. Each, when multiplied by its successor, yields an eight-digit palindrome. What is the smaller of the two? Three Steps Using a collaborative effort between human and machine, we can achieve more, while having fun solving problems and finding new patterns. Let’s try this in three steps. Step (I) is purely human effort. Step (II) is all brute power of the machine. Step (III) is a middle ground, a collaborative effort. Step I: Human Effort This section features human effort alone, without the use of the internet, or a computer program. To get started, we have to translate the puzzle from English to the language of Mathematics. Translation Let p and q be the two integers with q > p We are told p + 22 = q (p being smaller of the two) The successor of p is (p+1) and that of q is (q+1) Let the two palindromes be P = abcddcba and Q = wxyzzyxw Finding p is our objective Given these facts, we can write them as math equations: abcddcba and wxyzzyxw are the two palindromes with each letter denoting a digit — each different or with some overlap, we don’t know. The Basics At the risk of losing a few readers, I will try to be as accommodating as possible. If you know this already, you may skip to Spotting Patterns section. Integers are whole numbers, or counting numbers. Although not explicitly spelled out, this puzzle involves positive integers. -7, -8, -67 are a few negative integers56, 117, 2938 are a few positive integers Palindrome is a number that reads the same forwards and backwards. 25744752, 5665, 77, 8 are integers that are palindromes A successor to an integer is one that follows immediately after. Referred together, the two are consecutive integers. 7 is the successor to 6. 28 is a successor to 27. Decimal Expansion Any integer (in base-ten system or the decimal system) can be expressed as a sum of powers of 10. For example, the decimal expansion of 5665 is Prime Factors Any integer can be expressed as a product of its prime factors. A prime number is divisible by 1 and itself. 2 is an even prime. The rest of them are odd. Prime factors of 5665 are 5, 11 and 103 because 5665 = 5 x 11 x 103 Spotting Patterns (I) Test of divisibility by 11 A test of divisibility by 11 is interesting. For any integer, reading left to right, take the sums of alternate digits in that number. You must end up with two sums. If the sums differ by a multiple of 11, the number is divisible by 11. EXAMPLE: Take the number 947683 for instanceAlternate digits are S = {9, 7, 8} and T = {4, 6, 3}The sum of S = 9 + 7 + 8 = 24 The sum of T = 4 + 6 + 3 = 13Their difference is (T - S) = (13 - 24) = 11 x (-1), divisible by 11Therefore 947683 is divisible by 11In fact, 947683 = 86153 x 11 (II) Consecutive integers Even integers are divisible by 2. Odd integers are not. Two additional interesting facts about a pair of consecutive integers are: 1) One of them is odd, and the other even. Their product is even2) They share no prime factors. Their highest common factor is 1 (III) Perfect squares Two consecutive integers are of the form {2k, 2k+1} where k = {0, 1, 2, 3, …} A perfect square must be of the form {(2k)², (2k+1)²} = {4k², 4k(k+1) + 1}. A perfect square is exactly divisible by 4, OR It leaves 1 as remainder when divided by 4 (IV) Perfect square ending in 5 A curious fact about perfect squares ending in 5 is that the penultimate digit is always 2! If a number ends in 5, it’s square will always end in 25. EXAMPLES: Say you want to find 45². The only arithmetic we need is 4x5 = 20. Then attach 25 to it. The answer is 20\|25!45² = 4(4+1)\| 25 = 2025 Other examples:75² = 7(7+1)\| 25 = 5625115² = 11(12+1)\| 25 = 13225(10a+5)² = 100a² + 100a + 25 = 100a(a+1) + 25 If you look at the decimal expansion of a square ending in 5, this pattern is easy to spot. Notice that no matter what ‘a’ is, the result will end in 25. Also the other digits in the number are just product of two consecutive numbers (a) (a+1) shifted left by two digits to make way for 25! (V) Quadratic discriminant In a quadratic equation ax²+ bx + c = 0, the discriminant (Greek symbol Delta) given by Δ= ⎷(b²- 4ac) must be a perfect square. We want the solutions of x to be whole numbers! Δ is a perfect square for a quadratic equation with integral solutions (VI) Palindrome with even number of digits There are four pairs of integers that make up each palindrome. Let’s look at the decimal expansion of P for instance. Each term is divisible by 11. Notice how each digit is paired with another identical digit in a different location such that their sum {10000001, 100001, 1001, 11} is divisible by 11. P and Q must therefore, be divisible by 11. We can extend this to any palindrome with even number of digits. Using modular arithmetic and congruence relations, we can show they are divisible by 11. It is a unique pattern! A palindrome that has an even number of digits is divisible by 11 Detective Work Let’s start with digits {a, w} that can assume any of the values {0,1,2,...9}. But can they? Not really, because we have some enticing clues! Process of Elimination Using Pattern (II), both P and Q must be even. So they must begin (and end) with the following digits {2, 4, 6, 8}. I have excluded {0} because they cannot be zero if we want the palindromes to be eight-digits long! So we have eliminated {1, 3, 5, 7, 9}. That’s a reduction by 5/9 = 55.5%, right off the bat! {a,w} = {2, 4, 6, 8} P = p² + p is a quadratic equation with integer solutions. Using Pattern (V), the discriminant must be a perfect square. Comparing it with ax² + bx + c = 0 we have a = 1, b = 1, c = -P and Δ² = (b²- 4ac) = 1 + 4P Using Pattern (III), perfect squares are exactly divisible by 4 or leave a remainder 1 when divided by 4. We also know a set of rightmost digits of P and Q. Let’s check what the rightmost digits of 1 + 4P are. P = {2bcddcb2, 4bcddcb4, 6bcddcb6, 8bcddcb8}Δ²= 1 + 4 {2bcddcb2, 4bcddcb4, 6bcddcb6, 8bcddcb8}Δ² ends in digits {9, 7, 5, 3}There aren't any perfect squares that end in 7 or 3 because they are neither divisible by 4 nor leave a remainder 1 when divided by 4. We can apply the same logic to Q P cannot be { 4bcddcb4, 8bcddcb8}. And Q cannot be {4xyzzyz4, 8xyzzyx4}. Using this pattern, we managed to cut the possibilities in half again! {a,w} = {2, 6} Since the palindrome has eight-digits, both p and q must be four-digit numbers. Starting with {a} = {6} for P, we can show {w} = {0} for Q, which is a contradiction! Let’s understand how. Follow along using Table 1. Take row #5 for instance, where {a,w} = {6}. Two consecutive numbers (p, p+1) can end in (…2, …3) or (…7, …8) if the last digit of P is required to be 6. The leading dots denote digits we don’t yet know. Since q = p + 22, the digit endings for (q, q+1) will either be (…4, …5) or (…9, …0). Neither of those pairs, when multiplied give us a,w = {6} for Q. Using this logic, all pairs except the rows in green can be ruled out. {a, w} cannot be {6}. Get Venkat’s stories in your inbox Join Medium for free to get updates from this writer. We did it again, cut the solution space in half! When {a, w } = {2}, p ends in digits {1, 6} {a,w} = {2} which implies the following P = 2bcddcb2 and p = { ...1, ...6 } Q = 2xyzzyx2 and q = { ...3, ...8 } We know the largest and smallest eight-digit palindromes that begin (and end) with 2. They are bound by P(min, max) = {20000002, 29999992}. We can find out the leading (i.e, first) digit of p using an approximation that doesn’t alter its rightmost digit. If p² + p = P, approximately p² ~ P and p ~ ⎷P p is in the range p(min, max) = { ⎷20000002, ⎷29999992 } ~ [4500, 5500 ) I have used Pattern (IV) — technically we don’t need a calculator because 5500² =55² x 10⁴ = 30250000 and similarly 4500² = 45² x 10⁴ = 20250000 The first digit two digits of p are in this set {45, 46, 47,…, 53, 54}. Notice I have excluded 55 because 55² > 29999992 and included 45 because 45² > 20000002. We can write this more compactly: p = {4mn1, 4mn6, 5st1, 5st6} m = {5,6,7,8,9} and s={0,1,2,3,4} Using the Pattern (II) and (VI), {2, 11} are the prime factors of P = p (p+1) and Q = q(q+1). But p cannot have both 2 and 11 as its prime factors. That’s because they are consecutive, their highest common (prime) factor is 1. So we have to distribute {2, 11} between {p, p+1}. Case 1: p is odd If p is odd, 11 is its prime factor. In that case, p+1 is even and divisible by 2 p = {4mn1} m = {5,6,7,8,9} or p = {5st1} s={0,1,2,3,4} Consider p = 45n1 {m=5}: Apply the test of divisibility by 11. n + 4 = 5 + 1, which gives us n = 2. For the rest of {m, n} and {s,t} we can tabulate possible values of p. If n > 9, we discard the number because n is a digit! Case 2: p is even If p is even it is divisible by 2. In that case, p+1 is odd and 11, its factor. p = {4mn6} m = {5,6,7,8,9} or p = {5st6} s={0,1,2,3,4} Consider p = 45n6. In this case p+1 = 45n7 is divisible by 11. n + 4 = 7+ 5, which gives us n = 8. For the rest of {m, n} and {s,t} we can tabulate possible values of p. Again, if n > 9, we discard the number because n is a digit! If p > 5470 we discard it. Potential Suspects Our detective work is almost complete. We are down to a short list of suspects (sixteen candidates). There could be one or more culprits that fit the pattern! p = { 4521, 4631, 4741, 4851, 4961, 5071, 5181, 5291, 5401 }p = { 4586, 4696, 5026, 5136, 5246, 5356, 5466 } At this point, one could plug these numbers in a pocket calculator and check. For example, let’s test p = 4521 p = 4521 P = p(p+1) = 4521(4522) = 20443962We can save ourselves some time with long multiplication by spotting more patterns! If we multiply the last two digits 21 x 22 = 462 Similarly the first two digits 45 x 45 = 2025Notice the first two digits of P are 20, and last two are 62Since P isn't a palindrome, there is no need to check Q. We move on. I don’t know of any other patterns. I explored prime factors, digit endings of penultimate (tens place) digits and properties of consecutive integers and palindromes. But I ran out of ideas. I am interested in any other clever ideas to reduce the solution set further! Using simple numerical patterns for integers, we managed to reduce the solution space to scan. The bar chart shows how this reduction came about. Initial count was 9000, finally we are left with 16. The Culprit It turns out only p = 5291, q = 5313 have this unique property where they differ by 22 and each when multiplied by it’s successor yields an eight-digit palindrome. We have finally solved the problem by finding the values of p that have this property! p = 5291 = (11 ⨯ 13 ⨯ 37)p+1 = 5292 = (2² ⨯ 3² ⨯ 7²)P = p(p+1) = 5291(5292) = 27999972 Bingo! a palindrome ⏤⏤⏤⏤⏤⏤⏤⏤⏤⏤⏤⏤⏤⏤⏤⏤⏤⏤⏤⏤⏤⏤⏤q = p + 22 = 5313 = (3 ⨯ 7 ⨯ 11 ⨯ 23)q+1 = 5314 = (2 ⨯ 2657)Q = q(q+1) = 5313(5314) = 28233282 Another palindrome! Table 4: Candidates for palindromic product. Only p = 5291, q = 5313 fit the pattern Step II: Computing Power In this section, we will explore the power of gigahertz clock speed! Anyone with basic coding/programming experience can appreciate the difference in speed compared to a human with a pen/paper. I am using a MacBook with 16 GB RAM, 2.5 GHz Intel Core i7 processor. How Many Eight-Digit Palindromes? For curiosity’s sake, let’s find out how many eight-digit palindromes there are. The following code snippet provides the answer. It takes ~10.0 ± 0.2 s. Answer: There are 9000 eight-digit integer palindromes How Fast? This is extremely slow for a simple program. Modern laptops have processing speeds that easily exceed 10 billion steps or operations every second. A CPU clock cycle is roughly 0.3 ns (less than half a nanosecond). But to appreciate how fast the machine really operates, we need something we (as humans) can relate to. We understand and can relate to one second (One Mississippi). Let’s say we arbitrarily assign 1 CPU cycle ( 0.3 ns) to be one second. 1 CPU cycle = (0.3 ns or 0.0000000003 s) = 1 s Given this scale, how long is 10.0 seconds it took to generate all eight-digit palindromes? 10s / 0.0000000003 = 1056 years! We can do a lot better at generating and counting, if we know what a palindrome looks like. And we certainly do. We know half the digits (the first four or last four), we can construct the other half. Let us use this pattern to generate and count how many eight-digit palindromes are there. P = 11000i - 9900 (i//10) - 990 (i//100) - 99(i//1000)i = [1000, ..., 9999] Using this pattern, we can generate and count them all in 4.0 ± 0.2 milliseconds (ms) which is a significant (three orders of magnitude) improvement! Brute Force We can scan all eight-digit palindromes and test for this property. Let’s find out how long it takes to generate, check and find integers that have this property. The program is shown below. It takes about (7.1 ± 0.3 ms) Step III: Collaboration We went through the manual exercise in Step I. We managed to cut the solution space by half, in three steps. By the time we were done, we had a handful of candidates to verify. One could use any or all of the patterns to solve the problem at hand. I have chosen to start here: p = {4mn1, 4mn6, 5st1, 5st6} m = {5,6,7,8,9} and s={0,1,2,3,4} Let’s code this and find out how long it takes! The code snippet is shown below. It takes (55.6 ± 9 μs) which is a thousand-fold reduction! Now let’s compare how fast the execution is in human relatable time. Let’s recall we started with 1056 years to list all the eight-digit palindromes. If we calculate how long, we get about 2 days! That’s a remarkable reduction which went unnoticed. As humans it is hard to wrap our heads around both large and small numbers. 56 μs ~ 2 days Collaborate, Create Humans and machines are both powerful entities. We have evolved the capability to solve complex problems with our pattern recognition skills, capacity for abstract thought, and creative talents. If we collaborate with the machine to utilize its super-human strength and intelligence, the human-computer collaboration would make a winning combination! References Some Problems on the Prime Factors of Integers — P. Erdös, L. Selfridge, Illinois J. Math., Volume 11, Issue 3 (1967), 428–430. Link to PDF Divisibility by Eleven — Mudd Math Fun Facts Positive numbers k such that k(k+1) is a palindrome. A028336 — OEIS, Patrick De Geest Problem originally appeared in Mindsport, Sunday Times of India, By Mukul Sharma, Early 1990's The rise of human-computer cooperation, TED Talk 2012, Shyam Sankar Coding Horror blog post: The infinite space between words, 2014, Jeff Atwood ©️ Venkat Kaushik 2020. All Rights Reserved. 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15936	https://www.doubtnut.com/qna/646556792	Find the relation between γ (adiabatic constant) and degree of freedom (f) f=2γ−1 f=γγ−1 f=γ−12 f=γ−1γ To find the relation between the adiabatic constant γ and the degree of freedom f, we can follow these steps: 1. Understand the Definitions: - The adiabatic constant γ is defined as the ratio of the molar specific heat at constant pressure Cp to the molar specific heat at constant volume Cv: γ=CpCv 2. Express Cv in Terms of Degrees of Freedom: - The molar specific heat at constant volume Cv can be expressed in terms of the degree of freedom f: Cv=12fR - Here, R is the gas constant. 3. Use Mayer's Relation: - According to Mayer's relation, we have: Cp−Cv=R - Rearranging gives: Cp=Cv+R 4. Substitute Cv into Mayer's Relation: - Substitute the expression for Cv into Mayer's relation: Cp=12fR+R - This simplifies to: Cp=(12f+1)R 5. Express Cp in Terms of f: - Thus, we can express Cp as: Cp=(1+f2)R 6. Calculate γ: - Now, substituting Cp and Cv into the expression for γ: γ=CpCv=(1+f2)R12fR - The R cancels out: γ=1+f212f 7. Simplify the Expression: - This can be simplified further: γ=2(1+f2)f=2+ff=1+2f 8. Rearranging to Find f: - Rearranging gives: 2f=γ−1 - Therefore: f=2γ−1 Final Relation: The relation between the adiabatic constant γ and the degree of freedom f is: f=2γ−1 To find the relation between the adiabatic constant γ and the degree of freedom f, we can follow these steps: Understand the Definitions: The adiabatic constant γ is defined as the ratio of the molar specific heat at constant pressure Cp to the molar specific heat at constant volume Cv: γ=CpCv Express Cv in Terms of Degrees of Freedom: The molar specific heat at constant volume Cv can be expressed in terms of the degree of freedom f: Cv=12fR Here, R is the gas constant. Use Mayer's Relation: According to Mayer's relation, we have: Cp−Cv=R Rearranging gives: Cp=Cv+R Substitute Cv into Mayer's Relation: Substitute the expression for Cv into Mayer's relation: Cp=12fR+R This simplifies to: Cp=(12f+1)R Express Cp in Terms of f: Thus, we can express Cp as: Cp=(1+f2)R Calculate γ: Now, substituting Cp and Cv into the expression for γ: γ=CpCv=(1+f2)R12fR The R cancels out: γ=1+f212f Simplify the Expression: This can be simplified further: γ=2(1+f2)f=2+ff=1+2f Rearranging to Find f: Rearranging gives: 2f=γ−1 Therefore: f=2γ−1 Final Relation: The relation between the adiabatic constant γ and the degree of freedom f is: f=2γ−1 Topper's Solved these Questions Explore 80 Videos Explore 40 Videos Explore 473 Videos Explore 492 Videos Similar Questions What are degrees of freedom? Derive a relation between ionization constant and degree of ionization for acetic acid. Knowledge Check A gas that has 3 degrees of freedom is The correct relation between hydrolysis constant (Kb) and degree of hydrolysis (α) for the following equilibrium is The relation between the internal energy U adiabatic constant γ is Derive the relation between α,βandγ. What is the Relation between radians and degrees ? The adiabatic relation between P and T is The relation between βandγ of a solid is Assertion : The number of degrees of freedom of triatomic molecules is 6 Reason : Triatomic molecules have three translational degrees of freedom and three rotational degrees of freedom. JEE MAINS PREVIOUS YEAR-JEE MAIN 2021-PHYSICS SECTION B Find the relation between gamma (adiabatic constant) and degree of fre... A ball of mass 10 kg moving with a velocity 10sqrt3 m//s along the x-... As shown in the figure, a particle of mass 10 kg is placed at a point ... A particle performs simple harmonic motion with a period of 2 second. ... 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15937	https://ocw.mit.edu/courses/10-626-electrochemical-energy-systems-spring-2014/resources/mit10_626s14_lec24/	Browse Course Material Course Info Instructor Departments As Taught In Level Topics Learning Resource Types Electrochemical Energy Systems 10.626 Lecture Notes, Charge transfer at metal electrodes This resource contains the information regarding Charge transfer at metal electrodes. Course Info Instructor Departments As Taught In Level Topics Learning Resource Types You are leaving MIT OpenCourseWare
15938	https://en.wikibooks.org/wiki/Organic_Chemistry/Alkenes	Organic Chemistry/Alkenes - Wikibooks, open books for an open world Jump to content [x] Main menu Main menu move to sidebar hide Navigation Main Page Help Browse Cookbook Wikijunior Featured books Recent changes Special pages Random book Using Wikibooks Community Reading room forum Community portal Bulletin Board Help out! Policies and guidelines Contact us Search Search [x] Appearance Donations Create account Log in [x] Personal tools Donations Create account Log in [dismiss] The Wikibooks community is developing a policy on the use of generative AI. Please review the draft policy and provide feedback on its talk page. Contents move to sidebar hide Beginning 1 Naming AlkenesToggle Naming Alkenes subsection 1.1 EZ Notation 1.1.1 Comparison of E-Z with cis-trans 2 PropertiesToggle Properties subsection 2.1 Diastereomerism 2.1.1 Restricted rotation 3 Relative stability 4 ReactionsToggle Reactions subsection 4.1 Preparation 4.1.1 Dehydrohalogenation of Haloalkanes 4.1.2 Dehalogenation of Vicinal Dihalides 4.1.3 Dehydration of alcohols 4.1.4 Reduction of Alkynes 4.1.5 Wittig Reaction 4.2 Markovnikov's Rule 4.2.1 Examples 4.2.2 Markovnikov product 4.2.3 Markovnikov addition 4.2.4 Anti-Markovnikov addition 4.2.5 Why it works 4.2.6 Exceptions to the Rule 4.3 Addition reactions 4.3.1 Hydroboration 4.3.2 Hydroboration/Oxidation 4.3.2.1 Stereochemistry and Orientation 4.3.3 Oxymercuration/Reduction 4.3.4 Diels-Alder Reaction 4.3.5 Catalytic addition of hydrogen 4.3.6 Electrophilic addition 4.3.6.1 Halogenation 4.3.6.2 Hydrohalogenation 4.3.7 Oxidation 4.3.8 Polymerization 5 Substitution and Elimination Reaction MechanismsToggle Substitution and Elimination Reaction Mechanisms subsection 5.1 Nucleophilic Substitution Reactions 5.1.1 Note 5.1.2 S N 1 vs S N 2 5.1.3 S N 2 Reactions 5.1.4 Reactivity Due to Structure of S N 2 5.1.4.1 Nucleophilicity 5.1.5 List of descending nucleophilicities 5.1.5.1 Leaving Group 5.1.6 Relative Reactivity of Leaving Groups 5.1.6.1 Solvent 5.1.7 Relative Reactivity of Solvents 5.1.8 S N 1 Reactions 5.1.9 Reactivity Due to Structure of S N 1 5.1.9.1 Solvent 5.1.10 Summary 5.2 Elimination Reactions 5.2.1 Note 5.2.2 E1 vs E2 5.2.2.1 Reaction rates 5.2.2.2 Zaitsev's Rule 5.2.3 E2 5.2.4 Reactivity Due to Structure of E2 5.2.5 E1 6 References [x] Toggle the table of contents Organic Chemistry/Alkenes [x] 3 languages 日本語 Nederlands 中文 Edit links Book Discussion [x] English Read Latest draft Edit Edit source View history [x] Tools Tools move to sidebar hide Actions Read Latest draft Edit Edit source View history General What links here Related changes Upload file Permanent link Page information Cite this page Get shortened URL Download QR code Sister projects Wikipedia Wikiversity Wiktionary Wikiquote Wikisource Wikinews Wikivoyage Commons Wikidata MediaWiki Meta-Wiki Print/export Create a collection Download as PDF Printable version In other projects Wikimedia Commons Wikipedia Wikiquote Wikidata item Appearance move to sidebar hide From Wikibooks, open books for an open world <Organic Chemistry The latest reviewed version was checked on 25 March 2023. There are template/file changes awaiting review. << Haloalkanes \|Alkenes\| Alkynes >> Alkenes are aliphatichydrocarbons containing carbon-carbon double bonds and general formula C n H 2n. Naming Alkenes [edit \| edit source] Alkenes are named as if they were alkanes, but the "-ane" suffix is changed to "-ene". If the alkene contains only one double bond and that double bond is terminal (the double bond is at one end of the molecule or another) then it is not necessary to place any number in front of the name. butane: C 4 H 10 (CH 3 CH 2 CH 2 CH 3) butene: C 4 H 8 (CH 2=CHCH 2 CH 3) If the double bond is not terminal (if it is on a carbon somewhere in the center of the chain) then the carbons should be numbered in such a way as to give the first of the two double-bonded carbons the lowest possible number, and that number should precede the "ene" suffix with a dash, as shown below. correct: pent-2-ene (CH 3 CH=CHCH 2 CH 3) incorrect: pent-3-ene (CH 3 CH 2 CH=CHCH 3) The second one is incorrect because flipping the formula horizontally results in a lower number for the alkene. If there is more than one double bond in an alkene, all of the bonds should be numbered in the name of the molecule - even terminal double bonds. The numbers should go from lowest to highest, and be separated from one another by a comma. The IUPAC numerical prefixes are used to indicate the number of double bonds. octa-2,4-diene: CH 3 CH=CHCH=CHCH 2 CH 2 CH 3 deca-1,5-diene: CH 2=CHCH 2 CH 2 CH=CHCH 2 CH 2 CH 2 CH 3 Note that the numbering of "2-4" above yields a molecule with two double bonds separated by just one single bond. Double bonds in such a condition are called "conjugated", and they represent an enhanced stability of conformation, so they are energetically favored as reactants in many situations and combinations. EZ Notation [edit \| edit source] Earlier in stereochemistry, we discussed cis/trans notation where cis- means same side and trans- means opposite side. Alkenes can present a unique problem, however in that the cis/trans notation sometimes breaks down. The first thing to keep in mind is that alkenes are planar and there's no rotation of the bonds, as we'll discuss later. So when a substituent is on one side of the double-bond, it stays on that side. cis-but-2-ene and trans-but-2-ene The above example is pretty straight-forward. On the left, we have two methyl groups on the same side, so it's cis-but-2-ene. And on the right, we have them on opposite sides, so we have trans-but-2-ene. So in this situation, the cis/trans notation works and, in fact, these are the correct names. (E)-3-methylpent-2-ene and (Z)-3-methylpent-2-ene From the example above, how would you use cis and trans? Which is the same side and which is the opposite side? Whenever an alkene has 3 or 4 differing substituents, one must use the what's called the EZ nomenclature, coming from the German words, Entgegen (opposite) and Zusammen (same). E: Entgegen, opposite sides of double bond Z: Zusammen, same sides (zame zides) of double bond Let's begin with (Z)-3-methylpent-2-ene. We begin by dividing our alkene into left and right halves. On each side, we assign a substituent as being either a high priority or low priority substituent. The priority is based on the atomic number of the substituents. So on the left side, hydrogen is the lowest priority because its atomic number is 1 and carbon is higher because its atomic number is 6. On the right side, we have carbon substituents on both the top and bottom, so we go out to the next bond. On to the top, there's another carbon, but on the bottom, a hydrogen. So the top gets high priority and the bottom gets low priority. Because the high priorities from both sides are on the same side, they are Zusammen (as a mnemonic, think 'Zame Zide'). Now let's look at (E)-3-methylpent-2-ene. On the left, we have the same substituents on the same sides, so the priorities are the same as in the Zusammen version. However, the substituents are reversed on the right side with the high priority substituent on the bottom and the low priority substituent on the top. Because the High and Low priorities are opposite on the left and right, these are Entgegen, or opposite. The system takes a little getting used to and it's usually easier to name an alkene than it is to write one out given its name. But with a little practice, you'll find that it's quite easy. Comparison of E-Z with cis-trans [edit \| edit source] (Z)-but-2-ene(E)-but-2-ene cis-but-2-ene trans-but-2-ene To a certain extent, the Z configuration can be regarded as the cis- isomer and the E as the trans- isomers. This correspondence is exact only if the two carbon atoms are identically substituted. In general, cis-trans should only be used if each double-bonded carbon atom has a hydrogen atom (i.e. R-CH=CH-R'). IUPAC Gold book on cis-trans notation. IUPAC Gold book on E-Z notation. Properties [edit \| edit source] Alkenes are molecules with carbons bonded to hydrogens which contain at least two sp 2 hybridized carbon atoms. That is, to say, at least one carbon-to-carbon double bond, where the carbon atoms, in addition to an electron pair shared in a sigma (σ) bond, share one pair of electrons in a pi (π) bond between them. The general formula for an aliphatic alkene is: C n H 2n -- e.g. C 2 H 4 or C 3 H 6 Diastereomerism [edit \| edit source] Restricted rotation [edit \| edit source] Because of the characteristics of pi-bonds, alkenes have very limited rotation around the double bonds between two atoms. In order for the alkene structure to rotate the pi-bond would first have to be broken - which would require about 60 or 70 kcal of energy per mol. For this reason alkenes have different chemical properties based on which side of the bond each atom is located. For example, but-2-ene exists as two diastereomers: (Z)-but-2-ene(E)-But-2-ene cis-but-2-ene trans-but-2-ene Relative stability [edit \| edit source] Observing the reaction of the addition of hydrogen to 1-butene, (Z)-2-butene, and (E)-2-butene, we can see that all of the products are butane. The difference between the reactions is that each reaction has a different energy: -30.3 kcal/mol for 1-butene, -28.6 kcal/mol for (Z)-2-butene and -27.6 kcal/mol for (E)-2-butene. This illustrates that there are differences in the stabilities of the three species of butene isomers, due to the difference in how much energy can be released by reducing them. The relative stability of alkenes may be estimated based on the following concepts: An internal alkene (the double bond not on the terminal carbon) is more stable than a terminal alkene (the double bond is on a terminal carbon). Internal alkenes are more stable than terminal alkenes because they are connected to more carbons on the chain. Since a terminal alkene is located at the end of the chain, the double bond is only connected to one carbon, and is called primary (1°). Primary carbons are the least stable. In the middle of a chain, a double bond could be connected to two carbons. This is called secondary (2°). The most stable would be quaternary (4°). In general, the more and the bulkier the alkyl groups on a sp 2-hybridized carbon in the alkene, the more stable that alkene is. A trans double bond is more stable than a cis double bond. Reactions [edit \| edit source] Preparation [edit \| edit source] There are several methods for creating alkenes. Some of these methods, such as the Wittig reaction, we'll only describe briefly in this chapter and instead, cover them in more detail later in the book. For now, it's enough to know that they are ways of creating alkenes. Dehydrohalogenation of Haloalkanes [edit \| edit source] Synthesis of alkene by dehydrohalogenation Alkyl halides are converted into alkenes by dehydrohalogenation: elimination of the elements of hydrogen halide. Dehydrohalogenation involves removal of the halogen atom together with a hydrogen atom from a carbon adjacent to the one bearing the halogen. It uses the E2 elimination mechanism that we'll discuss in detail at the end of this chapter The haloalkane must have a hydrogen and halide 180° from each other on neighboring carbons. If there is no hydrogen 180° from the halogen on a neighboring carbon, the reaction will not take place. It is not surprising that the reagent required for the elimination of what amounts to a molecule of acid is a strong base for example: alcoholic KOH. In some cases this reaction yields a single alkene. and in other cases yield a mixture. n-Butyl chloride, for example, can eliminate hydrogen only from C-2 and hence yields only 1-butene. sec-Butyl chloride, on the other hand, can eliminate hydrogen from either C-l or C-3 and hence yields both 1-butene and 2-butene. Where the two alkenes can be formed, 2-butene is the chief product. Dehalogenation of Vicinal Dihalides [edit \| edit source] Synthesis of alkene via debromination of vicinal dihalides using Sodium Iodide Synthesis of alkene via debromination of vicinal dihalides using Zinc The dehalogenation of vicinal dihalides (halides on two neighboring carbons, think "vicinity") is another method for synthesizing alkenes. The reaction can take place using either sodium iodide in a solution of acetone, or it can be performed using zinc dust in a solution of either heated ethanol or acetic acid. This reaction can also be performed with magnesium in ether, though the mechanism is different as this actually produces, as an intermediate, a Grignard reagent that reacts with itself and causes an elimination, resulting in the alkene.[citation needed] Dehydration of alcohols [edit \| edit source] Synthesis of alkene by dehydration of an alcohol An alcohol is converted into an alkene by dehydration: elimination of a molecule of water. Dehydration requires the presence of an acid and the application of heat. It is generally carried out in either of two ways, heating the alcohol with sulfuric or phosphoric acid to temperatures as high as 200, or passing the alcohol vapor over alumina, Al 2 O 3 , at 350-400, alumina here serving as a Lewis acid. Ease of dehydration of alcohols: 3° > 2° > 1° Where isomeric alkenes can be formed, we again find the tendency for one isomer to predominate. Thus, sec-butyl alcohol, which might yield both 2-butene and 1-butene, actually yields almost exclusively the 2-isomer The formation of 2-butene from n-butyl alcohol illustrates a characteristic of dehydration that is not shared by dehydrohalogenalion: the double bond can be formed at a position remote from the carbon originally holding the -OH group. This characteristic is accounted for later. It is chiefly because of the greater certainty as to where the double bond will appear that dehydrohalogeation is often preferred over dehydration as a method of making alkenes. Reduction of Alkynes [edit \| edit source] Reduction of an alkyne to the double-bond stage can yield either a cis-alkene or a trans-alkene, unless the triple bond is at the end of a chain. Just which isomer predominates depends upon the choice of reducing agent. Predominantly trans-alkene is obtained by reduction of alkynes with sodium or lithium in liquid ammonia. Almost entirely cis-alkene (as high as 98%) is obtained by hydrogenation of alkynes with several different catalysts: a specially prepared palladium called Lindlar's catalyst; or a nickel boride called P-2 catalyst. Each of these reactions is, then, highly stereoselective. The stereoselectivity in the cis-reduction of alkynes is attributed, in a general way, to the attachment of two hydrogens to the same side of an alkyne sitting on the catalyst surface; presumably this same stereochemistry holds for the hydrogenation of terminal alkynes which cannot yield cis- and trans-alkenes. Wittig Reaction [edit \| edit source] Synthesis of alkene via Wittig reaction Markovnikov's Rule [edit \| edit source] Before we continue discussing reactions, we need to take a detour and discuss a subject that's very important in Alkene reactions, "Markovnikov's Rule." This is a simple rule stated by the Russian Vladmir Markovnikov in 1869, as he was showing the orientation of addition of HBr to alkenes. His rule states:"When an unsymmetrical alkene reacts with a hydrogen halide to give an alkyl halide, the hydrogen adds to the carbon of the alkene that has the greater number of hydrogen substituents, and the halogen to the carbon of the alkene with the fewer number of hydrogen substituents" (This rule is often compared to the phrase: "The rich get richer and the poor get poorer." Aka, the Carbon with the most Hydrogens gets another Hydrogen and the one with the least Hydrogens gets the halogen) This means that the nucleophile of the electrophile-nucleophile pair is bonded to the position most stable for a carbocation, or partial positive charge in the case of a transition state. Examples [edit \| edit source] C H 2=C H−C H 3+H−B r−>C H 3−C H B r−C H 3{\displaystyle CH_{2}=CH-CH_{3}+H-Br->CH3-CHBr-CH_{3}} Here the Br attaches to the middle carbon over the terminal carbon, because of Markovnikov's rule, and this is called a Markovnikov product. Markovnikov product [edit \| edit source] The product of a reaction that follows Markovnikov's rule is called a Markovnikov product. Markovnikov addition [edit \| edit source] Markovnikov addition is an addition reaction which follows Markovnikov's rule, producing a Markovnikov product. Anti-Markovnikov addition [edit \| edit source] Certain reactions produce the opposite of the Markovnikov product, yielding what is called anti-Markovnikov product. That is, hydrogen ends up on the more substituted carbon of the double bond. The hydroboration/oxidation reaction that we'll discuss shortly, is an example of this, as are reactions that are conducted in peroxides. A modernized version of Markovnikov's rule often explains the "anti-Markovnikov" behavior. The original Markovnikov rule predicts that the hydrogen (an electrophile) being added across a double bond will end up on the carbon with more hydrogens. Generalizing to all electrophiles, it is really the electrophile which ends up on the carbon with the greatest number of hydrogens. Usually hydrogen plays the role of the electrophile; however, hydrogen can also act as an nucleophile in some reactions. The following expansion of Markovnikov's rule is more versatile: "When an alkene undergoes electrophilic addition, the electrophile adds to the carbon with the greatest number of hydrogen substituents. The nucleophile adds to the more highly substituated carbon." Or more simply: "The species that adds first adds to the carbon with the greatest number of hydrogens." The fact that some reactions reliably produce anti-Markovnikov products is actually a powerful tool in organic chemistry. For example, in the reactions we discuss below, we'll show two different ways of creating alcohols from alkenes: Oxymercuration-Reduction and Hydroboration/Oxidation. Oxymercuration produces a Markovnikov product while Hydroboration produces an anti-Markovnikov product. This gives the organic chemist a choice in products without having to be stuck with a single product that might not be the most desired. Why it works [edit \| edit source] Markovnikov's rule works because of the stability of carbocation intermediates. Experiments tend to reveal that carbocations are planar molecules, with a carbon that has three substituents at 120° to each other and a vacant p orbital that is perpendicular to it in the 3rd plane. The p orbital extends above and below the trisubstituent plane. This leads to a stabilizing effect called hyperconjugation. Hyperconjugation is what happens when there is an unfilled (antibonding or vacant) C-C π orbital and a filled C-H σ bond orbital next to each other. The result is that the filled C-H σ orbital interacts with the unfilled C-C π orbital and stabilizes the molecule. The more highly substituted the molecule, the more chances there are for hyperconjugation and thus the more stable the molecule is. Another stabilizing effect is an inductive effect. Exceptions to the Rule [edit \| edit source] There are a few exceptions to the Markovnikov rule, and these are of tremendous importance to organic synthesis. HBr in Hydrogen Peroxide: Due to formation of free radicals, and the mechanism in which it reacts, the alkyl free radical forms at the middle atom, where it is most stable, and a hydrogen attaches itself here. Note here hydrogen addition is the second step, unlike in the above example. Addition reactions [edit \| edit source] Hydroboration [edit \| edit source] Hydroboration is a very useful reaction in Alkenes, not as an end product so much as an intermediate product for further reactions. The primary one we'll discuss below is the Hydroboration/Oxidation reaction which is actually a hydroboration reaction followed by a completely separate oxidation reaction. Hydroboration mechanism The addition of BH 3 is a concerted reaction in that several bonds are broken and formed at the same time. Hydroboration happens in what's called syn-addition because the boron and one of its hydrogens attach to the same side of the alkene at the same time. As you can see from the transition state in the center of the image, this produces a sort of box between the two alkene carbons and the boron and its hydrogen. In the final step, the boron, along with its other two hydrogens, remains attached to one carbon and the other hydrogen attaches to the adjacent carbon. This description is fairly adequate, however, the reaction actually continues to happen and the -BH 2 continue to react with other alkenes giving an R 2 BH and then again, until you end up with a complex of the boron atom attached to 3 alkyl groups, or R 3 B. This trialkyl-boron complex is then used in other reactions to produce various products. B 2 H 6 complex BH 3-THF complex Borane, in reality, is not stable as BH 3. Boron, in this configuration has only 6 electrons and wants 8, so in its natural state it actually creates the B 2 H 6 complex shown on the left. Furthermore, instead of using B 2 H 6 itself, BH 3 is often used in a complex with tetrahydrofuran (THF) as shown in the image on the right.In either situation, the result of the reactions are the same. Hydroboration/Oxidation [edit \| edit source] Hydroboration/Oxidation reaction With the reagent diborane, (BH 3)2, alkenes undergo hydroboration to yield alkylboranes, R 3 B, which on oxidation give alcohols.The reaction procedure is simple and convenient, the yields are exceedingly high, and the products are ones difficult to obtain from alkenes in anyother way. Diborane is the dimer of the hypothetical BH3 (borane) and, in the reactions that concern us, acts much as though it were BH 3 . Indeed, in tetrahydrofuran, one of the solvents used for hydroboration, the reagent exists as the monomer, in the form of an acid-base complex with the solvent. Hydroboration involves addition to the double bond of BH 3 (or, in following stages, BH 2 R and BHR 2), with hydrogen becoming attached to one doubly-bonded carbon, and boron to the other. The alkylborane can then undergo oxidation, in which the boron is replaced by -OH. Thus, the two-stage reaction process of hydroboration-oxidation permits, in effect, the addition to the carbon-carbon double bond of the elements of H-OH. Reaction is carried out in an ether, commonly tetrahydrofuran or "diglyme" (diethylene glycol methyl ether, CH 3 OCH 2 CH 2 OCH 2 CH 2 OCH 3). Diborane is commercially available in tetrahydrofuran solution. The alkylboranes are not isolated, but are simply treated in situ with alkaline hydrogen peroxide. Stereochemistry and Orientation [edit \| edit source] Hydroboration-oxidation, then, converts alkenes into alcohols. Addition is highly regiospecific; the preferred product here, however, is exactly opposite to the one formed by oxymercuration-demercuration or by direct acid-catalyzed hydration. The hydroboration-oxidation process gives products corresponding to anti-Markovnikov addition of water to the carbon-carbon double bond. The reaction of 3,3-dimethyl-l -butene illustrates a particular advantage of the method. Rearrangement does not occur in hydroboration evidently because carbonium ions are not intermediates and hence the method can be used without the complications that often accompany other addition reactions. The reaction of 1,2-dimethylcyclopentene illustrates the stereochemistry of the synthesis: hydroboration-oxidation involves overallsyn addition. Oxymercuration/Reduction [edit \| edit source] Oxymercuration/Reduction of 1-propene Alkenes react with mercuric acetate in the presence of water to give hydroxymercurial compounds which on reduction yield alcohols. The first stage, oxymercuration, involves addition to the carbon-carbon double bond of -OH and -HgOAc. Then, in reduction, the -HgOAc is replaced by -H. The reaction sequence amounts to hydration of the alkene, but is much more widely applicable than direct hydration. The two-stage process of oxymercuration/reduction is fast and convenient, takes place under mild conditions, and gives excellent yields often over 90%. The alkene is added at room temperature to an aqueous solution of mercuric acetate diluted with the solvent tetrahydrofuran. Reaction is generally complete within minutes. The organomercurial compound is not isolated but is simply reduced in situ by sodium borohydride, NaBH 4. (The mercury is recovered as a ball of elemental mercury.) Oxymercuration/reduction is highly regiospecific, and gives alcohols corresponding to Markovnikov addition of water to the carbon-carbon double bond. Oxymercuration involves electrophilic addition to the carbon-carbon double bond, with the mercuric ion acting as electrophile. The absence of rearrangement and the high degree of stereospecificity (typically anti) in the oxymercuration step argues against an open carbonium ion as intermediate. Instead, it has been proposed, there is formed a cyclic mercurinium ion, analogous to the bromonium and chloronium ions involved in the addition of halogens. In 1971, Olah reported spectroscopic evidence for the preparation of stable solutions of such mercurinium ions. The mercurinium ion is attacked by the nucleophilic solvent water, in the present case to yield the addition product. This attack is back-side (unless prevented by some structural feature) and the net result is anti addition, as in the addition of halogens. Attack is thus of the S N 2 type; yet the orientation of addition shows that the nucleophile becomes attached to the more highly substituted carbon as though there were a free carbonium ion intermediate. As we shall see, the transition state in reactions of such unstable threemembered rings has much S N 1 character. Reduction is generally not stereospecific and can, in certain special cases, be accompanied by rearrangement. Despite the stereospecificity of the first stage, then, the overall process is not,in general, stereospecific. Rearrangements can occur, but are not common. The reaction of 3,3-dimethyl-1-butene illustrates the absence of the rearrangements that are typical of intermediate carbonium ions. Diels-Alder Reaction [edit \| edit source] The Diels–Alder reaction is a reaction (specifically, a cycloaddition) between a conjugated diene and a substituted alkene, commonly termed the dienophile, to form a substituted cyclohexene system. The reaction can proceed even if some of the atoms in the newly formed ring are not carbon. Some of the Diels–Alder reactions are reversible; the decomposition reaction of the cyclic system is then called the retro-Diels–Alder. Diels-alder for 1,3-butadiene-Ethylene The Diels–Alder reaction is generally considered one of the more useful reactions in organic chemistry since it requires very little energy to create a cyclohexene ring, which is useful in many other organic reactions A concerted, single-step mechanism is almost certainly involved; both new carbon-carbon bonds are partly formed in the same transition state, although not necessarily to the same extent. The Diels-Alder reaction is the most important example of cycloaddition. Since reaction involves a system of 4 π electrons (the diene) and a system of 2 π it electrons (the dienophile), it is known as a [4 + 2] cycloaddition. Catalytic addition of hydrogen [edit \| edit source] Catalytic hydrogenation of alkenes produce the corresponding alkanes. The reaction is carried out under pressure in the presence of a metallic catalyst. Common industrial catalysts are based on platinum, nickel or palladium, but for laboratory syntheses, Raney nickel (formed from an alloy of nickel and aluminium) is often employed. The catalytic hydrogenation of ethylene to yield ethane proceeds thusly: CH 2=CH 2 + H 2 + catalyst → CH 3-CH 3 Electrophilic addition [edit \| edit source] Most addition reactions to alkenes follow the mechanism of electrophilic addition. An example is the Prins reaction, where the electrophile is a carbonyl group. Halogenation [edit \| edit source] Addition of elementary bromine or chlorine in the presence of an organic solvent to alkenes yield vicinal dibromo- and dichloroalkanes, respectively. The decoloration of a solution of bromine in water is an analytical test for the presence of alkenes: CH 2=CH 2 + Br 2 → BrCH 2-CH 2 Br The reaction works because the high electron density at the double bond causes a temporary shift of electrons in the Br-Br bond causing a temporary induced dipole. This makes the Br closest to the double bond slightly positive and therefore an electrophile. Hydrohalogenation [edit \| edit source] Addition of hydrohalic acids like HCl or HBr to alkenes yield the corresponding haloalkanes. an example of this type of reaction is: CH 3 CH=CH 2 + HBr → CH 3-CHBr-CH 3 If the two carbon atoms at the double bond are linked to a different number of hydrogen atoms, the halogen is found preferentially at the carbon with less hydrogen substituents (Markovnikov's rule). Addition of a carbene or carbenoid yields the corresponding cyclopropane Oxidation [edit \| edit source] Alkenes are oxidized with a large number of oxidizing agents. In the presence of oxygen, alkenes burn with a bright flame to form carbon dioxide and water. Catalytic oxidation with oxygen or the reaction with percarboxylic acids yields epoxides. Reaction with ozone in ozonolysis leads to the breaking of the double bond, yielding two aldehydes or ketones: R 1-CH=CH-R 2 + O 3 → R 1-CHO + R 2-CHO + H 2 O This reaction can be used to determine the position of a double bond in an unknown alkene. Polymerization [edit \| edit source] Polymerization of alkenes is an economically important reaction which yields polymers of high industrial value, such as the plastics polyethylene and polypropylene. Polymerization can either proceed via a free-radical or an ionic mechanism. Substitution and Elimination Reaction Mechanisms [edit \| edit source] Nucleophilic Substitution Reactions [edit \| edit source] Nucleophilic substitution reactions (S N 1 and S N 2) are very closely related to the E1 and E2 elimination reactions, discussed later in this section, and it is generally a good idea to learn the reactions together, as there are parallels in reaction mechanism, preferred substrates, and the reactions sometimes compete with each other. It's important to understand that substitution and elimination reactions are not associated with a specific compound or mixture so much as they're a representation of how certain reactions take place. At times, combinations of these mechanisms may occur together in the same reaction or may compete against each other, with influences such as solvent or nucleophile choice being the determining factor as to which reaction will dominate. Note In the notation S N 1 and S N 2, S stands for substitution (something takes the place of something else) N: stands for nucleophilic (a nucleophile displaces another nucleophile) 1: stands for unimolecular (the concentration of only one kind of molecule determines the rate of the reaction) 2: stands for bimolecular (the concentration of two types of molecules determine the rate of the reaction) In nucleophilic substitution, a nucleophile attacks a molecule and takes the place of another nucleophile, which then leaves. The nucleophile that leaves is called the leaving group. Nucleophilic substitutions require a nucleophile (such as a Lewis base) an electrophile with a leaving group A leaving group is a charged or neutral moiety (group) which breaks free. S N 1 vs S N 2 [edit \| edit source] One of the main differences between S N 1 and S N 2 is that the S N 1 reaction is a 2-step reaction, initiated by disassociation of the leaving group. The S N 2 reaction, on the other hand, is a 1-step reaction where the attacking nucleophile, because of its higher affinity for and stronger bonding with the carbon, forces the leaving group to leave. These two things happen in a single step. These two different mechanisms explain the difference in reaction rates between S N 1 and S N 2 reactions. S N 1 reactions are dependent on the leaving group disassociating itself from the carbon. It is the rate-limiting step and thus, the reaction rate is a first-order reaction whose rate depends solely on that step. R a t e=k[R X]{\displaystyle Rate=k[RX]} Alternatively, in S N 2 reactions, the single step of the nucleophile coming together with the reactant from the opposite side of the leaving group, is the key to its rate. Because of this, the rate is dependent on both the concentration of the nucleophile as well as the concentration of the reactant. The higher these two concentrations, the more frequent the collisions. Thus the reaction rate is a second-order reaction: R a t e=k[N u:][R X]{\displaystyle Rate=k[Nu:][RX]} (where Nu: is the attacking nucleophile) S N 2 Reactions [edit \| edit source] There are primarily 3 things that affect whether an S N 2 reaction will take place or not. The most important is structure. That is whether the alkyl halide is on a methyl, primary, secondary, or tertiary carbon. The other two components that determine whether an S N 2 reaction will take place or not, are the nucleophilicity of the nucleophile and the solvent used in the reaction. Reactivity Due to Structure of S N 2 CH 3 X > RCH 2 X > R 2 CHX >> R 3 CX The structure of the alkyl halide has a great effect on mechanism. CH 3 X & RCH 2 X are the preferred structures for S N 2. R 2 CHX can undergo the S N 2 under the proper conditions (see below), and R 3 CX rarely, if ever, is involved in S N 2 reactions. S N 2 nucleophilic substitution of bromine with a generic nucleophile The reaction takes place by the nucleophile attacking from the opposite side of the bromine atom. Notice that the other 3 bonds are all pointed away from the bromine and towards the attacking nucleophile. When these 3 bonds are hydrogen bonds, there's very little steric hinderance of the approaching nucleophile. However, as the number of R groups increases, so does the steric hinderance, making it more difficult for the nucleophile to get close enough to the α-carbon, to expel the bromine atom. In fact, tertiary carbons (R 3 CX) are so sterically hindered as to prevent the S N 2 mechanism from taking place at all. In the case of this example, a secondary α-carbon, there is still a great deal of steric hinderance and whether the S N 2 mechanism will happen will depend entirely on what the nucleophile and solvent are. S N 2 reactions are preferred for methyl halides and primary halides. Another important point to keep in mind, and this can be seen clearly in the example above, during an S N 2 reaction, the molecule undergoes an inversion. The bonds attached to the α-carbon are pushed away as the nucleophile approaches. During the transition state, these bonds become planar with the carbon and, as the bromine leaves and the nucleophile bonds to the α-carbon, the other bonds fold back away from the nucleophile. This is particularly important in chiral or pro-chiral molecules, where an R configuration will be converted into an S configuration and vice versa. As you'll see below, this is in contrast to the results of S N 1 reactions. Examples: OH- + CH 3—Cl → HO—CH 3 + Cl- OH- is the nucleophile, Cl is the electrophile, HOCH3 is the product, and Cl- is the leaving group. or, Na+I- + CH 3-Br → I-CH 3 + Na+Br- The above reaction, taking place in acetone as the solvent, sodium and iodide disassociate almost completely in the acetone, leaving the iodide ions free to attack the CH-Br molecules. The negatively charged iodide ion, a nucleophile, attacks the methyl bromide molecule, forcing off the negatively charged bromide ion and taking its place. The bromide ion is the leaving group. Nucleophilicity [edit \| edit source] Nucleophilicity is the rate at which a nucleophile displaces the leaving group in a reaction. Generally, nucleophilicity is stronger, the larger, more polarizable, and/or the less stable the nucleophile. No specific number or unit of measure is used. All other things being equal, nucleophiles are generally compared to each other in terms of relative reactivity. For example, a particular strong nucleophile might have a relative reactivity of 10,000 that of a particular weak nucleophile. These relationships are generalities as things like solvent and substrate can affect the relative rates, but they are generally good guidelines for which species make the best nucleophiles. All nucleophiles are Lewis bases. In S N 2 reactions, the preferred nucleophile is a strong nucleophile that is a weak base. Examples of these are N 3-, RS-, I-, Br-, and CN-. Alternatively, a strong nucleophile that's also a strong base can also work. However, as mentioned earlier in the text, sometimes reaction mechanisms compete and in the case of a strong nucleophile that's a strong base, the S N 2 mechanism will compete with the E2 mechanism. Examples of strong nucleophiles that are also strong bases, include RO- and OH-. List of descending nucleophilicities I-> Br-> Cl->> F-> -SeH > -OH > H 2 O Leaving Group [edit \| edit source] Leaving group is the group on the substrate that leaves. In the case of an alkyl halide, this is the halide ion that leaves the carbon atom when the nucleophile attacks. The tendency of the nucleophile to leave is Relative Reactivity of Leaving Groups I-> Br-> Cl->> F- Fluoride ions are very poor leaving groups because they bond very strongly and are very rarely used in alkyl halide substitution reactions. Reactivity of a leaving group is related to its basicity with stronger bases being poorer leaving groups. Solvent [edit \| edit source] The solvent can play an important role in S N 2 reactions, particularly in S N 2 involving secondary alkyl halide substrates, where it can be the determining factor in mechanism. Solvent can also have a great effect on reaction rate of S N 2 reactions. The S N 2 mechanism is preferred when the solvent is an aprotic, polar solvent. That is, a solvent that is polar, but without a polar hydrogen. Polar, protic solvents would include water, alcohols, and generally, solvents with polar NH or OH bonds. Good aprotic, polar solvents are HMPA, CH 3 CN, DMSO, and DMF. A polar solvent is preferred because it better allows the dissociation of the halide from the alkyl group. A protic solvent with a polar hydrogen, however, forms a 'cage' of hydrogen-bonded solvent around the nucleophile, hindering its approach to the substrate. Relative Reactivity of Solvents HMPA > CH 3 CN > DMF > DMSO >> H 2 O S N 1 Reactions [edit \| edit source] The S N 1 mechanism is very different from the S N 2 mechanism. In some of its preferences, its exactly the opposite and, in some cases, the results of the reaction can be significantly different. Like the S N 2 mechanism, structure plays an important role in the S N 1 mechanism. The role of structure in the S N 1 mechanism, however, is quite different and because of this, the reactivity of structures is more or less reversed. Reactivity Due to Structure of S N 1 CH 3 X < RCH 2 X << R 2 CHX < R 3 CX The S N 1 mechanism is preferred for tertiary alkyl halides and, depending on the solvent, may be preferred in secondary alkyl halides. The S N 1 mechanism does not operate on primary alkyl halides or methyl halides. To understand why this is so, let's take a look at how the S N 1 mechanism works. S N 1 nucleophilic substitution of a generic halide with a water molecule to produce an alcohol. At the top of the diagram, the first step is the spontaneous dissociation of the halide from the alkyl halide. Unlike the S N 2 mechanism, where the attacking nucleophile causes the halide to leave, the S N 1 mechanism depends on the ability of the halide to leave on its own. This requires certain conditions. In particular, the stability of the carbocation is crucial to the ability of the halide to leave. Since we know tertiary carbocations are the most stable, they are the best candidates for the S N 1 mechanism. And with appropriate conditions, secondary carbocations will also operate by the S N 1 mechanism. Primary and methyl carbocations however, are not stable enough to allow this mechanism to happen. Once the halide has dissociated, the water acts as a nucleophile to bond to the carbocation. In theS N 2 reactions, there is an inversion caused by the nucleophile attacking from the opposite side while the halide is still bonded to the carbon. In the S N 1 mechanism, since the halide has left, and the bonds off of the α-carbon have become planar, the water molecule is free to attack from either side. This results in, primarily, a racemic mixture. In the final step, one of the hydrogens of the bonded water molecule is attacked by another water molecule, leaving an alcohol. Note: Racemic mixtures imply entirely equal amounts of mixture, however this is rarely the case in S N 1. There is a slight tendency towards attack from the opposite side of the halide. This is the result some steric hinderence from the leaving halide which is sometimes close enough to the leaving side to block the nucleophile's approach from that side. Solvent [edit \| edit source] Like the S N 2 mechanism, the S N 1 is affected by solvent as well. As with structure, however, the reasons differ. In the S N 1 mechanism, a polar, protic solvent is used. The polarity of the solvent is associated with the dielectric constant of the solvent and solutions with high dielectric constants are better able to support separated ions in solution. In S N 2 reactions, we were concerned about polar hydrogen atoms "caging" our nucleophile. This still happens with a polar protic solvent in S N 1 reactions, so why don't we worry about it? You have to keep in mind the mechanism of the reaction. The first step, and more importantly, the rate-limiting step, of the S N 1 reaction, is the ability to create a stable carbocation by getting the halide anion to leave. With a polar protic solvent, just as with a polar aprotic solvent,we're creating a stable cation, however it's the polar hydrogens that stabilize the halide anion and make it better able to leave. Improving the rate-limiting step is always the goal. The "caging" of the nucleophile is unrelated to the rate-limiting step and even in its "caged" state, the second step, the attack of the nucleophile, is so much faster than the first step, that the "caging" can simply be ignored. Summary [edit \| edit source] S N 1, S N 2, E1, and E2, are all reaction mechanisms, not reactions themselves. They are mechanisms used by a number of different reactions. Usually in organic chemistry, the goal is to synthesize a product. In cases where you have possibly competing mechanisms, and this is particularly the case where an S N 1 and an E1 reaction are competing, the dominating mechanism is going to decide what your product is, so knowing the mechanisms and which conditions favor one over the other, will determine your product. In other cases, knowing the mechanism allows you to set up an environment favorable to that mechanism. This can mean the difference between having your product in a few minutes, or sometime around the next ice age. So when you're designing a synthesis for a product, you need to consider, I want to get product Y, so what are my options to get to Y? Once you know your options and you've decided on a reaction, then you need to consider the mechanism of the reaction and ask yourself, how do I create conditions that are going to make this happen correctly and happen quickly? Elimination Reactions [edit \| edit source] Nucleophilic substitution reactions and Elimination reactions share a lot of common characteristics, on top of which, the E1 and S N 1 as well as E2 and S N 2 reactions can sometimes compete and, since their products are different, it's important to understand them both. Without understanding both kinds of mechanisms, it would be difficult to get the product you desire from a reaction. In addition, the S N 1 and S N 2 reactions will be referenced quite a bit by way of comparison and contrast, so it's probably best to read that section first and then continue here. Elimination reactions are the mechanisms for creating alkene products from haloalkane reactants. E1 and E2 elimination, unlike S N 1 and S N 2 substitution, mechanisms do not occur with methyl halides because the reaction creates a double bond between two carbon atoms and methylhalides have only one carbon. Note In the notation E1 and E2, E stands for elimination 1: stands for unimolecular (the concentration of only one kind of molecule determines the rate of the reaction) 2: stands for bimolecular (the concentration of two types of molecules determine the rate of the reaction) E1 vs E2 [edit \| edit source] Reaction rates [edit \| edit source] E1 and E2 are two different pathways to creating alkenes from haloalkanes. As with S N 1 and S N 2 reactions, one of the key differences is in the reaction rate, as it provides great insight into the mechanisms. E1 reactions, like S N 1 reactions are 2-step reactions. Also like S N 1 reactions, the rate-limiting step is the dissociation of the halide from its alkane, making it a first-order reaction, depending on the concentration of the haloalkane, with a reaction rate of: R a t e=k[R X]{\displaystyle Rate=k[RX]} On the other hand, E2 reactions, like S N 2 reactions are 1-step reactions. And again, as with S N 2 reactions, the rate limiting step is the ability of a nucleophile to attach to the alkane and displace the halide. Thus it is a second-order reaction that depends on the concentrations of both the nucleophile and haloalkane, with a reaction rate of: R a t e=k[N u:][R X]{\displaystyle Rate=k[Nu:][RX]} (where Nu: is the attacking nucleophile) Zaitsev's Rule [edit \| edit source] Zaitsev's rule (sometimes spelled "Saytzeff") states that in an elimination reaction, when multiple products are possible, the most stable alkene is the major product. That is to say, the most highly substituted alkene (the alkene with the most non-hydrogen substituents) is the major product. Both E1 and E2 reactions produce a mixture of products, when possible, but generally follow Zaitsev's rule. We'll see below why E1 reactions follow Zaitsev's rule more reliably and tend to produce a purer product. Dehydrohalogenation reaction of (S)-2-bromo-3-methylbutane The above image represents two possible pathways for the dehydrohalogenation of (S)-2-bromo-3-methylbutane. The two potential products are 2-methylbut-2-ene and 3-methylbut-1-ene. The images on the right are simplified drawings of the molecular product shown in the images in the center. As you can see on the left, the bromine is on the second carbon and in an E1 or E2 reaction, the hydrogen could be removed from either the 1st or the 3rd carbon. Zaitsev's rule says that the hydrogen will be removed predominantly from the 3rd carbon. In reality, there will be a mixture, but most of the product will be 2-methylbut-2-ene by the E1 mechanism. By the E2 reaction, as we'll see later, this might not necessarily be the case. E2 [edit \| edit source] Reactivity Due to Structure of E2 RCH 2 X > R 2 CHX >> R 3 CX The E2 mechanism is concerted and highly stereospecific, because it can occur only when the H and the leaving group X are in an anti-coplanar position. That is, in a Newman projection, the H and X must be 180°, or in the anti-configuration. This behaviour stems from the best overlap of the 2 p orbitals of the adjacent carbons when the pi bond has to be formed. If the H and the leaving group cannot be brought into this position due to the structure of the molecule, the E2 mechanism will not take place. Mechanism of E2 elimination. Note the anti-coplanarity of the X-C-C-H atoms Therefore, only molecules having accessible H-X anti-coplanar conformations can react via this route. Furthermore, the E2 mechanism will operate contrary to Zaitsev's rule if the only anti-coplanar hydrogen from the leaving group results in the least stable alkene. A good example of how this can happen is by looking at how cyclohexane and cyclohexene derivatives might operate in E2 conditions. E2 with preferential elimination Let's look at the example above. The reactant we're using is 1-chloro-2-isopropylcyclohexane. The drawing at the top left is one conformation and the drawing below is after a ring flip. In the center are Newman projections of both conformations and the drawings on the right, the products. If we assume we're treating the 1-chloro-2-isopropylcyclohexane with a strong base, for example CH 3 CH 2 O- (ethanolate), the mechanism that dominates is E2. There are 3 hydrogens off of the carbons adjacent to our chlorinated carbon. The red and the green ones are two of them. The third would be hard to show but is attached to the same carbon as the red hydrogen, angled a little down from the plane and towards the viewer. The red hydrogen is the only hydrogen that's 180° from the chlorine atom, so it's the only one eligible for the E2 mechanism. Because of this, the product is going to be only 3-isopropylcylcohex-1-ene. Notice how this is contrary to Zaitsev's rule which says the most substituted alkene is preferred. By his rule, 1-isopropylcyclohexene should be our primary product, as that would leave the most substituted alkene. However it simply can't be produced because of the steric hindrance. The images below shows the molecule after a ring flip. In this conformation, no product is possible. As you can see from the Newman projection, there are no hydrogens 180° from the chlorine atom. So it's important, when considering the E2 mechanism, to understand the geometry of the molecule. Sometimes the geometry can be used to your advantage to preferentially get a single product. Other times it will prevent you from getting the product you want, and you'll need to consider a different mechanism to get your product. Note: Often the word periplanar is used instead of coplanar. Coplanar implies precisely 180 degree separation and "peri-", from Greek for "near", implies near 180 degrees. Periplanar may actually be more accurate. In the case of the 1-chloro-3-isopropylcyclohexane example, because of molecular forces, the chlorine atom is actually slightly less than 180 degrees from both the hydrogen and the isopropyl group, so in this case, periplanar might be a more correct term. E1 [edit \| edit source] E1 elimination of an alkyl halide by a base The E1 mechanism begins with the dissociation of the leaving group from an alkyl, producing a carbocation on the alkyl group and a leaving anion. This is the same way the S N 1 reaction begins, so the same thing that helps initiate that step in S N 1 reactions, help initiate the step in E1 reactions. More specifically, secondary and tertiary carbocations are preferred because they're more stable than primary carbocations. The choice of solvent is the same as S N 1 as well; a polar protic solvent is preferred because the polar aspect stabilizes the carbocation and the protic aspect stabilizes the anion. What makes the difference between whether the reaction takes the S N 1 or E1 pathway then, must depend on the second step; the action of the nucleophile. In S N 1 reactions, a strong nucleophile that's a weak base is preferred. The nucleophile will then attack and bond to the carbocation. In E1 reactions, a strong nucleophile is still preferred. The difference is that a strong nucleophile that's also a strong base, causes the nucleophile to attack the hydrogen at the β-carbon instead of the α-carbocation. The nucleophile/base then extracts the hydrogen causing the bonding electrons to fall in and produce a pi bond with the carbocation. Because the hydrogen and the leaving group are lost in two separate steps and the fact that it has no requirements as to geometry, the E1 mechanism more reliably produces products that follow Zaitsev's rule. References [edit \| edit source] ↑IIT Chemistry by Dr.O.P.Agrawal and Avinash Agrawal ↑ abOrganic Chemistry, John McMurry << Haloalkanes \|Alkenes\| Alkynes >> Retrieved from " Category: Book:Organic Chemistry This page was last edited on 25 March 2023, at 11:24. Text is available under the Creative Commons Attribution-ShareAlike License; additional terms may apply. By using this site, you agree to the Terms of Use and Privacy Policy. Privacy policy About Wikibooks Disclaimers Code of Conduct Developers Statistics Cookie statement Mobile view Search Search [x] Toggle the table of contents Organic Chemistry/Alkenes 3 languagesAdd topic
15939	https://www.sciencedirect.com/science/article/pii/B9780408012522500960	THE CRYOGENIC WIND TUNNEL FOR HIGH REYNOLDS NUMBER TESTING - ScienceDirect Typesetting math: 100% Skip to main contentSkip to article Journals & Books Access throughyour organization Purchase PDF Search ScienceDirect Article preview Abstract Cited by (2) Proceedings of the Ninth International Cryogenic Engineering Conference, Kobe, Japan, 11–14 May 1982 1982, Pages 389-394 THE CRYOGENIC WIND TUNNEL FOR HIGH REYNOLDS NUMBER TESTING Author links open overlay panel Robert A.Kilgore Show more Outline Add to Mendeley Share Cite rights and content An improved way to increase the Reynolds numbers capability of wind tunnels has been developed at the Langley Research Center. Cooling the test gas to cryogenic temperatures by spraying liquid nitrogen into the tunnel circuit increases Reynolds number with no increase in dynamic pressure and a reduction in drive power. In addition, the ability to vary the temperature of the test gas independently of pressure and Mach number allows for the first time the independent determination of Reynolds number, Mach number, and aeroelastic effects. A new fan-driven transonic cryogenic tunnel being built at the Langley Research Center will provide an order of magnitude increase in Reynolds number capability over existing transonic tunnels in the United States when it is completed later this year. Access through your organization Check access to the full text by signing in through your organization. Access through your organization Recommended articles References (0) Cited by (2) Research on Reynolds number effects and influencing factors of the wide-body transport aircraft standard model CHN-T2 2025, Aerospace Science and Technology Show abstract Current research on Reynolds number effects for large transport aircraft is often limited by conventional wind tunnels, leading to unreliable aerodynamic data. Low-temperature wind tunnel technology helps bridge the gap between test and flight Reynolds numbers by reducing gas temperatures. However, factors such as incoming turbulence and wall temperature variations can also significantly impact aerodynamic characteristics. This study combines wind tunnel tests and numerical simulations (γ−R e θ t−C F transition model)to investigate the effects of Reynolds number, wall temperature, and turbulence intensity on the aerodynamic characteristics, shock waves, transition, and boundary layer behavior of the wide-body transport aircraft standard model CHN-T2. Results show that higher Reynolds numbers lead to thinner boundary layers, downstream shock movement, reduced separation bubble size, earlier transition, and more stable aerodynamic characteristics. Higher wall temperatures lower lift and drag coefficients, thicken the boundary layer, delay transition, and shift shocks upstream. Increased turbulence intensity reduces the transition region, increases turbulent kinetic energy, and raises viscous drag by triggering earlier transition. This research provides a theoretical basis for wide-body aircraft aerodynamic design, clarifies the mechanisms of these factors on aerodynamic characteristics, and is significant for optimizing wind tunnel tests and improving prediction accuracy. ### The correlation between the running pressure ratio and state parameters in a cryogenic transonic wind tunnel 2021, Kongqi Donglixue Xuebao Acta Aerodynamica Sinica View full text Copyright © 1982 Butterworth & Co (Publishers) Ltd. Published by Elsevier Ltd. All rights reserved. Recommended articles Application of Box-Behnken design with RSM to predict the heat transfer performance of thermo-magnetic convection of hybrid nanofluid inside a novel oval-shaped annulus enclosure Case Studies in Thermal Engineering, Volume 61, 2024, Article 105010 Lotfi Ben Said, …, Mesfer Ahmad Mesfer Alwadai ### Fluid flow and heat transfer in microchannel heat sinks: Modelling review and recent progress Thermal Science and Engineering Progress, Volume 29, 2022, Article 101203 Jie Gao, …, Hongwei Wu About ScienceDirect Remote access Contact and support Terms and conditions Privacy policy Cookies are used by this site.Cookie settings All content on this site: Copyright © 2025 Elsevier B.V., 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. We use cookies that are necessary to make our site work. 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15940	http://www.jonblakely.com/wp-content/uploads/9_8.pdf	9.8 Graphing Rational Functions Lets begin with a definition. Definition: Rational Function A rational function is a function of the form    x Q x P x f  where P and Q are polynomials. An example of a simple rational function that we have seen before is  x x f 1  . As we can see, rational functions have limitations on their domains. The denominator can not be zero. So what happens when the denominator is zero? Lets investigate. We will consider our function  x x f 1  . Lets see what happens as we get closer and closer to zero from the positive and negative sides of zero. Here is a table of values. x f(x) 1 1 0.1 10 0.01 100 0.001 1000 0.0001 10,000 x f(x) -1 -1 -0.1 -10 -0.01 -100 -0.001 -1000 -0.0001 -10,000 As we can see, the functional values continue to increase as we get closer to zero from the positive side and they continue to become large negative from the negative side. We say that as x tends to zero, f(x) tends to infinity or negative infinity, respectively. This gives us the graph of When this situation happens (on one side or the other, or both) we have what is called a vertical asymptote. Property for Vertical Asymptotes A rational function, which has no common factors in the numerator and denominator, has vertical asymptotes at all points where the denominator is zero. So in order to find the vertical asymptotes we simply need to make sure there are no common factors and then find when the denominator is zero. These values would be the vertical asymptotes. The next thing we notice about our function  x x f 1  is that as the values of x increase, the values of f(x) get closer to zero. And as the values become increasingly large negatives the values of f(x) get closer to zero as well. This idea is one of a horizontal asymptote. A horizontal asymptote is any value that a function gets close to as the values of x get increasingly large in the positive or negative direction. We use the following property to determine the horizontal asymptote. The proof of this property is reserved for precalculus. Property for Horizontal Asymptotes Let    x Q x P x f  where the leading term of P is n ax and the leading term of Q is m bx . Then a. If n < m, then the horizontal asymptote is at y = 0. b. If n > m, then the function has no horizontal asymptote. c. If n = m, then the horizontal asymptote is at b a y  . So we need only to compare the leading terms of the numerator and denominator to determine the horizontal asymptotes. Example 1: Determine all the asymptotes of  6 5 2 5 3 2 2      x x x x x f . Solution: First we need to factor the numerator and denominator to see if we can cancel any common factors.        1 6 1 2 3 6 5 2 5 3 2 2           x x x x x x x x x f Now we will start with the vertical asymptotes. The vertical asymptotes occur when the denominator is zero. So set the denominator equal to zero and solve. We get    1 6 0 1 0 6 0 1 6 0 6 5 2              x x x x x x x x So the vertical asymptotes are at 6  x and 1   x . Now the horizontal asymptotes are found by using the property above. Since the degree of the numerator and denominator are the same, the horizontal asymptote occurs at the ratio of the leading coefficients. We get 3 1 3   y . Next we need to discuss finding the intercepts of a rational function. Recall, we find the y-intercept of an equation by letting x=0 and solve for y. We do the same thing in a rational function. Also, recall, we find the x-intercepts of an equation by letting y=0 and solve for x. In a rational function however, this is the same as setting the numerator equal to zero and solving. The reason for this is that a fraction can only be zero when its numerator is zero. Example 2: Find the intercepts of  6 5 2 5 3 2 2      x x x x x f . Solution: We will start with the y-intercept. So we let x=0 and get     3 1 6 2 6 0 5 0 2 0 5 0 3 2 2          y For the x-intercepts we set the numerator equal to zero and solve.    1 0 1 0 2 3 0 1 2 3 0 2 5 3 3 2 2             x x x x x x x x The last thing we need in order to graph a rational function is a sign chart. A sign chart is a chart used to find where an equation is entirely positive and negative. A sign chart tells us if the graph of a function is above the x-axis (positive) or below the x-axis (negative). Clearly, there are only a few ways in which a function can change from positive to negative. One way is to actually cross the x-axis, becoming an x –intercept, and the other way is if there is an asymptote. Once we have a sign chart we can use this information with our intercepts and asymptotes to generate the graph of a rational function. Here are the steps involved in generating a sign chart. Creating a Sign Chart for a Rational function 1. Find and plot on a number line, all x-intercepts and vertical asymptotes. Use an open circle for the vertical asymptotes and a closed circle for the x-intercepts. 2. Use the number line from step 1 to generate your intervals to be tested. We generally go from one asymptote or intercept to another to generate these intervals. 3. Test any value in the intervals from step 2. The sign of this value will characterize the sign of the entire interval. 4. Complete the sign chart by placing the sign of each interval directly above the interval on the graphed number line. Example 3: Construct the sign chart for  6 5 2 5 3 2 2      x x x x x f . Solution: From examples 1 and 2 we know the vertical asymptotes and x-intercepts for  6 5 2 5 3 2 2      x x x x x f . So we can put them on a number line as follows. Note: We need to use open circle on the asymptotes since those values are not in the domain and thus are not defined, and filled circles on the x-intercepts since that is the spots at which the graph would actually be touching the x-axis. So now we go from asymptote or intercept to asymptote to intercept to create our intervals. This would give us the following intervals:            , 6 , 6 , 1 , 1 , , , 1 , 1 , 3 2 3 2 We now test a value out of each interval. Since we are only concerned about the sign of the function in that interval, we only need to find if the test value makes the interval positive or negative. For the first interval   1 ,   , lets test –2.                           8 24 6 2 5 2 2 2 5 2 3 2 2 2 f So the first interval is positive. For the next interval   3 2 , 1  we can test 0. -1 3 2 1 6 O O                    6 2 6 0 5 0 2 0 5 0 3 2 2 2 f In a similar fashion we find that   1 , 3 2 is +,   6 , 1 is – and    , 6 is +. Summarizing this on the sign chart we get Now all we have to do is take all of the information we have found and put it all on a graph. Example 4: Graph  6 5 2 5 3 2 2      x x x x x f . Solution: From example 1 we see that we have vertical asymptotes of 6  x , 1   x and horizontal asymptote of 3  y . We start by graphing these on the rectangular grid with dashed lines. Then we plot the intercepts of    0 , 1 , 0 , 3 2 and   3 1 , 0  . Finally we use the sign chart and the behavior around the asymptotes to finish the graph. Plotting points can see the fact that the graph dips below the horizontal asymptote before getting close to it. Lets now put all of this together in our last example to graph a rational function. Example 5: -1 3 2 1 6 O O + - + - + Graph  6 1 2     x x x x f . Solution: First we should put the function in factored form.     2 3 1 6 1 2         x x x x x x x f From this we can see that the x-intercept is 1  x and the y-intercept is 6 1  y . The vertical asymptotes occur at 2   x and 3  x . Also, since the numerator has a smaller degree than the denominator, we have that the horizontal asymptote is 0  y . Now using the x-intercepts and vertical asymptotes we can construct our sign chart. So our intervals to test are           , 3 , 3 , 1 , 1 , 2 , 2 , . Testing a point in each interval will give us the following sign chart So putting it all together we get the graph. 9.8 Exercises Find all vertical and horizontal asymptotes of the following rational functions. 1.  1 1   x x f 2.  3 2   x x f 3.  1 3 2   x x x f 4.  1 2 1    x x x f 5.     2 1 5     x x x x f 6.    5 2   x x x f 7.  8 2 7 2 2     x x x x f 8.  5 4 4 2 2     x x x x f 9.  12 5 2 12 3 2 3     x x x x x f -2 1 3 O O -2 1 3 O O - + - + 10.  6 5 6 4 6 2 2     x x x x f 11.  x x x x x x f 24 24 6 1 4 2 2 3 2 4      12.  x x x x x f 16 22 3 25 2 3 2     13.  1 4  x x x f 14.  8 2 3 5  x x x f 15.  x x x x f    2 3 1 Find all x- and y-intercepts of the following rational functions. 16.  1 1   x x f 17.  3 2   x x f 18.  1 3 2   x x x f 19.  1 2 1    x x x f 20.     2 1 5     x x x x f 21.    5 1    x x x x f 22.     8 2 3 7 2 2      x x x x x f 23.  5 4 4 2 2     x x x x f 24.  12 2 2 27 3 2 3     x x x x x f 25.  6 5 6 4 2 3     x x x x x f 26.  x x x x x x f 24 24 6 9 10 2 3 2 4      27.  x x x x x x f 16 22 3 100 29 2 3 2 4      28.  x x x f 1 4   29.  8 40 16 2 3 3 4 5     x x x x x f 30.  x x x x f 2 1 2 6    Graph the following rational functions. Find all asymptotes and intercepts. 31.  1 1  x x f 32.  2 2  x x f 33.  1  x x x f 34.  2 2  x x x f 35.  2 1   x x x f 36.  1 2   x x x f 37.  4 3   x x x f 38.  x x x f   2 2 39.     1 2 1     x x x x f 40.     3 1    x x x x f 41.     1 1 2 2    x x x x f 42.     1 2 2     x x x x f 43.        3 2 2 1      x x x x x f 44.       2 2 1     x x x x x f 45.    2 2   x x x f 46.    2 2 1   x x x f 47.  3 2 2    x x x x f 48.  1 2 2 2    x x x x f 49.  3 2 1 2     x x x x f 50.  x x x x f    2 2 1 51.  x x x x x f 3 8 6 2 2     52.  3 4 8 2 2 2      x x x x x f 53.  1 2 3 4 2 2      x x x x x f 54.  3 2 1 2 2 2      x x x x x f 55.  4 3 2 3 2 2 2      x x x x x f 56.  2 3 2 6 2 2      x x x x x f 57.  4 18 2 2 2    x x x f 58.  5 4 4 8 4 2 2      x x x x x f 59.  x x x x f    3 1 60.  x x x x f 4 2 2 3 2    In the section we mentioned that the vertical asymptotes occurred where the denominator is zero, as long as the numerator and denominator have no common factors. In the case where they do have common factor it is a little different. For example     1 1 1     x x x x f clearly have a common factor. If we cancel it out we end up with  1 , 1    x x x f . Since the domain is still all real numbers except 1 the graph of f is just the line y=x-1 but with a hole at x=1 since the function is not defined there. Use this idea to graph the following functions. 61.  1 1 2 2     x x x x f 62.  x x x x f   2 63.  1 2 3 4 4 2     x x x x f 64.  x x x x x f    2 3 65.  6 2 2     x x x x f 66.  5 4 1 2     x x x x f
15941	https://artofproblemsolving.com/wiki/index.php/Combinatorial_identity?srsltid=AfmBOooQoF8JjSjHLRDGkBNEcU6Kn1phY1TWycl_0-9-FZpYuAO0DxgG	Art of Problem Solving Combinatorial identity - 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 Combinatorial identity Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search Combinatorial identity Contents [hide] 1 Pascal's Identity 1.1 Proof 1.2 Alternate Proofs 2 Vandermonde's Identity 2.1 Video Proof 2.2 Combinatorial Proof 2.3 Algebraic proof 3 Hockey-Stick Identity 3.1 Proof 4 Another Identity 4.1 Hat Proof 4.2 Proof 2 5 Even Odd Identity 6 Examples 7 See also Pascal's Identity Pascal's Identity is a very important formula for olympiad math and it states that for any positive integers and . Here, is the binomial coefficient . This result can be interpreted combinatorially as follows: the number of ways to choose things from things is equal to the number of ways to choose things from things added to the number of ways to choose things from things. Proof If then and so the result is trivial. So assume . Then Alternate Proofs Here, we prove this using committee forming. Consider picking one fixed object out of objects. Then, we can choose objects including that one in ways. Because our final group of objects either contains the specified one or doesn't, we can choose the group in ways. But we already know they can be picked in ways, so Also, we can look at Pascal's Triangle to see why this is. If we were to extend Pascal's Triangle to row n, we would see the term . Above that, we would see the terms and . Due to the definition of Pascal's Triangle, . Vandermonde's Identity Vandermonde's Identity states that , which can be proven combinatorially by noting that any combination of objects from a group of objects must have some objects from group and the remaining from group . Video Proof Combinatorial Proof Think of the right hand side as picking people from men and women. Think of the left hand side as picking men from the total men and picking women from the total women. The left hand side and right hand side are the same, thus Vandermonde's identity must be true. ~avn Algebraic proof For all The coefficients of on both sides must be the same, so using the Binomial Theorem, Hockey-Stick Identity For . This identity is known as the hockey-stick identity because, on Pascal's triangle, when the addends represented in the summation and the sum itself is highlighted, a hockey-stick shape is revealed. We can also flip the hockey stick because pascal's triangle is symmetrical. Proof Inductive Proof This identity can be proven by induction on . Base Case Let . . Inductive Step Suppose, for some , . Then . Algebraic Proof It can also be proven algebraically with Pascal's Identity, . Note that , which is equivalent to the desired result. Combinatorial Proof 1 Imagine that we are distributing indistinguishable candies to distinguishable children. By a direct application of Balls and Urns, there are ways to do this. Alternatively, we can first give candies to the oldest child so that we are essentially giving candies to kids and again, with Balls and Urns, , which simplifies to the desired result. Combinatorial Proof 2 We can form a committee of size from a group of people in ways. Now we hand out the numbers to of the people. We can divide this into disjoint cases. In general, in case , , person is on the committee and persons are not on the committee. This can be done in ways. Now we can sum the values of these disjoint cases, getting Algebraic Proof 2 Apply the finite geometric series formula to : Then expand with the Binomial Theorem and simplify: Finally, equate coefficients of on both sides: Since for , , this simplifies to the hockey stick identity. Algebraic Proof 3 Consider the number of solutions to the equation ++++++.......+ ≤ N where each is a non-negative integer for 1≤i≤r. METHOD 1 We know since all numbers on LHS are non-negative therefore 0≤N and N is a integer. Therfore, ++++++.......+ = 0,1,2......N. Consider each case seperately. ++++++.......+ =0 by Stars-and-bars the equation has solutions. ++++++.......+ =1 by Stars-and-bars the equation has solutions. ++++++.......+ =2 by Stars-and-bars the equation has solutions. ........... ++++++.......+ =N by Stars-and-bars the equation has solutions. Hence, the equation has + + +.... = (where k=N+r-1) SOLUTIONS. METHOD 2 Since, ++++++.......+ ≤ N Therefore we may say ++++++.......+ = N -m where m is another non-negative integer. 0 ≤++++++.......+ ⇒ 0≤ N-m ⇒ m≤ N So, we need not count this as an extra restriction. Now, ++++++.......+ +m = N. Again by Stars-and-bars this has solutions. Therefore, the equation has = solutions(As N+r-1 =k). Since, both methods yeild the same answer ⇒ = . Taking r-1= p redirects to the honeystick identity. ~SANSGANKRSNGUPTA Another Identity Hat Proof We have different hats. We split them into two groups, each with k hats: then we choose hats from the first group and hats from the second group. This may be done in ways. Evidently, to generate all possible choices of hats from the hats, we must choose hats from the first and the remaining hats from the second ; the sum over all such is the number of ways of choosing hats from . Therefore , as desired. Proof 2 This is a special case of Vandermonde's identity, in which we set . Even Odd Identity Examples 1986 AIME Problem 11 2000 AIME II Problem 7 2013 AIME I Problem 6 (Solution 2) 2015 AIME I Problem 12 2018 AIME I Problem 10 2020 AIME I Problem 7 2016 AMC 10A Problem 20 2021 AMC 12A Problem 15 1981 IMO Problem 2 2022 AIME I Problem 12 See also Pascal's Triangle Combinatorics Pascal's Identity Retrieved from " Categories: Theorems Combinatorics 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.
15942	https://physics.stackexchange.com/questions/78879/simple-explanation-of-relation-between-speed-of-sound-and-r-m-s-speed	Skip to main content Simple explanation of relation between speed of sound and r.m.s. speed? Ask Question Asked Modified 7 years, 2 months ago Viewed 13k times This question shows research effort; it is useful and clear 5 Save this question. Show activity on this post. In an ideal gas, the speed of sound vs is related to the r.m.s. molecular speed vm by vsvm=γ3−−√, where γ=Cp/Cv=7/5 for a diatomic gas. I understand how to prove this relation from first principles. However, it seems mysterious to me that it pops out like this in the end. I suppose it's inevitable for dimensional reasons that there will be some relation of this form, since if we want to describe a sample of an ideal gas, a sufficient set of unitful parameters is m and kT, and there is a unique way of combining these to give units of velocity. (In a solid or liquid, we have other parameters such as the Young's modulus.) However, this type of dimensional argument doesn't prove that the ratio of the speeds is of order unity. Is there any straightforward physical plausibility argument for the fact that this ratio of speeds is constant, and for the fact that the ratio is of order unity? I guess it's implausible to have vm≫vs, since it seems like then a sound wave would sort of get scrambled and lose its identity because of molecular motion, and this scrambling would take place in much less than a period. thermodynamics acoustics ideal-gas Share CC BY-SA 3.0 Improve this question Follow this question to receive notifications edited Sep 28, 2013 at 17:35 user4552 user4552 asked Sep 28, 2013 at 14:56 user4552user4552 7 2 Should they not be coupled through the mean time between collisions/mean-free-path? At the hand-waving level that alone would satisfy me as far as expecting a order unity coefficient. dmckee --- ex-moderator kitten – dmckee --- ex-moderator kitten 2013-09-28 15:00:44 +00:00 Commented Sep 28, 2013 at 15:00 1 @dmckee: Should they not be coupled through the mean time between collisions/mean-free-path? That's not immediately obvious to me, but if you can expand it into an answer, I'd love to read it. By reducing the pressure while maintaining constant temperature, you can make the mean free path go to infinity while the speed of sound is unchanged. user4552 – user4552 2013-09-28 17:33:45 +00:00 Commented Sep 28, 2013 at 17:33 @user23660: The ratio is constant and depends only on γ. This relation is valid only for ideal gases. user4552 – user4552 2013-09-28 17:34:47 +00:00 Commented Sep 28, 2013 at 17:34 Ben, On thinking a little I'd same I was way off the mark with that comment, and it should have been obvious: the ideal gas is nominally non-interacting. A better explanation for what I thought I was getting at is that pressure is intimately coupled to mean velocity in a gas (even a non-ideal one) because it is the expression of momentum transfer to a impinging object when the gas molecules collided with the surface. Sound being a variation in pressure there must be a relationship. If I get done with my work this weekend I'll work that up into a proper answer. dmckee --- ex-moderator kitten – dmckee --- ex-moderator kitten 2013-09-28 20:24:03 +00:00 Commented Sep 28, 2013 at 20:24 The simplest model that includes both sound wave propagation and molecule speeds would be kinetic theory. Consider Boltzmann equation ∂tf+(v⋅∇)f=St. Density function that corresponds to sound impulse propagating in direction x has the form f(x-v_s t, y, z, v_x, v_y, v_z) (with the only t dependence in the first argument). So, we will then have (−vs+v)∂xf=St. So there, if we assume that rms speed vm is increased α times, vs should also be increased accordingly. user23660 – user23660 2013-09-29 07:56:09 +00:00 Commented Sep 29, 2013 at 7:56 \| Show 2 more comments 1 Answer 1 Reset to default This answer is useful 3 Save this answer. Show activity on this post. Let us recap how we obtain the quantities in question. The easiest way to obtain the speed of sound is from continuous mechanics. Assuming isentropic flow and small perturbations of speed and density and linearizing Euler (or Navier-Stokes) and continuity equations we obtain v2s=(∂p∂ρ)s, where the derivative of pressure is taken at constant entropy. For an ideal gas that has adiabatic equation in the form p/ργ=const that means vs=γpρ−−−√=γkTm−−−−−√, where k is Boltzmann constant, T temperature and m is mass of molecule. The rms speed is obtained from statistical mechanics of gases using Maxwell–Boltzmann distribution: vm=3kTm−−−−−√, here the factor 3 under the square root is consequence of 3-dimensionality of our world. In order to clarify relationship between those two speeds we can consider the theory that includes both the sound wave propagation and statistical distributions of molecular speeds. That would be the kinetic theory and in particular Boltzmann equation: ∂∂tf+(v⋅∇)f=St, where f=f(r⃗ ,v⃗ ,t) is the density function, and St in the rhs (from Stosszahlansatz) is the integral collision operator. If we consider the sound wave propagating along the direction x, then its density function would have the form f(x−vst,vx,vy,vz) (with the only t dependence in the first argument and without dependence on y and z). Substituting this into Boltzmann equation we obtain −vs∂∂xf+vx∂∂xf=St. There, just from the form of equation we could see, that if we assume that rms speed vm is increased α times (by rescaling the density function f′=α−3⋅f(x−v′st,αv⃗ )), vs should also be increased accordingly to maintain the solution. So we can conclude that vs/vm=const. This is, of course, valid if we can assume that the collision term in rhs also transforms accordingly or if we could simply neglect the rhs at a first approximation, which should hold for rarefied gases. Another case is rigid-sphere approximation for the collisions that provides appropriate transformation, so vs/vm should also be independent on temperature there (but, obviously, if the density is large enough in this case it will be dependent on it). Here is random example of paper that considers sound-wave propagation in the framework of Boltzmann equation: R.D.M. Garcia, C.E. Siewert, 'The linearized Boltzmann equation: sound-wave propagation in a rarefied gas', DOI:10.1007/s00033-005-0007-8 (just the first from Google scholar results that has online source). Share CC BY-SA 3.0 Improve this answer Follow this answer to receive notifications answered Sep 30, 2013 at 5:30 user23660user23660 5,91711 gold badge2424 silver badges2626 bronze badges Add a comment \| Featured on Meta Community Asks Sprint Announcement - September 2025 stackoverflow.ai - rebuilt for attribution Linked 0 What is the microscopic equation to calculate sound speed in an ideal gas? 5 Why is the speed of sound wave in a gas always lesser than the r.m.s. speed at the same temperature? 0 Why does the amplitude not increase the speed of the sound wave? -1 How do the cell walls contain contents moving such high speeds? 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15943	https://pmt.physicsandmathstutor.com/download/Physics/A-level/Topic-Qs/Edexcel/02-Mechanics/Set-M/Hooke's%20Law%20(Multiple%20Choice)%20QP.pdf	Use the following information to answer questions 1 and 2. $ VSULQJ REH\V +RRNH¶V ODZ $ IRUFH RI 1 H[WHQGV WKH VSULQJ E\ P 1 A 6.0 N force will extend the spring by A P B 0.30 m C 0.60 m D 0.90 m (Total for Question = 1 mark) 2 7KH HQHUJ\ VWRUHG LQ WKH VSULQJ ZKHQ D IRUFH RI 1 LV DSSOLHG LV A 0.09 J B 0.30 J C 0.60 J D 0.90 J (Total for Question = 1 mark) PhysicsAndMathsTutor.com 3 A spring is suspended from a bar. When a load of 6.0 N is added to the bottom of the spring, its length changes from 0.040 m to 0.13 m. To find the spring constant of the spring you would use A . 6 0 . 0 13m N B . 6 0 0 13 . N m c . 6 0 0.090 N m D 0 090 . 6 0 . m N (total for Question = 1 mark) PhysicsAndMathsTutor.com 4 An apple is at rest on the ground. The diagram shows three forces of equal magnitude. P W = weight of apple P = push of apple on ground R = normal contact force of ground on apple Which row in the table shows Newton’s first and third laws being applied correctly. newton’s first law newton’s third law A P = W R = P B R = P W = R c W = R P = W D W = R R = P (total for Question = 1 mark) W R PhysicsAndMathsTutor.com 5. The graph shows how extension varies with applied force for a spring. 5 The stiffness of the spring in Nm–1 is A 1.5 B 54 C 67 D 150 (Total for Question = 1 mark) Force / N 6 ± 5 ± 4 ± 3 ± 2 ± 1 ± 0 ± ± ± ± ± ± ± ± ± ± ± 0 1 2 3 5 4 6 7 8 9 Extension / cm PhysicsAndMathsTutor.com Questions 6 and 7 refer to the graphs and information below. A force is applied to a spring and the spring extends. The new length of the spring is recorded. This procedure is repeated for different applied forces. Applied force Extension 0 P Applied force Length 0 Q Applied force Extension 0 Applied force Length 0 R S 6 Which of the above graphs could be obtained from this experiment? A P and Q B P and S C R and Q D R and S (Total for Question = 1 mark) 7 The graphs could show that the spring is A obeying Hooke’s law. B extending plastically. C extended beyond the limit of proportionality. D being compressed as well as extended. (Total for Question = 1 mark) PhysicsAndMathsTutor.com 8 A force is applied across the ends of a spring and the following force-extension graph is drawn. Three points, P, Q and R, are marked on the graph. At point Q the applied force is zero. Extension P Force Q R In the table below, the spring is represented using diagrams drawn to scale. The spring at Q is represented by . Select the row from the table that correctly represents the length of the spring at positions P, Q and R. P Q (no applied force) R A B C D (Total for Question = 1 mark) PhysicsAndMathsTutor.com
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Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Points on a Lattice Ask Question Asked 6 years, 6 months ago Modified5 years, 10 months ago Viewed 3k times This question shows research effort; it is useful and clear 2 Save this question. Show activity on this post. I got this question on a coding interview. Hanna moves in a lattice where every point can be represented by a pair of integers. She moves from point A to point B and then takes a turn 90 degrees right and starts moving till she reaches the first point on the lattice. Find what's the point she would reach? In essence the problem boils down to finding the first point where the perpendicular to a line will intersect. Can someone provide pseudo-code or code snippets as to how I can solve this? algorithm data-structures geometry coordinate-systems mathematical-lattices Share Share a link to this question Copy linkCC BY-SA 4.0 Improve this question Follow Follow this question to receive notifications asked Apr 2, 2019 at 20:17 Melissa StewartMelissa Stewart 3,625 15 15 gold badges 54 54 silver badges 91 91 bronze badges Add a comment\| 3 Answers 3 Sorted by: Reset to default This answer is useful 7 Save this answer. Show activity on this post. I'm assuming you mean that she moves in a straight line from A to B and then turns 90 degrees, and that the lattice is a Cartesian grid with the y axis pointing up and the x axis pointing right. Let (dx,dy) = (Bx,By)-(Ax,Ay), the vector from point A to point B. We can rotate this by 90 degrees to give (dy,-dx). After hanna turns right at B, she will head out along that rotated vector toward (Bx+dy,By-dx) Since she is moving in a straight line, her vector from B will follow (t.dy,-t.dx), and will hit another lattice point when both of those components are integers, i.e... She will hit another lattice point at: (Bx + dy/GCD(\|dx\|,\|dy\|), By - dx/GCD(\|dx\|,\|dy\|) ) Share Share a link to this answer Copy linkCC BY-SA 4.0 Improve this answer Follow Follow this answer to receive notifications edited Apr 2, 2019 at 20:59 answered Apr 2, 2019 at 20:23 Matt TimmermansMatt Timmermans 60.7k 3 3 gold badges 55 55 silver badges 105 105 bronze badges 2 Comments Add a comment Melissa Stewart Melissa StewartOver a year ago Can you please explain your reasoning a bit. 2019-04-02T20:30:07.5Z+00:00 0 Reply Copy link Matt Timmermans Matt TimmermansOver a year ago I made it more verbose 2019-04-02T20:35:33.237Z+00:00 2 Reply Copy link This answer is useful 1 Save this answer. Show activity on this post. const findNext = (Ax, Ay, Bx, By) => { // Move A to Origin const Cx = Bx - Ax; const Cy = By - Ay; // Rotate by 90 degree clockwise const Rx = Cy; const Ry = -Cx; // Normalize const norm = gcd(Math.abs(Rx), Math.abs(Ry)); const Nx = Rx / norm; const Ny = Ry / norm; return [Bx + Nx, By + Ny]; }; Here is gcd, ``` var gcd = function(a, b) { if (!b) { return a; } return gcd(b, a % b); } ``` Output: cons result = findNext(1,1,2,2); console.log(result); // [3, 1] Share Share a link to this answer Copy linkCC BY-SA 4.0 Improve this answer Follow Follow this answer to receive notifications answered Nov 18, 2019 at 19:52 Salmanul Faris KSalmanul Faris K 11 1 1 bronze badge Comments Add a comment This answer is useful 0 Save this answer. Show activity on this post. ```python A' . \| . \| . \| . . . A . \| . \| \| \| When you rotate clockwise a point A 90 degrees from origin, you get A' => f(x,y) = (-y, x) \| A \| . \| B \| . \| . ----------O------------- \| \| \| Based on a point A from origin, you can find point B by: By/Ay = Bx/Ax => By = Bx Ay/Ax \| A \| . \| . \| . \| ----------B-------------- . \| . \| C \| To make things easier, we can move the points to get point B on origin. After Hanna rotate 90 degrees to the right on point B, she will find eventually point C. Lets say that D is a point in a rect that is on B-C. Hanna will stop on a point on the lattice when the point (x,y) is integer So, from B we need to iterate Dx until we find a Dy integer def rotate_90_clockwise(A): return (-A, A) def find_B_y(A, x): return x A/A if A else A def find_next(A, B): # make B the origin Ao = (A - B, A - B) Bo = (0,0) # rotate Ao 90 clockwise C = rotate_90_clockwise(Ao) if C == 0: # C is on y axis x = 0 # Dy is one unit from origin y = C/abs(C) else: found = False # from origin x = 0 while not found: # inc x one unit x += C/abs(C) y = find_B_y(C, x) # check if y is integer found = y == round(y) # move back from origin x, y = (x + B, y + B) return x, y A = (-2, 3) B = (3, 2) D = find_next(A, B) print(D) B = (-4, 2) A = (-2, 2) D = find_next(A, B) print(D) B = (1, 20) A = (1, 5) D = find_next(A, B) print(D) ``` Output: python (2.0, -3.0) (-4, 3.0) (2.0, 20.0) Share Share a link to this answer Copy linkCC BY-SA 4.0 Improve this answer Follow Follow this answer to receive notifications edited Nov 3, 2019 at 20:56 answered Nov 3, 2019 at 20:51 RonivaldoRonivaldo 1 1 1 bronze badge Comments Add a comment Your Answer Thanks for contributing an answer to Stack Overflow! 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15945	https://zhuanlan.zhihu.com/p/553176988	关乎手机辐射安全的SAR测试是什么？ - 知乎关注推荐热榜专栏圈子 New付费咨询知学堂直答切换模式登录/注册关乎手机辐射安全的SAR测试是什么？切换模式关乎手机辐射安全的SAR测试是什么？李民Owen 国际国内外检测认证服务 1)什么是SAR测试？ SAR的英文全称为Specific Absorption Rate，中文一般称为电磁波吸收比值或比吸收率，是手机等无线产品的电磁波能量吸收比值。由于在外电磁场的作用下，人体将产生感应电磁场，而人体各种器官均为有耗介质，因此体内电磁场将会产生感应电流，导致吸收和耗散电磁能量，生物计量学中常用SAR来表征这一物理过程。通俗来讲，SAR是指单位时间内单位质量的人体组织所吸收或消耗的电磁波功率，单位为W/kg，或等效地，mW / g。 2)哪些产品需要进行SAR测试？电子产品的发射天线距离人体20cm以内，且无线发射功率大于SAR豁免功率限值需要进行SAR测试，常见的产品除手机外，还包括对讲机、平板电脑、笔记本电脑、USB Dongle、POS机、无绳电话等等。但是对于某些产品，例如医院用于病房内挂墙上使用的显示屏、电子展示屏等，虽然主体是平板电脑，且产品有无线发射功能，但产品的使用环境是固定在某一地点（属于固定式设备），使用时，发射天线不贴近人体，因此不需要做SAR测试。 3)SAR测试是如何进行的？哪些参数/因素可能影响SAR值的评估？测试是由人体模型、测量仪器、探针对及机械臂组成SAR测量系统，系统置于环境受控的测试室中，其中人体模型的内部是液态物质（模拟人体组织液），液体的电磁特性与人体组织的电磁特性一致，探针可在其内自由移动进行测试，最后通过公式计算出SAR值。受测无线产品的射频发射功率，发射天线的位置，测试环境的温度，模拟人体组织液的成分配比，探头的精度校准参数等等因素都可能影响SAR的测试结果。 4)世界各国对SAR测试有着怎样的管控规定？国内SAR测试标准为YD/T 1644《手持和身体佩戴使用的无线通信设备对人体的电磁照射人体模型，仪器和规程第一部分，靠近耳边使用的手持式无线通信设备的SAR评估规程（频率范围300MHz-3GHz）》。该标准对测量系统中的人体模型、测量仪器、探针、人体组织液、机械臂以及测量计算方法作了明确的描述和规定。限值要求：在进网测试中要求SAR限值取10g平均值，限值为2.0W/kg。 5）常规手机的SAR值是多少？和4G相比，5G设备的SAR值会有怎样的变化？下图是随机选择市面上10款5G手机进行4G频段和5G频段的SAR对比测试结果：（图源来自网络）从图中可以看出，5G手机的SAR测试值并不一定会比4G手机的SAR值大。被测手机在4G频段和5G的Sub-6GHz频段的SAR值相当;从发射功率来看,他们在相同的Class等级下允许的最大发射功率相同,但是可以根据需要、在容差范围内进行调整,也就导致了每款手机的实际功率在大部分情况下都是有区别的。一般在其他情况相同的条件下,功率的变化会导致SAR值不同。而在相同发射功率下,SAR值的大小ー般取决于天线电路的设计及相关技术和新材料的运用。也即是，5G技术相较于4G技术并不会引起手机SAR值的改变，无论是4G手机还是5G手机只要符合了本地法规的强制要求，它们都是安全的。（注：仅仅不同手机头部SAR值对比并不能说明手机辐射的大小。因为标准对测试位置有严格定义，手机摆放位置是固定的，如果天线在手机顶部则测出来头部SAR值大，天线在底部则测出来的头部SAR值小）。 6）对于上市产品来说，除了SAR测试，无线终端产品还需要通过哪些测试？无线产品销售的国家不同，需要满足的检测认证需求也是不一样的：国内销售：还需要做SRRC（国家无线电型号核准）、CTA（入网许可证）、CCC（中国强制性产品认证）、RoHs认证（关于限制在电子电气设备中使用某些有害成分的指令）等；欧盟成员国销售：还需要满足CE-RED指令、RoHs、REACH、WEE、CB、德国GS等法规管控，并通过相应的测试并取得证书；美国销售：还需要做FCC ID、UL、FDA等认证并取得证书，其中FCC ID认证必须提供当地代理商信息；加拿大销售：还需要做IC、CSA等认证并取得证书；日本销售：还需要做MIC、PSE等认证并取得证书；其他国家、地区无线认证包括：RCM（澳洲、新西兰）、NBTC（泰国）、ANATAL（巴西）、IMDA（新加坡）、ICASA（南非）、TRA（阿拉伯联合酋长国）、CITC（沙特阿拉伯）、WPC（印度）、NOM（墨西哥）、EAC/FAC（俄罗斯）… 发布于 2022-08-12 12:14 辐射（物理名词）手机辐射赞同添加评论分享喜欢收藏申请转载写下你的评论... 还没有评论，发表第一个评论吧关于作者李民Owen 国际国内外检测认证服务回答 840文章 826关注者 112 关注他发私信推荐阅读各国对SAR测试的限值和标准 ============== CTC华商检测SPI是什么？SPI检测是什么意思？SPI检测设备的作用是什么？ ================================ SPI是什么？SPI检测是什么意思？SPI检测设备的作用是什么？SPI是【Solder Paste Inspection】的简称，中文叫【锡膏检查】SPI(Solder Paste Inspection)到底有何用处？又可以帮我们做到检测… smt贴片...发表于smt行业...ict测试仪与fct测试原理,区别是什么? ===================== ict测试仪与fct测试原理,区别是什么? ICT测试仪是PCBA进行相对简单的模拟，主要用来检查元器件故障和焊接故障的，在板子焊接的下个流程进行，有问题的板子（比如器件焊反、短路等问题）直接… 曲淬 SAR测试是什么？ ========= 广电计量发表于检验检测想来知乎工作？请发送邮件到 jobs@zhihu.com 打开知乎App 在「我的页」右上角打开扫一扫其他扫码方式：微信下载知乎App 无障碍模式验证码登录密码登录开通机构号中国 +86 获取短信验证码获取语音验证码登录/注册其他方式登录未注册手机验证后自动登录，注册即代表同意《知乎协议》《隐私保护指引》扫码下载知乎 App 关闭二维码打开知乎App 在「我的页」右上角打开扫一扫其他扫码方式：微信下载知乎App 无障碍模式验证码登录密码登录开通机构号中国 +86 获取短信验证码获取语音验证码登录/注册其他方式登录未注册手机验证后自动登录，注册即代表同意《知乎协议》《隐私保护指引》扫码下载知乎 App 关闭二维码
15946	https://www.nagwa.com/en/videos/462171298740/	Question Video: Solving Cubic Equations over the Set of Real Numbers \| Nagwa Question Video: Solving Cubic Equations over the Set of Real Numbers \| 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: Solving Cubic Equations over the Set of Real Numbers Mathematics • Second Year of Preparatory School Given that 𝑥 ∈ ℝ and −𝑥/10 = 100/𝑥², determine the value of 𝑥. Pause Play % buffered 00:00 00:00 0:00 Unmute Mute Disable captions Enable captions Settings Captions English Quality 480p Speed Normal Captions Go back to previous menu Disabled English EN Quality Go back to previous menu 1080p HD 720p HD 480p SD 360p 240p Speed Go back to previous menu 0.5×0.75×Normal 1.25×1.5×1.75×2× Exit fullscreen Enter fullscreen Play 02:35 Video Transcript Given that 𝑥 is contained in all reals and negative 𝑥 over 10 is equal to 100 over 𝑥 squared, determine the value of 𝑥. The first thing I’m gonna do is just copy down this equation. To solve for 𝑥, I need to get both of my 𝑥s on the same side of the equation. And I’ll need them to be out of the denominator. To fix this, we can multiply the right side of the equation by 𝑥 squared over one. And if we multiply the right side of the equation by 𝑥 squared over one, we have to multiply the left side of the equation by 𝑥 squared over one. On the left, we’ll have 𝑥 squared times negative 𝑥 over 10. On the right, the two 𝑥 squareds cancel out, leaving us with 100. I can multiply 𝑥 squared times negative 𝑥. It equals negative 𝑥 cubed. And we’ll copy everything else down. Again, we’re trying to isolate 𝑥. We need to get the number 10 out of the denominator and onto the other side of the equation. So I multiply by 10 over one on both sides. On the left, we’re left with negative 𝑥 cubed equals 1000. I noticed that I have a negative 𝑥-value. So I want to multiply both sides of this equation by negative one. After I do that, I have an equation that says 𝑥 cubed equals negative 1000. After that, we need to take the cube root of both sides of the equation. The cube root of 𝑥 cubed equals 𝑥. And the cube root of negative 1000 equals negative 10. Let’s check and see if this makes sense in the context of our original equation. Negative negative 10 over 10, which is 10 over 10, is equal to 100 over negative 10 squared. Negative 10 squared is 100. 10 over 10 is equal to 100 over 100. So it seems we have a valid 𝑥-value. The value for 𝑥 that makes this statement true is negative 10. Lesson Menu Lesson Lesson Plan 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
15947	https://study.com/skill/learn/how-to-identify-opposite-adjacent-hypotenuse-sides-from-a-diagram-explanation.html	How to Identify Opposite, Adjacent & Hypotenuse Sides from a Diagram \| Trigonometry \| 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 Copyright How to Identify Opposite, Adjacent & Hypotenuse Sides from a Diagram Trigonometry Skills Practice Click for sound 4:36 You must c C reate an account to continue watching Register to access this and thousands of other videos Are you a student or a teacher? I am a student I am a teacher Try Study.com, risk-free As a member, you'll also get unlimited access to over 88,000 lessons in math, English, science, history, and more. Plus, get practice tests, quizzes, and personalized coaching to help you succeed. Get unlimited access to over 88,000 lessons. Try it risk-free It only takes a few minutes to setup and you can cancel any time. It only takes a few minutes. Cancel any time. Already registered? Log in here for access Back What teachers are saying about Study.com Try it risk-free for 30 days Already registered? Log in here for access 00:05 Introduction and… 00:36 First example on how… 01:53 Second example on how… 03:00 Third example on how… Jump to a specific example Speed Normal 0.5x Normal 1.25x 1.5x 1.75x 2x Speed Nidhi Agarwal, Kimberly Jones Instructors Nidhi Agarwal Nidhi holds a Bachelor's degree in Secondary Education with a teaching major Biological Sciences and Master's degree in education from Lucknow University and has taught middle and high school math. I have over 15 years of experience working with students of different ages including two years of teaching experience internationally. View bio Kimberly Jones Kimberly Jones has taught middle and high school mathematics for over 15 years and a Math Curriculum Specialist for 11 years. Kimberly has a Bachelor of Science in Mathematics from the University of Arkansas. View bio Example SolutionsPractice Questions Identifying Opposite, Adjacent & Hypotenuse Sides from a Diagram Step 1: Take a look at the given diagram and identify the right angle. The side opposite to the 90∘ angle. That side will be the hypotenuse. Step 2: Identify the angle in the diagram described by the given problem. Identify the side opposite to the angle. That will be the opposite side. Step 3: Identify the side adjacent to the angle other than the hypotenuse. That will be the adjacent side. Identifying Opposite, Adjacent & Hypotenuse Sides from a Diagram Vocabulary Opposite, adjacent, and hypotenuse sides: Opposite, adjacent, and hypotenuse sides refer to the sides of a right angle triangle. Hypotenuse is the side that is opposite to the right angle. it is also the longest side in the triangle. Adjacent and opposite sides are in respect to the angle asked and can vary accordingly. Opposite side is the side that is opposite to the angle asked for and the adjacent side is the side that is right by the angle that is asked for. Let's take a look at two diagrams of right angle triangles and identify the opposite, adjacent and hypotenuse sides from the diagram with the help of the steps explained above. Identifying Opposite, Adjacent & Hypotenuse Sides from a Diagram: Example 1 Identify the side opposite to ∠D, the side adjacent to ∠D, and the hypotenuse of right triangle △D C E in the given diagram. Step 1: Take a look at the given diagram and identify the right angle. The side opposite to the 90∘ angle. That side will be the hypotenuse. The right angle in the given triangle is ∠D E C. The side opposite the right angle ∠D E C is D C. Thus, D C is the hypotenuse. Step 2: Identify the angle in whose respect the opposite and adjacent sides are being asked for. The side opposite to that angle will be the opposite side. In the given problem the opposite and the adjacent sides are being asked in reference to ∠D. The side opposite of ∠D is C E. Thus, C E is the opposite side. Step 3: Identify the side adjacent to the given angle other than the hypotenuse. That will be the adjacent side. The side adjacent to ∠D other than the hypotenuse is D E. Thus, D E is the adjacent side. Identifying Opposite, Adjacent & Hypotenuse Sides from a Diagram: Example 2 Identify the side opposite to ∠F, the side adjacent to ∠F, and the hypotenuse of right triangle △F G H in the given diagram. Step 1: Take a look at the given diagram and identify the right angle. The side opposite to the 90∘ angle. That side will be the hypotenuse. The side opposite the right angle ∠F G H is F H. Thus, F H is the hypotenuse. Step 2: Identify the angle in whose respect the opposite and adjacent sides are being asked for. The side opposite to that angle will be the opposite side. The side opposite of ∠F is G H. Thus, G H is the opposite side. Step 3: Identify the side adjacent to the given angle other than the hypotenuse. That will be the adjacent side. The side adjacent to ∠F other than the hypotenuse is F G. Thus, F G is the adjacent side. Get access to thousands of practice questions and explanations! Create an account Table of Contents Identifying Opposite, Adjacent & Hypotenuse Sides from a Diagram Identifying Opposite, Adjacent & Hypotenuse Sides from a Diagram Vocabulary Example 1 Example 2 Test your current knowledge Practice Identifying Opposite, Adjacent & Hypotenuse sides from a Diagram Related Courses Math 102: College Mathematics High School Trigonometry: Tutoring Solution High School Geometry: Help and Review Study.com SAT Study Guide and Test Prep High School Geometry: Tutoring Solution Related Lessons Midpoint \| Formula & Examples Equilateral Triangle \| Definition, Properties & Measurements Isosceles Triangle \| Definition, Properties & Examples Triangles - Similar Triangles: Study.com SAT® Math Exam Prep Triangle Midsegment Theorem \| Definition, Formula & Examples Recently updated on Study.com Videos Courses Lessons Articles Quizzes Apps Concepts Teacher Resources Solving Problems Involving Systems of Equations The Great Leap Forward \| Overview, Definition & Purpose Henderson-Hasselbalch Equation \| Overview, Importance &... 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15948	https://patents.google.com/patent/US9685257B2/en	US9685257B2 - Electrical transmission cables with composite cores - Google Patents US9685257B2 - Electrical transmission cables with composite cores - Google Patents Electrical transmission cables with composite cores Download PDF Info Publication number US9685257B2 US9685257B2 US15/236,521 US201615236521A US9685257B2 US 9685257 B2 US9685257 B2 US 9685257B2 US 201615236521 A US201615236521 A US 201615236521A US 9685257 B2 US9685257 B2 US 9685257B2 Authority US United States Prior art keywords cable core composite range rovings Prior art date 2011-04-12 Legal status (The legal status is an assumption and is not a legal conclusion. Google has not performed a legal analysis and makes no representation as to the accuracy of the status listed.)Active Application number US15/236,521 Other versionsUS20160351300A1 (enInventor Allan Daniel Paul Springer Yuhsin Hawig Mark Lancaster David W. Eastep Sherri M. Nelson Tim Tibor Tim Regan Michael L. Wesley Current Assignee (The listed assignees may be inaccurate. Google has not performed a legal analysis and makes no representation or warranty as to the accuracy of the list.) Southwire Co LLC Original Assignee Southwire Co LLC Priority date (The priority date is an assumption and is not a legal conclusion. Google has not performed a legal analysis and makes no representation as to the accuracy of the date listed.)2011-04-12 Filing date 2016-08-15 Publication date 2017-06-20 2012-04-11 Priority claimed from US13/443,938 external-priority patent/US9012781B2/en 2016-08-15 Priority to US15/236,521 priority Critical patent/US9685257B2/en 2016-08-15 Application filed by Southwire Co LLC filed Critical Southwire Co LLC 2016-12-01 Publication of US20160351300A1 publication Critical patent/US20160351300A1/en 2017-05-16 Priority to US15/596,026 priority patent/US20170256338A1/en 2017-06-20 Publication of US9685257B2 publication Critical patent/US9685257B2/en 2017-06-20 Application granted granted Critical 2018-03-09 Priority to US15/916,561 priority patent/US20180197658A1/en 2018-05-17 Assigned to BANK OF AMERICA, N.A., AS COLLATERAL AGENT reassignment BANK OF AMERICA, N.A., AS COLLATERAL AGENT AMENDMENT NUMBER TWO TO GRANT OF SECURITY INTEREST IN PATENT RIGHTS Assignors: COLEMAN CABLE, LLC, SOUTHWIRE COMPANY, LLC, SUMNER MANUFACTURING COMPANY, LLC, TECHNOLOGY RESEARCH, LLC 2018-05-17 Assigned to WELLS FARGO BANK, NATIONAL ASSOCIATION, AS COLLATERAL AGENT reassignment WELLS FARGO BANK, NATIONAL ASSOCIATION, AS COLLATERAL AGENT SECURITY INTEREST (SEE DOCUMENT FOR DETAILS).Assignors: COLEMAN CABLE, LLC, SOUTHWIRE COMPANY, LLC, SUMNER MANUFACTURING COMPANY, LLC, TECHNOLOGY RESEARCH, LLC 2024-10-23 Assigned to WATTEREDGE, LLC, OBI PARTNERS, LLC, TECHNOLOGY RESEARCH, LLC (F/K/A TECHNOLOGY RESEARCH CORPORATION), TOPAZ LIGHTING COMPANY LLC, UNITED COPPER INDUSTRIES, LLC, SUMNER MANUFACTURING COMPANY, LLC, TAPPAN WIRE & CABLE, LLC, SOUTHWIRE COMPANY, LLC, COLEMAN CABLE, LLC (F/K/A COLEMAN CABLE, INC.), MADISON ELECTRIC PRODUCTS, LLC, NOVINIUM, LLC, WIIP, INC.reassignment WATTEREDGE, LLC TERMINATION AND RELEASE OF INTELLECTUAL PROPERTY SECURITY AGREEMENTS Assignors: BANK OF AMERICA, N.A., AS AGENT 2025-07-31 Assigned to SOUTHWIRE COMPANY, LLC, COLEMAN CABLE, LLC, TECHNOLOGY RESEARCH, LLC, SUMNER MANUFACTURING COMPANY, LLC, MADISON ELECTRIC PRODUCTS, LLC, NOVINIUM, LLC, NOVINIUM HOLDINGS, INC., OBI PARTNERS, LLC, TAPPAN WIRE & CABLE, LLC, TOPAZ LIGHTING COMPANY LLC, UNITED COPPER INDUSTRIES, LLC, WATTEREDGE, LLC, WIIP, INC.reassignment SOUTHWIRE COMPANY, LLC RELEASE BY SECURED PARTY (SEE DOCUMENT FOR DETAILS).Assignors: WELLS FARGO BANK, NATIONAL ASSOCIATION Status Active legal-status Critical Current 2032-04-11 Anticipated expiration legal-status Critical Links USPTO USPTO PatentCenter USPTO Assignment Espacenet Global Dossier Discuss Images Classifications H—ELECTRICITY H01—ELECTRIC ELEMENTS H01B—CABLES; CONDUCTORS; INSULATORS; SELECTION OF MATERIALS FOR THEIR CONDUCTIVE, INSULATING OR DIELECTRIC PROPERTIES H01B9/00—Power cables H01B9/04—Concentric cables B—PERFORMING OPERATIONS; TRANSPORTING B29—WORKING OF PLASTICS; WORKING OF SUBSTANCES IN A PLASTIC STATE IN GENERAL B29B—PREPARATION OR PRETREATMENT OF THE MATERIAL TO BE SHAPED; MAKING GRANULES OR PREFORMS; RECOVERY OF PLASTICS OR OTHER CONSTITUENTS OF WASTE MATERIAL CONTAINING PLASTICS B29B15/00—Pretreatment of the material to be shaped, not covered by groups B29B7/00-B29B13/00 B29B15/08—Pretreatment of the material to be shaped, not covered by groups B29B7/00-B29B13/00 of reinforcements or fillers B29B15/10—Coating or impregnating independently of the moulding or shaping step B29B15/12—Coating or impregnating independently of the moulding or shaping step of reinforcements of indefinite length B29B15/122—Coating or impregnating independently of the moulding or shaping step of reinforcements of indefinite length with a matrix in liquid form, e.g. as melt, solution or latex B—PERFORMING OPERATIONS; TRANSPORTING B29—WORKING OF PLASTICS; WORKING OF SUBSTANCES IN A PLASTIC STATE IN GENERAL B29C—SHAPING OR JOINING OF PLASTICS; SHAPING OF MATERIAL IN A PLASTIC STATE, NOT OTHERWISE PROVIDED FOR; AFTER-TREATMENT OF THE SHAPED PRODUCTS, e.g. REPAIRING B29C63/00—Lining or sheathing, i.e. applying preformed layers or sheathings of plastics; Apparatus therefor B29C63/02—Lining or sheathing, i.e. applying preformed layers or sheathings of plastics; Apparatus therefor using sheet or web-like material B29C63/04—Lining or sheathing, i.e. applying preformed layers or sheathings of plastics; Apparatus therefor using sheet or web-like material by folding, winding, bending or the like B29C63/08—Lining or sheathing, i.e. applying preformed layers or sheathings of plastics; Apparatus therefor using sheet or web-like material by folding, winding, bending or the like by winding helically B29C63/10—Lining or sheathing, i.e. applying preformed layers or sheathings of plastics; Apparatus therefor using sheet or web-like material by folding, winding, bending or the like by winding helically around tubular articles B29C63/105—Lining or sheathing, i.e. applying preformed layers or sheathings of plastics; Apparatus therefor using sheet or web-like material by folding, winding, bending or the like by winding helically around tubular articles continuously B—PERFORMING OPERATIONS; TRANSPORTING B29—WORKING OF PLASTICS; WORKING OF SUBSTANCES IN A PLASTIC STATE IN GENERAL B29C—SHAPING OR JOINING OF PLASTICS; SHAPING OF MATERIAL IN A PLASTIC STATE, NOT OTHERWISE PROVIDED FOR; AFTER-TREATMENT OF THE SHAPED PRODUCTS, e.g. REPAIRING B29C70/00—Shaping composites, i.e. plastics material comprising reinforcements, fillers or preformed parts, e.g. inserts B29C70/04—Shaping composites, i.e. plastics material comprising reinforcements, fillers or preformed parts, e.g. inserts comprising reinforcements only, e.g. self-reinforcing plastics B29C70/28—Shaping operations therefor B29C70/40—Shaping or impregnating by compression not applied B29C70/50—Shaping or impregnating by compression not applied for producing articles of indefinite length, e.g. prepregs, sheet moulding compounds [SMC] or cross moulding compounds [XMC] B29C70/52—Pultrusion, i.e. forming and compressing by continuously pulling through a die B29C70/523—Pultrusion, i.e. forming and compressing by continuously pulling through a die and impregnating the reinforcement in the die H—ELECTRICITY H01—ELECTRIC ELEMENTS H01B—CABLES; CONDUCTORS; INSULATORS; SELECTION OF MATERIALS FOR THEIR CONDUCTIVE, INSULATING OR DIELECTRIC PROPERTIES H01B1/00—Conductors or conductive bodies characterised by the conductive materials; Selection of materials as conductors H01B1/02—Conductors or conductive bodies characterised by the conductive materials; Selection of materials as conductors mainly consisting of metals or alloys H01B1/023—Alloys based on aluminium H—ELECTRICITY H01—ELECTRIC ELEMENTS H01B—CABLES; CONDUCTORS; INSULATORS; SELECTION OF MATERIALS FOR THEIR CONDUCTIVE, INSULATING OR DIELECTRIC PROPERTIES H01B1/00—Conductors or conductive bodies characterised by the conductive materials; Selection of materials as conductors H01B1/20—Conductive material dispersed in non-conductive organic material H01B1/24—Conductive material dispersed in non-conductive organic material the conductive material comprising carbon-silicon compounds, carbon or silicon H—ELECTRICITY H01—ELECTRIC ELEMENTS H01B—CABLES; CONDUCTORS; INSULATORS; SELECTION OF MATERIALS FOR THEIR CONDUCTIVE, INSULATING OR DIELECTRIC PROPERTIES H01B3/00—Insulators or insulating bodies characterised by the insulating materials; Selection of materials for their insulating or dielectric properties H01B3/18—Insulators or insulating bodies characterised by the insulating materials; Selection of materials for their insulating or dielectric properties mainly consisting of organic substances H01B3/30—Insulators or insulating bodies characterised by the insulating materials; Selection of materials for their insulating or dielectric properties mainly consisting of organic substances plastics; resins; waxes H01B3/42—Insulators or insulating bodies characterised by the insulating materials; Selection of materials for their insulating or dielectric properties mainly consisting of organic substances plastics; resins; waxes polyesters; polyethers; polyacetals H01B3/427—Polyethers H—ELECTRICITY H01—ELECTRIC ELEMENTS H01B—CABLES; CONDUCTORS; INSULATORS; SELECTION OF MATERIALS FOR THEIR CONDUCTIVE, INSULATING OR DIELECTRIC PROPERTIES H01B3/00—Insulators or insulating bodies characterised by the insulating materials; Selection of materials for their insulating or dielectric properties H01B3/18—Insulators or insulating bodies characterised by the insulating materials; Selection of materials for their insulating or dielectric properties mainly consisting of organic substances H01B3/48—Insulators or insulating bodies characterised by the insulating materials; Selection of materials for their insulating or dielectric properties mainly consisting of organic substances fibrous materials H—ELECTRICITY H01—ELECTRIC ELEMENTS H01B—CABLES; CONDUCTORS; INSULATORS; SELECTION OF MATERIALS FOR THEIR CONDUCTIVE, INSULATING OR DIELECTRIC PROPERTIES H01B5/00—Non-insulated conductors or conductive bodies characterised by their form H01B5/08—Several wires or the like stranded in the form of a rope H01B5/10—Several wires or the like stranded in the form of a rope stranded around a space, insulating material, or dissimilar conducting material H01B5/102—Several wires or the like stranded in the form of a rope stranded around a space, insulating material, or dissimilar conducting material stranded around a high tensile strength core H01B5/105—Several wires or the like stranded in the form of a rope stranded around a space, insulating material, or dissimilar conducting material stranded around a high tensile strength core composed of synthetic filaments, e.g. glass-fibres H—ELECTRICITY H01—ELECTRIC ELEMENTS H01B—CABLES; CONDUCTORS; INSULATORS; SELECTION OF MATERIALS FOR THEIR CONDUCTIVE, INSULATING OR DIELECTRIC PROPERTIES H01B7/00—Insulated conductors or cables characterised by their form H01B7/0009—Details relating to the conductive cores H—ELECTRICITY H01—ELECTRIC ELEMENTS H01B—CABLES; CONDUCTORS; INSULATORS; SELECTION OF MATERIALS FOR THEIR CONDUCTIVE, INSULATING OR DIELECTRIC PROPERTIES H01B7/00—Insulated conductors or cables characterised by their form H01B7/17—Protection against damage caused by external factors, e.g. sheaths or armouring H01B7/18—Protection against damage caused by wear, mechanical force or pressure; Sheaths; Armouring H—ELECTRICITY H01—ELECTRIC ELEMENTS H01B—CABLES; CONDUCTORS; INSULATORS; SELECTION OF MATERIALS FOR THEIR CONDUCTIVE, INSULATING OR DIELECTRIC PROPERTIES H01B9/00—Power cables H01B9/008—Power cables for overhead application B—PERFORMING OPERATIONS; TRANSPORTING B29—WORKING OF PLASTICS; WORKING OF SUBSTANCES IN A PLASTIC STATE IN GENERAL B29B—PREPARATION OR PRETREATMENT OF THE MATERIAL TO BE SHAPED; MAKING GRANULES OR PREFORMS; RECOVERY OF PLASTICS OR OTHER CONSTITUENTS OF WASTE MATERIAL CONTAINING PLASTICS B29B7/00—Mixing; Kneading B29B7/80—Component parts, details or accessories; Auxiliary operations B29B7/88—Adding charges, i.e. additives B29B7/90—Fillers or reinforcements, e.g. fibres B—PERFORMING OPERATIONS; TRANSPORTING B29—WORKING OF PLASTICS; WORKING OF SUBSTANCES IN A PLASTIC STATE IN GENERAL B29K—INDEXING SCHEME ASSOCIATED WITH SUBCLASSES B29B, B29C OR B29D, RELATING TO MOULDING MATERIALS OR TO MATERIALS FOR MOULDS, REINFORCEMENTS, FILLERS OR PREFORMED PARTS, e.g. INSERTS B29K2101/00—Use of unspecified macromolecular compounds as moulding material B29K2101/12—Thermoplastic materials B—PERFORMING OPERATIONS; TRANSPORTING B29—WORKING OF PLASTICS; WORKING OF SUBSTANCES IN A PLASTIC STATE IN GENERAL B29K—INDEXING SCHEME ASSOCIATED WITH SUBCLASSES B29B, B29C OR B29D, RELATING TO MOULDING MATERIALS OR TO MATERIALS FOR MOULDS, REINFORCEMENTS, FILLERS OR PREFORMED PARTS, e.g. INSERTS B29K2105/00—Condition, form or state of moulded material or of the material to be shaped B29K2105/06—Condition, form or state of moulded material or of the material to be shaped containing reinforcements, fillers or inserts B29K2105/08—Condition, form or state of moulded material or of the material to be shaped containing reinforcements, fillers or inserts of continuous length, e.g. cords, rovings, mats, fabrics, strands or yarns B—PERFORMING OPERATIONS; TRANSPORTING B29—WORKING OF PLASTICS; WORKING OF SUBSTANCES IN A PLASTIC STATE IN GENERAL B29K—INDEXING SCHEME ASSOCIATED WITH SUBCLASSES B29B, B29C OR B29D, RELATING TO MOULDING MATERIALS OR TO MATERIALS FOR MOULDS, REINFORCEMENTS, FILLERS OR PREFORMED PARTS, e.g. INSERTS B29K2995/00—Properties of moulding materials, reinforcements, fillers, preformed parts or moulds B29K2995/0037—Other properties B29K2995/0077—Yield strength; Tensile strength B—PERFORMING OPERATIONS; TRANSPORTING B29—WORKING OF PLASTICS; WORKING OF SUBSTANCES IN A PLASTIC STATE IN GENERAL B29K—INDEXING SCHEME ASSOCIATED WITH SUBCLASSES B29B, B29C OR B29D, RELATING TO MOULDING MATERIALS OR TO MATERIALS FOR MOULDS, REINFORCEMENTS, FILLERS OR PREFORMED PARTS, e.g. INSERTS B29K2995/00—Properties of moulding materials, reinforcements, fillers, preformed parts or moulds B29K2995/0037—Other properties B29K2995/0082—Flexural strength; Flexion stiffness B—PERFORMING OPERATIONS; TRANSPORTING B29—WORKING OF PLASTICS; WORKING OF SUBSTANCES IN A PLASTIC STATE IN GENERAL B29L—INDEXING SCHEME ASSOCIATED WITH SUBCLASS B29C, RELATING TO PARTICULAR ARTICLES B29L2031/00—Other particular articles B29L2031/34—Electrical apparatus, e.g. sparking plugs or parts thereof B29L2031/3462—Cables Definitions Composite wire structures are commonly used as transmission lines or cables for transmitting electricity to users. Examples of composite transmission wire constructions include, for instance, aluminum conductor steel reinforced (ACSR) cable, aluminum conductor steel supported (ACSS) cable, aluminum conductor composite reinforced (ACCR) cable, and aluminum conductor composite core (ACCC) cable. ACSR and ACSS cables include an aluminum outer conducting layer surrounding a steel inner core. the transmission lines or cables are designed not only to efficiently transmit electricity, but also to be strong and temperature resistant, especially when the transmission lines are strung on towers and stretched over long distances. Embodiments of the present invention may provide cables, e.g., electrical transmission cables for the overhead transmission of electricity, which may contain a cable core and conductive elements surrounding the cable core. the cable core may contain at least one composite core (the composite core also may be referred to as a composite strand or polymer composite strand). These core elements may serve as load-bearing members for the electrical transmission cable and, in some embodiments, these core elements may be non-conductive. a composite core for the electrical cable may extend in a longitudinal direction. the composite core may comprise at least one rod that comprises a continuous fiber component comprising a plurality of consolidated thermoplastic impregnated rovings (the rod also may be referred to as a fiber core). the rovings may contain continuous fibers oriented in the longitudinal direction, and a thermoplastic matrix that embeds the fibers. the fibers may have a ratio of ultimate tensile strength to mass per unit length of greater than about 1,000 Megapascals per gram per meter (MPa/g/m). the continuous fibers may constitute from about 25 wt. % to about 80 wt. the rod and the thermoplastic matrix may constitute from about 20 wt. % to about 75 wt. % of the rod. a capping layer may surround the rod, and this capping layer may be free of continuous fibers. the composite core may have a minimum flexural modulus of about 10 Gigapascals (GPa). a method for forming a composite core for an electrical transmission cable may comprise impregnating a plurality of rovings with a thermoplastic matrix and consolidating the rovings to form a ribbon, wherein the rovings may comprise continuous fibers oriented in the longitudinal direction. the fibers may have a ratio of ultimate tensile strength to mass per unit length of greater than about 1,000 MPa/g/m. the continuous fibers may constitute from about 25 wt. % to about 80 wt. % of the ribbon, and the thermoplastic matrix may constitute from about 20 wt. % to about 75 wt. % of the ribbon. the ribbon may be heated to a temperature at or above the softening temperature (or melting temperature) of the thermoplastic matrix and pulled through at least one forming die to compress and shape the ribbon into a rod. a capping layer may be applied to the rod to form the composite core. a method of making an electrical cable may comprise providing a cable core comprising at least one composite core, and surrounding the cable core with a plurality of conductive elements. the composite core may comprise at least one rod comprising a plurality of consolidated thermoplastic impregnated rovings. the rovings may comprise continuous fibers oriented in the longitudinal direction and a thermoplastic matrix that embeds the fibers. the fibers may have a ratio of ultimate tensile strength to mass per unit length of greater than about 1,000 MPa/g/m. the rod may comprise from about 25 wt. % to about 80 wt. % fibers, and from about 20 wt. % to about 75 wt. % thermoplastic matrix. a capping layer may surround the at least one rod, and this capping layer generally may be free of continuous fibers. the composite core may have a flexural modulus of greater than about 10 GPa. FIG. 1 is a perspective view of one embodiment of a consolidated ribbon for use in the present invention FIG. 2 is a cross-sectional view of another embodiment of a consolidated ribbon for use in the present invention. FIG. 3 is a schematic illustration of one embodiment of an impregnation system for use in the present invention. FIG. 4 is a cross-sectional view of the impregnation die shown in FIG. 3 ; FIG. 5 is an exploded view of one embodiment of a manifold assembly and gate passage for an impregnation die that may be employed in the present invention FIG. 6 is a perspective view of one embodiment of a plate at least partially defining an impregnation zone that may be employed in the present invention FIG. 7 is a schematic illustration of one embodiment of a pultrusion system that may be employed in the present invention. FIG. 8 is a perspective view of one embodiment of a composite core of the present invention. FIG. 9 is a perspective view of one embodiment of an electrical transmission cable of the present invention. FIG. 10 is a perspective view of another embodiment of an electrical transmission cable of the present invention. FIG. 11 is a top cross-sectional view of one embodiment of various calibration dies that may be employed in accordance with the present invention. FIG. 12 is a side cross-sectional view of one embodiment of a calibration die that may be employed in accordance with the present invention. FIG. 13 is a front view of a portion of one embodiment of a calibration die that may be employed in accordance with the present invention. FIG. 14 is a front view of one embodiment of forming rollers that may be employed in accordance with the present invention. FIG. 15 is a perspective view of the electrical cable of Examples 6-7; FIG. 16 is a stress-strain diagram for the electrical cable of Example 7. FIG. 17 is a perspective view of the electrical cable of Constructive Example 8. an electrical cable may comprise a cable core comprising at least one composite core (or composite strand), and a plurality of conductive elements surrounding the cable core. the composite core may contain a rod (or fiber core) comprising a continuous fiber component, surrounded by a capping layer. the rod may comprise a plurality of unidirectionally aligned fiber rovings embedded within a thermoplastic polymer matrix. While not wishing to be bound by theory, applicants believe that the degree to which the rovings are impregnated with the thermoplastic polymer matrix may be significantly improved through selective control over the impregnation process, and also through control over the degree of compression imparted to the rovings during formation and shaping of the rod, as well as the calibration of the final rod geometry. Such a well impregnated rod may have a very small void fraction, which may lead to excellent strength properties. Notably, the desired strength properties may be achieved without the need for different fiber types in the rod. the term “roving”generally refers to a bundle or tow of individual fibers. the fibers contained within the roving may be twisted or may be straight. different fibers may be used in individual or different rovings, it may be beneficial for each of the rovings to contain a single fiber type to minimize any adverse impact of using materials having different thermal expansion coefficients. the continuous fibers employed in the rovings may possess a high degree of tensile strength relative to their mass. the ultimate tensile strength of the fibers typically may be in a range from about 1,000 to about 15,000 Megapascals (MPa), in some embodiments from about 2,000 MPa to about 10,000 MPa, and in some embodiments, from about 3,000 MPa to about 6000 MPa. MPa Megapascals Such tensile strengths may be achieved even though the fibers are of a relatively light weight, such as a mass per unit length of from about 0.1 to about 2 grams per meter (g/m), in some embodiments from about 0.4 to about 1.5 g/m. the ratio of tensile strength to mass per unit length thus may be about 1,000 Megapascals per gram per meter (MPa/g/m) or greater, in some embodiments about 4,000 MPa/g/m or greater, and in some embodiments, from about 5,500 to about 20,000 MPa/g/m. Such high strength fibers may, for instance, be metal fibers, glass fibers (e.g., E-glass, A-glass, C-glass, D-glass, AR-glass, R-glass, S1-glass, S2-glass, etc.), carbon fibers (e.g., amorphous carbon, graphitic carbon, or metal-coated carbon, etc.), boron fibers, ceramic fibers (e.g., alumina or silica), aramid fibers (e.g., Kevlar® marketed by E. I. thermoplastic and/or thermoset compositions include synthetic organic fibers (e.g., polyamide, polyethylene, paraphenylene, terephthalamide, polyethylene terephthalate and polyphenylene sulfide), and various other natural or synthetic inorganic or organic fibrous materials known for reinforcing thermoplastic and/or thermoset compositions. Carbon fibers may be particularly suitable for use as the continuous fibers, which typically have a tensile strength to mass per unit length ratio in the range of from about 5,000 to about 7,000 MPa/g/m. the continuous fibers may have a nominal diameter of about 4 to about 35 micrometers ( ⁇ m), and in some embodiments, from about 5 to about 35 ⁇ m. the number of fibers contained in each roving may be constant or may vary from roving to roving. a roving may contain from about 1,000 fibers to about 100,000 individual fibers, and in some embodiments, from about 5,000 to about 50,000 fibers. thermoplastic polymers may be employed to form the thermoplastic matrix in which the continuous fibers are embedded. Suitable thermoplastic polymers for use in the present invention may include, for instance, polyolefins (e.g., polypropylene, propylene-ethylene copolymers, etc.), polyesters (e.g., polybutylene terephalate (PBT)), polycarbonates, polyam ides (e.g., NylonTM) polyether ketones (e.g., polyetherether ketone (PEEK)), polyetherimides, polyarylene ketones (e.g., polyphenylene diketone (PPDK)), liquid crystal polymers, polyarylene sulfides (e.g., polyphenylene sulfide (PPS), poly(biphenylene sulfide ketone), poly(phenylene sulfide diketone), poly(biphenylene sulfide), etc.), fluoropolymers (e.g. the properties of the thermoplastic matrix may be selected to achieve a desired combination of processability and end-use performance of the composite core. the melt viscosity of the thermoplastic matrix generally may be low enough so that the polymer may adequately impregnate the fibers and become shaped into the rod configuration. the melt viscosity typically may range from about 25 to about 2,000 Pascal-seconds (Pa-s), in some embodiments from 50 about 500 Pa-s, and in some embodiments, from about 60 to about 200 Pa-s, determined at the operating conditions used for the thermoplastic polymer (e.g., about 360° C.). thermoplastic polymer having a relatively high melting temperature may be employed. the melting temperature of such high temperature polymers may be in a range from about 200° C. to about 500° C., in some embodiments from about 225° C. to about 400° C., and in some embodiments, from about 250° C. to about 350° C. polyarylene sulfides may be used in the present invention as a high temperature matrix with the desired melt viscosity. Polyphenylene sulfide for example, is a semi-crystalline resin that generally includes repeating monomeric units represented by the following general formula: These monomeric units may constitute at least 80 mole %, and in some embodiments, at least 90 mole %, of the recurring units, in the polymer. the polyphenylene sulfide may contain additional recurring units, such as described in U.S. Pat. No. 5,075,381 to Gotoh, et al., which is incorporated herein in its entirety by reference thereto for all purposes. When employed, such additional recurring units typically may constitute less than about 20 mole % of the polymer. Commercially available high melt viscosity polyphenylene sulfides may include those available from Ticona, LLC (Florence, Ky.) under the trade designation FORTRON®. Such polymers may have a melting temperature of about 285° C. (determined according to ISO 11357-1,2,3) and a melt viscosity of from about 260 to about 320 Pa-s at 310° C. an extrusion device generally may be employed to impregnate the rovings with the thermoplastic matrix. the extrusion device may facilitate the application of the thermoplastic polymer to the entire surface of the fibers. the impregnated rovings also may have a very low void fraction, which may increase the resulting strength of the rod. the void fraction may be about 6% or less, in some embodiments about 4% or less, in some embodiments about 3% or less, in some embodiments about 2% or less, in some embodiments about 1% or less, and in some embodiments, about 0.5% or less. the void fraction may be measured using techniques well known to those skilled in the art. the void fraction may be measured using a “resin burn off” test in which samples are placed in an oven (e.g., at 600° C. for 3 hours) to burn off the resin. The mass of the remaining fibers may then be measured to calculate the weight and volume fractions. V f is the void fraction as a percentage ⁇ c is the density of the composite as measured using known techniques, such as with a liquid or gas pycnometer (e.g., helium pycnometer); ⁇ m is the density of the thermoplastic matrix (e.g., at the appropriate crystallinity); ⁇ f is the density of the fibers W f is the weight fraction of the fibers W m is the weight fraction of the thermoplastic matrix. the void fraction may be determined by chemically dissolving the resin in accordance with ASTM D 3171-09. the “burn off” and “dissolution” methods may be particularly suitable for glass fibers, which are generally resistant to melting and chemical dissolution. the void fraction may be indirectly calculated based on the densities of the thermoplastic polymer, fibers, and ribbon (or tape) in accordance with ASTM D 2734-09 (Method A), where the densities may be determined by ASTM D792-08 Method A. the void fraction also may be estimated using conventional microscopy equipment, or through the use of computed tomography (CT) scan equipment, such as a Metrotom 1500 (2k ⁇ 2k) high resolution detector. CT computed tomography the apparatus may include an extruder 120 containing a screw shaft 124 mounted inside a barrel 122 . a heater 130 e.g., an electrical resistance heater a thermoplastic polymer feedstock 127 may be supplied to the extruder 120 through a hopper 126 . the thermoplastic feedstock 127 may be conveyed inside the barrel 122 by the screw shaft 124 and heated by frictional forces inside the barrel 122 and by the heater 130 . the feedstock 127 may exit the barrel 122 through a barrel flange 128 and enter a die flange 132 of an impregnation die 150 . a continuous fiber roving 142 or a plurality of continuous fiber rovings 142 may be supplied from a reel or reels 144 to die 150 . the rovings 142 may be kept apart a certain distance before impregnation, such as at least about 4 mm, and in some embodiments, at least about 5 mm. the feedstock 127 may further be heated inside the die by heaters 133 mounted in or around the die 150 . the die generally may be operated at temperatures that are sufficient to cause melting and impregnation of the thermoplastic polymer. Typically, the operation temperatures of the die may be higher than the melt temperature of the thermoplastic polymer, such as at temperatures from about 200° C. to about 450° C. the continuous fiber rovings 142 may become embedded in the polymer matrix, which may be a resin 214 ( FIG. 4 ) processed from the feedstock 127 . the mixture then may be extruded from the impregnation die 150 to create an extrudate 152 . a pressure sensor 137 may monitor the pressure near the impregnation die 150 , so that the extruder 120 can be operated to deliver a correct amount of resin 214 for interaction with the fiber rovings 142 . the rate of extrusion may be varied by controlling the rotational speed of the screw shaft 124 and/or feed rate of the feedstock 127 . the extruder 120 may be operated to produce the extrudate 152 (impregnated fiber rovings), which after leaving the impregnation die 150 , may enter an optional pre-shaping or guiding section (not shown), before entering a nip formed between two adjacent rollers 190 . the rollers 190 may help to consolidate the extrudate 152 into the form of a ribbon (or tape), as well as to enhance fiber impregnation and to squeeze out any excess voids. other shaping devices also may be employed, such as a die system. the resulting consolidated ribbon 156 may be pulled by tracks 162 and 164 mounted on rollers. the tracks 162 and 164 also may pull the extrudate 152 from the impregnation die 150 and through the rollers 190 . the consolidated ribbon 156 may be wound up at a section 171 . the ribbons may be relatively thin and may have a thickness of from about 0.05 to about 1 millimeter (mm), in some embodiments from about 0.1 to about 0.8 mm, and in some embodiments, from about 0.2 to about 0.4 mm. the rovings 142 are traversed through an impregnation zone 250 to impregnate the rovings with the polymer resin 214 . the polymer resin may be forced generally transversely through the rovings by shear and pressure created in the impregnation zone 250 , which may significantly enhance the degree of impregnation. This may be particularly useful when forming a composite from ribbons of a high fiber content, such as about 35% weight fraction (Wf) or more, and in some embodiments, from about 40% Wf or more. the die 150 may include a plurality of contact surfaces 252 , such as, for example, at least 2, at least 3, from 4 to 7, from 2 to 20, from 2 to 30, from 2 to 40, from 2 to 50, or more contact surfaces 252 , to create a sufficient degree of penetration and pressure on the rovings 142 . the contact surfaces 252 typically may possess a curvilinear surface, such as a curved lobe, rod, etc. the contact surfaces 252 typically may be made of a metal material. FIG. 4 shows a cross-sectional view of an impregnation die 150 . the impregnation die 150 may include a manifold assembly 220 , a gate passage 270 , and an impregnation zone 250 . the manifold assembly 220 may be provided for flowing the polymer resin 214 therethrough. the manifold assembly 220 may include a channel 222 or a plurality of channels 222 . the resin 214 provided to the impregnation die 150 may flow through the channels 222 . the channels 222 may be curvilinear, and in exemplary embodiments, the channels 222 may have a symmetrical orientation along a central axis 224 . Further, in some embodiments, the channels may be a plurality of branched runners 222 , which may include first branched runner group 232 , second group 234 , third group 236 , and, if desired, more branched runner groups. Each group may include 2, 3, 4 or more runners 222 branching off from runners 222 in the preceding group, or from an initial channel 222 . the branched runners 222 and the symmetrical orientation thereof may evenly distribute the resin 214 , such that the flow of resin 214 exiting the manifold assembly 220 and coating the rovings 142 may be substantially uniformly distributed on the rovings 142 . Beneficially, this may result in generally uniform impregnation of the rovings 142 . the manifold assembly 220 may in some embodiments define an outlet region 242 , which generally encompasses at least a downstream portion of the channels or runners 222 from which the resin 214 exits. at least a portion of the channels or runners 222 disposed in the outlet region 242 may have an increasing area in a flow direction 244 of the resin 214 . The increasing area may permit diffusion and further distribution of the resin 214 as the resin 214 flows through the manifold assembly 220 , which may further result in substantially uniform distribution of the resin 214 on the rovings 142 . Gate passage 270 may be positioned between the manifold assembly 220 and the impregnation zone 250 , and may be configured for flowing the resin 214 from the manifold assembly 220 such that the resin 214 coats the rovings 142 . resin 214 exiting the manifold assembly 220 such as through outlet region 242 , may enter gate passage 270 and flow therethrough, as shown. the resin 214 may contact the rovings 142 being traversed through the die 150 . the resin 214 may substantially uniformly coat the rovings 142 , due to distribution of the resin 214 in the manifold assembly 220 and the gate passage 270 . the resin 214 may impinge on an upper surface of each of the rovings 142 , or on a lower surface of each of the rovings 142 , or on both an upper and lower surface of each of the rovings 142 . Initial impingement on the rovings 142 may provide for further impregnation of the rovings 142 with the resin 214 . the coated rovings 142 may traverse in run direction 282 through impregnation zone 250 , which is configured to impregnate the rovings 142 with the resin 214 . the rovings 142 may be traversed over contact surfaces 252 in the impregnation zone. Impingement of the rovings 142 on the contact surface 252 may create shear and pressure sufficient to impregnate the rovings 142 with the resin 214 , thereby coating the rovings 142 . the impregnation zone 250 may be defined between two spaced apart opposing plates 256 and 258 . First plate 256 may define a first inner surface 257 second plate 258 may define a second inner surface 259 . the contact surfaces 252 may be defined on or extend from both the first and second inner surfaces 257 and 259 , or only one of the first and second inner surfaces 257 and 259 . FIG. 6 illustrates the second plate 258 and the various contact surfaces thereon that may form at least a portion of the impregnation zone 250 according to these embodiments. FIG. 6 illustrates the second plate 258 and the various contact surfaces thereon that may form at least a portion of the impregnation zone 250 according to these embodiments. the contact surfaces 252 may be defined alternately on the first and second surfaces 257 and 259 such that the rovings alternately impinge on contact surfaces 252 on the first and second surfaces 257 and 259 . the rovings 142 may pass contact surfaces 252 in a waveform, tortuous, or sinusoidal-type pathway, which enhances shear. Angle 254 at which the rovings 142 traverse the contact surfaces 252 may be generally high enough to enhance shear, but not so high as to cause excessive forces that will break the fibers. the angle 254 may be in the range between approximately 1° and approximately 30°, and in some embodiments, between approximately 5° and approximately 25°. the impregnation zone 250 may include a plurality of pins (not shown), each pin having a contact surface 252 . the pins may be static, freely rotational, or rotationally driven. the contact surfaces 252 and impregnation zone 250 may comprise any suitable shapes and/or structures for impregnating the rovings 142 with the resin 214 as desired or required. the tension may, for example, range from about 5 to about 300 Newtons (N), in some embodiments from about 50 to about 250 N, and in some embodiments, from about 100 to about 200 N, per roving 142 or tow of fibers. a land zone 280 may be positioned downstream of the impregnation zone 250 in run direction 282 of the rovings 142 . the rovings 142 may traverse through the land zone 280 before exiting the die 150 . a faceplate 290 may adjoin the impregnation zone 250 . Faceplate 290 may be generally configured to meter excess resin 214 from the rovings 142 . apertures in the faceplate 290 through which the rovings 142 traverse, may be sized such that when the rovings 142 are traversed therethrough, the size of the apertures may cause excess resin 214 to be removed from the rovings 142 . the impregnation die shown and described above is but one of various possible configurations that may be employed in the present invention. the rovings may be introduced into a crosshead die that may be positioned at an angle relative to the direction of flow of the polymer melt. As the rovings move through the crosshead die and reach the point where the polymer exits from an extruder barrel, the polymer may be forced into contact with the rovings. Examples of such a crosshead die extruder are described, for instance, in U.S. Pat. No. 3,993,726 to Moyer, U.S. Pat. No. 4,588,538 to Chung, et al.; U.S. Pat. No. Such an assembly may include a supply of compressed air or another gas that may impinge in a generally perpendicular fashion on the moving rovings that pass across the exit ports. the spread rovings then may be introduced into a die for impregnation, such as described above. the continuous fibers may be oriented in the longitudinal direction (the machine direction “A” of the system of FIG. 3 ) to enhance tensile strength. other aspects of the pultrusion process also may be controlled to achieve the desired strength. a relatively high percentage of continuous fibers may be employed in the consolidated ribbon to provide enhanced strength properties. continuous fibers typically may constitute from about 25 wt. % to about 80 wt. %, in some embodiments from about 30 wt. % to about 75 wt. %, and in some embodiments, from about 35 wt. % to about 60 wt. % of the ribbon. thermoplastic polymer(s)typically may constitute from about 20 wt. the percentage of the fibers and thermoplastic matrix in the final rod also may be within the ranges noted above. the rovings may be consolidated into the form of one or more ribbons before being shaped into the desired rod configuration. the rovings When such a ribbon is subsequently compressed, the rovings may become distributed in a generally uniform manner about a longitudinal center of the rod. Such a uniform distribution enhances the consistency of the strength properties (e.g., flexural modulus, ultimate tensile strength, etc.) over the entire length of the rod. the number of consolidated ribbons used to form the rod may vary based on the desired thickness and/or cross-sectional area and strength of the rod, as well as the nature of the ribbons themselves. In most cases, however, the number of ribbons may be from 1 to 20, and in some embodiments, from 2 to 10. the number of rovings employed in each ribbon may likewise vary. Typically, however, a ribbon may contain from 2 to 10 rovings, and in some embodiments, from 3 to 5 rovings. To help achieve the symmetric distribution of the rovings in the final rod, it may be beneficial that they are spaced apart approximately the same distance from each other within the ribbon. Referring to FIG. 1 , for example, one embodiment of a consolidated ribbon 4 is shown that contains three (3) rovings 5 spaced equidistant from each other in the x direction. In other embodiments, however, it may be desired that the rovings are combined, such that the fibers of the rovings are generally evenly distributed throughout the ribbon 4 . the rovings may be generally indistinguishable from each other. FIG. 2 for example, one embodiment of a consolidated ribbon 4 is shown that contains rovings that are combined such that the fibers are generally evenly distributed throughout. two ribbons 12 initially may be provided in a wound package on a creel 20 . the creel 20 may be an unreeling creel that includes a frame provided with horizontal spindles 22 , each supporting a package. a pay-out creel also may be employed, particularly if desired to induce a twist into the fibers, such as when using raw fibers in a one-step configuration. the ribbons also may be formed in-line with the formation of the rod. the extrudate 152 exiting the impregnation die 150 from FIG. 3 may be supplied directly to the system used to form a rod. a tension-regulating device 40 also may be employed to help control the degree of tension in the ribbons 12 . the device 40 may include inlet plate 30 that lies in a vertical plane parallel to the rotating spindles 22 of the creel 20 and/or perpendicular to the incoming ribbons. the tension-regulating device 40 may contain cylindrical bars 41 arranged in a staggered configuration so that the ribbon 12 may pass over and under these bars to define a wave pattern. The height of the bars may be adjusted to modify the amplitude of the wave pattern and control tension. the ribbons 12 may be heated in an oven 45 before entering a consolidation die 50 . Heating may be conducted using any known type of oven, such as an infrared oven, a convection oven, etc. During heating, the fibers in the ribbon may be unidirectionally oriented to optimize the exposure to the heat and maintain even heat across the entire ribbon. the temperature to which the ribbons 12 are heated generally may be high enough to soften the thermoplastic polymer to an extent that the ribbons may bond together. However, the temperature may not be so high as to destroy the integrity of the material. the temperature may, for example, range from about 100° C. to about 500° C., in some embodiments from about 200° C. to about 400° C., and in some embodiments, from about 250° C. to about 350° C. PPS polyphenylene sulfide the ribbons may be heated to or above the melting point of PPS, which may be about 285° C. the ribbons 12 may be provided to a consolidation die 50 that may compress them together into a preform 14 , as well as may align and form the initial shape of the rod. the ribbons 12 may be guided through a flow passage 51 of the die 50 in a direction “A” from an inlet 53 to an outlet 55 . the passage 51 may have any of a variety of shapes and/or sizes to achieve the rod configuration. the channel and rod configuration may be circular, elliptical, parabolic, trapezoidal, rectangular, etc. the ribbons generally may be maintained at a temperature at or above the melting point of the thermoplastic matrix used in the ribbon to ensure adequate consolidation. the desired heating, compression, and shaping of the ribbons 12 may be accomplished through the use of a die 50 having one or multiple sections. the consolidation die 50 may possess multiple sections that function together to compress and shape the ribbons 12 into the desired configuration. a first section of the passage 51 may be a tapered zone that initially may shape the material as it flows into the die 50 . the tapered zone generally may possess a cross-sectional area that is larger at its inlet than at its outlet. the cross-sectional area of the passage 51 at the inlet of the tapered zone may be about 2% or more, in some embodiments about 5% or more, and in some embodiments, from about 10% to about 20% greater than the cross-sectional area at the outlet of the tapered zone. the cross-section of the flow passage typically may change gradually and smoothly within the tapered zone so that a balanced flow of the composite material through the die may be maintained. a shaping zone may follow the tapered zone, and may compress the material and provide a generally homogeneous flow therethrough. the shaping zone also may pre-shape the material into an intermediate shape that is similar to that of the rod, but typically of a larger cross-sectional area to allow for expansion of the thermoplastic polymer while heated to minimize the risk of backup within the die 50 . the shaping zone also may include one or more surface features that impart a directional change to the preform. The directional change may force the material to be redistributed, resulting in a more even distribution of the fiber/resin in the final shape. This also may reduce the risk of dead spots in the die that may cause burning of the resin. the cross-sectional area of the passage 51 at a shaping zone may be about 2% or more, in some embodiments about 5% or more, and in some embodiments, from about 10% to about 20% greater than the width of the preform 14 . a die land also may follow the shaping zone to serve as an outlet for the passage 51 . the shaping zone, tapered zone, and/or die land may be heated to a temperature at or above that of the glass transition temperature or melting point of the thermoplastic matrix. a second die 60 also may be employed to compress the preform 14 into the final shape of the rod. a second die 60 e.g., a calibration die cooling may reduce the temperature of the exterior of the rod below the melting point temperature of the thermoplastic matrix to minimize and substantially prevent the occurrence of melt fracture on the exterior surface of the rod. the internal section of the rod may remain molten to ensure compression when the rod enters the calibration die body. Such cooling may be accomplished by simply exposing the preform 14 to the ambient atmosphere (e.g., room temperature) or through the use of active cooling techniques (e.g., water bath or air cooling) as is known in the art. active cooling techniques e.g., water bath or air cooling air may blown onto the preform 14 (e.g., with an air ring). the cooling between these stages generally may occur over a small period of time to ensure that the preform 14 still may be soft enough to be further shaped. the preform 14 may be exposed to the ambient environment for only from about 1 to about 20 seconds, and in some embodiments, from about 2 to about 10 seconds, before entering the second die 60 . the preform generally may be kept at a temperature below the melting point of the thermoplastic matrix used in the ribbon so that the shape of the rod can be maintained. the dies 50 and 60 may in fact be formed from multiple individual dies (e.g., face plate dies). multiple individual dies 60 may be utilized to gradually shape the material into the desired configuration. the dies 60 may be placed in series, and provide for gradual decreases in the dimensions of the material. Such gradual decreases may allow for shrinkage during and between the various steps. a first die 60 may include one or more inlets 62 and corresponding outlets 64 , as shown. Any number of inlets 62 and corresponding outlets 64 may be included in a die 60 , such as four as shown, or one, two, three, five, six, or more. An inlet 62 in some embodiments may be generally oval or circular shaped. In other embodiments, the inlet 62 may have a curved rectangular shape, i.e., a rectangular shape with curved corners or a rectangular shape with straight longer sidewalls and curved shorter sidewalls. an outlet 64 may be generally oval or circular shaped, or may have a curved rectangular shape. the inlet 62 may have a major axis length 66 to minor axis length 68 ratio in a range between approximately 3:1 and approximately 5:1. the outlet 64 may have a major axis length 66 to minor axis length 68 ratio in a range between approximately 1:1 and approximately 3:1. the inlet and outlet may have major axis length 66 to minor axis length 66 ratios (aspect ratios) between approximately 2:1 and approximately 7:1, and the outlet 64 ratio may be less than the inlet 62 ratio. the cross-sectional area of an inlet 62 and the cross-sectional area of a corresponding outlet 64 of the first die 60 may have a ratio in a range between approximately 1.5:1 and 6:1. the first die 60 thus may provide a generally smooth transformation of polymer impregnated fiber material to a shape that is relatively similar to a final shape of the resulting rod, which in exemplary embodiments has a circular or oval shaped cross-section. Subsequent dies such as a second die 60 and third die 60 as shown in FIG. 11 , may provide for further gradual decreases and/or changes in the dimensions of the material, such that the shape of the material is converted to a final cross-sectional shape of the rod. These subsequent dies 60 may both shape and cool the material. each subsequent die 60 may be maintained at a lower temperature than the previous dies. all dies 60 may be maintained at temperatures that are higher than a softening point temperature for the material. dies 60 having relatively long land lengths 69 may be desired, due to, for example, proper cooling and solidification, which may be important in achieving a desired rod shape and size. Relatively long land lengths 69 may reduce stresses and provide smooth transformations to desired shapes and sizes, and with minimal void fraction and bow characteristics. a ratio of land length 69 at an outlet 64 to major axis length 66 at the outlet 64 for a die 60 may be in the range between 0 and approximately 20, such as between approximately 2 and approximately 6. calibration dies 60 may provide for gradual changes in material cross-section, as discussed. These gradual changes may in exemplary embodiments ensure that the resulting product, such as a rod or other suitable product, has a generally uniform fiber distribution with relatively minimal void fraction. any suitable number of dies 60 may be utilized to gradually form the material into a profile having any suitable cross-sectional shape, as desired or as required by various end-use applications. rollers 90 may be employed between the consolidation die 50 and the calibration die 60 , between the various calibration dies 60 , and/or after the calibration dies 60 to further compress the preform 14 before it is converted into its final shape. the rollers may have any configuration, such as pinch rollers, overlapping rollers, etc., and may be vertical as shown or horizontal rollers. the surfaces of the rollers 90 may be machined to impart the dimensions of the final product, such as the rod, core, profile, or other suitable product, to the preform 14 . the pressure of the rollers 90 may be adjustable to optimize the quality of the final product. the rollers 90 in exemplary embodiments, such as at least the portions contacting the material,may have generally smooth surfaces. relatively hard, polished surfaces may be beneficial in many embodiments. the surface of the rollers may be formed from a relatively smooth chrome or other suitable material. This may allow the rollers 90 to manipulate the preform 14 without damaging or undesirably altering the preform 14 . such surfaces may prevent the material from sticking to the rollers, and the rollers may impart smooth surfaces onto the materials. the temperature of the rollers 90 may be controlled. This may be accomplished by heating of the rollers 90 themselves, or by placing the rollers 90 in a temperature controlled environment. surface features 92 may be provided on the rollers 90 . the surface features 92 may guide and/or control the preform 14 in one or more directions as it is passed through the rollers. surface features 92 may be provided to prevent the preform 14 from folding over on itself as it is passed through the rollers 90 . the surface features 92 may guide and control deformation of the preform 14 in the cross-machine direction relative to the machine direction A as well as in the vertical direction relative to the machine direction A. The preform 14 thus may be pushed together in the cross-machine direction, rather than folded over on itself, as it is passed through the rollers 90 in the machine direction A. tension regulation devices may be provided in communication with the rollers. These devices may be utilized with the rollers to apply tension to the preform 14 in the machine direction, cross-machine direction, and/or vertical direction to further guide and/or control the preform. the resulting rod also may be applied with a capping layer to protect it from environmental conditions and/or to improve wear resistance. a capping layer may be applied via an extruder oriented at any desired angle to introduce a thermoplastic resin into a capping die 72 . the capping material may have a dielectric strength of at least about 1 kV per millimeter (kV/mm), in some embodiments at least about 2 kV/mm, in some embodiments from about 3 kV/mm to about 50 kV/mm, and in some embodiments, from about 4 kV/mm to about 30 kV/mm, such as determined in accordance with ASTM D149-09. Suitable thermoplastic polymers for this purpose may include, for instance, polyolefins (e.g., polypropylene, propylene-ethylene copolymers, etc.), polyesters (e.g., polybutylene terephalate (PBT)), polycarbonates, polyamides (e.g., NylonTM) polyether ketones (e.g., polyetherether ketone (PEEK)), polyetherimides, polyarylene ketones (e.g., polyphenylene diketone (PPDK)), liquid crystal polymers, polyarylene sulfides (e.g., polyphenylene sulfide (PPS), poly(biphenylene sulfide ketone), poly(phenylene sulfide diketone), poly(biphenylene sulfide), etc.), fluoropolymers (e.g., polytetrafluoroethylene-perfluoromethylvinylether polymer, perfluoro-al the capping layer generally may be “free” of continuous fibers. That is, the capping layer may contain “less than about 10 wt. %” of continuous fibers, in some embodiments about 5 wt. % or less of continuous fibers, and in some embodiments, about 1 wt. % or less of continuous fibers (e.g., 0 wt. %). Nevertheless, the capping layer may contain other additives for improving the final properties of the composite core. Additive materials employed at this stage may include those that are not suitable for incorporating into the continuous fiber material. For instance, it may be beneficial to add pigments to reduce finishing labor, or it may be beneficial to add flame retardant agents to enhance the flame retardancy of the core. Additive materials may include, for instance, mineral reinforcing agents, lubricants, flame retardants, blowing agents, foaming agents, ultraviolet light resistant agents, thermal stabilizers, pigments, and combinations thereof. Suitable mineral reinforcing agents may include, for instance, calcium carbonate, silica, mica, clays, talc, calcium silicate, graphite, calcium silicate, alumina trihydrate, barium ferrite, and combinations thereof. the capping die 72 may include various features known in the art to help achieve the desired application of the capping layer. the capping die 72 may include an entrance guide that aligns the incoming rod. the capping die also may include a heating mechanism (e.g., heated plate) that pre-heats the rod before application of the capping layer to help ensure adequate bonding. the shaped part 15 then may be finally cooled using a cooling system 80 as is known in the art. the cooling system 80 may, for instance, be a sizing system that includes one or more blocks (e.g., aluminum blocks) that may completely encapsulate the composite core while a vacuum pulls the hot shape out against its walls as it cools. a cooling medium may be supplied to the sizer, such as air or water, to solidify the composite core in the correct shape. the composite core may be beneficial to cool the composite core after it exits the capping die (or the consolidation or calibration die, if capping is not applied). Cooling may occur using any technique known in the art, such a water tank, cool air stream or air jet, cooling jacket, an internal cooling channel, cooling fluid circulation channels, etc. Regardless, the temperature at which the material is cooled may be controlled to achieve certain mechanical properties, part dimensional tolerances, good processing, and an aesthetically pleasing composite. For instance, if the temperature of the cooling station is too high, the material might swell in the tool and interrupt the process. a water tank may be employed at a temperature of from about 0° C. to about 30° C., in some embodiments from about 1° C. to about 20° C., and in some embodiments, from about 2° C. to about 15° C. one or more sizing blocks also may be employed, such as after capping. Such blocks may contain openings that are cut to the exact core shape, graduated from oversized at first to the final core shape. As the composite core passes therethrough, any tendency for it to move or sag may be counteracted, and it may be pushed back (repeatedly) to its correct shape. the composite core may be cut to the desired length at a cutting station (not shown), such as with a cut-off saw capable of performing cross-sectional cuts, or the composite core may be wound on a reel in a continuous process. the length of rod and/or the composite core may be limited to the length of the fiber tow. the temperature of the rod or composite core as it advances through any section of the system of the present invention may be controlled to yield certain manufacturing and final composite properties. Any or all of the assembly sections may be temperature controlled utilizing electrical cartridge heaters, circulated fluid cooling, etc., or any other temperature controlling device known to those skilled in the art. a pulling device 82 may be positioned downstream from the cooling system 80 to pull the finished composite core 16 through the system for final sizing of the composite. the pulling device 82 may be any device capable of pulling the core through the process system at a desired rate. Typical pulling devices may include, for example, caterpillar pullers and reciprocating pullers. the composite core 516 may have a substantially circular shape and may include a rod (or fiber core) 514 comprising one or more consolidated ribbons (a continuous fiber component). substantially circular it is generally meant that the aspect ratio of the core (height divided by the width) is typically from about 1.0 to about 1.5, and in some embodiments, about 1.0. Due to the selective control over the process used to impregnate the rovings and form a consolidated ribbon, as well the process for compressing and shaping the ribbon, the composite core may comprise a relatively even distribution of the thermoplastic matrix across along its entire length. the continuous fibers may be distributed in a generally uniform manner about a longitudinal central axis “L” of the composite core 516 . the rod 514 of the composite core 516 may include continuous fibers 526 embedded within a thermoplastic matrix 528 . the fibers 526 may be distributed generally uniformly about the longitudinal axis “L.” It should be understood that only a few fibers are shown in FIG. 8 , and that the composite core typically may contain a substantially greater number of uniformly distributed fibers. a capping layer 519 also may extend around the perimeter of the rod 514 and define an external surface of the composite core 516 . the cross-sectional thickness of the rod 514 may be selected strategically to help achieve a particular strength for the composite core. the rod 514 may have a thickness (e.g., diameter) of from about 0.1 to about 40 mm, in some embodiments from about 0.5 to about 30 mm, and in some embodiments, from about 1 to about 10 mm. the thickness of the capping layer 519 may depend on the intended function of the part, but typically may be from about 0.01 to about 10 mm, and in some embodiments, from about 0.02 to about 5 mm. the total cross-sectional thickness, or height, of the composite core 516 also may range from about 0.1 to about 50 mm, in some embodiments from about 0.5 to about 40 mm, and in some embodiments, from about 1 to about 20 mm (e.g., diameter, if a circular cross-section). While the composite core may be substantially continuous in length, the length of the composite core may be limited in practice by the spool onto which it will be wound and stored and/or by the length of the continuous fibers. For example, the length often may range from about 1,000 m to about 5,000 m, although even greater lengths are certainly possible. the composite cores may exhibit a relatively high flexural modulus. flexural modulus generally refers to the ratio of stress to strain in flexural deformation (units of force per unit area), or the tendency for a material to bend. It is determined from the slope of a stress-strain curve produced by a “three point flexural” test (such as ASTM D790-10, Procedure A, room temperature). the composite core of the present invention may exhibit a flexural modulus of from about 10 GPa or more, in some embodiments from about 12 to about 400 GPa, in some embodiments from about 15 to about 200 GPa, and in some embodiments, from about 20 to about 150 GPa. Composite cores used to produce electrical cables consistent with certain embodiments disclosed herein may have ultimate tensile strengths over about 300 MPa, such as, for instance, in a range from about 400 MPa to about 5,000 MPa, or from about 500 MPa to about 3,500 MPa. Further, suitable composite cores may have an ultimate tensile strength in a range from about 700 MPa to about 3,000 MPa; alternatively, from about 900 MPa to about 1,800 MPa; or alternatively, from about 1,100 MPa to about 1,500 MPa. the term “ultimate tensile strength”generally refers to the maximum stress that a material can withstand while being stretched or pulled before breaking, and is the maximum stress reached on a stress-strain curve produced by a tensile test (such as ASTM D3916-08) at room temperature. the composite core may have a tensile modulus of elasticity, or elastic modulus, in a range from about 50 GPa to about 500 GPa, from about 70 GPa to about 400 GPa, from about 70 GPa to about 300 GPa, or from about 70 GPa to about 250 GPa. the composite core may have an elastic modulus in a range from about 70 GPa to about 200 GPa; alternatively, from about 70 GPa to about 150 GPa; or alternatively, from about 70 GPa to about 130 GPa. tensile modulus of elasticity generally refers to the ratio of tensile stress over tensile strain and is the slope of a stress-strain curve produced by a tensile test (such as ASTM 3916-08) at room temperature. Composite cores made according to the present disclosure may further have relatively high flexural fatigue life, and may exhibit relatively high residual strength. Flexural fatigue life and residual flexural strength may be determined based on a “three point flexural fatigue” test (such as ASTM D790, typically at room temperature). For example, the cores of the present invention may exhibit residual flexural strength after one million cycles at 160 Newtons (“N”) or 180 N loads of from about 60 kilograms per square inch (“ksi”) to about 115 ksi, in some embodiments from about 70 ksi to about 115 ksi, and in some embodiments from about 95 ksi to about 115 ksi. Further, the cores may exhibit relatively minimal reductions in flexural strength. N three point flexural fatigue cores having void fractions of about 4% or less, in some embodiments about 3% or less may exhibit reductions in flexural strength after three point flexural fatigue testing of about 1% (for example, from a maximum pristine flexural strength of about 106 ksi to a maximum residual flexural strength of about 105 ksi). Flexural strength may be tested before and after fatigue testing using, for example, a three point flexural test as discussed above. the composite core may have a density or specific gravity of less than about 2.5 g/cc, less than about 2.2 g/cc, less than about 2 g/cc, or less than about 1.8 g/cc. the composite core density may be in a range from about 1 g/cc to about 2.5 g/cc; alternatively, from about 1.1 g/cc to about 2.2 g/cc; alternatively, from about 1.1 g/cc to about 2 g/cc; alternatively, from about 1.1 g/cc to about 1.9 g/cc; alternatively, from about 1.2 g/cc to about 1.8 g/cc; or alternatively, from about 1.3 g/cc to about 1.7 g/cc. the strength to weight ratio of the composite core may be important. the ratio may be quantified by the ratio of the tensile strength of the core material to the density of the core material (in units of MPa/(g/cc)). Illustrative and non-limiting strength to weight ratios of composite cores in accordance with embodiments of the present invention may be in a range from about 400 to about 1,300, from about 400 to about 1,200, from about 500 to about 1,100, from about 600 to about 1,100, from about 700 to about 1,100, from about 700 to about 1,000, or from about 750 to about 1,000. Again, the ratios are based on the tensile strength in MPa, and the composite core density in g/cc. the percent elongation at break for the composite core may be less than 4%, less than 3%, or less than 2%, while in other embodiments, the elongation at break may be in a range from about 0.5% to about 2.5%, from about 1% to about 2.5%, or from about 1% to about 2%. the linear thermal expansion coefficient of the composite core may be less than about 5 ⁇ 10 ⁇ 6 /° C., less than about 4 ⁇ 10 ⁇ 6 /° C., less than about 3 ⁇ 10 ⁇ 6 /° C., or less than about 2 ⁇ 10 ⁇ 6 /° C. (or in units of m/m/° C.). the linear thermal expansion coefficient may be, on a ppm basis per ° C., less than about 5, less than about 4, less than about 3, or less than about 2. the coefficient (ppm/° C.)may be in a range from about ⁇ 0.4 to about 5; alternatively, from about ⁇ 0.2 to about 4; alternatively, from about 0.4 to about 4; or alternatively, from about 0.2 to about 2. the temperature range contemplated for this linear thermal expansion coefficient may be generally in the ⁇ 50° C. to 200° C. range, the 0° C. to 200° C. range, the 0° C. to 175° C. range, or the 25° C. to 150° C. range. the linear thermal expansion coefficient is measured in the longitudinal direction, i.e., along the length of the fibers. the composite core also may exhibit a relatively small “bending radius”, which is the minimum radius that the rod can be bent without damage and is measured to the inside curvature of the composite core or composite strand. a smaller bend radius means that the composite core may be more flexible and may be spooled onto a smaller diameter bobbin. This property also may permit easier substitution of the composite core in cables that currently use metal cores, and allow for the use of tools and installation methods presently in use in conventional overhead transmission cables. the bending radius for the composite core may, in some embodiments, be in a range from about 1 cm to about 60 cm, from about 1 cm to about 50 cm, from about 1 cm to about 50 cm, or from about 2 cm to about 45 cm, as determined at a temperature of about 25° C. the bending radius may be in a range from about 2 cm to about 40 cm, or from about 3 cm to about 40 cm in certain embodiments contemplated herein. In other embodiments, bending radiuses may be achieved that are less than about 40 times the outer diameter of the composite core, in some embodiments from about 1 to about 30 times the outer diameter of the composite core, and in some embodiments, from about 2 to about 25 times the outer diameter of the composite core, determined at a temperature of about 25° C. the strength, physical, and thermal properties of the composite core referenced above also may be maintained over a relatively wide temperature range, such as from about ⁇ 50° C. to about 300° C., from about 100° C. to about 300° C., from about 110° C. to about 250° C., from about 120° C. to about 200° C., from about 150° C. to about 200° C., or from about 180° C. to about 200° C. the composite core also may have a low void fraction, such as about 6% or less, in some embodiments about 3% or less, in some embodiments about 2% or less, in some embodiments about 1% or less, and in some embodiments, about 0.5% or less. the void fraction may be determined in the manner described above, such as using a “resin burn off” test in accordance with ASTM D 2584-08 or through the use of computed tomography (CT) scan equipment, such as a Metrotom 1500 (2k ⁇ 2k) high resolution detector. CT computed tomography a composite core of the present invention may be characterized by the following properties: an ultimate tensile strength in a range from about 700 MPa to about 3,500 MPa; an elastic modulus from about 70 GPa to about 300 GPa; and a linear thermal expansion coefficient (in units of ppm per ° C.) in a range from about ⁇ 0.4 to about 5. the composite core may have a density of less than about 2.5 g/cc and/or a strength to weight ratio (in units of MPa/(g/cc)) in a range from about 500 to about 1,100. the composite core may have a bending radius in a range from about 1 cm to about 50 cm. Still further, the composite core may have a percent elongation at break of less than about 3%. a composite core of the present invention may be characterized by the following properties: an ultimate tensile strength in a range from about 1,100 MPa to about 1,500 MPa; an elastic modulus in a range about 70 GPa to about 130 GPa; and a linear thermal expansion coefficient (in units of ppm per ° C.) in a range from about 0.2 to about 2. the composite core may have a density in a range from about 1.2 g/cc to about 1.8 g/cc and/or a strength to weight ratio (in units of MPa/(g/cc)) in a range from about 700 to about 1,100. the composite core may have a bending radius in a range from about 2 cm to about 40 cm. the composite core may have a percent elongation at break in a range from about 1% to about 2.5%. the particular composite core embodiments described above are merely exemplary of the numerous designs that may be within the scope of the present invention. additional layers of material may be employed in addition to those described above. Such multi-component cores may be particularly useful in increasing overall strength without requiring the need for more expensive high strength materials for the entire core. the lower and/or higher strength components may comprise ribbon(s) that contain continuous fibers embedded within a thermoplastic matrix. the composite cores may contain various other components depending on the desired application and its required properties. the additional components may be formed from a continuous fiber ribbon, such as described herein, as well as other types of materials. the composite core may contain a layer of discontinuous fibers (e.g., short fibers, long fibers, etc.) to improve its transverse strength. the discontinuous fibers may be oriented so that at least a portion of these fibers may be positioned at an angle relative to the direction in which the continuous fibers extend. electrical cables of the present invention may comprise a cable core comprising at least one composite core, and a plurality of conductive elements surrounding the cable core. the cable core may be a single composite core, incorporating any composite core design and accompanying physical and thermal properties provided above. the cable core may comprise two or more composite cores, or composite strands, having either the same or different designs, and either the same or different physical and thermal properties. These two or more composite cores may be assembled parallel to each other (straight), or stranded, e.g., about a central composite core member. an electrical cable may comprise a cable core comprising one composite core surrounded by a plurality of conductive elements an electrical cable may comprise a cable core comprising two or more composite cores, the cable core surrounded by a plurality of conductive elements. the cable core may comprise, for instance, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, or 20 composite cores, or more (e.g., 37 composite cores), each of which may incorporate any composite core design and accompanying physical and thermal properties provided above. the composite cores may be arranged, bundled, or oriented in any suitable fashion, as would be Recognized by one of skill in the art. the composite cores can be stranded, such as a cable core comprising 7 stranded composite cores or 19 stranded composite cores. the composite cores can be parallel, such as a cable core comprising a bundle of 7 composite cores aligned parallel to each other. the electrical cable may comprise a plurality of conductive elements surrounding the cable core (e.g., a single composite core, a plurality of stranded composite cores). the conductive elements may be of any geometric shape, and may be round/circular wires or trapezoidal wires, among others, and including combinations thereof. the conductive elements may be in one layer, or 2 layers, or 3 layers, or 4 layers, and so forth, around the cable core. the conductive elements may be configured parallel to the cable core, or wrapped helically, or in any other suitable arrangement. Any number of conductive elements (e.g., wires) may be used, but a typical number of conductive elements in a cable may be up to 84 conductive elements, and often in a range from 2 to about 50. For instance, in some common conductor arrangements, 7, 19, 26, or 37 wires may be employed. the electrical cable 420 may include a plurality of conductive elements 422 (e.g., aluminum or an alloy thereof) radially disposed about a substantially cylindrical cable core 400 , which is illustrated as a single composite core but could be a plurality of stranded composite cores. the conductive elements may be arranged in a single layer or in multiple layers. the conductive elements 422 are arranged to form a first concentric layer 426 and a second concentric layer 428 . the shape of the conductive elements 422 also may be varied around the cable core 400 . the conductive elements 422 have a generally trapezoidal cross-sectional shape. Other shapes also may be employed, such as circular, elliptical, rectangular, square, etc. the conductive elements 422 also may be twisted or wrapped around the cable core 400 in any desired geometrical configuration, such as in a helical manner. the electrical transmission cable 420 may include a plurality of conductive elements 422 (e.g., aluminum or an alloy thereof) radially disposed about a bundle of generally cylindrical composite cores 400 , which may be formed in accordance with the present invention. FIG. 10 illustrates six composite cores 400 surrounding a single core 400 , although any suitable number of composite cores 400 in any suitable arrangement is within the scope and spirit of the present disclosure and may be used as the cable core. a capping layer 519 also may extend around the perimeter of and define an external surface of each rod. the conductive elements may be arranged in a single layer or in multiple layers. the conductive elements 422 are arranged to form a first concentric layer 426 and a second concentric layer 428 . the shape of the conductive elements 422 also may be varied to optimize the number of elements that can be disposed about the cable core. the conductive elements 422 have a generally trapezoidal cross sectional shape. Other shapes also may be employed, such as circular, elliptical, rectangular, square, etc. the conductive elements 422 also may be twisted or wrapped around the cable core containing the bundle of composite cores 400 in any desired geometrical configuration, such as in a helical manner. the cross-sectional area of individual conductive elements may vary considerably, but generally the cross-sectional area of individual elements may be in a range from about 10 to about 50 mm 2 , or from about 15 to about 45 mm 2 . the overall conductor area may range (in kcmil), for instance, from about 167 to about 3500 kcmil, from about 210 to about 2700 kcmil, from about 750 to about 3500 kcmil, or from about 750 to about 3000 kcmil. Overall conductor areas of about 795, about 825, about 960, and about 1020 kcmil often may be employed in many end-uses for electrical cables, such as in overheard power transmission lines. a common aluminum conductor steel reinforced cable known in the industry is often referred to as the 795 kcmil ACSR “Drake” conductor cable. the outside diameter of cables in accordance with the present invention is not limited to any particular range. However, typical cable outside diameters may be within a range, for example, of from about 7 to about 50 mm, from about 10 to about 48 mm, from about 20 to about 40 mm, from about 25 to about 35 mm, or from about 28 to about 30 mm. the cross-sectional area of a composite core in the cable is not limited to any particular range. However, typical cross-sectional areas of the composite core may be within a range from about 20 to about 140 mm 2 , or from about 30 to about 120 mm 2 . the conductive elements may be made from any suitable conductive or metal material, including various alloys. the conductive elements may comprise copper, a copper alloy, aluminum, an aluminum alloy, or combinations thereof. the term “aluminum or an aluminum alloy”is meant to collectively refer to grades of aluminum or aluminum alloys having at least 97% aluminum by weight, at least 98% aluminum by weight, or at least 99% aluminum by weight, including pure or substantially pure aluminum. Aluminum alloys or grades of aluminum having an IACS electrical conductivity of at least 57%, at least 58%, at least 59%, at least 60%, or at least 61% (e.g., 59% to 65%)may be employed in embodiments disclosed herein, and this is inclusive of any method that could produce such conductivities (e.g., annealing, tempering, etc.). aluminum 1350 alloy may be employed as the aluminum or aluminum alloy in certain embodiments of this invention. Aluminum 1350, its composition, and its minimum IACS,are described in ASTM B233, the disclosure of which is incorporated herein by reference in its entirety. the sag of the electrical cable may be an important feature. Sag is generally considered to be the distance that a cable departs from a straight line between the end points of a span. the sag across a span of towers may affect the ground clearance, and subsequently, the tower height and/or the number of towers needed. Sag generally may increase with the square of the span length, but may be reduced by an increase in tensile strength of the cable and/or a decrease in weight of the cable. Electrical cables in some embodiments of the present invention may have a sag (at rated temperature (180° C.), and for a 300-meter level span) with a NESC light loading of from about 3 to about 9.5 m, from about 4.5 to about 9.5 m, from about 5.5 to about 8 m, or from about 6 to about 7.5 m. the sag may be in a range from about 3 to about 9.5 m, from about 3 to about 7.5 m, from about 4.5 to about 7.5 m, or from about 5 to about 7 m. the cable also may be characterized as having a stress parameter of about 10 MPa or more, in some embodiments about 15 MPa or more, and in some embodiments, from about 20 to about 50 MPa. the method for determining the stress parameter is described in more detail in U.S. Pat. No. 7,093,416 to Johnson, et al., which is incorporated herein in its entirety by reference thereto for all purposes. sag and temperature may be measured and plotted as a graph of sag versus temperature. a calculated curve may be fitted to the measured data using an Alcoa Sag10 graphic method available in a software program from Southwire Company (Carrollton, Ga.) under the trade designation SAG10 (version 3.0 update 3.10.10). the stress parameter is a fitting parameter in SAG10 labeled as the “built-in aluminum stress”, which may be altered to fit other parameters, if a material other than aluminum is used (e.g., an aluminum alloy), and which adjusts the position of the knee-point on the predicted graph and also the amount of sag in the high temperature, post-knee-point regime. a description of the stress parameter also may be provided in the Sag10 Users Manual (Version 2.0), incorporated herein by reference in its entirety. creep is generally considered to be the permanent elongation of a cable under load over a long period of time. the amount of creep of a length of cable may be impacted by the length of time in service, the load on the cable, the tension on the cable, the encountered temperature conditions, amongst other factors. cables disclosed herein may have 10-year creep values at 15%, 20%, 25%, and/or 30% RBS (rated breaking stress) of less than about 0.25%, less than about 0.2%, or less than about 0.175%. the 10-year creep value at 15% RBS may be less than about 0.25%; alternatively, less than about 0.2%; alternatively, less than about 0.15%; alternatively, less than about 0.1%; or alternatively, less than about 0.075%. the 10-year creep value at 30% RBS may be less than about 0.25%; alternatively, less than about 0.225%; alternatively, less than about 0.2; or alternatively, less than about 0.175%. Electrical cables in accordance with embodiments of this invention may have a maximum operating temperature up to about 300° C., up to about 275° C., or up to about 250° C. Certain cables provided herein may have maximum operating temperatures that may be up to about 225° C.; alternatively, up to 200° C.; alternatively, up to 180° C.; or alternatively, up to 175° C. Maximum operating temperatures may be in a range from about 100 to about 300° C., from about 100 to about 250° C., from about 110 to about 250° C., from about 120 to about 200° C., or from about 120 to about 180° C., in various embodiments of the present invention. the electrical cable may have certain fatigue and/or vibrational resistance properties. the electrical cable may pass (meet or exceed) the Aeolian vibration test specified in IEEE 1138, incorporated herein by reference, at 100 million cycles. the electrical cable may comprise a partial or complete layer of a material between the cable core and the conductive elements. the material may be conductive or non-conductive, and may be a tape that partially or completely wraps/covers the cable core. the material may be configured to hold or secure the individual composite core elements of a cable core together. the material may comprise a metal or aluminum foil tape, a polymer tape (e.g., a polypropylene tape, a polyester tape, a Teflon tape, etc.), a tape with glass-reinforcement, and the like. a polymer tape e.g., a polypropylene tape, a polyester tape, a Teflon tape, etc. the thickness of the material may be in a range from about 0.025 mm to about 0.25 mm, although the thickness is not limited only to this range. the tape or other material may be applied so that each subsequent wrap overlaps the previous wrap. In another embodiment, the tape or other material may be applied so that each subsequent wrap leaves a gap between the previous wrap. In yet another embodiment, the tape or other material may be applied so that abuts the previous wrap with no overlap and no gap. In these and other embodiments, the tape or other material may be applied helically around the cable core. the electrical cable may comprise a partial or complete coating of a material between the cable core and the conductive elements. the material may be, or may comprise, a polymer. Suitable polymers may include, but are not limited to, a polyolefin (e.g., polyethylene and polypropylene homopolymers, copolymers, etc.), a polyester (e.g., polybutylene terephalate (PBT)), a polycarbonate, a polyamide (e.g., NylonTM), a polyether ketone (e.g., polyetherether ketone (PEEK)), a polyetherimide, a polyarylene ketone (e.g., polyphenylene diketone (PPDK)), a liquid crystal polymer, a polyarylene sulfide (e.g., polyphenylene sulfide (PPS), poly(biphenylene sulfide ketone), poly(phenylene s the polymer may be an elastomeric polymer. the coating may be conductive or non-conductive, and may contain various additives typically employed in wire and cable applications. the coating may serve, in some embodiments, as a protective coating for the cable core. the coating may be used in instances where the composite core does not contain a capping layer, and the coating partially or completely covers the rod (or fiber core), for example, as a protective coating for the rod. the coating may partially or completely fill the spaces between the individual core elements. the present invention also encompasses methods of making an electrical cable comprising a cable core and a plurality of conductive elements surrounding the cable core. electrical cables using various cable core configurations and conductor element configurations disclosed herein may be produced by any suitable method known to those of skill in the art. a rigid-frame strander which can rotate spools of composite cores or strands to assemble a cable core, may be employed. the rigid-frame strander may impart one twist per machine revolution into all composite cores or strands, except for the center composite core, which is not twisted. Each successive layer over the center composite core may be closed by a round die. the cable core containing the composite cores or strands may be secured with a tape or other material. If tape is employed, it may be applied using a concentric taping machine. The resulting cable core with tape may be taken-up onto a reel. The cable core then may be fed back through the same rigid-frame strander for the application of a plurality of conductive elements around the cable core. One such method of transmitting electricity may comprise (i) installing an electrical cable as disclosed herein, e.g., comprising a cable core and a plurality of conductive elements surrounding the cable core, and (ii) transmitting electricity across the electrical cable. Another method of transmitting electricity may comprise (i) providing an electrical cable as disclosed herein, e.g., comprising a cable core and a plurality of conductive elements surrounding the cable core, and (ii) transmitting electricity across the electrical cable. the electrical cable, cable core, and conductive elements may be any electrical cable, cable core, and conductive elements described herein. the cable core may comprise any composite core described herein, i.e., one or more composite cores or strands. Carbon fiber rovings(Toray T700SC, which contained 12,000 carbon filaments having a tensile strength of 4,900 MPa and a mass per unit length of 0.8 g/m) were employed for the continuous fibers with each individual ribbon containing 4 rovings. the thermoplastic polymer used to impregnate the fibers was polyphenylene sulfide (PPS) (FORTRON® PPS 205, available from Ticona, LLC), which has a melting point of about 280° C. PPS polyphenylene sulfide Each ribbon contained 50 wt. % carbon fibers and 50 wt. % PPS. the ribbons had a thickness of about 0.18 mm and a void fraction of less than 1.0%. the ribbons were then fed to a pultrusion line operating at a speed of 20 ft/min. Before shaping, the ribbons were heated within an infrared oven (power setting of 305). The heated ribbons were then supplied to a consolidation die having a circular-shaped channel that received the ribbons and compressed them together while forming the initial shape of the rod. Within the die, the ribbons remained at a temperature of about 177° C. the resulting preform was then briefly cooled with an air ring/tunnel device that supplied ambient air at a pressure of 1 psig. the preform was then passed through a nip formed between two rollers, and then to a calibration die for final shaping. Within the calibration die, the preform remained at a temperature of about 140° C. After exiting this die the profile was capped with a polyether ether ketone (PEEK), which had a melting point of 350° C. The capping layer had an average thickness of about 0.1-0.15 mm. The resulting part was then cooled with an air stream. The resulting composite core had an average outside diameter of about 3.4-3.6 mm, and contained 45 wt. % carbon fibers, 50 wt. % PPS, and 5 wt. % capping material. PEEK polyether ether ketone Two (2) continuous fiber ribbons were initially formed using an extrusion system as substantially described above. Carbon fiber rovings (Toray T700SC) were employed for the continuous fibers with each individual ribbon containing 4 rovings. the thermoplastic polymer used to impregnate the fibers was FORTRON® PPS 205. Each ribbon contained 50 wt. % carbon fibers and 50 wt. % PPS. the ribbons had a thickness of about 0.18 mm and a void fraction of less than 1.0%. the ribbons were then fed to a pultrusion line operating at a speed of 20 ft/min. Before shaping, the ribbons were heated within an infrared oven (power setting of 305). the heated ribbons were then supplied to a consolidation die having a circular-shaped channel that received the ribbons and compressed them together while forming the initial shape of the rod. Within the die, the ribbons remained at a temperature of about 343° C. the resulting preform was then briefly cooled with an air ring/tunnel device that supplied ambient air at a pressure of 1 psig. the preform was then passed through a nip formed between two rollers, and then to a calibration die for final shaping. Within the calibration die, the preform remained at a temperature of about 140° C. After exiting this die the profile was capped with FORTRON® PPS 320, which had a melting point of 280° C. the capping layer had an average thickness of about 0.1-0.15 mm. the resulting part was then cooled with an air stream. the resulting composite core had an average outside diameter of about 3.4-3.6 mm, and contained 45 wt. % carbon fibers, 50 wt. % PPS, and 5 wt. % capping material. Two (2) continuous fiber ribbons were initially formed using an extrusion system as substantially described above. Glass fiber rovings(TUFRov® 4588 from PPG, which contained E-glass filaments having a tensile strength of 2,599 MPa and a mass per unit length of 0.0044 lb/yd (2.2 g/m) were employed for the continuous fibers with each individual ribbon containing 2 rovings. the thermoplastic polymer used to impregnate the fibers was polyphenylene sulfide (PPS) (FORTRON® 205, available from Ticona, LLC), which has a melting point of about 280° C. PPS polyphenylene sulfide Each ribbon contained 56 wt. % glass fibers and 44 wt. % PPS. the ribbons had a thickness of about 0.18 mm and a void fraction of less than 1.0%. the ribbons were then fed to a pultrusion line operating at a speed of 20 ft/min. Before shaping, the ribbons were heated within an infrared oven (power setting of 330). The heated ribbons were then supplied to a consolidation die having a circular-shaped channel that received the ribbons and compressed them together while forming the initial shape of the rod. Upon consolidation, the resulting preform was then briefly cooled with ambient air. The preform was then passed through a nip formed between two rollers, and then to a calibration die for final shaping. Within the calibration die, the preform remained at a temperature of about 275° C. the profile was capped with FORTRON® 205. the capping layer had an average thickness of about 0.1-0.15 mm. the resulting part was then cooled with an air stream. the resulting composite core had an average outside diameter of about 3.4-3.6 mm, and contained 50 wt. % glass fibers and 50 wt. % PPS. Two (2) continuous fiber ribbons were initially formed using an extrusion system as substantially described above. Glass fiber rovings (TUFRov® 4588) were employed for the continuous fibers with each individual ribbon containing 2 rovings. the thermoplastic polymer used to impregnate the fibers was Nylon 66 (PA66), which has a melting point of about 250° C. PA66 Nylon 66 Each ribbon contained 60 wt. % glass fibers and 40 wt. % Nylon 66. the ribbons had a thickness of about 0.18 mm and a void fraction of less than 1.0%. Once formed, the ribbons were then fed to a pultrusion line operating at a speed of 10 ft/min. the ribbons were heated within an infrared oven (power setting of 320). the heated ribbons were then supplied to a consolidation die having a circular-shaped channel that received the ribbons and compressed them together while forming the initial shape of the rod. the resulting preform was then briefly cooled with ambient air. the preform was then passed through a nip formed between two rollers, and then to a calibration die for final shaping. Within the calibration die, the preform remained at a temperature of about 170° C. the profile was capped with Nylon 66. the capping layer had an average thickness of about 0.1-0.15 mm. the resulting part was then cooled with an air stream. the resulting composite core had an average outside diameter of about 3.4-3.6 mm, and contained 53 wt. % glass fibers, 40 wt. % Nylon 66, and 7 wt. % capping material. Three (3) batches of eight (8) cores were formed having different void fraction levels. two (2) continuous fiber ribbons were initially formed using an extrusion system as substantially described above. Carbon fiber rovings(Toray T700SC, which contained 12,000 carbon filaments having a tensile strength of 4,900 MPa and a mass per unit length of 0.8 g/m) were employed for the continuous fibers with each individual ribbon containing 4 rovings. the thermoplastic polymer used to impregnate the fibers was polyphenylene sulfide (“PPS”) (FORTRON® PPS 205, available from Ticona, LLC), which had a melting point of about 280° C. Each ribbon contained 50 wt. % carbon fibers and 50 wt. the ribbons had a thickness of about 0.18 mm and a void fraction of less than 1.0%. Once formed, the ribbons were then fed to a pultrusion line operating at a speed of 20 ft/min. Before shaping, the ribbons were heated within an infrared oven (power setting of 305). The heated ribbons were then supplied to a consolidation die having a circular-shaped channel that received the ribbons and compressed them together while forming the initial shape of the rod. Within the die, the ribbons remained at a temperature of about 177° C. Upon consolidation, the resulting preform was then briefly cooled with an air ring/tunnel device that supplied ambient air at a pressure of 1 psi. the preform was then passed through a nip formed between two rollers, and then to a calibration die for final shaping. Within the calibration die, the preform remained at a temperature of about 140° C. After exiting this die, the profile was capped with a polyether ether ketone (“PEEK”), which had a melting point of 350° C. The capping layer had a thickness of about 0.1 mm. PEEK polyether ether ketone the resulting composite core was then cooled with an air stream. the resulting composite core had a diameter of about 3.5 mm, and contained 45 wt. % carbon fibers, 50 wt. % PPS, and 5 wt. % capping material. a first batch of composite cores had a mean void fraction of 2.78%. a second batch of composite cores had a mean void fraction of 4.06%. a third batch of composite cores had a mean void fraction of 8.74%. Void fraction measurements were performed using CT scanning. a Metrotom 1500 (2k ⁇ 2k) high resolution detector was used to scan the core specimens. Detection was done using an enhanced analysis mode with a low probability threshold. Once the specimens were scanned for void fraction, Volume Graphics software was used to interpret the data from the 3D scans, and calculate the void levels in each specimen. FIG. 15 illustrates the electrical cable 520 produced in Example 6. the 26 conductive elements 522 formed a first layer 526 and a second layer 528 . the cable core 500 was a strand of 7 composite cores. a tape 530 between the cable core 500 and the conductive elements 522 partially covered the cable core 500 in a helical arrangement. Electrical cable was produced as follows. Seven (7) composite cores having a diameter of about 3.5 mm were stranded to form a stranded cable core with a 508-mm lay length. The composite cores were similar to those produced in Example 1 above. The cable core was secured with an aluminum foil tape laminated to a fiberglass scrim and a silicone based adhesive. 26 conductor wires were placed above and around the cable core and tape in two layers as shown in FIG. 15 . The conductor wires had a diameter of about 4.5 mm, and were fabricated from fully annealed 1350 aluminum. The ultimate tensile strength of the cable was approximately 19,760 psi (136 MPa). FIG. 15 illustrates the electrical cable 520 produced in Example 7. the 26 conductive elements 522 formed a first layer 526 and a second layer 528 . the cable core 500 was a strand of 7 composite cores. a tape 530 between the cable core 500 and the conductive elements 522 partially covered the cable core 500 in a helical arrangement. FIG. 16 illustrates the stress-strain data for the electrical cable of Example 7. the electrical cable of Example 7 was tested for its fatigue and/or vibrational resistance properties in accordance with the Aeolian vibration test specified in IEEE 1138. the electrical cable of Example 7 passed the Aeolian vibration test at 100 million cycles. Example 7 Using mathematical modeling based on overhead transmission cables similar to Example 7, the 10-year creep (elongation) values for the electrical cable of Example 7 were estimated. The calculated 10-year creep values at 15%, 20%, 25%, and 30% RBS (rated breaking stress) were approximately 0.054%, approximately 0.081%, approximately 0.119%, and approximately 0.163%, respectively. FIG. 17 illustrates an electrical cable 620 that can be produced in Constructive Example 8. the 26 conductive elements 622 can form a first layer 626 and a second layer 628 . the cable core 600 can be a strand of 7 composite cores. a tape 630 between the cable core 600 and the conductive elements 622 can partially cover the cable core 600 in a helical arrangement. the electrical cable of FIG. 17 can be produced as follows. Seven (7) composite cores having a diameter of about 3.5 mm can be stranded to form a stranded cable core with a 508-mm lay length. The composite cores can be similar to those produced in Example 1 above. the cable core can be secured with an aluminum foil tape laminated to a fiberglass scrim and a silicone based adhesive. 26 conductor wires can be placed above and around the cable core and tape in two layers as shown in FIG. 17 . the conductors can be trapezoidal wires having a cross-sectional area of about 15-17 mm 2 , and can be fabricated from annealed 1350 aluminum (or alternatively, an aluminum alloy containing zirconium). Landscapes Spectroscopy & Molecular Physics (AREA) Physics & Mathematics (AREA) Engineering & Computer Science (AREA) Chemical & Material Sciences (AREA) Mechanical Engineering (AREA) Composite Materials (AREA) Dispersion Chemistry (AREA) Manufacturing & Machinery (AREA) Reinforced Plastic Materials (AREA) Insulated Conductors (AREA) Conductive Materials (AREA) Laminated Bodies (AREA) Ropes Or Cables (AREA) Organic Insulating Materials (AREA) Communication Cables (AREA) Abstract The present invention discloses electrical cables containing a cable core and a plurality of conductive elements surrounding the cable core. The cable core contains at least one composite core, and each composite core contains a rod which contains a plurality of unidirectionally aligned fiber rovings embedded within a thermoplastic polymer matrix, and surrounded by a capping layer. Description REFERENCE TO RELATED APPLICATIONS This application is a continuation application of U.S. patent application Ser. No. 14/661,780, filed on Mar. 18, 2015, now U.S. Pat. No. 9,443,635, which is a continuation application of U.S. patent application Ser. No. 13/443,938, filed on Apr. 11, 2012, now U.S. Pat. No. 9,012,781, which claims the benefit of U.S. Provisional Application No. 61/474,423, filed on Apr. 12, 2011, and relates to U.S. Provisional Application No. 61/474,458, filed on Apr. 12, 2011, all of which are incorporated herein by reference in their entirety. BACKGROUND OF THE INVENTION Composite wire structures are commonly used as transmission lines or cables for transmitting electricity to users. Examples of composite transmission wire constructions include, for instance, aluminum conductor steel reinforced (ACSR) cable, aluminum conductor steel supported (ACSS) cable, aluminum conductor composite reinforced (ACCR) cable, and aluminum conductor composite core (ACCC) cable. ACSR and ACSS cables include an aluminum outer conducting layer surrounding a steel inner core. The transmission lines or cables are designed not only to efficiently transmit electricity, but also to be strong and temperature resistant, especially when the transmission lines are strung on towers and stretched over long distances. It would be beneficial to produce cables with a composite core that are capable of achieving the desired strength, durability, and temperature performance demanded by applications such as overhead power transmission cables. Accordingly, it is to these ends that the present disclosure is directed. SUMMARY OF THE INVENTION This summary is provided to introduce a selection of concepts in a simplified form that are further described below in the detailed description. This summary is not intended to identify required or essential features of the claimed subject matter. Nor is this summary intended to be used to limit the scope of the claimed subject matter. Embodiments of the present invention may provide cables, e.g., electrical transmission cables for the overhead transmission of electricity, which may contain a cable core and conductive elements surrounding the cable core. The cable core may contain at least one composite core (the composite core also may be referred to as a composite strand or polymer composite strand). These core elements may serve as load-bearing members for the electrical transmission cable and, in some embodiments, these core elements may be non-conductive. In accordance with one embodiment of the present invention, a composite core for the electrical cable is disclosed. Generally, the cables and cores disclosed herein may extend in a longitudinal direction. The composite core may comprise at least one rod that comprises a continuous fiber component comprising a plurality of consolidated thermoplastic impregnated rovings (the rod also may be referred to as a fiber core). The rovings may contain continuous fibers oriented in the longitudinal direction, and a thermoplastic matrix that embeds the fibers. The fibers may have a ratio of ultimate tensile strength to mass per unit length of greater than about 1,000 Megapascals per gram per meter (MPa/g/m). The continuous fibers may constitute from about 25 wt. % to about 80 wt. % of the rod and the thermoplastic matrix may constitute from about 20 wt. % to about 75 wt. % of the rod. A capping layer may surround the rod, and this capping layer may be free of continuous fibers. The composite core may have a minimum flexural modulus of about 10 Gigapascals (GPa). In accordance with another embodiment of the present invention, a method for forming a composite core for an electrical transmission cable is disclosed. The method may comprise impregnating a plurality of rovings with a thermoplastic matrix and consolidating the rovings to form a ribbon, wherein the rovings may comprise continuous fibers oriented in the longitudinal direction. The fibers may have a ratio of ultimate tensile strength to mass per unit length of greater than about 1,000 MPa/g/m. The continuous fibers may constitute from about 25 wt. % to about 80 wt. % of the ribbon, and the thermoplastic matrix may constitute from about 20 wt. % to about 75 wt. % of the ribbon. The ribbon may be heated to a temperature at or above the softening temperature (or melting temperature) of the thermoplastic matrix and pulled through at least one forming die to compress and shape the ribbon into a rod. A capping layer may be applied to the rod to form the composite core. In accordance with yet another embodiment of the present invention, a method of making an electrical cable is disclosed. This method may comprise providing a cable core comprising at least one composite core, and surrounding the cable core with a plurality of conductive elements. The composite core may comprise at least one rod comprising a plurality of consolidated thermoplastic impregnated rovings. The rovings may comprise continuous fibers oriented in the longitudinal direction and a thermoplastic matrix that embeds the fibers. The fibers may have a ratio of ultimate tensile strength to mass per unit length of greater than about 1,000 MPa/g/m. Typically, the rod may comprise from about 25 wt. % to about 80 wt. % fibers, and from about 20 wt. % to about 75 wt. % thermoplastic matrix. A capping layer may surround the at least one rod, and this capping layer generally may be free of continuous fibers. In these and other embodiments, the composite core may have a flexural modulus of greater than about 10 GPa. Both the foregoing summary and the following detailed description provide examples and are explanatory only. Accordingly, the foregoing summary and the following detailed description should not be considered to be restrictive. Further, features or variations may be provided in addition to those set forth herein. For example, certain aspects and embodiments may be directed to various feature combinations and sub-combinations described in the detailed description. BRIEF DESCRIPTION OF THE DRAWINGS The accompanying drawings, which are incorporated in and constitute a part of this disclosure, illustrate various aspects and embodiments of the present invention. In the drawings: FIG. 1 is a perspective view of one embodiment of a consolidated ribbon for use in the present invention; FIG. 2 is a cross-sectional view of another embodiment of a consolidated ribbon for use in the present invention; FIG. 3 is a schematic illustration of one embodiment of an impregnation system for use in the present invention; FIG. 4 is a cross-sectional view of the impregnation die shown in FIG. 3; FIG. 5 is an exploded view of one embodiment of a manifold assembly and gate passage for an impregnation die that may be employed in the present invention; FIG. 6 is a perspective view of one embodiment of a plate at least partially defining an impregnation zone that may be employed in the present invention; FIG. 7 is a schematic illustration of one embodiment of a pultrusion system that may be employed in the present invention; FIG. 8 is a perspective view of one embodiment of a composite core of the present invention; and FIG. 9 is a perspective view of one embodiment of an electrical transmission cable of the present invention; FIG. 10 is a perspective view of another embodiment of an electrical transmission cable of the present invention; FIG. 11 is a top cross-sectional view of one embodiment of various calibration dies that may be employed in accordance with the present invention; FIG. 12 is a side cross-sectional view of one embodiment of a calibration die that may be employed in accordance with the present invention; FIG. 13 is a front view of a portion of one embodiment of a calibration die that may be employed in accordance with the present invention; FIG. 14 is a front view of one embodiment of forming rollers that may be employed in accordance with the present invention; FIG. 15 is a perspective view of the electrical cable of Examples 6-7; FIG. 16 is a stress-strain diagram for the electrical cable of Example 7; and FIG. 17 is a perspective view of the electrical cable of Constructive Example 8. DETAILED DESCRIPTION OF THE INVENTION The following detailed description refers to the accompanying drawings. Wherever possible, the same or similar reference numbers are used in the drawings and the following description to refer to the same or similar elements or features. While aspects and embodiments of the invention may be described, modifications, adaptations, and other implementations are possible. For example, substitutions, additions, or modifications may be made to the elements illustrated in the drawings, and the methods described herein may be modified by substituting, reordering, or adding stages to the disclosed methods. Accordingly, the following detailed description and its exemplary embodiments do not limit the scope of the invention. The present invention is directed generally to electrical cables, such as high voltage overhead transmission lines, and to the composite cores contained within these electrical cables. In certain embodiments of the invention, an electrical cable may comprise a cable core comprising at least one composite core (or composite strand), and a plurality of conductive elements surrounding the cable core. Composite Core The composite core may contain a rod (or fiber core) comprising a continuous fiber component, surrounded by a capping layer. The rod may comprise a plurality of unidirectionally aligned fiber rovings embedded within a thermoplastic polymer matrix. While not wishing to be bound by theory, applicants believe that the degree to which the rovings are impregnated with the thermoplastic polymer matrix may be significantly improved through selective control over the impregnation process, and also through control over the degree of compression imparted to the rovings during formation and shaping of the rod, as well as the calibration of the final rod geometry. Such a well impregnated rod may have a very small void fraction, which may lead to excellent strength properties. Notably, the desired strength properties may be achieved without the need for different fiber types in the rod. As used herein, the term “roving” generally refers to a bundle or tow of individual fibers. The fibers contained within the roving may be twisted or may be straight. Although different fibers may be used in individual or different rovings, it may be beneficial for each of the rovings to contain a single fiber type to minimize any adverse impact of using materials having different thermal expansion coefficients. The continuous fibers employed in the rovings may possess a high degree of tensile strength relative to their mass. For example, the ultimate tensile strength of the fibers typically may be in a range from about 1,000 to about 15,000 Megapascals (MPa), in some embodiments from about 2,000 MPa to about 10,000 MPa, and in some embodiments, from about 3,000 MPa to about 6000 MPa. Such tensile strengths may be achieved even though the fibers are of a relatively light weight, such as a mass per unit length of from about 0.1 to about 2 grams per meter (g/m), in some embodiments from about 0.4 to about 1.5 g/m. The ratio of tensile strength to mass per unit length thus may be about 1,000 Megapascals per gram per meter (MPa/g/m) or greater, in some embodiments about 4,000 MPa/g/m or greater, and in some embodiments, from about 5,500 to about 20,000 MPa/g/m. Such high strength fibers may, for instance, be metal fibers, glass fibers (e.g., E-glass, A-glass, C-glass, D-glass, AR-glass, R-glass, S1-glass, S2-glass, etc.), carbon fibers (e.g., amorphous carbon, graphitic carbon, or metal-coated carbon, etc.), boron fibers, ceramic fibers (e.g., alumina or silica), aramid fibers (e.g., Kevlar® marketed by E. I. duPont de Nemours, Wilmington, Del.), synthetic organic fibers (e.g., polyamide, polyethylene, paraphenylene, terephthalamide, polyethylene terephthalate and polyphenylene sulfide), and various other natural or synthetic inorganic or organic fibrous materials known for reinforcing thermoplastic and/or thermoset compositions. Carbon fibers may be particularly suitable for use as the continuous fibers, which typically have a tensile strength to mass per unit length ratio in the range of from about 5,000 to about 7,000 MPa/g/m. Often, the continuous fibers may have a nominal diameter of about 4 to about 35 micrometers (μm), and in some embodiments, from about 5 to about 35 μm. The number of fibers contained in each roving may be constant or may vary from roving to roving. Typically, a roving may contain from about 1,000 fibers to about 100,000 individual fibers, and in some embodiments, from about 5,000 to about 50,000 fibers. Any of a variety of thermoplastic polymers may be employed to form the thermoplastic matrix in which the continuous fibers are embedded. Suitable thermoplastic polymers for use in the present invention may include, for instance, polyolefins (e.g., polypropylene, propylene-ethylene copolymers, etc.), polyesters (e.g., polybutylene terephalate (PBT)), polycarbonates, polyam ides (e.g., Nylon™) polyether ketones (e.g., polyetherether ketone (PEEK)), polyetherimides, polyarylene ketones (e.g., polyphenylene diketone (PPDK)), liquid crystal polymers, polyarylene sulfides (e.g., polyphenylene sulfide (PPS), poly(biphenylene sulfide ketone), poly(phenylene sulfide diketone), poly(biphenylene sulfide), etc.), fluoropolymers (e.g., polytetrafluoroethylene-perfluoromethylvinylether polymer, perfluoro-alkoxyalkane polymer, petrafluoroethylene polymer, ethylene-tetrafluoroethylene polymer, etc.), polyacetals, polyurethanes, polycarbonates, styrenic polymers (e.g., acrylonitrile butadiene styrene (ABS)), and the like, or combinations thereof. Generally, the properties of the thermoplastic matrix may be selected to achieve a desired combination of processability and end-use performance of the composite core. For example, the melt viscosity of the thermoplastic matrix generally may be low enough so that the polymer may adequately impregnate the fibers and become shaped into the rod configuration. In this regard, the melt viscosity typically may range from about 25 to about 2,000 Pascal-seconds (Pa-s), in some embodiments from 50 about 500 Pa-s, and in some embodiments, from about 60 to about 200 Pa-s, determined at the operating conditions used for the thermoplastic polymer (e.g., about 360° C.). Likewise, because the core may be used at high temperatures (e.g., in high voltage transmission cables), a thermoplastic polymer having a relatively high melting temperature may be employed. For example, the melting temperature of such high temperature polymers may be in a range from about 200° C. to about 500° C., in some embodiments from about 225° C. to about 400° C., and in some embodiments, from about 250° C. to about 350° C. In particular embodiments contemplated herein, polyarylene sulfides may be used in the present invention as a high temperature matrix with the desired melt viscosity. Polyphenylene sulfide, for example, is a semi-crystalline resin that generally includes repeating monomeric units represented by the following general formula: These monomeric units may constitute at least 80 mole %, and in some embodiments, at least 90 mole %, of the recurring units, in the polymer. It should be understood, however, that the polyphenylene sulfide may contain additional recurring units, such as described in U.S. Pat. No. 5,075,381 to Gotoh, et al., which is incorporated herein in its entirety by reference thereto for all purposes. When employed, such additional recurring units typically may constitute less than about 20 mole % of the polymer. Commercially available high melt viscosity polyphenylene sulfides may include those available from Ticona, LLC (Florence, Ky.) under the trade designation FORTRON®. Such polymers may have a melting temperature of about 285° C. (determined according to ISO 11357-1,2,3) and a melt viscosity of from about 260 to about 320 Pa-s at 310° C. According to the present invention, an extrusion device generally may be employed to impregnate the rovings with the thermoplastic matrix. Among other things, the extrusion device may facilitate the application of the thermoplastic polymer to the entire surface of the fibers. The impregnated rovings also may have a very low void fraction, which may increase the resulting strength of the rod. For instance, the void fraction may be about 6% or less, in some embodiments about 4% or less, in some embodiments about 3% or less, in some embodiments about 2% or less, in some embodiments about 1% or less, and in some embodiments, about 0.5% or less. The void fraction may be measured using techniques well known to those skilled in the art. For example, the void fraction may be measured using a “resin burn off” test in which samples are placed in an oven (e.g., at 600° C. for 3 hours) to burn off the resin. The mass of the remaining fibers may then be measured to calculate the weight and volume fractions. Such “burn off” testing may be performed in accordance with ASTM D 2584-08 to determine the weights of the fibers and the thermoplastic matrix, which may then be used to calculate the “void fraction” based on the following equations: V f=100(ρ t−ρ c)/ρ t where, V f is the void fraction as a percentage; ρ c is the density of the composite as measured using known techniques, such as with a liquid or gas pycnometer (e.g., helium pycnometer); ρ t is the theoretical density of the composite as is determined by the following equation: ρ t=1/[W f/ρ f +W m/ρ m] ρ m is the density of the thermoplastic matrix (e.g., at the appropriate crystallinity); ρ f is the density of the fibers; W f is the weight fraction of the fibers; and W m is the weight fraction of the thermoplastic matrix. Alternatively, the void fraction may be determined by chemically dissolving the resin in accordance with ASTM D 3171-09. The “burn off” and “dissolution” methods may be particularly suitable for glass fibers, which are generally resistant to melting and chemical dissolution. In other cases, however, the void fraction may be indirectly calculated based on the densities of the thermoplastic polymer, fibers, and ribbon (or tape) in accordance with ASTM D 2734-09 (Method A), where the densities may be determined by ASTM D792-08 Method A. Of course, the void fraction also may be estimated using conventional microscopy equipment, or through the use of computed tomography (CT) scan equipment, such as a Metrotom 1500 (2k×2k) high resolution detector. Referring to FIG. 3, one embodiment of an extrusion device is shown. More particularly, the apparatus may include an extruder120 containing a screw shaft124 mounted inside a barrel122. A heater 130 (e.g., an electrical resistance heater) may be mounted outside the barrel122. During use, a thermoplastic polymer feedstock127 may be supplied to the extruder120 through a hopper126. The thermoplastic feedstock127 may be conveyed inside the barrel122 by the screw shaft124 and heated by frictional forces inside the barrel122 and by the heater130. Upon being heated, the feedstock127 may exit the barrel122 through a barrel flange128 and enter a die flange132 of an impregnation die150. A continuous fiber roving 142 or a plurality of continuous fiber rovings142 may be supplied from a reel or reels144 to die 150. Generally, the rovings142 may be kept apart a certain distance before impregnation, such as at least about 4 mm, and in some embodiments, at least about 5 mm. The feedstock127 may further be heated inside the die by heaters133 mounted in or around the die150. The die generally may be operated at temperatures that are sufficient to cause melting and impregnation of the thermoplastic polymer. Typically, the operation temperatures of the die may be higher than the melt temperature of the thermoplastic polymer, such as at temperatures from about 200° C. to about 450° C. When processed in this manner, the continuous fiber rovings142 may become embedded in the polymer matrix, which may be a resin 214 (FIG. 4) processed from the feedstock127. The mixture then may be extruded from the impregnation die 150 to create an extrudate152. A pressure sensor 137 (FIG. 3) may monitor the pressure near the impregnation die 150, so that the extruder120 can be operated to deliver a correct amount of resin214 for interaction with the fiber rovings142. The rate of extrusion may be varied by controlling the rotational speed of the screw shaft124 and/or feed rate of the feedstock127. The extruder120 may be operated to produce the extrudate 152 (impregnated fiber rovings), which after leaving the impregnation die 150, may enter an optional pre-shaping or guiding section (not shown), before entering a nip formed between two adjacent rollers190. The rollers190 may help to consolidate the extrudate152 into the form of a ribbon (or tape), as well as to enhance fiber impregnation and to squeeze out any excess voids. In addition to the rollers190, other shaping devices also may be employed, such as a die system. The resulting consolidated ribbon156 may be pulled by tracks162 and 164 mounted on rollers. The tracks162 and 164 also may pull the extrudate152 from the impregnation die 150 and through the rollers190. If desired, the consolidated ribbon156 may be wound up at a section171. Generally speaking, the ribbons may be relatively thin and may have a thickness of from about 0.05 to about 1 millimeter (mm), in some embodiments from about 0.1 to about 0.8 mm, and in some embodiments, from about 0.2 to about 0.4 mm. Within the impregnation die, it may be beneficial that the rovings142 are traversed through an impregnation zone250 to impregnate the rovings with the polymer resin214. In the impregnation zone250, the polymer resin may be forced generally transversely through the rovings by shear and pressure created in the impregnation zone250, which may significantly enhance the degree of impregnation. This may be particularly useful when forming a composite from ribbons of a high fiber content, such as about 35% weight fraction (Wf) or more, and in some embodiments, from about 40% Wf or more. Typically, the die150 may include a plurality of contact surfaces 252, such as, for example, at least 2, at least 3, from 4 to 7, from 2 to 20, from 2 to 30, from 2 to 40, from 2 to 50, or more contact surfaces 252, to create a sufficient degree of penetration and pressure on the rovings142. Although their particular form may vary, the contact surfaces 252 typically may possess a curvilinear surface, such as a curved lobe, rod, etc. The contact surfaces 252 typically may be made of a metal material. FIG. 4 shows a cross-sectional view of an impregnation die150. As shown, the impregnation die 150 may include a manifold assembly220, a gate passage270, and an impregnation zone250. The manifold assembly220 may be provided for flowing the polymer resin214 therethrough. For example, the manifold assembly220 may include a channel222 or a plurality of channels222. The resin214 provided to the impregnation die 150 may flow through the channels222. As shown in FIG. 5, some portions of the channels222 may be curvilinear, and in exemplary embodiments, the channels222 may have a symmetrical orientation along a central axis224. Further, in some embodiments, the channels may be a plurality of branched runners222, which may include first branched runner group232, second group234, third group236, and, if desired, more branched runner groups. Each group may include 2, 3, 4 or more runners222 branching off from runners222 in the preceding group, or from an initial channel222. The branched runners222 and the symmetrical orientation thereof may evenly distribute the resin214, such that the flow of resin214 exiting the manifold assembly220 and coating the rovings142 may be substantially uniformly distributed on the rovings142. Beneficially, this may result in generally uniform impregnation of the rovings142. Further, the manifold assembly220 may in some embodiments define an outlet region242, which generally encompasses at least a downstream portion of the channels or runners222 from which the resin214 exits. In some embodiments, at least a portion of the channels or runners222 disposed in the outlet region242 may have an increasing area in a flow direction244 of the resin214. The increasing area may permit diffusion and further distribution of the resin214 as the resin214 flows through the manifold assembly220, which may further result in substantially uniform distribution of the resin214 on the rovings142. As further illustrated in FIGS. 4 and 5, after flowing through the manifold assembly220, the resin214 may flow through gate passage270. Gate passage270 may be positioned between the manifold assembly220 and the impregnation zone250, and may be configured for flowing the resin214 from the manifold assembly220 such that the resin214 coats the rovings142. Thus, resin214 exiting the manifold assembly220, such as through outlet region242, may enter gate passage270 and flow therethrough, as shown. Upon exiting the manifold assembly220 and the gate passage270 of the die 150 as shown in FIG. 4, the resin214 may contact the rovings142 being traversed through the die150. As discussed above, the resin214 may substantially uniformly coat the rovings142, due to distribution of the resin214 in the manifold assembly220 and the gate passage270. Further, in some embodiments, the resin214 may impinge on an upper surface of each of the rovings142, or on a lower surface of each of the rovings142, or on both an upper and lower surface of each of the rovings142. Initial impingement on the rovings142 may provide for further impregnation of the rovings142 with the resin214. As shown in FIG. 4, the coated rovings142 may traverse in run direction282 through impregnation zone250, which is configured to impregnate the rovings142 with the resin214. For example, as shown in FIGS. 4 and 6, the rovings142 may be traversed over contact surfaces 252 in the impregnation zone. Impingement of the rovings142 on the contact surface252 may create shear and pressure sufficient to impregnate the rovings142 with the resin214, thereby coating the rovings142. In some embodiments, as shown in FIG. 4, the impregnation zone250 may be defined between two spaced apart opposing plates256 and 258. First plate256 may define a first inner surface257, while second plate258 may define a second inner surface259. The contact surfaces 252 may be defined on or extend from both the first and second inner surfaces257 and 259, or only one of the first and second inner surfaces257 and 259. FIG. 6 illustrates the second plate258 and the various contact surfaces thereon that may form at least a portion of the impregnation zone250 according to these embodiments. In exemplary embodiments, as shown in FIG. 4, the contact surfaces 252 may be defined alternately on the first and second surfaces257 and 259 such that the rovings alternately impinge on contact surfaces252 on the first and second surfaces257 and 259. Thus, the rovings142 may pass contact surfaces 252 in a waveform, tortuous, or sinusoidal-type pathway, which enhances shear. Angle254 at which the rovings142 traverse the contact surfaces 252 may be generally high enough to enhance shear, but not so high as to cause excessive forces that will break the fibers. Thus, for example, the angle254 may be in the range between approximately 1° and approximately 30°, and in some embodiments, between approximately 5° and approximately 25°. In alternative embodiments, the impregnation zone250 may include a plurality of pins (not shown), each pin having a contact surface252. The pins may be static, freely rotational, or rotationally driven. In further alternative embodiments, the contact surfaces 252 and impregnation zone250 may comprise any suitable shapes and/or structures for impregnating the rovings142 with the resin214 as desired or required. To further facilitate impregnation of the rovings142, they also may be kept under tension while present within the impregnation die. The tension may, for example, range from about 5 to about 300 Newtons (N), in some embodiments from about 50 to about 250 N, and in some embodiments, from about 100 to about 200 N, per roving 142 or tow of fibers. As shown in FIG. 4, in some embodiments, a land zone280 may be positioned downstream of the impregnation zone250 in run direction282 of the rovings142. The rovings142 may traverse through the land zone280 before exiting the die150. As further shown in FIG. 4, in some embodiments, a faceplate290 may adjoin the impregnation zone250. Faceplate290 may be generally configured to meter excess resin214 from the rovings142. Thus, apertures in the faceplate290, through which the rovings142 traverse, may be sized such that when the rovings142 are traversed therethrough, the size of the apertures may cause excess resin214 to be removed from the rovings142. The impregnation die shown and described above is but one of various possible configurations that may be employed in the present invention. In alternative embodiments, for example, the rovings may be introduced into a crosshead die that may be positioned at an angle relative to the direction of flow of the polymer melt. As the rovings move through the crosshead die and reach the point where the polymer exits from an extruder barrel, the polymer may be forced into contact with the rovings. Examples of such a crosshead die extruder are described, for instance, in U.S. Pat. No. 3,993,726 to Moyer, U.S. Pat. No. 4,588,538 to Chung, et al.; U.S. Pat. No. 5,277,566 to Augustin, et al.; and U.S. Pat. No. 5,658,513 to Amaike, et al., which are incorporated herein in their entirety by reference thereto for all purposes. It should also be understood that any other extruder design also may be employed, such as a twin screw extruder. Still further, other components optionally may be employed to assist in the impregnation of the fibers. For example, a “gas jet” assembly may be employed in certain embodiments to help uniformly spread a roving of individual fibers, which may each contain up to as many as 24,000 fibers, across the entire width of the merged tow. This may help achieve uniform distribution of strength properties. Such an assembly may include a supply of compressed air or another gas that may impinge in a generally perpendicular fashion on the moving rovings that pass across the exit ports. The spread rovings then may be introduced into a die for impregnation, such as described above. Regardless of the technique employed, the continuous fibers may be oriented in the longitudinal direction (the machine direction “A” of the system of FIG. 3) to enhance tensile strength. Besides fiber orientation, other aspects of the pultrusion process also may be controlled to achieve the desired strength. For example, a relatively high percentage of continuous fibers may be employed in the consolidated ribbon to provide enhanced strength properties. For instance, continuous fibers typically may constitute from about 25 wt. % to about 80 wt. %, in some embodiments from about 30 wt. % to about 75 wt. %, and in some embodiments, from about 35 wt. % to about 60 wt. % of the ribbon. Likewise, thermoplastic polymer(s) typically may constitute from about 20 wt. % to about 75 wt. %, in some embodiments from about 25 wt. % to about 70 wt. %, and in some embodiments, from about 40 wt. % to about 65 wt. % of the ribbon. The percentage of the fibers and thermoplastic matrix in the final rod also may be within the ranges noted above. As noted above, the rovings may be consolidated into the form of one or more ribbons before being shaped into the desired rod configuration. When such a ribbon is subsequently compressed, the rovings may become distributed in a generally uniform manner about a longitudinal center of the rod. Such a uniform distribution enhances the consistency of the strength properties (e.g., flexural modulus, ultimate tensile strength, etc.) over the entire length of the rod. When employed, the number of consolidated ribbons used to form the rod may vary based on the desired thickness and/or cross-sectional area and strength of the rod, as well as the nature of the ribbons themselves. In most cases, however, the number of ribbons may be from 1 to 20, and in some embodiments, from 2 to 10. The number of rovings employed in each ribbon may likewise vary. Typically, however, a ribbon may contain from 2 to 10 rovings, and in some embodiments, from 3 to 5 rovings. To help achieve the symmetric distribution of the rovings in the final rod, it may be beneficial that they are spaced apart approximately the same distance from each other within the ribbon. Referring to FIG. 1, for example, one embodiment of a consolidated ribbon4 is shown that contains three (3) rovings5 spaced equidistant from each other in the x direction. In other embodiments, however, it may be desired that the rovings are combined, such that the fibers of the rovings are generally evenly distributed throughout the ribbon4. In these embodiments, the rovings may be generally indistinguishable from each other. Referring to FIG. 2, for example, one embodiment of a consolidated ribbon4 is shown that contains rovings that are combined such that the fibers are generally evenly distributed throughout. The specific manner in which the rovings are shaped also may be carefully controlled to ensure that rod may be formed with an adequate degree of compression and strength properties. Referring to FIG. 7, for example, one particular embodiment of a system and method for forming a rod is shown. In this embodiment, two ribbons12 initially may be provided in a wound package on a creel20. The creel20 may be an unreeling creel that includes a frame provided with horizontal spindles22, each supporting a package. A pay-out creel also may be employed, particularly if desired to induce a twist into the fibers, such as when using raw fibers in a one-step configuration. It should also be understood that the ribbons also may be formed in-line with the formation of the rod. In one embodiment, for example, the extrudate152 exiting the impregnation die 150 from FIG. 3 may be supplied directly to the system used to form a rod. A tension-regulating device40 also may be employed to help control the degree of tension in the ribbons12. The device40 may include inlet plate30 that lies in a vertical plane parallel to the rotating spindles22 of the creel20 and/or perpendicular to the incoming ribbons. The tension-regulating device40 may contain cylindrical bars41 arranged in a staggered configuration so that the ribbon12 may pass over and under these bars to define a wave pattern. The height of the bars may be adjusted to modify the amplitude of the wave pattern and control tension. The ribbons12 may be heated in an oven45 before entering a consolidation die50. Heating may be conducted using any known type of oven, such as an infrared oven, a convection oven, etc. During heating, the fibers in the ribbon may be unidirectionally oriented to optimize the exposure to the heat and maintain even heat across the entire ribbon. The temperature to which the ribbons12 are heated generally may be high enough to soften the thermoplastic polymer to an extent that the ribbons may bond together. However, the temperature may not be so high as to destroy the integrity of the material. The temperature may, for example, range from about 100° C. to about 500° C., in some embodiments from about 200° C. to about 400° C., and in some embodiments, from about 250° C. to about 350° C. In one particular embodiment, for example, polyphenylene sulfide (“PPS”) may be used as the polymer, and the ribbons may be heated to or above the melting point of PPS, which may be about 285° C. Upon being heated, the ribbons12 may be provided to a consolidation die 50 that may compress them together into a preform14, as well as may align and form the initial shape of the rod. As shown generally in FIG. 7, for example, the ribbons12 may be guided through a flow passage51 of the die 50 in a direction “A” from an inlet53 to an outlet55. The passage51 may have any of a variety of shapes and/or sizes to achieve the rod configuration. For example, the channel and rod configuration may be circular, elliptical, parabolic, trapezoidal, rectangular, etc. Within the die50, the ribbons generally may be maintained at a temperature at or above the melting point of the thermoplastic matrix used in the ribbon to ensure adequate consolidation. The desired heating, compression, and shaping of the ribbons12 may be accomplished through the use of a die 50 having one or multiple sections. For instance, although not shown in detail herein, the consolidation die 50 may possess multiple sections that function together to compress and shape the ribbons12 into the desired configuration. For instance, a first section of the passage51 may be a tapered zone that initially may shape the material as it flows into the die50. The tapered zone generally may possess a cross-sectional area that is larger at its inlet than at its outlet. For example, the cross-sectional area of the passage51 at the inlet of the tapered zone may be about 2% or more, in some embodiments about 5% or more, and in some embodiments, from about 10% to about 20% greater than the cross-sectional area at the outlet of the tapered zone. Regardless, the cross-section of the flow passage typically may change gradually and smoothly within the tapered zone so that a balanced flow of the composite material through the die may be maintained. A shaping zone may follow the tapered zone, and may compress the material and provide a generally homogeneous flow therethrough. The shaping zone also may pre-shape the material into an intermediate shape that is similar to that of the rod, but typically of a larger cross-sectional area to allow for expansion of the thermoplastic polymer while heated to minimize the risk of backup within the die50. The shaping zone also may include one or more surface features that impart a directional change to the preform. The directional change may force the material to be redistributed, resulting in a more even distribution of the fiber/resin in the final shape. This also may reduce the risk of dead spots in the die that may cause burning of the resin. For example, the cross-sectional area of the passage51 at a shaping zone may be about 2% or more, in some embodiments about 5% or more, and in some embodiments, from about 10% to about 20% greater than the width of the preform14. A die land also may follow the shaping zone to serve as an outlet for the passage51. The shaping zone, tapered zone, and/or die land may be heated to a temperature at or above that of the glass transition temperature or melting point of the thermoplastic matrix. If desired, a second die 60 (e.g., a calibration die) also may be employed to compress the preform14 into the final shape of the rod. When employed, it may be beneficial to allow the preform14 to cool briefly after exiting the consolidation die 50 and before entering the optional second die60. This may allow the consolidated preform14 to retain its initial shape before progressing further through the system. Typically, cooling may reduce the temperature of the exterior of the rod below the melting point temperature of the thermoplastic matrix to minimize and substantially prevent the occurrence of melt fracture on the exterior surface of the rod. The internal section of the rod, however, may remain molten to ensure compression when the rod enters the calibration die body. Such cooling may be accomplished by simply exposing the preform14 to the ambient atmosphere (e.g., room temperature) or through the use of active cooling techniques (e.g., water bath or air cooling) as is known in the art. In one embodiment, for example, air may blown onto the preform 14 (e.g., with an air ring). The cooling between these stages, however, generally may occur over a small period of time to ensure that the preform14 still may be soft enough to be further shaped. For example, after exiting the consolidation die 50, the preform14 may be exposed to the ambient environment for only from about 1 to about 20 seconds, and in some embodiments, from about 2 to about 10 seconds, before entering the second die60. Within the die60, the preform generally may be kept at a temperature below the melting point of the thermoplastic matrix used in the ribbon so that the shape of the rod can be maintained. Although referred to above as single dies, it should be understood that the dies 50 and 60 may in fact be formed from multiple individual dies (e.g., face plate dies). Thus, in some embodiments, multiple individual dies 60 may be utilized to gradually shape the material into the desired configuration. The dies 60 may be placed in series, and provide for gradual decreases in the dimensions of the material. Such gradual decreases may allow for shrinkage during and between the various steps. For example, as shown in FIGS. 11 through 13, a first die60 may include one or more inlets62 and corresponding outlets64, as shown. Any number of inlets62 and corresponding outlets64 may be included in a die60, such as four as shown, or one, two, three, five, six, or more. An inlet62 in some embodiments may be generally oval or circular shaped. In other embodiments, the inlet62 may have a curved rectangular shape, i.e., a rectangular shape with curved corners or a rectangular shape with straight longer sidewalls and curved shorter sidewalls. Further, an outlet64 may be generally oval or circular shaped, or may have a curved rectangular shape. In some embodiments wherein an oval shaped inlet is utilized, the inlet62 may have a major axis length66 to minor axis length68 ratio in a range between approximately 3:1 and approximately 5:1. In some embodiments wherein an oval or circular shaped inlet is utilized, the outlet64 may have a major axis length66 to minor axis length68 ratio in a range between approximately 1:1 and approximately 3:1. In embodiments wherein a curved rectangular shape is utilized, the inlet and outlet may have major axis length66 to minor axis length66 ratios (aspect ratios) between approximately 2:1 and approximately 7:1, and the outlet64 ratio may be less than the inlet62 ratio. In further embodiments, the cross-sectional area of an inlet62 and the cross-sectional area of a corresponding outlet64 of the first die60 may have a ratio in a range between approximately 1.5:1 and 6:1. The first die 60 thus may provide a generally smooth transformation of polymer impregnated fiber material to a shape that is relatively similar to a final shape of the resulting rod, which in exemplary embodiments has a circular or oval shaped cross-section. Subsequent dies, such as a second die60 and third die 60 as shown in FIG. 11, may provide for further gradual decreases and/or changes in the dimensions of the material, such that the shape of the material is converted to a final cross-sectional shape of the rod. These subsequent dies 60 may both shape and cool the material. For example, in some embodiments, each subsequent die60 may be maintained at a lower temperature than the previous dies. In exemplary embodiments, all dies 60 may be maintained at temperatures that are higher than a softening point temperature for the material. In further exemplary embodiments, dies 60 having relatively long land lengths69 may be desired, due to, for example, proper cooling and solidification, which may be important in achieving a desired rod shape and size. Relatively long land lengths69 may reduce stresses and provide smooth transformations to desired shapes and sizes, and with minimal void fraction and bow characteristics. In some embodiments, for example, a ratio of land length69 at an outlet64 to major axis length66 at the outlet64 for a die 60 may be in the range between 0 and approximately 20, such as between approximately 2 and approximately 6. The use of calibration dies 60 according to the present disclosure may provide for gradual changes in material cross-section, as discussed. These gradual changes may in exemplary embodiments ensure that the resulting product, such as a rod or other suitable product, has a generally uniform fiber distribution with relatively minimal void fraction. It should be understood that any suitable number of dies 60 may be utilized to gradually form the material into a profile having any suitable cross-sectional shape, as desired or as required by various end-use applications. In addition to the use of one or more dies, other mechanisms also may be employed to help compress the preform14 into the shape of a rod. For example, forming rollers90, as shown in FIG. 14, may be employed between the consolidation die 50 and the calibration die 60, between the various calibration dies 60, and/or after the calibration dies 60 to further compress the preform14 before it is converted into its final shape. The rollers may have any configuration, such as pinch rollers, overlapping rollers, etc., and may be vertical as shown or horizontal rollers. Depending on the roller90 configuration, the surfaces of the rollers90 may be machined to impart the dimensions of the final product, such as the rod, core, profile, or other suitable product, to the preform14. In an exemplary embodiment, the pressure of the rollers90 may be adjustable to optimize the quality of the final product. The rollers90 in exemplary embodiments, such as at least the portions contacting the material, may have generally smooth surfaces. For example, relatively hard, polished surfaces may be beneficial in many embodiments. For example, the surface of the rollers may be formed from a relatively smooth chrome or other suitable material. This may allow the rollers90 to manipulate the preform14 without damaging or undesirably altering the preform14. For example, such surfaces may prevent the material from sticking to the rollers, and the rollers may impart smooth surfaces onto the materials. In some embodiments, the temperature of the rollers90 may be controlled. This may be accomplished by heating of the rollers90 themselves, or by placing the rollers90 in a temperature controlled environment. Further, in some embodiments, surface features 92 may be provided on the rollers90. The surface features 92 may guide and/or control the preform14 in one or more directions as it is passed through the rollers. For example, surface features 92 may be provided to prevent the preform14 from folding over on itself as it is passed through the rollers90. Thus, the surface features 92 may guide and control deformation of the preform14 in the cross-machine direction relative to the machine direction A as well as in the vertical direction relative to the machine direction A. The preform14 thus may be pushed together in the cross-machine direction, rather than folded over on itself, as it is passed through the rollers90 in the machine direction A. In some embodiments, tension regulation devices may be provided in communication with the rollers. These devices may be utilized with the rollers to apply tension to the preform14 in the machine direction, cross-machine direction, and/or vertical direction to further guide and/or control the preform. As indicated above, the resulting rod also may be applied with a capping layer to protect it from environmental conditions and/or to improve wear resistance. Referring again to FIG. 7, for example, such a capping layer may be applied via an extruder oriented at any desired angle to introduce a thermoplastic resin into a capping die72. To help prevent a galvanic response, it may be beneficial for the capping material to have a dielectric strength of at least about 1 kV per millimeter (kV/mm), in some embodiments at least about 2 kV/mm, in some embodiments from about 3 kV/mm to about 50 kV/mm, and in some embodiments, from about 4 kV/mm to about 30 kV/mm, such as determined in accordance with ASTM D149-09. Suitable thermoplastic polymers for this purpose may include, for instance, polyolefins (e.g., polypropylene, propylene-ethylene copolymers, etc.), polyesters (e.g., polybutylene terephalate (PBT)), polycarbonates, polyamides (e.g., Nylon™) polyether ketones (e.g., polyetherether ketone (PEEK)), polyetherimides, polyarylene ketones (e.g., polyphenylene diketone (PPDK)), liquid crystal polymers, polyarylene sulfides (e.g., polyphenylene sulfide (PPS), poly(biphenylene sulfide ketone), poly(phenylene sulfide diketone), poly(biphenylene sulfide), etc.), fluoropolymers (e.g., polytetrafluoroethylene-perfluoromethylvinylether polymer, perfluoro-alkoxyalkane polymer, petrafluoroethylene polymer, ethylene-tetrafluoroethylene polymer, etc.), polyacetals, polyurethanes, polycarbonates, styrenic polymers (e.g., acrylonitrile butadiene styrene (ABS)), acrylic polymers, polyvinyl chloride (PVC), etc. Particularly suitable high dielectric strength capping layer materials may include polyketone (e.g., polyetherether ketone (PEEK)), polysulfide (e.g., polyarylene sulfide), or a mixture thereof. The capping layer generally may be “free” of continuous fibers. That is, the capping layer may contain “less than about 10 wt. %” of continuous fibers, in some embodiments about 5 wt. % or less of continuous fibers, and in some embodiments, about 1 wt. % or less of continuous fibers (e.g., 0 wt. %). Nevertheless, the capping layer may contain other additives for improving the final properties of the composite core. Additive materials employed at this stage may include those that are not suitable for incorporating into the continuous fiber material. For instance, it may be beneficial to add pigments to reduce finishing labor, or it may be beneficial to add flame retardant agents to enhance the flame retardancy of the core. Because many additive materials may be heat sensitive, an excessive amount of heat may cause them to decompose and produce volatile gases. Therefore, if a heat sensitive additive material is extruded with an impregnation resin under high heating conditions, the result may be a complete degradation of the additive material. Additive materials may include, for instance, mineral reinforcing agents, lubricants, flame retardants, blowing agents, foaming agents, ultraviolet light resistant agents, thermal stabilizers, pigments, and combinations thereof. Suitable mineral reinforcing agents may include, for instance, calcium carbonate, silica, mica, clays, talc, calcium silicate, graphite, calcium silicate, alumina trihydrate, barium ferrite, and combinations thereof. While not shown in detail herein, the capping die 72 may include various features known in the art to help achieve the desired application of the capping layer. For instance, the capping die 72 may include an entrance guide that aligns the incoming rod. The capping die also may include a heating mechanism (e.g., heated plate) that pre-heats the rod before application of the capping layer to help ensure adequate bonding. Following capping, the shaped part15 then may be finally cooled using a cooling system80 as is known in the art. The cooling system80 may, for instance, be a sizing system that includes one or more blocks (e.g., aluminum blocks) that may completely encapsulate the composite core while a vacuum pulls the hot shape out against its walls as it cools. A cooling medium may be supplied to the sizer, such as air or water, to solidify the composite core in the correct shape. Even if a sizing system is not employed, it may be beneficial to cool the composite core after it exits the capping die (or the consolidation or calibration die, if capping is not applied). Cooling may occur using any technique known in the art, such a water tank, cool air stream or air jet, cooling jacket, an internal cooling channel, cooling fluid circulation channels, etc. Regardless, the temperature at which the material is cooled may be controlled to achieve certain mechanical properties, part dimensional tolerances, good processing, and an aesthetically pleasing composite. For instance, if the temperature of the cooling station is too high, the material might swell in the tool and interrupt the process. For semi-crystalline materials, too low of a temperature likewise may cause the material to cool down too rapidly and not allow complete crystallization, thereby adversely affecting the mechanical and chemical resistance properties of the composite. Multiple cooling die sections with independent temperature control may be utilized to impart a beneficial balance of processing and performance attributes. In one particular embodiment, for example, a water tank may be employed at a temperature of from about 0° C. to about 30° C., in some embodiments from about 1° C. to about 20° C., and in some embodiments, from about 2° C. to about 15° C. If desired, one or more sizing blocks (not shown) also may be employed, such as after capping. Such blocks may contain openings that are cut to the exact core shape, graduated from oversized at first to the final core shape. As the composite core passes therethrough, any tendency for it to move or sag may be counteracted, and it may be pushed back (repeatedly) to its correct shape. Once sized, the composite core may be cut to the desired length at a cutting station (not shown), such as with a cut-off saw capable of performing cross-sectional cuts, or the composite core may be wound on a reel in a continuous process. The length of rod and/or the composite core may be limited to the length of the fiber tow. As will be appreciated, the temperature of the rod or composite core as it advances through any section of the system of the present invention may be controlled to yield certain manufacturing and final composite properties. Any or all of the assembly sections may be temperature controlled utilizing electrical cartridge heaters, circulated fluid cooling, etc., or any other temperature controlling device known to those skilled in the art. Referring again to FIG. 7, a pulling device82 may be positioned downstream from the cooling system80 to pull the finished composite core16 through the system for final sizing of the composite. The pulling device82 may be any device capable of pulling the core through the process system at a desired rate. Typical pulling devices may include, for example, caterpillar pullers and reciprocating pullers. One embodiment of the composite core (or composite strand) formed from the method described above is shown in more detail in FIG. 8 as element516. As illustrated, the composite core516 may have a substantially circular shape and may include a rod (or fiber core) 514 comprising one or more consolidated ribbons (a continuous fiber component). By “substantially circular,” it is generally meant that the aspect ratio of the core (height divided by the width) is typically from about 1.0 to about 1.5, and in some embodiments, about 1.0. Due to the selective control over the process used to impregnate the rovings and form a consolidated ribbon, as well the process for compressing and shaping the ribbon, the composite core may comprise a relatively even distribution of the thermoplastic matrix across along its entire length. This also means that the continuous fibers may be distributed in a generally uniform manner about a longitudinal central axis “L” of the composite core516. As shown in FIG. 8, for example, the rod514 of the composite core516 may include continuous fibers526 embedded within a thermoplastic matrix528. The fibers526 may be distributed generally uniformly about the longitudinal axis “L.” It should be understood that only a few fibers are shown in FIG. 8, and that the composite core typically may contain a substantially greater number of uniformly distributed fibers. A capping layer519 also may extend around the perimeter of the rod514 and define an external surface of the composite core516. The cross-sectional thickness of the rod514 may be selected strategically to help achieve a particular strength for the composite core. For example, the rod514 may have a thickness (e.g., diameter) of from about 0.1 to about 40 mm, in some embodiments from about 0.5 to about 30 mm, and in some embodiments, from about 1 to about 10 mm. The thickness of the capping layer519 may depend on the intended function of the part, but typically may be from about 0.01 to about 10 mm, and in some embodiments, from about 0.02 to about 5 mm. The total cross-sectional thickness, or height, of the composite core516 also may range from about 0.1 to about 50 mm, in some embodiments from about 0.5 to about 40 mm, and in some embodiments, from about 1 to about 20 mm (e.g., diameter, if a circular cross-section). While the composite core may be substantially continuous in length, the length of the composite core may be limited in practice by the spool onto which it will be wound and stored and/or by the length of the continuous fibers. For example, the length often may range from about 1,000 m to about 5,000 m, although even greater lengths are certainly possible. Through control over the various parameters mentioned above, cores having very high strengths may be formed. For example, the composite cores may exhibit a relatively high flexural modulus. The term “flexural modulus” generally refers to the ratio of stress to strain in flexural deformation (units of force per unit area), or the tendency for a material to bend. It is determined from the slope of a stress-strain curve produced by a “three point flexural” test (such as ASTM D790-10, Procedure A, room temperature). For example, the composite core of the present invention may exhibit a flexural modulus of from about 10 GPa or more, in some embodiments from about 12 to about 400 GPa, in some embodiments from about 15 to about 200 GPa, and in some embodiments, from about 20 to about 150 GPa. Composite cores used to produce electrical cables consistent with certain embodiments disclosed herein may have ultimate tensile strengths over about 300 MPa, such as, for instance, in a range from about 400 MPa to about 5,000 MPa, or from about 500 MPa to about 3,500 MPa. Further, suitable composite cores may have an ultimate tensile strength in a range from about 700 MPa to about 3,000 MPa; alternatively, from about 900 MPa to about 1,800 MPa; or alternatively, from about 1,100 MPa to about 1,500 MPa. The term “ultimate tensile strength” generally refers to the maximum stress that a material can withstand while being stretched or pulled before breaking, and is the maximum stress reached on a stress-strain curve produced by a tensile test (such as ASTM D3916-08) at room temperature. Additionally or alternatively, the composite core may have a tensile modulus of elasticity, or elastic modulus, in a range from about 50 GPa to about 500 GPa, from about 70 GPa to about 400 GPa, from about 70 GPa to about 300 GPa, or from about 70 GPa to about 250 GPa. In certain embodiments, the composite core may have an elastic modulus in a range from about 70 GPa to about 200 GPa; alternatively, from about 70 GPa to about 150 GPa; or alternatively, from about 70 GPa to about 130 GPa. The term “tensile modulus of elasticity” or “elastic modulus” generally refers to the ratio of tensile stress over tensile strain and is the slope of a stress-strain curve produced by a tensile test (such as ASTM 3916-08) at room temperature. Composite cores made according to the present disclosure may further have relatively high flexural fatigue life, and may exhibit relatively high residual strength. Flexural fatigue life and residual flexural strength may be determined based on a “three point flexural fatigue” test (such as ASTM D790, typically at room temperature). For example, the cores of the present invention may exhibit residual flexural strength after one million cycles at 160 Newtons (“N”) or 180 N loads of from about 60 kilograms per square inch (“ksi”) to about 115 ksi, in some embodiments from about 70 ksi to about 115 ksi, and in some embodiments from about 95 ksi to about 115 ksi. Further, the cores may exhibit relatively minimal reductions in flexural strength. For example, cores having void fractions of about 4% or less, in some embodiments about 3% or less, may exhibit reductions in flexural strength after three point flexural fatigue testing of about 1% (for example, from a maximum pristine flexural strength of about 106 ksi to a maximum residual flexural strength of about 105 ksi). Flexural strength may be tested before and after fatigue testing using, for example, a three point flexural test as discussed above. In some embodiments, the composite core may have a density or specific gravity of less than about 2.5 g/cc, less than about 2.2 g/cc, less than about 2 g/cc, or less than about 1.8 g/cc. In other embodiments, the composite core density may be in a range from about 1 g/cc to about 2.5 g/cc; alternatively, from about 1.1 g/cc to about 2.2 g/cc; alternatively, from about 1.1 g/cc to about 2 g/cc; alternatively, from about 1.1 g/cc to about 1.9 g/cc; alternatively, from about 1.2 g/cc to about 1.8 g/cc; or alternatively, from about 1.3 g/cc to about 1.7 g/cc. In some cable applications, such as in overhead transmission lines, the strength to weight ratio of the composite core may be important. The ratio may be quantified by the ratio of the tensile strength of the core material to the density of the core material (in units of MPa/(g/cc)). Illustrative and non-limiting strength to weight ratios of composite cores in accordance with embodiments of the present invention may be in a range from about 400 to about 1,300, from about 400 to about 1,200, from about 500 to about 1,100, from about 600 to about 1,100, from about 700 to about 1,100, from about 700 to about 1,000, or from about 750 to about 1,000. Again, the ratios are based on the tensile strength in MPa, and the composite core density in g/cc. In some embodiments, the percent elongation at break for the composite core may be less than 4%, less than 3%, or less than 2%, while in other embodiments, the elongation at break may be in a range from about 0.5% to about 2.5%, from about 1% to about 2.5%, or from about 1% to about 2%. The linear thermal expansion coefficient of the composite core may be less than about 5×10−6/° C., less than about 4×10−6/° C., less than about 3×10−6/° C., or less than about 2×10−6/° C. (or in units of m/m/° C.). Stated another way, the linear thermal expansion coefficient may be, on a ppm basis per ° C., less than about 5, less than about 4, less than about 3, or less than about 2. For instance, the coefficient (ppm/° C.) may be in a range from about −0.4 to about 5; alternatively, from about −0.2 to about 4; alternatively, from about 0.4 to about 4; or alternatively, from about 0.2 to about 2. The temperature range contemplated for this linear thermal expansion coefficient may be generally in the −50° C. to 200° C. range, the 0° C. to 200° C. range, the 0° C. to 175° C. range, or the 25° C. to 150° C. range. The linear thermal expansion coefficient is measured in the longitudinal direction, i.e., along the length of the fibers. The composite core also may exhibit a relatively small “bending radius”, which is the minimum radius that the rod can be bent without damage and is measured to the inside curvature of the composite core or composite strand. A smaller bend radius means that the composite core may be more flexible and may be spooled onto a smaller diameter bobbin. This property also may permit easier substitution of the composite core in cables that currently use metal cores, and allow for the use of tools and installation methods presently in use in conventional overhead transmission cables. The bending radius for the composite core may, in some embodiments, be in a range from about 1 cm to about 60 cm, from about 1 cm to about 50 cm, from about 1 cm to about 50 cm, or from about 2 cm to about 45 cm, as determined at a temperature of about 25° C. The bending radius may be in a range from about 2 cm to about 40 cm, or from about 3 cm to about 40 cm in certain embodiments contemplated herein. In other embodiments, bending radiuses may be achieved that are less than about 40 times the outer diameter of the composite core, in some embodiments from about 1 to about 30 times the outer diameter of the composite core, and in some embodiments, from about 2 to about 25 times the outer diameter of the composite core, determined at a temperature of about 25° C. Notably, the strength, physical, and thermal properties of the composite core referenced above also may be maintained over a relatively wide temperature range, such as from about −50° C. to about 300° C., from about 100° C. to about 300° C., from about 110° C. to about 250° C., from about 120° C. to about 200° C., from about 150° C. to about 200° C., or from about 180° C. to about 200° C. The composite core also may have a low void fraction, such as about 6% or less, in some embodiments about 3% or less, in some embodiments about 2% or less, in some embodiments about 1% or less, and in some embodiments, about 0.5% or less. The void fraction may be determined in the manner described above, such as using a “resin burn off” test in accordance with ASTM D 2584-08 or through the use of computed tomography (CT) scan equipment, such as a Metrotom 1500 (2k×2k) high resolution detector. In one embodiment, a composite core of the present invention may be characterized by the following properties: an ultimate tensile strength in a range from about 700 MPa to about 3,500 MPa; an elastic modulus from about 70 GPa to about 300 GPa; and a linear thermal expansion coefficient (in units of ppm per ° C.) in a range from about −0.4 to about 5. Additionally, the composite core may have a density of less than about 2.5 g/cc and/or a strength to weight ratio (in units of MPa/(g/cc)) in a range from about 500 to about 1,100. Further, in certain embodiments, the composite core may have a bending radius in a range from about 1 cm to about 50 cm. Still further, the composite core may have a percent elongation at break of less than about 3%. In another embodiment, a composite core of the present invention may be characterized by the following properties: an ultimate tensile strength in a range from about 1,100 MPa to about 1,500 MPa; an elastic modulus in a range about 70 GPa to about 130 GPa; and a linear thermal expansion coefficient (in units of ppm per ° C.) in a range from about 0.2 to about 2. Additionally, the composite core may have a density in a range from about 1.2 g/cc to about 1.8 g/cc and/or a strength to weight ratio (in units of MPa/(g/cc)) in a range from about 700 to about 1,100. Further, in certain embodiments, the composite core may have a bending radius in a range from about 2 cm to about 40 cm. Still further, the composite core may have a percent elongation at break in a range from about 1% to about 2.5%. As will be appreciated, the particular composite core embodiments described above are merely exemplary of the numerous designs that may be within the scope of the present invention. Among the various possible composite core designs, it should be understood that additional layers of material may be employed in addition to those described above. In certain embodiments, for example, it may be beneficial to form a multi-component core in which one component comprises a higher strength material and another component comprises from a lower strength material. Such multi-component cores may be particularly useful in increasing overall strength without requiring the need for more expensive high strength materials for the entire core. The lower and/or higher strength components may comprise ribbon(s) that contain continuous fibers embedded within a thermoplastic matrix. Further, it should be understood that the scope of the present invention is by no means limited to the embodiments described above. For example, the composite cores may contain various other components depending on the desired application and its required properties. The additional components may be formed from a continuous fiber ribbon, such as described herein, as well as other types of materials. In one embodiment, for example, the composite core may contain a layer of discontinuous fibers (e.g., short fibers, long fibers, etc.) to improve its transverse strength. The discontinuous fibers may be oriented so that at least a portion of these fibers may be positioned at an angle relative to the direction in which the continuous fibers extend. Electrical Cable Consistent with embodiments disclosed herein, electrical cables of the present invention, such as high voltage overhead transmission lines, may comprise a cable core comprising at least one composite core, and a plurality of conductive elements surrounding the cable core. The cable core may be a single composite core, incorporating any composite core design and accompanying physical and thermal properties provided above. Alternatively, the cable core may comprise two or more composite cores, or composite strands, having either the same or different designs, and either the same or different physical and thermal properties. These two or more composite cores may be assembled parallel to each other (straight), or stranded, e.g., about a central composite core member. Accordingly, in some embodiments, an electrical cable may comprise a cable core comprising one composite core surrounded by a plurality of conductive elements, while in other embodiments, an electrical cable may comprise a cable core comprising two or more composite cores, the cable core surrounded by a plurality of conductive elements. For example, the cable core may comprise, for instance, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, or 20 composite cores, or more (e.g., 37 composite cores), each of which may incorporate any composite core design and accompanying physical and thermal properties provided above. The composite cores may be arranged, bundled, or oriented in any suitable fashion, as would be Recognized by one of skill in the art. For instance, the composite cores can be stranded, such as a cable core comprising 7 stranded composite cores or 19 stranded composite cores. Alternatively, the composite cores can be parallel, such as a cable core comprising a bundle of 7 composite cores aligned parallel to each other. The electrical cable may comprise a plurality of conductive elements surrounding the cable core (e.g., a single composite core, a plurality of stranded composite cores). The conductive elements may be of any geometric shape, and may be round/circular wires or trapezoidal wires, among others, and including combinations thereof. The conductive elements may be in one layer, or 2 layers, or 3 layers, or 4 layers, and so forth, around the cable core. The conductive elements may be configured parallel to the cable core, or wrapped helically, or in any other suitable arrangement. Any number of conductive elements (e.g., wires) may be used, but a typical number of conductive elements in a cable may be up to 84 conductive elements, and often in a range from 2 to about 50. For instance, in some common conductor arrangements, 7, 19, 26, or 37 wires may be employed. Exemplary transmission cable designs with composites cores which may be employed in various embodiments of the present invention are described in U.S. Pat. No. 7,211,319 to Heil, et al., which is incorporated herein in its entirety by reference thereto for all purposes. Referring now to FIG. 9, one embodiment of an electrical cable420 is shown. As illustrated, the electrical cable420 may include a plurality of conductive elements 422 (e.g., aluminum or an alloy thereof) radially disposed about a substantially cylindrical cable core400, which is illustrated as a single composite core but could be a plurality of stranded composite cores. The conductive elements may be arranged in a single layer or in multiple layers. In the illustrated embodiment, the conductive elements422 are arranged to form a first concentric layer426 and a second concentric layer428. The shape of the conductive elements422 also may be varied around the cable core400. In the illustrated embodiment, the conductive elements422 have a generally trapezoidal cross-sectional shape. Other shapes also may be employed, such as circular, elliptical, rectangular, square, etc. The conductive elements422 also may be twisted or wrapped around the cable core400 in any desired geometrical configuration, such as in a helical manner. Referring to FIG. 10, for instance, another embodiment of an electrical transmission cable420 is shown. As illustrated, the electrical transmission cable420 may include a plurality of conductive elements 422 (e.g., aluminum or an alloy thereof) radially disposed about a bundle of generally cylindrical composite cores400, which may be formed in accordance with the present invention. FIG. 10 illustrates six composite cores400 surrounding a single core400, although any suitable number of composite cores400 in any suitable arrangement is within the scope and spirit of the present disclosure and may be used as the cable core. A capping layer519 also may extend around the perimeter of and define an external surface of each rod. The conductive elements may be arranged in a single layer or in multiple layers. In the illustrated embodiment, for example, the conductive elements422 are arranged to form a first concentric layer426 and a second concentric layer428. Of course, any number of concentric layers may be employed. The shape of the conductive elements422 also may be varied to optimize the number of elements that can be disposed about the cable core. In the illustrated embodiment, for example, the conductive elements422 have a generally trapezoidal cross sectional shape. Other shapes also may be employed, such as circular, elliptical, rectangular, square, etc. The conductive elements422 also may be twisted or wrapped around the cable core containing the bundle of composite cores400 in any desired geometrical configuration, such as in a helical manner. The cross-sectional area of individual conductive elements may vary considerably, but generally the cross-sectional area of individual elements may be in a range from about 10 to about 50 mm 2, or from about 15 to about 45 mm 2. The overall conductor area may range (in kcmil), for instance, from about 167 to about 3500 kcmil, from about 210 to about 2700 kcmil, from about 750 to about 3500 kcmil, or from about 750 to about 3000 kcmil. Overall conductor areas of about 795, about 825, about 960, and about 1020 kcmil, often may be employed in many end-uses for electrical cables, such as in overheard power transmission lines. For instance, a common aluminum conductor steel reinforced cable known in the industry is often referred to as the 795 kcmil ACSR “Drake” conductor cable. The outside diameter of cables in accordance with the present invention is not limited to any particular range. However, typical cable outside diameters may be within a range, for example, of from about 7 to about 50 mm, from about 10 to about 48 mm, from about 20 to about 40 mm, from about 25 to about 35 mm, or from about 28 to about 30 mm. Likewise, the cross-sectional area of a composite core in the cable is not limited to any particular range. However, typical cross-sectional areas of the composite core may be within a range from about 20 to about 140 mm 2, or from about 30 to about 120 mm 2. The conductive elements may be made from any suitable conductive or metal material, including various alloys. The conductive elements may comprise copper, a copper alloy, aluminum, an aluminum alloy, or combinations thereof. As used herein, the term “aluminum or an aluminum alloy” is meant to collectively refer to grades of aluminum or aluminum alloys having at least 97% aluminum by weight, at least 98% aluminum by weight, or at least 99% aluminum by weight, including pure or substantially pure aluminum. Aluminum alloys or grades of aluminum having an IACS electrical conductivity of at least 57%, at least 58%, at least 59%, at least 60%, or at least 61% (e.g., 59% to 65%) may be employed in embodiments disclosed herein, and this is inclusive of any method that could produce such conductivities (e.g., annealing, tempering, etc.). For example, aluminum 1350 alloy may be employed as the aluminum or aluminum alloy in certain embodiments of this invention. Aluminum 1350, its composition, and its minimum IACS, are described in ASTM B233, the disclosure of which is incorporated herein by reference in its entirety. In some applications, such as in overhead transmission lines, the sag of the electrical cable may be an important feature. Sag is generally considered to be the distance that a cable departs from a straight line between the end points of a span. The sag across a span of towers may affect the ground clearance, and subsequently, the tower height and/or the number of towers needed. Sag generally may increase with the square of the span length, but may be reduced by an increase in tensile strength of the cable and/or a decrease in weight of the cable. Electrical cables in some embodiments of the present invention may have a sag (at rated temperature (180° C.), and for a 300-meter level span) with a NESC light loading of from about 3 to about 9.5 m, from about 4.5 to about 9.5 m, from about 5.5 to about 8 m, or from about 6 to about 7.5 m. Likewise, with a NESC heavy loading, under similar conditions, the sag may be in a range from about 3 to about 9.5 m, from about 3 to about 7.5 m, from about 4.5 to about 7.5 m, or from about 5 to about 7 m. In some embodiments, the cable also may be characterized as having a stress parameter of about 10 MPa or more, in some embodiments about 15 MPa or more, and in some embodiments, from about 20 to about 50 MPa. The method for determining the stress parameter is described in more detail in U.S. Pat. No. 7,093,416 to Johnson, et al., which is incorporated herein in its entirety by reference thereto for all purposes. For example, sag and temperature may be measured and plotted as a graph of sag versus temperature. A calculated curve may be fitted to the measured data using an Alcoa Sag10 graphic method available in a software program from Southwire Company (Carrollton, Ga.) under the trade designation SAG10 (version 3.0 update 3.10.10). The stress parameter is a fitting parameter in SAG10 labeled as the “built-in aluminum stress”, which may be altered to fit other parameters, if a material other than aluminum is used (e.g., an aluminum alloy), and which adjusts the position of the knee-point on the predicted graph and also the amount of sag in the high temperature, post-knee-point regime. A description of the stress parameter also may be provided in the Sag10 Users Manual (Version 2.0), incorporated herein by reference in its entirety. In relation to the subject matter disclosed herein, creep is generally considered to be the permanent elongation of a cable under load over a long period of time. The amount of creep of a length of cable may be impacted by the length of time in service, the load on the cable, the tension on the cable, the encountered temperature conditions, amongst other factors. It is contemplated that cables disclosed herein may have 10-year creep values at 15%, 20%, 25%, and/or 30% RBS (rated breaking stress) of less than about 0.25%, less than about 0.2%, or less than about 0.175%. For instance, the 10-year creep value at 15% RBS may be less than about 0.25%; alternatively, less than about 0.2%; alternatively, less than about 0.15%; alternatively, less than about 0.1%; or alternatively, less than about 0.075%. The 10-year creep value at 30% RBS may be less than about 0.25%; alternatively, less than about 0.225%; alternatively, less than about 0.2; or alternatively, less than about 0.175%. These creep values are determined in accordance with the 10-year ACSR conductor creep test (Aluminum Association Creep Test rev. 1999), incorporated herein by reference in its entirety. Electrical cables in accordance with embodiments of this invention may have a maximum operating temperature up to about 300° C., up to about 275° C., or up to about 250° C. Certain cables provided herein may have maximum operating temperatures that may be up to about 225° C.; alternatively, up to 200° C.; alternatively, up to 180° C.; or alternatively, up to 175° C. Maximum operating temperatures may be in a range from about 100 to about 300° C., from about 100 to about 250° C., from about 110 to about 250° C., from about 120 to about 200° C., or from about 120 to about 180° C., in various embodiments of the present invention. In accordance with some embodiments, it may be beneficial for the electrical cable to have certain fatigue and/or vibrational resistance properties. For instance, the electrical cable may pass (meet or exceed) the Aeolian vibration test specified in IEEE 1138, incorporated herein by reference, at 100 million cycles. In some embodiments, the electrical cable may comprise a partial or complete layer of a material between the cable core and the conductive elements. For instance, the material may be conductive or non-conductive, and may be a tape that partially or completely wraps/covers the cable core. The material may be configured to hold or secure the individual composite core elements of a cable core together. In some embodiments, the material may comprise a metal or aluminum foil tape, a polymer tape (e.g., a polypropylene tape, a polyester tape, a Teflon tape, etc.), a tape with glass-reinforcement, and the like. Often, the thickness of the material (e.g., the tape) may be in a range from about 0.025 mm to about 0.25 mm, although the thickness is not limited only to this range. In one embodiment, the tape or other material may be applied so that each subsequent wrap overlaps the previous wrap. In another embodiment, the tape or other material may be applied so that each subsequent wrap leaves a gap between the previous wrap. In yet another embodiment, the tape or other material may be applied so that abuts the previous wrap with no overlap and no gap. In these and other embodiments, the tape or other material may be applied helically around the cable core. In some embodiments, the electrical cable may comprise a partial or complete coating of a material between the cable core and the conductive elements. For instance, the material may be, or may comprise, a polymer. Suitable polymers may include, but are not limited to, a polyolefin (e.g., polyethylene and polypropylene homopolymers, copolymers, etc.), a polyester (e.g., polybutylene terephalate (PBT)), a polycarbonate, a polyamide (e.g., Nylon™), a polyether ketone (e.g., polyetherether ketone (PEEK)), a polyetherimide, a polyarylene ketone (e.g., polyphenylene diketone (PPDK)), a liquid crystal polymer, a polyarylene sulfide (e.g., polyphenylene sulfide (PPS), poly(biphenylene sulfide ketone), poly(phenylene sulfide diketone), poly(biphenylene sulfide), etc.), a fluoropolymer (e.g., polytetrafluoroethylene-perfluoromethylvinylether polymer, perfluoro-alkoxyalkane polymer, petrafluoroethylene polymer, ethylene-tetrafluoroethylene polymer, etc.), a polyacetal, a polyurethane, a styrenic polymer (e.g., acrylonitrile butadiene styrene (ABS)), an acrylic polymer, a polyvinyl chloride polymer (PVC), and the like, including combinations thereof. Moreover, the polymer may be an elastomeric polymer. The coating may be conductive or non-conductive, and may contain various additives typically employed in wire and cable applications. The coating may serve, in some embodiments, as a protective coating for the cable core. Additionally, the coating may be used in instances where the composite core does not contain a capping layer, and the coating partially or completely covers the rod (or fiber core), for example, as a protective coating for the rod. In circumstances where the cable core comprises two or more composite cores (e.g., composite strands), the coating may partially or completely fill the spaces between the individual core elements. The present invention also encompasses methods of making an electrical cable comprising a cable core and a plurality of conductive elements surrounding the cable core. Generally, electrical cables using various cable core configurations and conductor element configurations disclosed herein may be produced by any suitable method known to those of skill in the art. For instance, a rigid-frame strander, which can rotate spools of composite cores or strands to assemble a cable core, may be employed. In some embodiments, the rigid-frame strander may impart one twist per machine revolution into all composite cores or strands, except for the center composite core, which is not twisted. Each successive layer over the center composite core may be closed by a round die. After the final layer is applied, the cable core containing the composite cores or strands may be secured with a tape or other material. If tape is employed, it may be applied using a concentric taping machine. The resulting cable core with tape may be taken-up onto a reel. The cable core then may be fed back through the same rigid-frame strander for the application of a plurality of conductive elements around the cable core. Consistent with embodiments of the present invention, methods of transmitting electricity are provided herein. One such method of transmitting electricity may comprise (i) installing an electrical cable as disclosed herein, e.g., comprising a cable core and a plurality of conductive elements surrounding the cable core, and (ii) transmitting electricity across the electrical cable. Another method of transmitting electricity may comprise (i) providing an electrical cable as disclosed herein, e.g., comprising a cable core and a plurality of conductive elements surrounding the cable core, and (ii) transmitting electricity across the electrical cable. In these and other embodiments, the electrical cable, cable core, and conductive elements may be any electrical cable, cable core, and conductive elements described herein. For instance, the cable core may comprise any composite core described herein, i.e., one or more composite cores or strands. EXAMPLES The invention is further illustrated by the following examples, which are not to be construed in any way as imposing limitations to the scope of this invention. Various other aspects, embodiments, modifications, and equivalents thereof which, after reading the description herein, may suggest themselves to one of ordinary skill in the art without departing from the spirit of the present invention or the scope of the appended claims. Example 1 Two (2) continuous fiber ribbons were initially formed using an extrusion system as substantially described above. Carbon fiber rovings (Toray T700SC, which contained 12,000 carbon filaments having a tensile strength of 4,900 MPa and a mass per unit length of 0.8 g/m) were employed for the continuous fibers with each individual ribbon containing 4 rovings. The thermoplastic polymer used to impregnate the fibers was polyphenylene sulfide (PPS) (FORTRON® PPS 205, available from Ticona, LLC), which has a melting point of about 280° C. Each ribbon contained 50 wt. % carbon fibers and 50 wt. % PPS. The ribbons had a thickness of about 0.18 mm and a void fraction of less than 1.0%. Once formed, the ribbons were then fed to a pultrusion line operating at a speed of 20 ft/min. Before shaping, the ribbons were heated within an infrared oven (power setting of 305). The heated ribbons were then supplied to a consolidation die having a circular-shaped channel that received the ribbons and compressed them together while forming the initial shape of the rod. Within the die, the ribbons remained at a temperature of about 177° C. Upon consolidation, the resulting preform was then briefly cooled with an air ring/tunnel device that supplied ambient air at a pressure of 1 psig. The preform was then passed through a nip formed between two rollers, and then to a calibration die for final shaping. Within the calibration die, the preform remained at a temperature of about 140° C. After exiting this die the profile was capped with a polyether ether ketone (PEEK), which had a melting point of 350° C. The capping layer had an average thickness of about 0.1-0.15 mm. The resulting part was then cooled with an air stream. The resulting composite core had an average outside diameter of about 3.4-3.6 mm, and contained 45 wt. % carbon fibers, 50 wt. % PPS, and 5 wt. % capping material. To determine the strength properties of the composite core, three-point flexural testing was performed in accordance with ASTM D790-10, Procedure A. The support and nose radius was 0.25 inch, the support span was 30 mm, the specimen length was 2 in, and the test speed was 2 mm/min. The resulting flexural modulus was about 31 GPa and the flexural strength was about 410 MPa. The density of the part was 1.48 g/cm 3 and the void content was less than about 3%. The bend radius was 3.27 cm. Example 2 Two (2) continuous fiber ribbons were initially formed using an extrusion system as substantially described above. Carbon fiber rovings (Toray T700SC) were employed for the continuous fibers with each individual ribbon containing 4 rovings. The thermoplastic polymer used to impregnate the fibers was FORTRON® PPS 205. Each ribbon contained 50 wt. % carbon fibers and 50 wt. % PPS. The ribbons had a thickness of about 0.18 mm and a void fraction of less than 1.0%. Once formed, the ribbons were then fed to a pultrusion line operating at a speed of 20 ft/min. Before shaping, the ribbons were heated within an infrared oven (power setting of 305). The heated ribbons were then supplied to a consolidation die having a circular-shaped channel that received the ribbons and compressed them together while forming the initial shape of the rod. Within the die, the ribbons remained at a temperature of about 343° C. Upon consolidation, the resulting preform was then briefly cooled with an air ring/tunnel device that supplied ambient air at a pressure of 1 psig. The preform was then passed through a nip formed between two rollers, and then to a calibration die for final shaping. Within the calibration die, the preform remained at a temperature of about 140° C. After exiting this die the profile was capped with FORTRON® PPS 320, which had a melting point of 280° C. The capping layer had an average thickness of about 0.1-0.15 mm. The resulting part was then cooled with an air stream. The resulting composite core had an average outside diameter of about 3.4-3.6 mm, and contained 45 wt. % carbon fibers, 50 wt. % PPS, and 5 wt. % capping material. To determine the strength properties of the composite core, three-point flexural testing was performed in accordance with ASTM D790-10, Procedure A. The support and nose radius was 0.25 inch, the support span was 30 mm, the specimen length was 2 in, and the test speed was 2 mm/min. The resulting flexural modulus was 20.3 GPa and the flexural strength was about 410 MPa. The density of the part was 1.48 g/cm 3 and the void content was less than about 3%. The bend radius was 4.37 cm. Example 3 Two (2) continuous fiber ribbons were initially formed using an extrusion system as substantially described above. Glass fiber rovings (TUFRov® 4588 from PPG, which contained E-glass filaments having a tensile strength of 2,599 MPa and a mass per unit length of 0.0044 lb/yd (2.2 g/m) were employed for the continuous fibers with each individual ribbon containing 2 rovings. The thermoplastic polymer used to impregnate the fibers was polyphenylene sulfide (PPS) (FORTRON® 205, available from Ticona, LLC), which has a melting point of about 280° C. Each ribbon contained 56 wt. % glass fibers and 44 wt. % PPS. The ribbons had a thickness of about 0.18 mm and a void fraction of less than 1.0%. Once formed, the ribbons were then fed to a pultrusion line operating at a speed of 20 ft/min. Before shaping, the ribbons were heated within an infrared oven (power setting of 330). The heated ribbons were then supplied to a consolidation die having a circular-shaped channel that received the ribbons and compressed them together while forming the initial shape of the rod. Upon consolidation, the resulting preform was then briefly cooled with ambient air. The preform was then passed through a nip formed between two rollers, and then to a calibration die for final shaping. Within the calibration die, the preform remained at a temperature of about 275° C. After exiting this die, the profile was capped with FORTRON® 205. The capping layer had an average thickness of about 0.1-0.15 mm. The resulting part was then cooled with an air stream. The resulting composite core had an average outside diameter of about 3.4-3.6 mm, and contained 50 wt. % glass fibers and 50 wt. % PPS. To determine the strength properties of the composite core, three-point flexural testing was performed in accordance with ASTM D790-10, Procedure A. The support and nose radius was 0.25 inch, the support span was 30 mm, the specimen length was 2 in, and the test speed was 2 mm/min. The resulting flexural modulus was about 18 GPa and the flexural strength was about 590 MPa. The void content was about 0%, and the bend radius was 1.87 cm. Example 4 Two (2) continuous fiber ribbons were initially formed using an extrusion system as substantially described above. Glass fiber rovings (TUFRov® 4588) were employed for the continuous fibers with each individual ribbon containing 2 rovings. The thermoplastic polymer used to impregnate the fibers was Nylon 66 (PA66), which has a melting point of about 250° C. Each ribbon contained 60 wt. % glass fibers and 40 wt. % Nylon 66. The ribbons had a thickness of about 0.18 mm and a void fraction of less than 1.0%. Once formed, the ribbons were then fed to a pultrusion line operating at a speed of 10 ft/min. Before shaping, the ribbons were heated within an infrared oven (power setting of 320). The heated ribbons were then supplied to a consolidation die having a circular-shaped channel that received the ribbons and compressed them together while forming the initial shape of the rod. Upon consolidation, the resulting preform was then briefly cooled with ambient air. The preform was then passed through a nip formed between two rollers, and then to a calibration die for final shaping. Within the calibration die, the preform remained at a temperature of about 170° C. After exiting this die, the profile was capped with Nylon 66. The capping layer had an average thickness of about 0.1-0.15 mm. The resulting part was then cooled with an air stream. The resulting composite core had an average outside diameter of about 3.4-3.6 mm, and contained 53 wt. % glass fibers, 40 wt. % Nylon 66, and 7 wt. % capping material. To determine the strength properties of the composite core, three-point flexural testing was performed in accordance with ASTM D790-10, Procedure A. The support and nose radius was 0.25 inch, the support span was 30 mm, the specimen length was 2 in, and the test speed was 2 mm/min. The resulting flexural modulus was about 19 GPa and the flexural strength was about 549 MPa. The void content was about 0%, and the bend radius was 2.34 cm. Example 5 Three (3) batches of eight (8) cores were formed having different void fraction levels. For each rod, two (2) continuous fiber ribbons were initially formed using an extrusion system as substantially described above. Carbon fiber rovings (Toray T700SC, which contained 12,000 carbon filaments having a tensile strength of 4,900 MPa and a mass per unit length of 0.8 g/m) were employed for the continuous fibers with each individual ribbon containing 4 rovings. The thermoplastic polymer used to impregnate the fibers was polyphenylene sulfide (“PPS”) (FORTRON® PPS 205, available from Ticona, LLC), which had a melting point of about 280° C. Each ribbon contained 50 wt. % carbon fibers and 50 wt. % PPS. The ribbons had a thickness of about 0.18 mm and a void fraction of less than 1.0%. Once formed, the ribbons were then fed to a pultrusion line operating at a speed of 20 ft/min. Before shaping, the ribbons were heated within an infrared oven (power setting of 305). The heated ribbons were then supplied to a consolidation die having a circular-shaped channel that received the ribbons and compressed them together while forming the initial shape of the rod. Within the die, the ribbons remained at a temperature of about 177° C. Upon consolidation, the resulting preform was then briefly cooled with an air ring/tunnel device that supplied ambient air at a pressure of 1 psi. The preform was then passed through a nip formed between two rollers, and then to a calibration die for final shaping. Within the calibration die, the preform remained at a temperature of about 140° C. After exiting this die, the profile was capped with a polyether ether ketone (“PEEK”), which had a melting point of 350° C. The capping layer had a thickness of about 0.1 mm. The resulting composite core was then cooled with an air stream. The resulting composite core had a diameter of about 3.5 mm, and contained 45 wt. % carbon fibers, 50 wt. % PPS, and 5 wt. % capping material. A first batch of composite cores had a mean void fraction of 2.78%. A second batch of composite cores had a mean void fraction of 4.06%. A third batch of composite cores had a mean void fraction of 8.74%. Void fraction measurements were performed using CT scanning. A Metrotom 1500 (2k×2k) high resolution detector was used to scan the core specimens. Detection was done using an enhanced analysis mode with a low probability threshold. Once the specimens were scanned for void fraction, Volume Graphics software was used to interpret the data from the 3D scans, and calculate the void levels in each specimen. To determine the flexural fatigue life and residual flexural strength of the rods, three-point flexural fatigue testing was performed in accordance with ASTM D790. The support span was 2.2 in and the specimen length was 3 in. Four (4) composite cores from each batch were tested at a loading level of 160 Newtons (“N”) and four (4) composite cores from each batch were tested at a loading level of 180 N, respectively, representing about 50% and 55% of the pristine (static) flexural strength of the cores. Each specimen was tested to one million cycles at a frequency of 10 Hertz (Hz). Before and after fatigue testing, to determine the respective pristine and residual flexural strength properties of the rods, three-point flexural testing was performed in accordance with ASTM D790-10, Procedure A. The average pristine and residual flexural strengths of each batch at each loading level were recorded. The resulting pristine flexural strength for the third batch was 107 ksi, and the resulting residual flexural strength for the third batch was 75 ksi, thus resulting in a reduction of about 29%. The resulting pristine flexural strength for the second batch was 108 ksi, and the resulting residual flexural strength for the second batch was 72 ksi, thus resulting in a reduction of about 33%. The resulting pristine flexural strength for the first batch was 106 ksi, and the resulting residual flexural strength for the first batch was 105 ksi, thus resulting in a reduction of about 1%. Example 6 FIG. 15 illustrates the electrical cable520 produced in Example 6. The 26 conductive elements522 formed a first layer526 and a second layer528. The cable core500 was a strand of 7 composite cores. A tape530 between the cable core500 and the conductive elements522 partially covered the cable core500 in a helical arrangement. Electrical cable was produced as follows. Seven (7) composite cores having a diameter of about 3.5 mm were stranded to form a stranded cable core with a 508-mm lay length. The composite cores were similar to those produced in Example 1 above. The cable core was secured with an aluminum foil tape laminated to a fiberglass scrim and a silicone based adhesive. 26 conductor wires were placed above and around the cable core and tape in two layers as shown in FIG. 15. The conductor wires had a diameter of about 4.5 mm, and were fabricated from fully annealed 1350 aluminum. The ultimate tensile strength of the cable was approximately 19,760 psi (136 MPa). Example 7 FIG. 15 illustrates the electrical cable520 produced in Example 7. The 26 conductive elements522 formed a first layer526 and a second layer528. The cable core500 was a strand of 7 composite cores. A tape530 between the cable core500 and the conductive elements522 partially covered the cable core500 in a helical arrangement. Electrical cable was produced as follows. Seven (7) composite cores having a diameter of about 3.5 mm were stranded to form a stranded cable core with a 508-mm lay length. The composite cores were similar to those produced in Example 1 above. The cable core was secured with an aluminum foil tape laminated to a fiberglass scrim and a silicone based adhesive. 26 conductor wires were placed above and around the cable core and tape in two layers as shown in FIG. 15. The conductor wires had a diameter of about 4.5 mm, and were fabricated from an aluminum alloy containing zirconium (approximately 0.2-0.33% zirconium). FIG. 16 illustrates the stress-strain data for the electrical cable of Example 7. The electrical cable of Example 7 was tested for its fatigue and/or vibrational resistance properties in accordance with the Aeolian vibration test specified in IEEE 1138. The electrical cable of Example 7 passed the Aeolian vibration test at 100 million cycles. Using mathematical modeling based on overhead transmission cables similar to Example 7, the 10-year creep (elongation) values for the electrical cable of Example 7 were estimated. The calculated 10-year creep values at 15%, 20%, 25%, and 30% RBS (rated breaking stress) were approximately 0.054%, approximately 0.081%, approximately 0.119%, and approximately 0.163%, respectively. Constructive Example 8 FIG. 17 illustrates an electrical cable620 that can be produced in Constructive Example 8. The 26 conductive elements622 can form a first layer626 and a second layer628. The cable core600 can be a strand of 7 composite cores. A tape630 between the cable core600 and the conductive elements622 can partially cover the cable core600 in a helical arrangement. The electrical cable of FIG. 17 can be produced as follows. Seven (7) composite cores having a diameter of about 3.5 mm can be stranded to form a stranded cable core with a 508-mm lay length. The composite cores can be similar to those produced in Example 1 above. The cable core can be secured with an aluminum foil tape laminated to a fiberglass scrim and a silicone based adhesive. 26 conductor wires can be placed above and around the cable core and tape in two layers as shown in FIG. 17. The conductors can be trapezoidal wires having a cross-sectional area of about 15-17 mm 2, and can be fabricated from annealed 1350 aluminum (or alternatively, an aluminum alloy containing zirconium). Claims (20) We claim: An electrical cable comprising: (a) a cable core comprising at least one composite core, the composite core comprising: (i) at least one rod comprising a plurality of consolidated thermoplastic impregnated rovings, the rovings comprising continuous carbon fibers oriented in the longitudinal direction and a thermoplastic matrix that embeds the carbon fibers, the carbon fibers having a ratio of ultimate tensile strength to mass per unit length of greater than about 1,000 MPa/g/m, wherein the thermoplastic matrix comprises a polyphenylene sulfide, and wherein the rod comprises from about 30 wt. % to about 75 wt. % carbon fibers and from about 25 wt. % to about 70 wt. % thermoplastic matrix; and (ii) a capping layer surrounding the at least one rod, wherein the capping layer comprises a polyether ether ketone and contains less than 5 wt. % of continuous carbon fibers; wherein the composite core has a flexural modulus of from about 15 to about 200 GPa; and (b) a plurality of conductive elements surrounding the cable core, wherein the conductive elements comprise aluminum or an aluminum alloy. The cable of claim 1, further comprising a partial or complete layer of a tape or a coating between the cable core and the plurality of conductive elements. The cable of claim 1, wherein the cable core comprises two or more composite cores. The cable of claim 1, wherein the cable core comprises from 2to 37 composite cores. The cable of claim 1, wherein the cable comprises 7, 19, 26, or 37 conductive elements. The cable of claim 1, wherein the cable comprises from 2 to about 50 conductive elements. The cable of claim 1, wherein: the cable core is a stranded core comprising 7 composite cores; and the cable comprises 26 conductive elements arranged in 2 layers around the cable core. The cable of claim 7, wherein the composite cores have a density in a range from about 1.3 g/cc to about 1.7 g/cc. The cable of claim 1, wherein the cable has: a 10-year creep value at 15% RBS (rated breaking stress) of less than about 0.2%; and a 10-year creep value at 30% RBS (rated breaking stress) of less than about 0.25%. The cable of claim 1, wherein the capping layer has a thickness in a range from about 0.02 mm to about 5 mm and contains less than about 1 wt. % of continuous carbon fibers. The cable of claim 1, wherein the capping layer contains 0 wt. % of continuous carbon fibers. The cable of claim 1, wherein the cable has: a sag, at rated temperature of 180° C., for a 300-meter level span with a NESC light loading, in a range from about 3 m to about 9.5 m; and a sag, at rated temperature of 180° C., for a 300-meter level span with a NESC heavy loading, in a range from about 3 m to about 7.5 m. The cable of claim 1, wherein the cable passes an Aeolian vibration test at 100 million cycles. The cable of claim 1, wherein the capping layer has a thickness in a range from about 0.01 mm to about 10 mm. The cable of claim 1, wherein the composite core has: a bending radius in a range from about 1 cm to about 50 cm; and a void fraction of about 6% or less. The cable of claim 15, wherein the composite core has a flexural modulus in a range from about 20 to about 150 GPa. The cable of claim 15, wherein the composite core has a density in a range from about 1.2 g/cc to about 1.8 g/cc. The cable of claim 15, wherein the composite core has a bending radius in a range from about 3 cm to about 40 cm. The cable of claim 1, wherein the composite core has a void fraction of about 4% or less. The cable of claim 1, wherein the composite core has: a flexural modulus of from about 20 to about 150 GPa; a density from about 1.3 g/cc to about 1.7 g/cc; and a void fraction of about 3% or less. US15/236,521 2011-04-12 2016-08-15 Electrical transmission cables with composite cores ActiveUS9685257B2 (en) Priority Applications (3) \| Application Number \| Priority Date \| Filing Date \| Title \| --- --- \| \| US15/236,521US9685257B2 (en) \| 2011-04-12 \| 2016-08-15 \| Electrical transmission cables with composite cores \| \| US15/596,026US20170256338A1 (en) \| 2011-04-12 \| 2017-05-16 \| Electrical Transmission Cables With Composite Cores \| \| US15/916,561US20180197658A1 (en) \| 2011-04-12 \| 2018-03-09 \| Electrical Transmission Cables With Composite Cores \| Applications Claiming Priority (4) \| Application Number \| Priority Date \| Filing Date \| Title \| --- --- \| \| US201161474423P \| 2011-04-12 \| 2011-04-12 \| \| \| US13/443,938US9012781B2 (en) \| 2011-04-12 \| 2012-04-11 \| Electrical transmission cables with composite cores \| \| US14/661,780US9443635B2 (en) \| 2011-04-12 \| 2015-03-18 \| Electrical transmission cables with composite cores \| \| US15/236,521US9685257B2 (en) \| 2011-04-12 \| 2016-08-15 \| Electrical transmission cables with composite cores \| Related Parent Applications (1) \| Application Number \| Title \| Priority Date \| Filing Date \| --- --- \| \| US14/661,780 ContinuationUS9443635B2 (en) \| 2011-04-12 \| 2015-03-18 \| Electrical transmission cables with composite cores \| Related Child Applications (1) \| Application Number \| Title \| Priority Date \| Filing Date \| --- --- \| \| US15/596,026 ContinuationUS20170256338A1 (en) \| 2011-04-12 \| 2017-05-16 \| Electrical Transmission Cables With Composite Cores \| Publications (2) \| Publication Number \| Publication Date \| --- \| \| US20160351300A1US20160351300A1 (en) \| 2016-12-01 \| \| US9685257B2 trueUS9685257B2 (en) \| 2017-06-20 \| Family ID=46022657 Family Applications (3) \| Application Number \| Title \| Priority Date \| Filing Date \| --- --- \| \| US15/236,521 ActiveUS9685257B2 (en) \| 2011-04-12 \| 2016-08-15 \| Electrical transmission cables with composite cores \| \| US15/596,026 AbandonedUS20170256338A1 (en) \| 2011-04-12 \| 2017-05-16 \| Electrical Transmission Cables With Composite Cores \| \| US15/916,561 AbandonedUS20180197658A1 (en) \| 2011-04-12 \| 2018-03-09 \| Electrical Transmission Cables With Composite Cores \| Family Applications After (2) \| Application Number \| Title \| Priority Date \| Filing Date \| --- --- \| \| US15/596,026 AbandonedUS20170256338A1 (en) \| 2011-04-12 \| 2017-05-16 \| Electrical Transmission Cables With Composite Cores \| \| US15/916,561 AbandonedUS20180197658A1 (en) \| 2011-04-12 \| 2018-03-09 \| Electrical Transmission Cables With Composite Cores \| Country Status (17) \| Country \| Link \| --- \| \| US (3) \| US9685257B2 (en) \| \| EP (1) \| EP2697800B1 (en) \| \| CN (2) \| CN107742542B (en) \| \| AR (1) \| AR085999A1 (en) \| \| AU (1) \| AU2012242930B2 (en) \| \| BR (1) \| BR112013026310B1 (en) \| \| CA (1) \| CA2832453C (en) \| \| CL (1) \| CL2013002932A1 (en) \| \| DK (1) \| DK2697800T3 (en) \| \| ES (1) \| ES2617596T3 (en) \| \| HU (1) \| HUE033251T2 (en) \| \| MX (1) \| MX346917B (en) \| \| PL (1) \| PL2697800T3 (en) \| \| RU (1) \| RU2594016C2 (en) \| \| TW (3) \| TWI609383B (en) \| \| WO (1) \| WO2012142129A1 (en) \| \| ZA (1) \| ZA201308410B (en) \| Cited By (7) Cited by examiner, † Cited by third party\| Publication number \| Priority date \| Publication date \| Assignee \| Title \| --- --- \| USD830311S1 (en) \| 2014-09-25 \| 2018-10-09 \| Conway Electric, LLC \| Overbraided electrical cord with X pattern \| \| WO2019147838A1 (en) \| 2018-01-24 \| 2019-08-01 \| Ctc Global Corporation \| Termination arrangement for an overhead electrical cable \| \| US20200126686A1 (en) \| 2018-10-18 \| 2020-04-23 \| Saudi Arabian Oil Company \| Power cable with non-conductive armor \| \| WO2021127659A1 (en) \| 2019-12-20 \| 2021-06-24 \| Ctc Global Corporation \| Ported hardware for overhead electrical cables \| \| US20210249160A1 (en) \| 2020-02-12 \| 2021-08-12 \| Jonathan Jan \| Wire having a hollow micro-tubing and method therefor \| \| WO2021222663A1 (en) \| 2020-04-29 \| 2021-11-04 \| Ctc Global Corporation \| Strength member assemblies and overhead electrical cables incorporating optical fibers \| \| EP4273892A2 (en) \| 2018-02-27 \| 2023-11-08 \| CTC Global Corporation \| Systems, methods and tools for the interrogation of composite strength members \| Families Citing this family (22) Cited by examiner, † Cited by third party\| Publication number \| Priority date \| Publication date \| Assignee \| Title \| --- --- \| MX346917B (en) \| 2011-04-12 \| 2017-04-05 \| Southwire Co \| Electrical transmission cables with composite cores. \| \| KR20140027252A (en) \| 2011-04-12 \| 2014-03-06 \| 티코나 엘엘씨 \| Composite core for electrical transmission cables \| \| US20150017437A1 (en) \| 2013-07-10 \| 2015-01-15 \| Ticona Llc \| Composite Rod Having an Abrasion Resistant Capping Layer \| \| DE102014010777A1 (en) \| 2014-01-30 \| 2015-07-30 \| Dürr Systems GmbH \| High voltage cables \| \| EP3129991B1 (en) \| 2014-04-09 \| 2021-03-03 \| De Angeli Prodotti S.r.l. \| Conductor for bare overhead electric lines, especially for middle-high thermal limit, and low expansion at high electronic loads \| \| US10767421B2 (en) \| 2014-04-29 \| 2020-09-08 \| Halliburton Energy Services, Inc. \| Composite cables \| \| CN106626444A (en) \| 2016-07-27 \| 2017-05-10 \| 山东极威新材料科技有限公司 \| Production technology of continuous fiber reinforced thermoplastic composites (CFRTP) pultruded profiles \| \| CN111655403B (en) \| 2017-10-31 \| 2022-10-14 \| 梅尔德制造公司 \| Solid additive manufacturing system and composition and structure of material \| \| RU2683252C1 (en) \| 2017-12-11 \| 2019-03-27 \| Виктор Александрович Фокин \| Insulated steel-aluminum wire \| \| JP7451500B2 (en) \| 2018-09-11 \| 2024-03-18 \| メルドマニュファクチュアリングコーポレイション \| Solid additive production method for synthesizing conductive polymer composites, production of conductive plastic parts and conductive coatings \| \| RU2695317C1 (en) \| 2018-10-15 \| 2019-07-23 \| Общество с ограниченной ответственностью "Метсбытсервис" \| Steel-aluminium high-strength, high-temperature insulated wire for overhead power transmission line \| \| US11441001B1 (en) \| 2018-11-29 \| 2022-09-13 \| University Of Tennessee Research Foundation \| Process to manufacture carbon fiber intermediate products in-line with carbon fiber production \| \| EP3696214A1 (en) \| 2019-02-13 \| 2020-08-19 \| Borealis AG \| Reinforcement of moulded parts with pp unidirectional long glass fiber (lgf) rods \| \| RU2706957C1 (en) \| 2019-03-21 \| 2019-11-21 \| Виктор Александрович Фокин \| Non-insulated steel-aluminum high-temperature high-strength wire \| \| JP6987824B2 (en) \| 2019-10-25 \| 2022-01-05 \| 矢崎総業株式会社 \| Communication cable and wire harness \| \| MX2022013850A (en) \| 2020-05-02 \| 2023-04-14 \| Terry Mcquarrie \| Electrical conductor and composite core for an electrical conductor having a nanoparticle modified resin. \| \| US10964447B1 (en) \| 2020-07-01 \| 2021-03-30 \| International Business Machines Corporation \| Periodic transmission line cable filtering \| \| CN112102981B (en) \| 2020-09-21 \| 2021-04-16 \| 江苏易鼎复合技术有限公司 \| Metal-clad composite molded line stranded reinforced core overhead conductor and manufacturing method thereof \| \| RU209892U1 (en) \| 2021-08-17 \| 2022-03-23 \| Общество с ограниченной ответственностью "Кабельный Завод "ЭКСПЕРТ-КАБЕЛЬ" \| CABLE PRODUCT THAT CHANGES COLOR DEPENDING ON TEMPERATURE \| \| CN113715323B (en) \| 2021-11-01 \| 2022-01-04 \| 四川久远特种高分子材料技术有限公司 \| Assembling equipment for cable connector \| \| CN115609962A (en) \| 2022-10-10 \| 2023-01-17 \| 江苏合东新材料科技有限公司 \| Glass fiber reinforced pultruded composite solar module frame and preparation method thereof \| \| CN117542581B (en) \| 2023-11-11 \| 2024-08-06 \| 重庆特发信息光缆有限公司 \| Wire cable core stranding machine and process \| Citations (187) Cited by examiner, † Cited by third party\| Publication number \| Priority date \| Publication date \| Assignee \| Title \| --- --- \| US2589507A (en) \| 1947-01-11 \| 1952-03-18 \| Aluminum Co Of America \| Expanded electrical transmission cable \| \| US3244784A (en) \| 1960-01-15 \| 1966-04-05 \| Universal Moulded Fiber Glass \| Method for forming fibre reinforced resin articles \| \| US3509269A (en) \| 1968-06-11 \| 1970-04-28 \| Western Electric Co \| Thermal barriers for cables \| \| US3706216A (en) \| 1970-12-16 \| 1972-12-19 \| Joseph L Weingarten \| Process for reinforcing extruded articles \| \| US3795524A (en) \| 1971-03-01 \| 1974-03-05 \| Minnesota Mining & Mfg \| Aluminum borate and aluminum borosilicate articles \| \| US3823255A (en) \| 1972-04-20 \| 1974-07-09 \| Cyprus Mines Corp \| Flame and radiation resistant cable \| \| US3980808A (en) \| 1974-09-19 \| 1976-09-14 \| The Furukawa Electric Co., Ltd. \| Electric cable \| \| US3993726A (en) \| 1974-01-16 \| 1976-11-23 \| Hercules Incorporated \| Methods of making continuous length reinforced plastic articles \| \| US4047965A (en) \| 1976-05-04 \| 1977-09-13 \| Minnesota Mining And Manufacturing Company \| Non-frangible alumina-silica fibers \| \| US4228641A (en) \| 1978-09-28 \| 1980-10-21 \| Exxon Research & Engineering Co. \| Thermoplastic twines \| \| US4454709A (en) \| 1982-03-12 \| 1984-06-19 \| General Electric Company \| Method of forming concentric cable layer and article formed \| \| US4499716A (en) \| 1983-06-13 \| 1985-02-19 \| E. I. Du Pont De Nemours And Company \| Reinforcement structure \| \| US4549920A (en) \| 1981-07-28 \| 1985-10-29 \| Imperial Chemical Industries, Plc \| Method for impregnating filaments with thermoplastic \| \| US4552432A (en) \| 1983-04-21 \| 1985-11-12 \| Cooper Industries, Inc. \| Hybrid cable \| \| US4559262A (en) \| 1981-01-21 \| 1985-12-17 \| Imperial Chemical Industries, Plc \| Fibre reinforced compositions and methods for producing such compositions \| \| US4588538A (en) \| 1984-03-15 \| 1986-05-13 \| Celanese Corporation \| Process for preparing tapes from thermoplastic polymers and carbon fibers \| \| US4649060A (en) \| 1984-03-22 \| 1987-03-10 \| Agency Of Industrial Science & Technology \| Method of producing a preform wire, sheet or tape fiber reinforced metal composite \| \| US4657990A (en) \| 1984-01-19 \| 1987-04-14 \| Imperial Chemical Industries Plc \| Aromatic polyetherketones \| \| US4677818A (en) \| 1984-07-11 \| 1987-07-07 \| Toho Beslon Co., Ltd. \| Composite rope and manufacture thereof \| \| US4680224A (en) \| 1984-03-06 \| 1987-07-14 \| Phillips Petroleum Company \| Reinforced plastic \| \| US4767841A (en) \| 1987-02-24 \| 1988-08-30 \| Phillips Petroleum Company \| Arylene sulfide polymer preparation from dehydrated admixture comprising sulfur source, cyclic amide solvent and water \| \| US4792481A (en) \| 1986-11-28 \| 1988-12-20 \| Phillips Petroleum Company \| Reinforced plastic \| \| US4843696A (en) \| 1987-05-11 \| 1989-07-04 \| Southwire Company \| Method and apparatus for forming a stranded conductor \| \| US4861947A (en) \| 1987-04-13 \| 1989-08-29 \| Schweizerische Isola-Werke \| Communication or control cable with supporting element \| \| US4877643A (en) \| 1988-03-24 \| 1989-10-31 \| Director General Agency Of Industrial Science And Technology \| Process for producing preformed wire from silicon carbide fiber-reinforced aluminum \| \| US4954462A (en) \| 1987-06-05 \| 1990-09-04 \| Minnesota Mining And Manufacturing Company \| Microcrystalline alumina-based ceramic articles \| \| CN2071207U (en) \| 1990-06-16 \| 1991-02-13 \| 辽宁省农业工程研究设计所 \| Steel and plastic composite rope \| \| US5010209A (en) \| 1988-12-20 \| 1991-04-23 \| Pirelli Cable Corp. \| Power cable with water swellable agents and elongated metal elements outside cable insulation \| \| US5017643A (en) \| 1990-03-20 \| 1991-05-21 \| Phillips Petroleum Company \| Composition and process for making poly (arylene sulfide) resins reinforced with glass fibers \| \| JPH03119188A (en) \| 1989-10-03 \| 1991-05-21 \| Mitsubishi Kasei Corp \| Fiber-reinforced plastic composite material \| \| JPH03129606A (en) \| 1989-07-27 \| 1991-06-03 \| Hitachi Cable Ltd \| Aerial power cable \| \| US5039572A (en) \| 1989-07-24 \| 1991-08-13 \| Phillips Petroleum Company \| High strength thermoplastic resin/carbon fiber composites and methods \| \| GB2240997A (en) \| 1990-02-19 \| 1991-08-21 \| Bridon Plc \| Strand or rope product of composite rods \| \| US5060466A (en) \| 1988-10-31 \| 1991-10-29 \| Tokyo Rope Mfg. Co. Ltd. \| Composite rope and manufacturing method for the same \| \| US5066536A (en) \| 1987-12-10 \| 1991-11-19 \| Imperial Chemical Industries Plc \| Fibre reinforced thermoplastic composite structures \| \| US5068142A (en) \| 1989-01-31 \| 1991-11-26 \| Teijin Limited \| Fiber-reinforced polymeric resin composite material and process for producing same \| \| US5073413A (en) \| 1990-05-31 \| 1991-12-17 \| American Composite Technology, Inc. \| Method and apparatus for wetting fiber reinforcements with matrix materials in the pultrusion process using continuous in-line degassing \| \| US5075381A (en) \| 1989-10-21 \| 1991-12-24 \| Mitsubishi Petrochemical Company Limited \| Thermoplastic resin composition \| \| US5080175A (en) \| 1990-03-15 \| 1992-01-14 \| Williams Jerry G \| Use of composite rod-stiffened wireline cable for transporting well tool \| \| US5126167A (en) \| 1990-06-13 \| 1992-06-30 \| Ube-Nitto Kasei Co., Ltd. \| Process of manufacturing a fiber reinforced plastic armored cable \| \| US5130193A (en) \| 1988-11-10 \| 1992-07-14 \| Nippon Oil Co., Ltd. \| Fiber-reinforced composite cable \| \| US5149749A (en) \| 1990-05-31 \| 1992-09-22 \| Phillips Petroleum Company \| Poly(phenylene sulfide) composition and articles having improved thermal stability at high temperatures \| \| US5149581A (en) \| 1989-11-21 \| 1992-09-22 \| Idemitsu Kosan Co., Ltd. \| Polyether copolymers, resin compositions containing them, and molded articles formed from them \| \| US5153381A (en) \| 1990-03-20 \| 1992-10-06 \| Alcan Aluminum Corporation \| Metal clad cable and method of making \| \| US5159157A (en) \| 1989-09-12 \| 1992-10-27 \| Kabelwerke Reinshagen Gmbh \| Electrical cable with element of high tensile strength \| \| US5171942A (en) \| 1991-02-28 \| 1992-12-15 \| Southwire Company \| Oval shaped overhead conductor and method for making same \| \| US5176775A (en) \| 1989-10-20 \| 1993-01-05 \| Montsinger Lawrence V \| Apparatus and method for forming fiber filled thermoplastic composite materials \| \| JPH0511729A (en) \| 1991-07-01 \| 1993-01-22 \| Seiko Epson Corp \| Information processing equipment \| \| JPH0533278A (en) \| 1991-07-19 \| 1993-02-09 \| Toray Ind Inc \| Rope comprising carbon fiber-reinforced composite material and production thereof \| \| US5198621A (en) \| 1991-12-31 \| 1993-03-30 \| The Furukawa Electric Co., Ltd. \| Twisted cable \| \| US5207850A (en) \| 1990-07-17 \| 1993-05-04 \| General Electric Company \| Process for making thermoplastic composites with cyclics oligomers and composites made thereby \| \| US5210128A (en) \| 1991-07-24 \| 1993-05-11 \| Phillips Petroleum Company \| Poly(arylene sulfide) compositions, composites, and methods of production \| \| US5211500A (en) \| 1989-04-06 \| 1993-05-18 \| Tokyo Rope Mfg. Co., Ltd. \| Composite rope having molded-on fixing member at end portion thereof \| \| JPH05148780A (en) \| 1991-11-28 \| 1993-06-15 \| Toray Ind Inc \| Production of rope composed of fiber-reinforced composite material \| \| US5277566A (en) \| 1988-10-19 \| 1994-01-11 \| Hoechst Aktiengesellschaft \| Extrusion impregnating device \| \| JPH067112A (en) \| 1992-06-23 \| 1994-01-18 \| Iharaya Shoten:Kk \| Frame for device for squeezing unrefined sake @(3754/24)rice wine) \| \| US5298318A (en) \| 1990-03-21 \| 1994-03-29 \| Phillips Petroleum Company \| Poly(arylene sulfide) resins reinforced with glass fibers \| \| US5319003A (en) \| 1992-09-30 \| 1994-06-07 \| Union Carbide Chemicals & Plastics Technology Corporation \| Method for improving the mechanical performance of composite articles \| \| US5324563A (en) \| 1990-08-08 \| 1994-06-28 \| Bell Helicopter Textron Inc. \| Unidirectional carbon fiber reinforced pultruded composite material having improved compressive strength \| \| US5325457A (en) \| 1991-09-20 \| 1994-06-28 \| Bottoms Jack Jr \| Field protected self-supporting fiber optic cable \| \| US5364657A (en) \| 1990-04-06 \| 1994-11-15 \| The University Of Akron \| Method of depositing and fusing polymer particles onto moistened continuous filaments \| \| US5374694A (en) \| 1989-08-09 \| 1994-12-20 \| Ici Composites Inc. \| Curable blends of cyanate esters and polyarylsulphones \| \| US5424388A (en) \| 1993-06-24 \| 1995-06-13 \| Industrial Technology Research Institute \| Pultrusion process for long fiber-reinforced nylon composites \| \| US5439632A (en) \| 1993-02-05 \| 1995-08-08 \| Ici Composites, Inc. \| Process for pretreatment of non-isotropic cylindrical products \| \| JPH07279940A (en) \| 1994-03-31 \| 1995-10-27 \| Nippon Cable Syst Inc \| High bending withstanding rope \| \| US5462618A (en) \| 1993-03-23 \| 1995-10-31 \| Bell Helicopter Textron Inc. \| Continuous process of making unidirectional graphite fiber reinforced pultruded rods having minimal fiber waviness \| \| US5468327A (en) \| 1994-01-24 \| 1995-11-21 \| University Of Massachusetts Lowell \| Method and device for continuous formation of braid reinforced thermoplastic structural and flexible members \| \| US5501906A (en) \| 1994-08-22 \| 1996-03-26 \| Minnesota Mining And Manufacturing Company \| Ceramic fiber tow reinforced metal matrix composite \| \| US5525003A (en) \| 1993-12-29 \| 1996-06-11 \| Conoco Inc. \| Connection termination for composite rods \| \| US5529652A (en) \| 1987-07-11 \| 1996-06-25 \| Kabushiki Kaisha Kobe Seiko Sho \| Method of manufacturing continuous fiber-reinforced thermoplastic prepregs \| \| US5542230A (en) \| 1993-04-02 \| 1996-08-06 \| Deutsche Forschungsanstalt F. Luft-Und Raumfahrt E.V. \| Connecting rods \| \| US5554826A (en) \| 1992-06-25 \| 1996-09-10 \| Southwire Company \| Overhead transmission conductor \| \| US5614139A (en) \| 1993-05-18 \| 1997-03-25 \| Eniricerche S.P.A. \| Process of making a composite article \| \| US5658513A (en) \| 1994-10-18 \| 1997-08-19 \| Polyplastics Co., Ltd. \| Cross-head die and method for manufacturing a resin structure reinforced with long fibers \| \| US5700417A (en) \| 1995-01-27 \| 1997-12-23 \| Kabushiki Kaisha Kobe Seiko Sho \| Pultrusion process for preparing fiber-reinforced composite rod \| \| US5705241A (en) \| 1995-07-01 \| 1998-01-06 \| Deutsche Forschungsanstalt Fur Luft-Und Raumfahrt E.V. \| Fibre-composite rod \| \| WO1998006109A1 (en) \| 1996-08-03 \| 1998-02-12 \| Bicc Public Limited Company \| Electrical and optical cable \| \| US5725954A (en) \| 1995-09-14 \| 1998-03-10 \| Montsinger; Lawrence V. \| Fiber reinforced thermoplastic composite with helical fluted surface and method of producing same \| \| US5727357A (en) \| 1996-05-22 \| 1998-03-17 \| Owens-Corning Fiberglas Technology, Inc. \| Composite reinforcement \| \| US5780154A (en) \| 1994-03-22 \| 1998-07-14 \| Tokuyama Corporation \| Boron nitride fiber and process for production thereof \| \| US5783129A (en) \| 1993-08-17 \| 1998-07-21 \| Polyplastics Co., Ltd. \| Apparatus, method, and coating die for producing long fiber-reinforced thermoplastic resin composition \| \| US5788908A (en) \| 1994-08-19 \| 1998-08-04 \| Polyplastics Co., Ltd. \| Method for producing long fiber-reinforced thermoplastic resin composition \| \| US5792527A (en) \| 1994-06-17 \| 1998-08-11 \| Chisso Corporation \| Products in a continuous length formed from fiber-reinforced resin and process for preparing the same \| \| US5840424A (en) \| 1993-02-19 \| 1998-11-24 \| Fiberite, Inc. \| Curable composite materials \| \| JPH10321048A (en) \| 1997-05-16 \| 1998-12-04 \| Furukawa Electric Co Ltd:The \| Tension member, lightweight low-loose overhead wire using it \| \| US5888609A (en) \| 1990-12-18 \| 1999-03-30 \| Valtion Teknillinen Tutkimuskeskus \| Planar porous composite structure and method for its manufacture \| \| US5895808A (en) \| 1996-01-25 \| 1999-04-20 \| Ems-Inventa Ag \| Process for producing compound materials with a polylactam matrix which can be thermally postformed \| \| CN1220330A (en) \| 1997-12-17 \| 1999-06-23 \| 沈阳机电研究设计院 \| Steel plastic composite rope \| \| US5935508A (en) \| 1996-07-30 \| 1999-08-10 \| Kabushiki Kaisha Kobe Seiko Sho \| Fibre-reinforced compositions and methods for their production \| \| US6003356A (en) \| 1997-01-23 \| 1999-12-21 \| Davinci Technology Corporation \| Reinforced extruded products and process of manufacture \| \| US6015953A (en) \| 1994-03-11 \| 2000-01-18 \| Tohoku Electric Power Co., Inc. \| Tension clamp for stranded conductor \| \| US6048598A (en) \| 1997-12-17 \| 2000-04-11 \| Balaba Concrete Supply, Inc. \| Composite reinforcing member \| \| US6117551A (en) \| 1996-12-18 \| 2000-09-12 \| Toray Industries, Inc. \| Carbon fiber prepreg, and a production process thereof \| \| JP3119188B2 (en) \| 1997-01-28 \| 2000-12-18 \| 日本電気株式会社 \| Semiconductor device \| \| US6180232B1 (en) \| 1995-06-21 \| 2001-01-30 \| 3M Innovative Properties Company \| Overhead high power transmission cable comprising fiber reinforced aluminum matrix composite wire \| \| US6244014B1 (en) \| 1999-07-22 \| 2001-06-12 \| Andrew Barmakian \| Steel rod-reinforced plastic piling \| \| US6248262B1 (en) \| 2000-02-03 \| 2001-06-19 \| General Electric Company \| Carbon-reinforced thermoplastic resin composition and articles made from same \| \| US6258453B1 (en) \| 1996-09-19 \| 2001-07-10 \| Lawrence V. Montsinger \| Thermoplastic composite materials made by rotational shear \| \| US6260343B1 (en) \| 1998-05-01 \| 2001-07-17 \| Wire Rope Corporation Of America, Incorporated \| High-strength, fatigue resistant strands and wire ropes \| \| US6270851B1 (en) \| 1997-07-14 \| 2001-08-07 \| Daelim Industrial Co., Ltd. \| Process for manufacturing resin-coated fibers composite and an application thereof \| \| US6329056B1 (en) \| 2000-07-14 \| 2001-12-11 \| 3M Innovative Properties Company \| Metal matrix composite wires, cables, and method \| \| US6334293B1 (en) \| 1999-03-04 \| 2002-01-01 \| N.V. Bekaert S.A. \| Steel cord with polymer core \| \| US6344270B1 (en) \| 2000-07-14 \| 2002-02-05 \| 3M Innovative Properties Company \| Metal matrix composite wires, cables, and method \| \| US6346325B1 (en) \| 1999-07-01 \| 2002-02-12 \| The Dow Chemical Company \| Fiber-reinforced composite encased in a thermoplastic and method of making same \| \| US20020019182A1 (en) \| 1996-11-06 \| 2002-02-14 \| Toray Industries, Inc. \| Molding material and production process \| \| US20020041049A1 (en) \| 2000-09-29 \| 2002-04-11 \| Mccullough Kevin A. \| Nozzle insert for long fiber compounding \| \| US6391959B1 (en) \| 1997-12-03 \| 2002-05-21 \| Toray Industries, Inc. \| Phenolic resin composition for fiber-reinforced composite material, prepreg for fiber-reinforced composite material, and process for producing prepreg for fiber-reinforced composite material \| \| US6485796B1 (en) \| 2000-07-14 \| 2002-11-26 \| 3M Innovative Properties Company \| Method of making metal matrix composites \| \| US20020189845A1 (en) \| 2001-06-14 \| 2002-12-19 \| Gorrell Brian E. \| High voltage cable \| \| US20030037529A1 (en) \| 2001-04-27 \| 2003-02-27 \| Conoco Inc. \| Composite tether and methods for manufacturing, transporting, and installing same \| \| US6528729B1 (en) \| 1999-09-30 \| 2003-03-04 \| Yazaki Corporation \| Flexible conductor of high strength and light weight \| \| US20030082380A1 (en) \| 2001-10-31 \| 2003-05-01 \| Hager Thomas P. \| Compact, hybrid fiber reinforced rods for optical cable reinforcements and method for making same \| \| US6559385B1 (en) \| 2000-07-14 \| 2003-05-06 \| 3M Innovative Properties Company \| Stranded cable and method of making \| \| US6568072B2 (en) \| 2001-06-13 \| 2003-05-27 \| Jerry W. Wilemon \| Reinforced utility cable and method for producing the same \| \| FR2836591A1 (en) \| 2002-02-27 \| 2003-08-29 \| Pierre Robert Gouniot \| Composite conductive wire, used in the manufacture of overhead electricity conductors, comprises a reinforcing core of organic or inorganic material, coated with one or more braided aluminum wire layers \| \| US6656316B1 (en) \| 1997-07-09 \| 2003-12-02 \| Joel A. Dyksterhouse \| Method of prepregging with resin and novel prepregs produced by such method \| \| US6658836B2 (en) \| 2001-03-14 \| 2003-12-09 \| The Goodyear Tire & Rubber Company \| Hybrid cord \| \| US6723451B1 (en) \| 2000-07-14 \| 2004-04-20 \| 3M Innovative Properties Company \| Aluminum matrix composite wires, cables, and method \| \| US20040098963A1 (en) \| 2001-02-15 \| 2004-05-27 \| Jan Calleeuw \| Metal rope and fabric comprising such a metal rope \| \| CN2637534Y (en) \| 2003-09-12 \| 2004-09-01 \| 江苏法尔胜技术开发中心 \| Armoured carbon fiber rope \| \| US20040182597A1 (en) \| 2003-03-20 \| 2004-09-23 \| Smith Jack B. \| Carbon-core transmission cable \| \| JP2004300609A (en) \| 2003-03-31 \| 2004-10-28 \| Tokyo Seiko Co Ltd \| Fiber rope for moving rope \| \| US20040224590A1 (en) \| 2003-03-31 \| 2004-11-11 \| George Rawa \| Thermoplastic/fiber material composites, composite/metallic articles and methods for making composite/metallic articles \| \| US20040265558A1 (en) \| 2001-05-29 \| 2004-12-30 \| Berard Steven O. \| Thermally conductive carbon fiber extrusion compounder \| \| US6846857B1 (en) \| 1999-07-06 \| 2005-01-25 \| Fact Future Advanced Composites & Technology, Gmbh \| Long fiber-reinforced thermoplastics material and a method for producing the same \| \| US6861131B2 (en) \| 2000-12-06 \| 2005-03-01 \| Complastik Corp. \| Hybrid composite articles and methods for their production \| \| US20050061538A1 (en) \| 2001-12-12 \| 2005-03-24 \| Blucher Joseph T. \| High voltage electrical power transmission cable having composite-composite wire with carbon or ceramic fiber reinforcement \| \| US6872343B2 (en) \| 2000-01-13 \| 2005-03-29 \| Fulcrum Composites, Inc. \| Process for in-line forming of pultruded composites \| \| US20050181228A1 (en) \| 2004-02-13 \| 2005-08-18 \| 3M Innovative Properties Company \| Metal-cladded metal matrix composite wire \| \| US20050186410A1 (en) \| 2003-04-23 \| 2005-08-25 \| David Bryant \| Aluminum conductor composite core reinforced cable and method of manufacture \| \| US20050205287A1 (en) \| 2004-03-17 \| 2005-09-22 \| Raymond Browning \| Electrical conductor cable and method for forming the same \| \| US20050244231A1 (en) \| 2004-04-13 \| 2005-11-03 \| Deepwater Marine Technology L.L.C. \| Hybrid composite steel tendon for offshore platform \| \| US20060021729A1 (en) \| 2004-07-29 \| 2006-02-02 \| 3M Innovative Properties Company \| Metal matrix composites, and methods for making the same \| \| US20060024490A1 (en) \| 2004-07-29 \| 2006-02-02 \| 3M Innovative Properties Company \| Metal matrix composites, and methods for making the same \| \| US20060024489A1 (en) \| 2004-07-29 \| 2006-02-02 \| 3M Innovative Properties Company \| Metal matrix composites, and methods for making the same \| \| US20060049541A1 (en) \| 2002-10-15 \| 2006-03-09 \| Sutton Tonja R \| Articles comprising a fiber-reinforced thermoplastic polymer composition \| \| US7015395B2 (en) \| 2000-02-08 \| 2006-03-21 \| Gift Technologies, Lp \| Composite reinforced electrical transmission conductor \| \| US7019217B2 (en) \| 2002-04-23 \| 2006-03-28 \| Ctc Cable Corporation \| Collet-type splice and dead end use with an aluminum conductor composite core reinforced cable \| \| US7041909B2 (en) \| 2002-04-23 \| 2006-05-09 \| Compsite Technology Corporation \| Methods of installing and apparatuses to install an aluminum conductor composite core reinforced cable \| \| US7059091B2 (en) \| 2000-05-31 \| 2006-06-13 \| Aker Kvaerner Subsea As \| Tension member \| \| US7093416B2 (en) \| 2004-06-17 \| 2006-08-22 \| 3M Innovative Properties Company \| Cable and method of making the same \| \| US20060204739A1 (en) \| 2003-04-30 \| 2006-09-14 \| Ticona Gmbh \| Pultrusion method and an article produced by said method \| \| US7131308B2 (en) \| 2004-02-13 \| 2006-11-07 \| 3M Innovative Properties Company \| Method for making metal cladded metal matrix composite wire \| \| EP1724306A1 (en) \| 2004-03-02 \| 2006-11-22 \| Toray Industries, Inc. \| Epoxy resin composition for fiber-reinforced composite material, prepreg and fiber-reinforced composite material \| \| US20060283323A1 (en) \| 2005-06-15 \| 2006-12-21 \| Gas Technology Institute \| Polyoxometalate material for gaseous stream purification at high temperature \| \| US7179522B2 (en) \| 2002-04-23 \| 2007-02-20 \| Ctc Cable Corporation \| Aluminum conductor composite core reinforced cable and method of manufacture \| \| US7220492B2 (en) \| 2003-12-18 \| 2007-05-22 \| 3M Innovative Properties Company \| Metal matrix composite articles \| \| US20070128435A1 (en) \| 2002-04-23 \| 2007-06-07 \| Clement Hiel \| Aluminum conductor composite core reinforced cable and method of manufacture \| \| US20070193767A1 (en) \| 2006-02-01 \| 2007-08-23 \| Daniel Guery \| Electricity transport conductor for overhead lines \| \| US20070202331A1 (en) \| 2006-02-24 \| 2007-08-30 \| Davis Gregory A \| Ropes having improved cyclic bend over sheave performance \| \| US7291263B2 (en) \| 2003-08-21 \| 2007-11-06 \| Filtrona Richmond, Inc. \| Polymeric fiber rods for separation applications \| \| US20070269645A1 (en) \| 2006-04-05 \| 2007-11-22 \| Venkat Raghavendran \| Lightweight thermoplastic composite including reinforcing skins \| \| US7326854B2 (en) \| 2005-06-30 \| 2008-02-05 \| Schlumberger Technology Corporation \| Cables with stranded wire strength members \| \| US20080141614A1 (en) \| 2006-12-14 \| 2008-06-19 \| Knouff Brian J \| Flexible fiber reinforced composite rebar \| \| US7402753B2 (en) \| 2005-01-12 \| 2008-07-22 \| Schlumberger Technology Corporation \| Enhanced electrical cables \| \| US20080250631A1 (en) \| 2007-04-14 \| 2008-10-16 \| Buckley David L \| Method and device for handling elongate strength members \| \| US7438971B2 (en) \| 2003-10-22 \| 2008-10-21 \| Ctc Cable Corporation \| Aluminum conductor composite core reinforced cable and method of manufacture \| \| US20080282666A1 (en) \| 2007-05-19 \| 2008-11-20 \| Chia-Te Chou \| Composite rope structures and systems and methods for fabricating cured composite rope structures \| \| US20080282664A1 (en) \| 2007-05-18 \| 2008-11-20 \| Chia-Te Chou \| Composite rope structures and systems and methods for making composite rope structures \| \| US7516051B2 (en) \| 2006-05-19 \| 2009-04-07 \| 3M Innovative Properties Company \| Overhead power transmission line conductor selection \| \| US7547843B2 (en) \| 2006-12-28 \| 2009-06-16 \| 3M Innovative Properties Company \| Overhead electrical power transmission line \| \| US7563983B2 (en) \| 2002-04-23 \| 2009-07-21 \| Ctc Cable Corporation \| Collet-type splice and dead end for use with an aluminum conductor composite core reinforced cable \| \| US20090229452A1 (en) \| 2005-06-09 \| 2009-09-17 \| Markus Milwich \| Rod-shaped fibre composite, and method and device for the production thereof \| \| WO2009130525A1 (en) \| 2008-04-24 \| 2009-10-29 \| Szaplonczay Pal \| Process and equipment for producing composite core with thermoplastic matrix for recyclable and thermally stable electrical transmission line conductor \| \| WO2010002878A1 (en) \| 2008-07-01 \| 2010-01-07 \| Dow Global Technologies Inc. \| Fiber-polymer composite \| \| US7650742B2 (en) \| 2004-10-19 \| 2010-01-26 \| Tokyo Rope Manufacturing Co., Ltd. \| Cable made of high strength fiber composite material \| \| US20100038112A1 (en) \| 2008-08-15 \| 2010-02-18 \| 3M Innovative Properties Company \| Stranded composite cable and method of making and using \| \| US7683262B2 (en) \| 2006-12-01 \| 2010-03-23 \| Nexans \| Power transmission conductor for an overhead line \| \| US7705242B2 (en) \| 2007-02-15 \| 2010-04-27 \| Advanced Technology Holdings Ltd. \| Electrical conductor and core for an electrical conductor \| \| US20100181012A1 (en) \| 2002-04-23 \| 2010-07-22 \| Ctc Cable Corporation \| Method for the manufacture of a composite core for an electrical cable \| \| CN101908391A (en) \| 2010-07-01 \| 2010-12-08 \| 嘉兴中宝碳纤维有限责任公司 \| Carbon fiber-resin composite core for overhead cable \| \| AU2010315935A1 (en) \| 2009-11-03 \| 2012-06-14 \| Calvin T. Coffey \| Folding exercise rowing machine \| \| CA2823637A1 (en) \| 2011-01-05 \| 2012-07-12 \| Alcan Products Corporation \| Aluminum alloy conductor composite reinforced for high voltage overhead power lines \| \| US8250845B2 (en) \| 2010-02-09 \| 2012-08-28 \| Tokyo Rope Manufacturing Co., Ltd. \| Fiber composite twisted cable \| \| JP5033278B2 (en) \| 1998-04-29 \| 2012-09-26 \| エリコン・トレーディング・アクチェンゲゼルシャフト，トリュープバッハ \| Tools or machine parts and methods for increasing the wear resistance of such parts \| \| WO2012142098A2 (en) \| 2011-04-12 \| 2012-10-18 \| Ticona Llc \| Umbilical for use in subsea applications \| \| WO2012142129A1 (en) \| 2011-04-12 \| 2012-10-18 \| Daniel Allan \| Electrical transmission cables with composite cores \| \| WO2012142107A1 (en) \| 2011-04-12 \| 2012-10-18 \| Ticona Llc \| Continious fiber reinforced thermoplastic rod and pultrusion method for its manufacture \| \| WO2012142096A1 (en) \| 2011-04-12 \| 2012-10-18 \| Ticona Llc \| Composite core for electrical transmission cables \| \| JP5148780B2 (en) \| 2010-07-13 \| 2013-02-20 \| クロリンエンジニアズ株式会社 \| Electrolyzer for producing chlorine and sodium hydroxide and method for producing chlorine and sodium hydroxide \| \| KR20130074165A (en) \| 2011-12-26 \| 2013-07-04 \| 한국과학기술원 \| Power amplifier using transmission line transformer \| \| KR20130085780A (en) \| 2012-01-20 \| 2013-07-30 \| 주식회사 비송 \| Pan tilt system by wire power transmission machnism for under water condition \| \| CA2768065A1 (en) \| 2012-02-24 \| 2013-08-24 \| Silbert S. Barrett \| Thermal energy generation from underground power transmission cables (double circuit 750kv to 120kv) \| \| US8519664B2 (en) \| 2010-02-08 \| 2013-08-27 \| Abb As \| Method for controlling a machine or an electrical load supplied with electric power over a long line \| \| USD688625S1 (en) \| 2012-07-26 \| 2013-08-27 \| AEP Transmission Holding Company, LLC \| Power transmission line \| \| US20130221747A1 (en) \| 2010-11-09 \| 2013-08-29 \| Telefonaktiebolaget Lm Ericsson (Publ) \| Subscriber Line Power Distribution System \| \| US9093194B2 (en) \| 2009-07-16 \| 2015-07-28 \| 3M Innovative Properties Company \| Insulated composite power cable and method of making and using same \| Family Cites Families (4) Cited by examiner, † Cited by third party\| Publication number \| Priority date \| Publication date \| Assignee \| Title \| --- --- \| CN101404190A (en) \| 2008-11-21 \| 2009-04-08 \| 蔡浩田 \| Multifunctional composite core aluminum stranded wire and cable \| \| RU2386183C1 (en) \| 2008-12-04 \| 2010-04-10 \| Дмитрий Григорьевич Сильченков \| Composite bearing core for external current-conducting strands of overhead high-voltage power transmission line wires and method of its production \| \| FR2941812A1 (en) \| 2009-02-03 \| 2010-08-06 \| Nexans \| ELECTRICAL TRANSMISSION CABLE WITH HIGH VOLTAGE. \| \| CN101740161A (en) \| 2009-12-29 \| 2010-06-16 \| 上海电缆研究所 \| Novel electric cable reinforced core and preparation method thereof \| 2012 2012-04-11 MX MX2013011923Apatent/MX346917B/enactive IP Right Grant 2012-04-11 CA CA2832453Apatent/CA2832453C/enactive Active 2012-04-11 AU AU2012242930Apatent/AU2012242930B2/ennot_active Ceased 2012-04-11 RU RU2013145605/07Apatent/RU2594016C2/ennot_active IP Right Cessation 2012-04-11 WO PCT/US2012/033077patent/WO2012142129A1/enactive Application Filing 2012-04-11 HU HUE12717953Apatent/HUE033251T2/enunknown 2012-04-11 PL PL12717953Tpatent/PL2697800T3/enunknown 2012-04-11 DK DK12717953.9Tpatent/DK2697800T3/enactive 2012-04-11 CN CN201710960995.XApatent/CN107742542B/enactive Active 2012-04-11 ES ES12717953.9Tpatent/ES2617596T3/enactive Active 2012-04-11 CN CN201280022956.0Apatent/CN103534763B/enactive Active 2012-04-11 BR BR112013026310-5Apatent/BR112013026310B1/ennot_active IP Right Cessation 2012-04-11 EP EP12717953.9Apatent/EP2697800B1/enactive Active 2012-04-12 AR ARP120101255Apatent/AR085999A1/enactive IP Right Grant 2012-04-12 TW TW101113063Apatent/TWI609383B/ennot_active IP Right Cessation 2012-04-12 TW TW107146951Apatent/TWI681412B/enactive 2012-04-12 TW TW106137348Apatent/TWI654626B/enactive 2013 2013-10-11 CL CL2013002932Apatent/CL2013002932A1/enunknown 2013-11-08 ZA ZA2013/08410Apatent/ZA201308410B/enunknown 2016 2016-08-15 US US15/236,521patent/US9685257B2/enactive Active 2017 2017-05-16 US US15/596,026patent/US20170256338A1/ennot_active Abandoned 2018 2018-03-09 US US15/916,561patent/US20180197658A1/ennot_active Abandoned Patent Citations (208) Cited by examiner, † Cited by third party\| Publication number \| Priority date \| Publication date \| Assignee \| Title \| --- --- \| US2589507A (en) \| 1947-01-11 \| 1952-03-18 \| Aluminum Co Of America \| Expanded electrical transmission cable \| \| US3244784A (en) \| 1960-01-15 \| 1966-04-05 \| Universal Moulded Fiber Glass \| Method for forming fibre reinforced resin articles \| \| US3509269A (en) \| 1968-06-11 \| 1970-04-28 \| Western Electric Co \| Thermal barriers for cables \| \| US3706216A (en) \| 1970-12-16 \| 1972-12-19 \| Joseph L Weingarten \| Process for reinforcing extruded articles \| \| US3795524A (en) \| 1971-03-01 \| 1974-03-05 \| Minnesota Mining & Mfg \| Aluminum borate and aluminum borosilicate articles \| \| US3823255A (en) \| 1972-04-20 \| 1974-07-09 \| Cyprus Mines Corp \| Flame and radiation resistant cable \| \| US3993726A (en) \| 1974-01-16 \| 1976-11-23 \| Hercules Incorporated \| Methods of making continuous length reinforced plastic articles \| \| US3980808A (en) \| 1974-09-19 \| 1976-09-14 \| The Furukawa Electric Co., Ltd. \| Electric cable \| \| US4047965A (en) \| 1976-05-04 \| 1977-09-13 \| Minnesota Mining And Manufacturing Company \| Non-frangible alumina-silica fibers \| \| US4228641A (en) \| 1978-09-28 \| 1980-10-21 \| Exxon Research & Engineering Co. \| Thermoplastic twines \| \| US4559262A (en) \| 1981-01-21 \| 1985-12-17 \| Imperial Chemical Industries, Plc \| Fibre reinforced compositions and methods for producing such compositions \| \| US4549920A (en) \| 1981-07-28 \| 1985-10-29 \| Imperial Chemical Industries, Plc \| Method for impregnating filaments with thermoplastic \| \| US4454709A (en) \| 1982-03-12 \| 1984-06-19 \| General Electric Company \| Method of forming concentric cable layer and article formed \| \| US4552432A (en) \| 1983-04-21 \| 1985-11-12 \| Cooper Industries, Inc. \| Hybrid cable \| \| US4499716A (en) \| 1983-06-13 \| 1985-02-19 \| E. I. Du Pont De Nemours And Company \| Reinforcement structure \| \| US4657990A (en) \| 1984-01-19 \| 1987-04-14 \| Imperial Chemical Industries Plc \| Aromatic polyetherketones \| \| US4680224A (en) \| 1984-03-06 \| 1987-07-14 \| Phillips Petroleum Company \| Reinforced plastic \| \| US4680224B1 (en) \| 1984-03-06 \| 1992-04-07 \| Phillips Petroleum Co \| \| \| US4588538A (en) \| 1984-03-15 \| 1986-05-13 \| Celanese Corporation \| Process for preparing tapes from thermoplastic polymers and carbon fibers \| \| US4649060A (en) \| 1984-03-22 \| 1987-03-10 \| Agency Of Industrial Science & Technology \| Method of producing a preform wire, sheet or tape fiber reinforced metal composite \| \| US4779563A (en) \| 1984-03-22 \| 1988-10-25 \| Agency Of Industrial Science & Technology \| Ultrasonic wave vibration apparatus for use in producing preform wire, sheet or tape for a fiber reinforced metal composite \| \| US4677818A (en) \| 1984-07-11 \| 1987-07-07 \| Toho Beslon Co., Ltd. \| Composite rope and manufacture thereof \| \| US4792481A (en) \| 1986-11-28 \| 1988-12-20 \| Phillips Petroleum Company \| Reinforced plastic \| \| US4767841A (en) \| 1987-02-24 \| 1988-08-30 \| Phillips Petroleum Company \| Arylene sulfide polymer preparation from dehydrated admixture comprising sulfur source, cyclic amide solvent and water \| \| US4861947A (en) \| 1987-04-13 \| 1989-08-29 \| Schweizerische Isola-Werke \| Communication or control cable with supporting element \| \| US4843696A (en) \| 1987-05-11 \| 1989-07-04 \| Southwire Company \| Method and apparatus for forming a stranded conductor \| \| US4954462A (en) \| 1987-06-05 \| 1990-09-04 \| Minnesota Mining And Manufacturing Company \| Microcrystalline alumina-based ceramic articles \| \| US5529652A (en) \| 1987-07-11 \| 1996-06-25 \| Kabushiki Kaisha Kobe Seiko Sho \| Method of manufacturing continuous fiber-reinforced thermoplastic prepregs \| \| US5066536A (en) \| 1987-12-10 \| 1991-11-19 \| Imperial Chemical Industries Plc \| Fibre reinforced thermoplastic composite structures \| \| US4877643A (en) \| 1988-03-24 \| 1989-10-31 \| Director General Agency Of Industrial Science And Technology \| Process for producing preformed wire from silicon carbide fiber-reinforced aluminum \| \| US5277566A (en) \| 1988-10-19 \| 1994-01-11 \| Hoechst Aktiengesellschaft \| Extrusion impregnating device \| \| US5060466A (en) \| 1988-10-31 \| 1991-10-29 \| Tokyo Rope Mfg. Co. Ltd. \| Composite rope and manufacturing method for the same \| \| US5130193A (en) \| 1988-11-10 \| 1992-07-14 \| Nippon Oil Co., Ltd. \| Fiber-reinforced composite cable \| \| US5010209A (en) \| 1988-12-20 \| 1991-04-23 \| Pirelli Cable Corp. \| Power cable with water swellable agents and elongated metal elements outside cable insulation \| \| US5068142A (en) \| 1989-01-31 \| 1991-11-26 \| Teijin Limited \| Fiber-reinforced polymeric resin composite material and process for producing same \| \| US5211500A (en) \| 1989-04-06 \| 1993-05-18 \| Tokyo Rope Mfg. Co., Ltd. \| Composite rope having molded-on fixing member at end portion thereof \| \| US5039572A (en) \| 1989-07-24 \| 1991-08-13 \| Phillips Petroleum Company \| High strength thermoplastic resin/carbon fiber composites and methods \| \| JPH03129606A (en) \| 1989-07-27 \| 1991-06-03 \| Hitachi Cable Ltd \| Aerial power cable \| \| US5374694A (en) \| 1989-08-09 \| 1994-12-20 \| Ici Composites Inc. \| Curable blends of cyanate esters and polyarylsulphones \| \| US5159157A (en) \| 1989-09-12 \| 1992-10-27 \| Kabelwerke Reinshagen Gmbh \| Electrical cable with element of high tensile strength \| \| JPH03119188A (en) \| 1989-10-03 \| 1991-05-21 \| Mitsubishi Kasei Corp \| Fiber-reinforced plastic composite material \| \| US5176775A (en) \| 1989-10-20 \| 1993-01-05 \| Montsinger Lawrence V \| Apparatus and method for forming fiber filled thermoplastic composite materials \| \| US5075381A (en) \| 1989-10-21 \| 1991-12-24 \| Mitsubishi Petrochemical Company Limited \| Thermoplastic resin composition \| \| US5149581A (en) \| 1989-11-21 \| 1992-09-22 \| Idemitsu Kosan Co., Ltd. \| Polyether copolymers, resin compositions containing them, and molded articles formed from them \| \| GB2240997A (en) \| 1990-02-19 \| 1991-08-21 \| Bridon Plc \| Strand or rope product of composite rods \| \| US5080175A (en) \| 1990-03-15 \| 1992-01-14 \| Williams Jerry G \| Use of composite rod-stiffened wireline cable for transporting well tool \| \| US5153381A (en) \| 1990-03-20 \| 1992-10-06 \| Alcan Aluminum Corporation \| Metal clad cable and method of making \| \| US5017643A (en) \| 1990-03-20 \| 1991-05-21 \| Phillips Petroleum Company \| Composition and process for making poly (arylene sulfide) resins reinforced with glass fibers \| \| US5298318A (en) \| 1990-03-21 \| 1994-03-29 \| Phillips Petroleum Company \| Poly(arylene sulfide) resins reinforced with glass fibers \| \| US5364657A (en) \| 1990-04-06 \| 1994-11-15 \| The University Of Akron \| Method of depositing and fusing polymer particles onto moistened continuous filaments \| \| US5073413A (en) \| 1990-05-31 \| 1991-12-17 \| American Composite Technology, Inc. \| Method and apparatus for wetting fiber reinforcements with matrix materials in the pultrusion process using continuous in-line degassing \| \| US5149749A (en) \| 1990-05-31 \| 1992-09-22 \| Phillips Petroleum Company \| Poly(phenylene sulfide) composition and articles having improved thermal stability at high temperatures \| \| US5126167A (en) \| 1990-06-13 \| 1992-06-30 \| Ube-Nitto Kasei Co., Ltd. \| Process of manufacturing a fiber reinforced plastic armored cable \| \| CN2071207U (en) \| 1990-06-16 \| 1991-02-13 \| 辽宁省农业工程研究设计所 \| Steel and plastic composite rope \| \| US5207850A (en) \| 1990-07-17 \| 1993-05-04 \| General Electric Company \| Process for making thermoplastic composites with cyclics oligomers and composites made thereby \| \| US5324563A (en) \| 1990-08-08 \| 1994-06-28 \| Bell Helicopter Textron Inc. \| Unidirectional carbon fiber reinforced pultruded composite material having improved compressive strength \| \| US5888609A (en) \| 1990-12-18 \| 1999-03-30 \| Valtion Teknillinen Tutkimuskeskus \| Planar porous composite structure and method for its manufacture \| \| US5171942A (en) \| 1991-02-28 \| 1992-12-15 \| Southwire Company \| Oval shaped overhead conductor and method for making same \| \| JPH0511729A (en) \| 1991-07-01 \| 1993-01-22 \| Seiko Epson Corp \| Information processing equipment \| \| JPH0533278A (en) \| 1991-07-19 \| 1993-02-09 \| Toray Ind Inc \| Rope comprising carbon fiber-reinforced composite material and production thereof \| \| US5210128A (en) \| 1991-07-24 \| 1993-05-11 \| Phillips Petroleum Company \| Poly(arylene sulfide) compositions, composites, and methods of production \| \| US5286561A (en) \| 1991-07-24 \| 1994-02-15 \| Phillips Petroleum Company \| Poly(arylene sulfide)composites and methods of production \| \| US5325457A (en) \| 1991-09-20 \| 1994-06-28 \| Bottoms Jack Jr \| Field protected self-supporting fiber optic cable \| \| JPH05148780A (en) \| 1991-11-28 \| 1993-06-15 \| Toray Ind Inc \| Production of rope composed of fiber-reinforced composite material \| \| US5198621A (en) \| 1991-12-31 \| 1993-03-30 \| The Furukawa Electric Co., Ltd. \| Twisted cable \| \| JPH067112A (en) \| 1992-06-23 \| 1994-01-18 \| Iharaya Shoten:Kk \| Frame for device for squeezing unrefined sake @(3754/24)rice wine) \| \| US5554826A (en) \| 1992-06-25 \| 1996-09-10 \| Southwire Company \| Overhead transmission conductor \| \| US5319003A (en) \| 1992-09-30 \| 1994-06-07 \| Union Carbide Chemicals & Plastics Technology Corporation \| Method for improving the mechanical performance of composite articles \| \| US5439632A (en) \| 1993-02-05 \| 1995-08-08 \| Ici Composites, Inc. \| Process for pretreatment of non-isotropic cylindrical products \| \| US5840424A (en) \| 1993-02-19 \| 1998-11-24 \| Fiberite, Inc. \| Curable composite materials \| \| US5462618A (en) \| 1993-03-23 \| 1995-10-31 \| Bell Helicopter Textron Inc. \| Continuous process of making unidirectional graphite fiber reinforced pultruded rods having minimal fiber waviness \| \| US5542230A (en) \| 1993-04-02 \| 1996-08-06 \| Deutsche Forschungsanstalt F. Luft-Und Raumfahrt E.V. \| Connecting rods \| \| US5614139A (en) \| 1993-05-18 \| 1997-03-25 \| Eniricerche S.P.A. \| Process of making a composite article \| \| US5424388A (en) \| 1993-06-24 \| 1995-06-13 \| Industrial Technology Research Institute \| Pultrusion process for long fiber-reinforced nylon composites \| \| US5783129A (en) \| 1993-08-17 \| 1998-07-21 \| Polyplastics Co., Ltd. \| Apparatus, method, and coating die for producing long fiber-reinforced thermoplastic resin composition \| \| US5525003A (en) \| 1993-12-29 \| 1996-06-11 \| Conoco Inc. \| Connection termination for composite rods \| \| US5468327A (en) \| 1994-01-24 \| 1995-11-21 \| University Of Massachusetts Lowell \| Method and device for continuous formation of braid reinforced thermoplastic structural and flexible members \| \| US6015953A (en) \| 1994-03-11 \| 2000-01-18 \| Tohoku Electric Power Co., Inc. \| Tension clamp for stranded conductor \| \| US5780154A (en) \| 1994-03-22 \| 1998-07-14 \| Tokuyama Corporation \| Boron nitride fiber and process for production thereof \| \| JPH07279940A (en) \| 1994-03-31 \| 1995-10-27 \| Nippon Cable Syst Inc \| High bending withstanding rope \| \| US5792527A (en) \| 1994-06-17 \| 1998-08-11 \| Chisso Corporation \| Products in a continuous length formed from fiber-reinforced resin and process for preparing the same \| \| US5788908A (en) \| 1994-08-19 \| 1998-08-04 \| Polyplastics Co., Ltd. \| Method for producing long fiber-reinforced thermoplastic resin composition \| \| US5501906A (en) \| 1994-08-22 \| 1996-03-26 \| Minnesota Mining And Manufacturing Company \| Ceramic fiber tow reinforced metal matrix composite \| \| US5658513A (en) \| 1994-10-18 \| 1997-08-19 \| Polyplastics Co., Ltd. \| Cross-head die and method for manufacturing a resin structure reinforced with long fibers \| \| US5700417A (en) \| 1995-01-27 \| 1997-12-23 \| Kabushiki Kaisha Kobe Seiko Sho \| Pultrusion process for preparing fiber-reinforced composite rod \| \| US6245425B1 (en) \| 1995-06-21 \| 2001-06-12 \| 3M Innovative Properties Company \| Fiber reinforced aluminum matrix composite wire \| \| US6180232B1 (en) \| 1995-06-21 \| 2001-01-30 \| 3M Innovative Properties Company \| Overhead high power transmission cable comprising fiber reinforced aluminum matrix composite wire \| \| US6447927B1 (en) \| 1995-06-21 \| 2002-09-10 \| 3M Innovative Properties Company \| Fiber reinforced aluminum matrix composite \| \| US6544645B1 (en) \| 1995-06-21 \| 2003-04-08 \| 3M Innovative Properties Company \| Fiber reinforced aluminum matrix composite wire \| \| US6460597B1 (en) \| 1995-06-21 \| 2002-10-08 \| 3M Innovative Properties Company \| Method of making fiber reinforced aluminum matrix composite \| \| US6336495B1 (en) \| 1995-06-21 \| 2002-01-08 \| 3M Innovative Properties Company \| Method of making fiber reinforced aluminum matrix composite wire \| \| US5705241A (en) \| 1995-07-01 \| 1998-01-06 \| Deutsche Forschungsanstalt Fur Luft-Und Raumfahrt E.V. \| Fibre-composite rod \| \| US5725954A (en) \| 1995-09-14 \| 1998-03-10 \| Montsinger; Lawrence V. \| Fiber reinforced thermoplastic composite with helical fluted surface and method of producing same \| \| US5895808A (en) \| 1996-01-25 \| 1999-04-20 \| Ems-Inventa Ag \| Process for producing compound materials with a polylactam matrix which can be thermally postformed \| \| US5727357A (en) \| 1996-05-22 \| 1998-03-17 \| Owens-Corning Fiberglas Technology, Inc. \| Composite reinforcement \| \| US5935508A (en) \| 1996-07-30 \| 1999-08-10 \| Kabushiki Kaisha Kobe Seiko Sho \| Fibre-reinforced compositions and methods for their production \| \| WO1998006109A1 (en) \| 1996-08-03 \| 1998-02-12 \| Bicc Public Limited Company \| Electrical and optical cable \| \| US6258453B1 (en) \| 1996-09-19 \| 2001-07-10 \| Lawrence V. Montsinger \| Thermoplastic composite materials made by rotational shear \| \| US6455143B1 (en) \| 1996-11-06 \| 2002-09-24 \| Toray Industries Inc. \| Molding material and production process \| \| US20020019182A1 (en) \| 1996-11-06 \| 2002-02-14 \| Toray Industries, Inc. \| Molding material and production process \| \| US6117551A (en) \| 1996-12-18 \| 2000-09-12 \| Toray Industries, Inc. \| Carbon fiber prepreg, and a production process thereof \| \| US6003356A (en) \| 1997-01-23 \| 1999-12-21 \| Davinci Technology Corporation \| Reinforced extruded products and process of manufacture \| \| JP3119188B2 (en) \| 1997-01-28 \| 2000-12-18 \| 日本電気株式会社 \| Semiconductor device \| \| JPH10321048A (en) \| 1997-05-16 \| 1998-12-04 \| Furukawa Electric Co Ltd:The \| Tension member, lightweight low-loose overhead wire using it \| \| US6656316B1 (en) \| 1997-07-09 \| 2003-12-02 \| Joel A. Dyksterhouse \| Method of prepregging with resin and novel prepregs produced by such method \| \| US6270851B1 (en) \| 1997-07-14 \| 2001-08-07 \| Daelim Industrial Co., Ltd. \| Process for manufacturing resin-coated fibers composite and an application thereof \| \| US6391959B1 (en) \| 1997-12-03 \| 2002-05-21 \| Toray Industries, Inc. \| Phenolic resin composition for fiber-reinforced composite material, prepreg for fiber-reinforced composite material, and process for producing prepreg for fiber-reinforced composite material \| \| US6048598A (en) \| 1997-12-17 \| 2000-04-11 \| Balaba Concrete Supply, Inc. \| Composite reinforcing member \| \| CN1220330A (en) \| 1997-12-17 \| 1999-06-23 \| 沈阳机电研究设计院 \| Steel plastic composite rope \| \| JP5033278B2 (en) \| 1998-04-29 \| 2012-09-26 \| エリコン・トレーディング・アクチェンゲゼルシャフト，トリュープバッハ \| Tools or machine parts and methods for increasing the wear resistance of such parts \| \| US6260343B1 (en) \| 1998-05-01 \| 2001-07-17 \| Wire Rope Corporation Of America, Incorporated \| High-strength, fatigue resistant strands and wire ropes \| \| US6334293B1 (en) \| 1999-03-04 \| 2002-01-01 \| N.V. Bekaert S.A. \| Steel cord with polymer core \| \| US6346325B1 (en) \| 1999-07-01 \| 2002-02-12 \| The Dow Chemical Company \| Fiber-reinforced composite encased in a thermoplastic and method of making same \| \| US6846857B1 (en) \| 1999-07-06 \| 2005-01-25 \| Fact Future Advanced Composites & Technology, Gmbh \| Long fiber-reinforced thermoplastics material and a method for producing the same \| \| US6244014B1 (en) \| 1999-07-22 \| 2001-06-12 \| Andrew Barmakian \| Steel rod-reinforced plastic piling \| \| US6528729B1 (en) \| 1999-09-30 \| 2003-03-04 \| Yazaki Corporation \| Flexible conductor of high strength and light weight \| \| US6872343B2 (en) \| 2000-01-13 \| 2005-03-29 \| Fulcrum Composites, Inc. \| Process for in-line forming of pultruded composites \| \| US6248262B1 (en) \| 2000-02-03 \| 2001-06-19 \| General Electric Company \| Carbon-reinforced thermoplastic resin composition and articles made from same \| \| US7752754B2 (en) \| 2000-02-08 \| 2010-07-13 \| Gift Technologies, Inc. \| Method for increasing the current carried between two high voltage conductor support towers \| \| US7015395B2 (en) \| 2000-02-08 \| 2006-03-21 \| Gift Technologies, Lp \| Composite reinforced electrical transmission conductor \| \| US20100293783A1 (en) \| 2000-02-08 \| 2010-11-25 \| Gift Technologies, Llc \| Method for increasing the current carried between two high voltage conductor support towers \| \| US8371028B2 (en) \| 2000-02-08 \| 2013-02-12 \| Gift Technologies, Llc \| Method for increasing the current carried between two high voltage conductor support towers \| \| US7059091B2 (en) \| 2000-05-31 \| 2006-06-13 \| Aker Kvaerner Subsea As \| Tension member \| \| US6329056B1 (en) \| 2000-07-14 \| 2001-12-11 \| 3M Innovative Properties Company \| Metal matrix composite wires, cables, and method \| \| US6485796B1 (en) \| 2000-07-14 \| 2002-11-26 \| 3M Innovative Properties Company \| Method of making metal matrix composites \| \| US6723451B1 (en) \| 2000-07-14 \| 2004-04-20 \| 3M Innovative Properties Company \| Aluminum matrix composite wires, cables, and method \| \| US6344270B1 (en) \| 2000-07-14 \| 2002-02-05 \| 3M Innovative Properties Company \| Metal matrix composite wires, cables, and method \| \| US6559385B1 (en) \| 2000-07-14 \| 2003-05-06 \| 3M Innovative Properties Company \| Stranded cable and method of making \| \| US20020041049A1 (en) \| 2000-09-29 \| 2002-04-11 \| Mccullough Kevin A. \| Nozzle insert for long fiber compounding \| \| US6861131B2 (en) \| 2000-12-06 \| 2005-03-01 \| Complastik Corp. \| Hybrid composite articles and methods for their production \| \| US20040098963A1 (en) \| 2001-02-15 \| 2004-05-27 \| Jan Calleeuw \| Metal rope and fabric comprising such a metal rope \| \| US6658836B2 (en) \| 2001-03-14 \| 2003-12-09 \| The Goodyear Tire & Rubber Company \| Hybrid cord \| \| US20070271897A1 (en) \| 2001-04-27 \| 2007-11-29 \| Conocophillips Company \| Composite tether and methods for manufacturing, transporting, and installing same \| \| US20030037529A1 (en) \| 2001-04-27 \| 2003-02-27 \| Conoco Inc. \| Composite tether and methods for manufacturing, transporting, and installing same \| \| US20040265558A1 (en) \| 2001-05-29 \| 2004-12-30 \| Berard Steven O. \| Thermally conductive carbon fiber extrusion compounder \| \| US6568072B2 (en) \| 2001-06-13 \| 2003-05-27 \| Jerry W. Wilemon \| Reinforced utility cable and method for producing the same \| \| US20020189845A1 (en) \| 2001-06-14 \| 2002-12-19 \| Gorrell Brian E. \| High voltage cable \| \| US20030082380A1 (en) \| 2001-10-31 \| 2003-05-01 \| Hager Thomas P. \| Compact, hybrid fiber reinforced rods for optical cable reinforcements and method for making same \| \| US20050061538A1 (en) \| 2001-12-12 \| 2005-03-24 \| Blucher Joseph T. \| High voltage electrical power transmission cable having composite-composite wire with carbon or ceramic fiber reinforcement \| \| FR2836591A1 (en) \| 2002-02-27 \| 2003-08-29 \| Pierre Robert Gouniot \| Composite conductive wire, used in the manufacture of overhead electricity conductors, comprises a reinforcing core of organic or inorganic material, coated with one or more braided aluminum wire layers \| \| US20100181012A1 (en) \| 2002-04-23 \| 2010-07-22 \| Ctc Cable Corporation \| Method for the manufacture of a composite core for an electrical cable \| \| US7179522B2 (en) \| 2002-04-23 \| 2007-02-20 \| Ctc Cable Corporation \| Aluminum conductor composite core reinforced cable and method of manufacture \| \| US7368162B2 (en) \| 2002-04-23 \| 2008-05-06 \| Ctc Cable Corporation \| Aluminum conductor composite core reinforced cable and method of manufacture \| \| US20070128435A1 (en) \| 2002-04-23 \| 2007-06-07 \| Clement Hiel \| Aluminum conductor composite core reinforced cable and method of manufacture \| \| US7019217B2 (en) \| 2002-04-23 \| 2006-03-28 \| Ctc Cable Corporation \| Collet-type splice and dead end use with an aluminum conductor composite core reinforced cable \| \| US7041909B2 (en) \| 2002-04-23 \| 2006-05-09 \| Compsite Technology Corporation \| Methods of installing and apparatuses to install an aluminum conductor composite core reinforced cable \| \| US7211319B2 (en) \| 2002-04-23 \| 2007-05-01 \| Ctc Cable Corporation \| Aluminum conductor composite core reinforced cable and method of manufacture \| \| US7060326B2 (en) \| 2002-04-23 \| 2006-06-13 \| Composite Technology Corporation \| Aluminum conductor composite core reinforced cable and method of manufacture \| \| US7563983B2 (en) \| 2002-04-23 \| 2009-07-21 \| Ctc Cable Corporation \| Collet-type splice and dead end for use with an aluminum conductor composite core reinforced cable \| \| US20060049541A1 (en) \| 2002-10-15 \| 2006-03-09 \| Sutton Tonja R \| Articles comprising a fiber-reinforced thermoplastic polymer composition \| \| US20040182597A1 (en) \| 2003-03-20 \| 2004-09-23 \| Smith Jack B. \| Carbon-core transmission cable \| \| US20040224590A1 (en) \| 2003-03-31 \| 2004-11-11 \| George Rawa \| Thermoplastic/fiber material composites, composite/metallic articles and methods for making composite/metallic articles \| \| JP2004300609A (en) \| 2003-03-31 \| 2004-10-28 \| Tokyo Seiko Co Ltd \| Fiber rope for moving rope \| \| US7608783B2 (en) \| 2003-04-23 \| 2009-10-27 \| Ctc Cable Corporation \| Collet-type splice and dead end for use with an aluminum conductor composite core reinforced cable \| \| US20050186410A1 (en) \| 2003-04-23 \| 2005-08-25 \| David Bryant \| Aluminum conductor composite core reinforced cable and method of manufacture \| \| US20060204739A1 (en) \| 2003-04-30 \| 2006-09-14 \| Ticona Gmbh \| Pultrusion method and an article produced by said method \| \| US7291263B2 (en) \| 2003-08-21 \| 2007-11-06 \| Filtrona Richmond, Inc. \| Polymeric fiber rods for separation applications \| \| CN2637534Y (en) \| 2003-09-12 \| 2004-09-01 \| 江苏法尔胜技术开发中心 \| Armoured carbon fiber rope \| \| US7438971B2 (en) \| 2003-10-22 \| 2008-10-21 \| Ctc Cable Corporation \| Aluminum conductor composite core reinforced cable and method of manufacture \| \| US20100163275A1 (en) \| 2003-10-22 \| 2010-07-01 \| Ctc Cable Corporation \| Composite core for an electrical cable \| \| US7220492B2 (en) \| 2003-12-18 \| 2007-05-22 \| 3M Innovative Properties Company \| Metal matrix composite articles \| \| US7131308B2 (en) \| 2004-02-13 \| 2006-11-07 \| 3M Innovative Properties Company \| Method for making metal cladded metal matrix composite wire \| \| US20050181228A1 (en) \| 2004-02-13 \| 2005-08-18 \| 3M Innovative Properties Company \| Metal-cladded metal matrix composite wire \| \| EP1724306A1 (en) \| 2004-03-02 \| 2006-11-22 \| Toray Industries, Inc. \| Epoxy resin composition for fiber-reinforced composite material, prepreg and fiber-reinforced composite material \| \| US20050205287A1 (en) \| 2004-03-17 \| 2005-09-22 \| Raymond Browning \| Electrical conductor cable and method for forming the same \| \| US20050244231A1 (en) \| 2004-04-13 \| 2005-11-03 \| Deepwater Marine Technology L.L.C. \| Hybrid composite steel tendon for offshore platform \| \| US7093416B2 (en) \| 2004-06-17 \| 2006-08-22 \| 3M Innovative Properties Company \| Cable and method of making the same \| \| US20060021729A1 (en) \| 2004-07-29 \| 2006-02-02 \| 3M Innovative Properties Company \| Metal matrix composites, and methods for making the same \| \| US20060024489A1 (en) \| 2004-07-29 \| 2006-02-02 \| 3M Innovative Properties Company \| Metal matrix composites, and methods for making the same \| \| US20060024490A1 (en) \| 2004-07-29 \| 2006-02-02 \| 3M Innovative Properties Company \| Metal matrix composites, and methods for making the same \| \| US7650742B2 (en) \| 2004-10-19 \| 2010-01-26 \| Tokyo Rope Manufacturing Co., Ltd. \| Cable made of high strength fiber composite material \| \| US7402753B2 (en) \| 2005-01-12 \| 2008-07-22 \| Schlumberger Technology Corporation \| Enhanced electrical cables \| \| US20090229452A1 (en) \| 2005-06-09 \| 2009-09-17 \| Markus Milwich \| Rod-shaped fibre composite, and method and device for the production thereof \| \| US20060283323A1 (en) \| 2005-06-15 \| 2006-12-21 \| Gas Technology Institute \| Polyoxometalate material for gaseous stream purification at high temperature \| \| US7326854B2 (en) \| 2005-06-30 \| 2008-02-05 \| Schlumberger Technology Corporation \| Cables with stranded wire strength members \| \| US20070193767A1 (en) \| 2006-02-01 \| 2007-08-23 \| Daniel Guery \| Electricity transport conductor for overhead lines \| \| US20070202331A1 (en) \| 2006-02-24 \| 2007-08-30 \| Davis Gregory A \| Ropes having improved cyclic bend over sheave performance \| \| US20070269645A1 (en) \| 2006-04-05 \| 2007-11-22 \| Venkat Raghavendran \| Lightweight thermoplastic composite including reinforcing skins \| \| US7516051B2 (en) \| 2006-05-19 \| 2009-04-07 \| 3M Innovative Properties Company \| Overhead power transmission line conductor selection \| \| US7683262B2 (en) \| 2006-12-01 \| 2010-03-23 \| Nexans \| Power transmission conductor for an overhead line \| \| US20080141614A1 (en) \| 2006-12-14 \| 2008-06-19 \| Knouff Brian J \| Flexible fiber reinforced composite rebar \| \| US7547843B2 (en) \| 2006-12-28 \| 2009-06-16 \| 3M Innovative Properties Company \| Overhead electrical power transmission line \| \| US7705242B2 (en) \| 2007-02-15 \| 2010-04-27 \| Advanced Technology Holdings Ltd. \| Electrical conductor and core for an electrical conductor \| \| US20100206606A1 (en) \| 2007-02-15 \| 2010-08-19 \| Winterhalter Michael A \| Electrical Conductor and Core for an Electrical Conductor \| \| US20080250631A1 (en) \| 2007-04-14 \| 2008-10-16 \| Buckley David L \| Method and device for handling elongate strength members \| \| US20080282664A1 (en) \| 2007-05-18 \| 2008-11-20 \| Chia-Te Chou \| Composite rope structures and systems and methods for making composite rope structures \| \| US20080282666A1 (en) \| 2007-05-19 \| 2008-11-20 \| Chia-Te Chou \| Composite rope structures and systems and methods for fabricating cured composite rope structures \| \| WO2009130525A1 (en) \| 2008-04-24 \| 2009-10-29 \| Szaplonczay Pal \| Process and equipment for producing composite core with thermoplastic matrix for recyclable and thermally stable electrical transmission line conductor \| \| WO2010002878A1 (en) \| 2008-07-01 \| 2010-01-07 \| Dow Global Technologies Inc. \| Fiber-polymer composite \| \| US20100038112A1 (en) \| 2008-08-15 \| 2010-02-18 \| 3M Innovative Properties Company \| Stranded composite cable and method of making and using \| \| US9093194B2 (en) \| 2009-07-16 \| 2015-07-28 \| 3M Innovative Properties Company \| Insulated composite power cable and method of making and using same \| \| AU2010315935A1 (en) \| 2009-11-03 \| 2012-06-14 \| Calvin T. Coffey \| Folding exercise rowing machine \| \| US8519664B2 (en) \| 2010-02-08 \| 2013-08-27 \| Abb As \| Method for controlling a machine or an electrical load supplied with electric power over a long line \| \| US8250845B2 (en) \| 2010-02-09 \| 2012-08-28 \| Tokyo Rope Manufacturing Co., Ltd. \| Fiber composite twisted cable \| \| CN101908391A (en) \| 2010-07-01 \| 2010-12-08 \| 嘉兴中宝碳纤维有限责任公司 \| Carbon fiber-resin composite core for overhead cable \| \| JP5148780B2 (en) \| 2010-07-13 \| 2013-02-20 \| クロリンエンジニアズ株式会社 \| Electrolyzer for producing chlorine and sodium hydroxide and method for producing chlorine and sodium hydroxide \| \| US20130221747A1 (en) \| 2010-11-09 \| 2013-08-29 \| Telefonaktiebolaget Lm Ericsson (Publ) \| Subscriber Line Power Distribution System \| \| CA2823637A1 (en) \| 2011-01-05 \| 2012-07-12 \| Alcan Products Corporation \| Aluminum alloy conductor composite reinforced for high voltage overhead power lines \| \| WO2012142098A2 (en) \| 2011-04-12 \| 2012-10-18 \| Ticona Llc \| Umbilical for use in subsea applications \| \| WO2012142096A1 (en) \| 2011-04-12 \| 2012-10-18 \| Ticona Llc \| Composite core for electrical transmission cables \| \| WO2012142107A1 (en) \| 2011-04-12 \| 2012-10-18 \| Ticona Llc \| Continious fiber reinforced thermoplastic rod and pultrusion method for its manufacture \| \| US9012781B2 (en) \| 2011-04-12 \| 2015-04-21 \| Southwire Company, Llc \| Electrical transmission cables with composite cores \| \| WO2012142129A1 (en) \| 2011-04-12 \| 2012-10-18 \| Daniel Allan \| Electrical transmission cables with composite cores \| \| US9443635B2 (en) \| 2011-04-12 \| 2016-09-13 \| Southwire Company, Llc \| Electrical transmission cables with composite cores \| \| KR20130074165A (en) \| 2011-12-26 \| 2013-07-04 \| 한국과학기술원 \| Power amplifier using transmission line transformer \| \| KR20130085780A (en) \| 2012-01-20 \| 2013-07-30 \| 주식회사 비송 \| Pan tilt system by wire power transmission machnism for under water condition \| \| CA2768065A1 (en) \| 2012-02-24 \| 2013-08-24 \| Silbert S. Barrett \| Thermal energy generation from underground power transmission cables (double circuit 750kv to 120kv) \| \| USD688625S1 (en) \| 2012-07-26 \| 2013-08-27 \| AEP Transmission Holding Company, LLC \| Power transmission line \| Non-Patent Citations (12) Cited by examiner, † Cited by third party\| Title \| \| ASTM B 233-92 Designation, entitled "Standard Specification for Aluminum 1350 Drawing Stock for Electrical Purposes" published Oct. 1992; pp. 222-225. \| \| Gambone, L. R. entitled "Alternative Materials for Overhead Conductors," Dec. 1991, 80 pages. \| \| Hiel, Clem, entitled "Development of a composite reinforced aluminum conductor," California Energy Commission Report, Nov. 2000, 41 pages. \| \| High-Temperature, Low-Sag Transmission Conductors, EPRI, Palo Alto, CA, 2002, 1001811, 60 pages. \| \| International Search Report and the Written Opinion in PCT/US2012/033077 dated Aug. 7, 2012. 12 pages. \| \| Johnson et al., entitled "A New Generation of High Performance Conductors," IEEE, 2001, pp. 175-179. \| \| Karbhari, entitled "Use of Composite Materials in Civil Infrastructure in Japan," ITRI, WTEC Division, Aug. 1998, 211 pages. \| \| Sato, et al. entitled "Development of a Low Sag Aluminum Conductor Carbon Fiber Reinforced for Transmission Lines," CIGRE, May 2002, 6 pages. \| \| SBIR Phase I: Advanced Carbon Composite Transmission Conductor Development, www.sbir.gov/sbirsearch/detail/94623, 2001, 2 pages. \| \| Song, et al., "Mechanical and thermal properties of poly(arylene disulfide) derived from cyclic(4,4′-oxybis(benzene)disulfide) via ring-opening polymerization," Springer Science, Jan. 23, 2007, vol. 42, © Springer Science+Business Media, LLC 2007; pp. 1156-1161. \| \| Technical Data on Aluminum Conductor Carbon Fiber Reinforced, Tokyo Rope Mfg. Co., Ltd, 2002, 41 pages. \| \| Zimmerman, et al., "Polymerization of poly(p-phenylene sulfide) from a cyclic precursor," Polymer, vol. 37, No. 14, 1996, Copyright © 1996 Published by Elsevier Science Ltd.; pp. 3111-3116. \| Cited By (10) Cited by examiner, † Cited by third party\| Publication number \| Priority date \| Publication date \| Assignee \| Title \| --- --- \| USD830311S1 (en) \| 2014-09-25 \| 2018-10-09 \| Conway Electric, LLC \| Overbraided electrical cord with X pattern \| \| WO2019147838A1 (en) \| 2018-01-24 \| 2019-08-01 \| Ctc Global Corporation \| Termination arrangement for an overhead electrical cable \| \| US11329467B2 (en) \| 2018-01-24 \| 2022-05-10 \| Ctc Global Corporation \| Termination arrangement for an overhead electrical cable \| \| US12136804B2 (en) \| 2018-01-24 \| 2024-11-05 \| Ctc Global Corporation \| Termination arrangement for an overhead electrical cable \| \| EP4273892A2 (en) \| 2018-02-27 \| 2023-11-08 \| CTC Global Corporation \| Systems, 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2014-09-05 \| \| TW201308357A (en) \| 2013-02-16 \| \| HUE033251T2 (en) \| 2017-11-28 \| \| TW201921385A (en) \| 2019-06-01 \| \| RU2594016C2 (en) \| 2016-08-10 \| \| WO2012142129A1 (en) \| 2012-10-18 \| \| CN103534763B (en) \| 2017-11-14 \| \| EP2697800A1 (en) \| 2014-02-19 \| \| CN107742542A (en) \| 2018-02-27 \| \| TWI654626B (en) \| 2019-03-21 \| \| TWI609383B (en) \| 2017-12-21 \| \| RU2013145605A (en) \| 2015-05-20 \| \| CN103534763A (en) \| 2014-01-22 \| \| US20180197658A1 (en) \| 2018-07-12 \| \| AR085999A1 (en) \| 2013-11-13 \| \| TW201805956A (en) \| 2018-02-16 \| \| AU2012242930A1 (en) \| 2013-05-02 \| \| DK2697800T3 (en) \| 2017-02-27 \| \| CN107742542B (en) \| 2019-10-01 \| \| CA2832453C (en) \| 2019-09-10 \| \| US20170256338A1 (en) \| 2017-09-07 \| \| US20160351300A1 (en) \| 2016-12-01 \| \| PL2697800T3 (en) \| 2017-07-31 \| \| HK1251351A1 (en) \| 2019-01-25 \| \| MX2013011923A (en) \| 2014-03-27 \| \| TWI681412B (en) \| 2020-01-01 \| \| ZA201308410B (en) \| 2014-08-27 \| 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15949	https://www.youtube.com/watch?v=rBLm0GvuCSE	Code Challenge - Python \| CodeWars - Alphametics Solver The Technology Sandbox 24800 subscribers 1 likes Description 108 views Posted: 26 Nov 2024 Challenge URL: Software development is about problem solving. It’s not just about writing code, but about understanding the problem at hand, breaking it down into manageable pieces, and finding the most efficient and effective solutions. In this series of videos, we will exercise our problem solving skills. technologysandbox #codechallenge #codewars Transcript: hey welcome back to another code challenge so this one I found interesting it took me more than a couple of hours to actually complete and I know sometimes it's informative to you know watch people struggle and how they solve problems um but it was just getting a little long so I had recorded I was about an hour and a half and then I just stopped that recording so I already have a working solution and what I'm going to do is I'm going to erase that now and then we can kind of fill it back in together and I'll give you my thought processing what I [Music] did so I'm just going to reset this solution here at the very bottom and we're going to go from scratch and I guess the problem here is you know were're given some formula in words you know send plus more equals money and each digit has to be replaced with a number and then you have to make the equation still be true and there's some constraints so it's only uppercase you know every letter can only assign one digit there will be maximum of 10 unique letters so presumably 0 1 2 3 4 5 6 7 8 9 are the number of characters and then there's a couple other things that are come important here like no leading zeros and then finally this addition thing so let's go and take a look the first thing I always do is I print out you know what the puzzle looks like you know what what are your inputs and this is informative just because I like to visually you know although we can see it over here I still want to just kind of confirm what things look like okay so we are getting exactly what they said everything is fine in this case I wanted to have a kind of objectoriented solution so there's a lot of um Pathways and a lot of algorithm here that we're going to need and it just serves our purpose a little bit better I think so I'm going to create a cipher class and we're going to feed that puzzle in and then down here how I want to use this thing eventually is I'm going to create an instance of that cyer class and I want to be able to say things like cipher solve and then down here I want to be able to return cipher result so I need a solve and a result here so we're going to go ahead and fill those [Music] in and for now I am going to we don't know what the solution is so we're just going to set it to none but I do know that when this is completed we are going to return that solution all right so if we take one of these things so we take like this example let's just talk about the algorithm here um we'll put this in the notes maybe Sobe we can just kind of up here at the very top let's just make a quick note so we're given something like this and if we were going to traditionally do this math um we would have like send more and then there would be like a plus I guess it's uncore like that and then it would have to add up to money all right so that's kind of tricky Additionally you know what we're really looking at so the algorithm that I'm going to implement and I'll show you this in a second is we're going to start with the ones digit and I'm going to just guess at what I want for y and I'll start with zero so zero will be the the digit I'm going to start with I'm going to then I'm going to pick all values of d and e from the remaining digits so everything but zero um 1 through n in this case that add up to zero so in that case it would be like 1 and n it would be 2 and 8 um 3 and 7even four and six right so that's the first pass through we're going to take a look at the ones digit and we're going to assume that Y is equal to zero and then all values of d& e must be so whatever D and E are they must be something like this it must be in one of these it must be uh like we talked about one and N but it could also be 9 and one um it could be like 2 and 8 CU those add up to 10 or it could also be 8 and two and it could be 3 and 7 or 7 and three and then finally it could be um six and four or four and six and then right above this maybe we should just say well the remaining digits are everything except for zero right cuz we just peled Zero from the the thing so then the next level I'm going to choose one of these so let's say d& is equal to um 1 and N so then I'm going to remove 1 and N from my list and then I'm going to move over to the 10 digit and then I'm going to guess what e is going to be right so so the next pass is I'm going to have to iterate through all of these here um but let's say we're going to choose one and [Music] N then our next pass is we computer remaining digits [Music] again so we'll have 2 3 4 five 6 7 and 8 we're going to select the value for e now e is already set over here so we said e was nine so we have to have n and R have to add up to nine right just like here D plus e had to be something we don't know what that first digit is but the last digit had to be a zero based upon this so what we know now is n + r has got to be something that ends in e and we already have a value for E because we chose it to be [Music] nine so then we're going to go through all the solutions again right so from the remaining digits what adds up to nine so two and seven 7 comma 2 anyways we're getting a little long with I think you get this idea this is the algorithm that we're going to do and it is just kind of figuring out what I do working backwards from the one digit to the 10 digits to the hundreds digit to the thousands all the way back up until I get to a solution the first one that I find that works anyways I thought maybe they said there could be more than one solution but maybe I miss that there but anyways soon we find one solution then we're out of here so that's what we have to code up we got to code that up so in my solve what I'm going to have is I'm going to have a start [Music] digit and I'm going to use the word total because this is on the total side so over here that's the total um these are going to be the terms so term one is send term two is more the total is money and so that start total digit we're going to start with zero and then we're going to have a while our solution is none I'm going to call some method that I don't have yet and I'm just going to call that um kind of bubble the digits up that's what I'm kind of named this algorithm is we're kind of taking those solutions that are possible we're going to Bubble those digits up and it's going to take into it the start total digit we're going to feed in the current solution puzzle and we're going to feed in the index that we're working on so minus one is the last digit on this total and the total is money so the last digit is minus one that's a y and we're in the one spot and when we want to do the 10 spot we'll switch this to minus two but I will have to have that function and it's going to have to do some Workforce in a little bit right now it doesn't do anything so back to my solve I'm G to write this whole function and then we're going to go and start worrying about the bubble digits up so I'm going to start with this total digit I'm going to come back through here I'm going to have to increase my start total digit if this is equal to 10 then I've gone too far and I'm going to break out of this loop I don't need pass anymore okay so that makes sense so this solve is a totally complete function now the only thing I'm going to do is I'm going to copy and paste in a couple of functions that we're going to use to help us out a little bit so you watching me type all of this I'm going to copy and paste in a print function um it's undor print it's going to use a self is debug so I'm going to set this self that is debug to true for now and then I'm going to copy in the actual solve function because it has a couple of prints in there all right so the only thing it changed I think is it added this line here so we're going to run this let's run this real quick we won't have a solution yet but we'll see um cyber solve does not like something with indentation that's a bit weird there's something going on there all right so there was something going on with some spacing down here but but you can see how this thing this algorithm works now solve is now starting with the Zero's digit it's going to ones twoos 3es four it's going to run through all the start digits and we're going to try to Bubble Up then we're going to start with zero here then we're going to go to one then two then three and then we'll just keep bubbling it up and doing these passes down here in this um kind of bubble digits up algorithm I'm going to copy and paste in a couple of more functions here just so we can save a little bit of time on here um we're going to need a function to validate whether our current solution so given a solution and again a solution looks like this um so when all the digits or all the characters have been replaced we're going to have a function that looks like that so this validate solution we're going to split apart into a right left hand and a right hand side side using the equal sign we're going to get all the terms by splitting apart by the plus sign so could be multiple terms like down here we have three and so we'll split this up so it'll be like 57 870 94041 if any term starts with a zero or if the total starts with a zero we're going to return false because those are not Solutions um and that's an explicit thing here so no leading zeros then we're going to eval the left hand side so the left-and side then would be this whole thing where it would say you know the number plus another number plus another number plus another number we're going to eval that and get a number for it and if that is equal to the value on the right hand side then we have a solution so that's just a validate solution function that'll come in handy in a little bit there's another function I'm going to just drop and put in place here that we're going to use in this bubble a little bit here and this is going to be what's going on I'm missing it's complaining of something here but I'm not sure if it really is still functions even though it's highlighted as red me it's because of the past um but I have a function called analyze solution so again given this solution and it could be halfway done so you know maybe we've replaced you know the Y with a zero and we have replaced the D and the E with a one and a 9 um so you're going to have some digits replaced so what analyze solution does is it takes the solution and it's going to figure out how many digits are used all right so it's going to go every single character in this you know solution and if it's a digit it's going to add it to digits use and it's going to remove it from digits free so we'll we'll know how many or which digits are free and which are used and they're also going to create a list of remaining characters so you know if we have for example replaced everything but this s it'll say that there's one remaining character left and it's an S so this analysis will be important later on we already kind of saw how we were going to use this digits free up here in our algorithm so remaining digits that's digits free all right so let's talk about this bubble digits up now a little bit and this is a bit lengthy so I'm just going to copy and paste a little bit as it time we're going to put a little note in here put this in so here's our note still doesn't like the how I get rid of that that bugs me there we go okay so just a little note this is recursive we're going to call this bubble digits up later on down below and we're going to keep incrementing this digit index to the left so it starts with the ones digit and then we're going to go to the T and then the hundreds and the thousands so this is going to incrementally get I guess more negative starts at negative 1 then it goes to -23 -4 and we'll always carry with it all the context that we need for the solution so this solution will be everything like this you know that's the first pass but it second pass it'll have some of those characters replaced um third pass it'll have even more characters replaced this digit index like we just talked about this is -1 - 2 - 3 -4 that's going to be the digit we're replacing this total digit is the next digit that we're going to try to replace so up here you know we said we just guessed Y is equal to zero and we fed that in so that's the we put a zero in there and so that's the next digit that we're guessing so that's what those mean and we just have a little um helper here this debug again we looked at this already I think I talked about this but that little print statement is just a flag if we're running in debug mode we'll do print otherwise we're not going to print and then there is a case in the code later on you're going to see you know if the solution was set then we're done so we can just break out early and this because this is a recursive function there may be a bunch of recursive calls that are still kind of in the stack that need to happen and so this is just a way for all those recursive calls that are just there already to exit early once we have a solution anything else is just extra work we don't need to do the extra work so we're out of there all right so the next thing we have to do is we have this total digit which is kind of zero and we have our function here and we need to replace uh this y with a zero so that is a bit of easy code here just drop place it put in place and so not just you know we're starting with this y we're going to replace that but if there's y's over here then we're going to replace all occurrences of Y with that same value so we're replace it with total digit so the character be to replace right now is we're looking at the solution and this is a minus one so this is the whole thing we're looking at the last character there and then I have an if statement here so if this character is a digit later on when we get to recursive calls you know for example e on the next layer e will already be replaced cuz we'll have replaced it with whatever value we're trying here for this e and so if it's already been replaced I don't need to do a replacement so I'm not trying to do a replacement again this is only if the uh it hasn't been replaced so if it's not a digit and this character isn't in a space Plus or equals you know there is a case where maybe I run all the way back and I I get the space I don't want to be back in that area so I don't need to replace anything at that point either otherwise I go ahead and I replace uh that character and then again I'm just kind of printing out the solution so if we look and see where we we got right now you know you can see here's that first print line right up there it's telling me you know here's my solution there's no digits replace we're at the minus1 index and the digit that we're going to use and replace it is the zeros which is what this big line indicates so I'm going to replace that y with a zero and I end up with that's my solution so far we're just guessing and we're going to Bubble kind of these digits back up until we end up with either with something that doesn't work anymore or we end up with the solution so next thing I want to do is I want to split this solution back into kind of a total um which is on the right hand side that's the total and then I also want to get the terms that are on the left hand side and so this is just doing a little bit of splitting based upon the equal sign and also for the terms based upon the plus sign and then I do this exit early against if I ever run into a case where the very first number in my term or my total is a zero that is not a solution and again that's based upon uh this no leading zeros technical detail there okay so now we need to do a little bit of the algorithm like we talked about at the very top um so at the very top here you know I just replaced this with a zero and I have to go up and see what the characters are up here now that are in this ones column so given my index which which is my I'm at the ones column I'm going to have a d and an e and I also could have I mean if you get back here further this e is going to be replaced eventually so that's not necessarily e but maybe it's a number and when I'm adding up these columns I need to know how many characters I have but I also need to know what the total is already and so that's what we're doing here is we're collecting the total so we're adding up the total we're also going to collect up the total number of characters I find and this is going to be a dictionary because not only do I need to know how many characters I find so maybe I found an e but I also need to know that hey there was actually three e in that column and so sometimes the words that they chose you know we're going to not just have a single instance of an e but it could be like three and so what we're going to do is we're going to Loop over all of the terms and we're looking at this digit index if that length of that term has a value at that index so if that length say you know if we're looking at minus1 and there's at least a single digit in the term we'll get either a character or we're going to get um the actual value if that character is not a digit then what we're going to do is we're going to add to our term character dictionary here um that character and we're going to set the number of Curren is to zero and then we're just going to increment it by one so now I'll know that hey there's one e and then I've see another E from a different variable oh there's two e's and so all the different terms will contribute to this uh term characters if it was a digit we're just going to track the total and we need to know that because again we try to add up these columns I need to know what the total is going to be and it has to be whatever we're going to predict for our next value so let's go and print here and see what we get now and so you can see here so here's send more money I put we replace the Y with a zero and it's going to see that there's a d and a e in that ones column there's one instance of a d and there's one instance of an e and then there's no more terms so that's zero let's see if we can find one down here [Music] where so here's one where it replaced you know it's guessing nine on the last digit and then what it's going to do is going to find an e and then it found another e so there's two e in that ones column so you can see that there and so far we don't have any uh instances where the total is um greater than zero I didn't see anything there all right so that's kind of an example of what we're doing next now we know how many characters that we're working with we have a DNA that we have to you know figure out all the different combinations next now for this next bit I'm going to import a library ital iter tools we're going to use that eventually to help us with all the combinations and stuff like that but let's see where we're at here at this point what I need to do next is I need to analyze my current solution and figure out you my remaining digits right so just like I did up here when we did um I chose zero I manually computed the remaining digits and so I'm going to just use this anal analyze solution a helper that we already kind of went over and that's kind of right there and it's going to return these three things one of them is the um digits that are still free and I guess with this let me just add um this print statement so if we run this again now we should see you know here's d and e and here's the digits that we've used so far we've only used the zero 1 through 9 are free and the remaining characters that we still need to be replaced are you know send M um and here's the E for it so those are the remaining characters we have send more m o n e you know these are all number or letters that already in here so the S is in there the E is in there the N the d The M the O the r this e is already there on the other side m is already there there o is there n is there and then the E is already there so these are the remaining characters that I have to to replace eventually in fact at some point we may end up that in a place where there are no remaining characters so if the length of this remaining characters array if it's zero then I want to check for a possible solution right I I don't need to go any deeper I don't need to try to replace anything else there's nothing else replace but we just need to stop at this point and see if our solution is valid and so we're going to use this validate solution helper which is down here and we already talked about this essentially it's parsing off the left and the right side making sure that the total in our equation you know we'll end up with something like this um and a value on this side we got to make sure that that equality is really there and so if we do find a valid solution we're just going to print it out so in this case so far we should never end up in this Loop or in this if condition so we're not going to see any of that now the next part is probably one of the trickiest bits um if we go back up to our example once we start to fill this in and I'm say back here on say the the hundreds digits so the E the O I'm going to choose an N I'm going to have some values that are already replaced so that e is already going to exist and I need to know not just the values here which is what we just did but I also need to know the previous digits that have already been replaced so that' be the tens and the ones they should already have numbers in them if I add all those up what's the carry digit you know am I carrying a two into my column or is it a one into my column or maybe not carrying any digit in my column so that was a bit of trickiness that I kind of missed the first pass through but I eventually saw it and I added a bit of code here so we're going just copy and paste that code in and so what this code does is now this calculates the carry value for all the terms that we've already kind of replaced so if we're on the hundreds you know we're looking in the hundreds column that means the ones and the tens digits have already been replaced so now here I'm going to kind of loop over all those terms and I'm going to look for those indexes that are to the right of us and so what you can see here then is and for every single term I'm going to build up this term number it's going to be a string and I'm just grabbing all the characters from digit index so if we're looking at the hundreds column that'll be -1 - 2 - 3+ one so we're going to from minus 2 to minus1 to I won't include zero so we're just going to go minus 2 and minus one that's the tens and the hundreds and we're going to grab the digits that are there so if they're digits then we're going to add that to our term and so all those things that are filled out I'm going to end up with some term number it'll be like a maybe 15 um but as a string so we just going toate as a string if I end up with something that's greater than zero then I'm going to convert this term number into an integer and we're going to pin this to our list of term numbers so depending on how many terms you have I'm going to have maybe a 10 a 15 25 whatever they are and I'm going to have this whole list in fact let me go ahead and grab I have a print statement here that I didn't copy and paste in let me put this here so now we can kind of see what everything's going to look like we'll test this out again so just incrementally we're just kind of going through all the solutions so the very first pass here isn't going to be very interesting all these term numbers because there aren't going to be any terms we're on the very first one you know we just replaced the Y with a zero I have a d& e there's no column to the right of this that I can possibly add up so until we get to a spot where we've starting to do some recursion we're just going to see this information the way it is so the current sum is zero um you can see the current sum so that value is just being computed here so you can see current sum is zero once I convert this to an INT we can just keep adding up that current sum um eventually we're going to figure out what the carry digit is from that current sum if the length of it is going to spill over into my next column I'll know what the digit is and so we're going to go and grab that digit and I'm going to put it into this value so that that's the previous digits carry value right there and it'll be an integer and so I'll know what that number is all right so the next bits this is where we're starting to set ourselves up for recursion and again this is what we did manually up here you know we figured out that hey my digits a zero I need D+ e to be something that adds up to you know that has a that least common digit or least common value of zero and I figured out what all of these um possible combinations are and so this is the code that's going to do that next is I'm going to create a list at the end we're going to have this total possible combinations list right here um and we can see what they are and here's where we're using our iter tools so using itter tools give me the combinations and what this is going to spit out is I have the number of free digits so 1 through nine in this case how many am I looking for I need to take them two at a time so I'm looking for a d and a e so given that two at a time um I'm going to get out of this then all com Combs combinations is going to be an example like you know one and two and one and three and two and three so if the total digits three are these three numbers and I'm taking them two at a time then it's all combinations of just two at a time 1 comma 2 1 comma 3 2 comma 3 now what it isn't is the reverse of it so the 2 comma 1 the 3 comma 1 and the two or the 3 comma 2 for that I need to also run this through all permutations and so then I'll end up with all the different possible combinations now just because I have this combination this is kind of a theoretical hey let's just you know I have five numbers I need to take them two at a time this isn't take into account what they add up to so now what I'm going to do is for every single one of those combinations I have to ensure that that column adds up to the value that I'm looking for so in our case it's a zero so this so I need to compute a total and this little for Loop is all it's doing is adding up but it's multiplying by the number of occurrences remember we found this one down here where there was multiple e so if there was multiple this is just the term the number of those characters and so I'm going to take whatever the value is in that placeholder and I'm going to multiply by the number of those kind of characters and I'm going to add that to my total and then I can't forget these are just the total of the digits that I selected in the combination but I also have to add to this you know here's my solution total um the total that we computed from the term you know here is the term total and this particular case it was Zero because there hadn't been any digits replaced and then I Al also have to add in the uh carry digit so if the carry was Zero then I don't have anything to add but if the other columns are going to contribute a two to this one then I add in the total number of digits this is probably another one of the tricky spots here is just to make sure that we're accounting for all the numbers and then making sure that we're filtering out all those solutions that don't meet our requirements so the least common digit um that's the remainder if I divide this by 10 has to be equal to the total digit which is the number that we are replacing it with so if we go all the way back to the top that's this number so that's the total digit I'm going to make one more line here I'm going to print out also just so you can see them all now we'll run this again we'll see the possible combinations you know like we manually did so we we took our our starting with our zero and we replaced the Y we saw that we had a d and a e and then we figured out all the possible combinations 1 9 91 28 so we did this manually up above and there's a total of eight of those so so far we're just implementing this we did this manually next thing is you know we're going to run each of these solutions back through our um our function that we're running here here and so we can start to see this for Loop here so now we have all these possible combinations so for every possible combination um we are going to now run this for Loop and there's going to be a lot more in here that we're going to add so we're essentially going to start trying all of those and that'll be kind of the next little bits here you're going to see okay I'm trying this solution 1 n I'm trying 91 I'm trying 28 trying 82 but there's more obviously that we have to do here first thing is you know this context that we pull in you know at the very top we're going to try to we need to call our bubble digits up function and we need to set this up so that we have a new context um a new digit index and a new Total digit and that's what this next bit of code is going to do so the first thing is we're going to go ahead and copy our um current solution and get a totally different copy of it because I don't want to ruin this me this is going to be recursive this has to stand Zone but I need to carry that context back up into the bubble um function when I call that again so next thing I'm going to do is again I'm I've got a possible combination this has some digits in it now I have to replace in this solution copy all right so I have that solution copy now I'm going to have to go over all the characters that are that we possibly have and I'm going to replace those with their value so given that this is a a possible combination or possible solution I'm going to go back in and we're going to replace it so let's take a look and see what this looks like now and so you can see here you know there's eight possible combinations we're going to try one n um here our we start off with a zero and then what we're saying is d and e so e is going to be equal to nine so every time you see an e you're going to replace that with a nine so now you have nine here um there's a nine and here's a nine one is going to be the D so you can see every time we see a d we're going to replace that with a one so here's our solution copy right now so we're just kind of incrementally replacing characters with digits as we guess the solution all right the next part is you know we need to do the analysis that we had before you know we're going to take this new solution we're going to feed it into our analyze solution function and we're going to get the number of free digit and the remaining characters if there's no remaining characters then I'm going to test and see if there's a possible solution at this point I'm never going to get into here but this is um eventually you know hopefully when we are set you know we'll have no more characters to replace so let's go ahead and test and see if it's a possible solution all right next thing is we need to start setting ourselves up for call in our function recursively so this this is going to be our new digit index so this is the current one I'm at say minus one I'm going subtract one so now I'm going to move over to the 10's digit so when I call the bubble function here in a little bit now we're going to be in the 10's digit it's not at this point still just a simple call back to the bubble function there's a little bit more logic here um and again this is one of the tricky areas that I kind of got caught up on is this if statement so let's go ahead and and go through this so what this is saying is if our solution copy if the digit so our new now we're at the 10 um if it's already a digit that means it's already been replaced so in this particular case that's not true um oh I'm sorry in this particular case it is true so you can see here you know we had an e we had an e over here so we replace that e with a nine and so you can see here is the digit nine so it already is true so we're going to end up in here so the digit total which is what we're going to feed back into the bubble digits up is just going to be a nine so all we're doing is we're just saying hey that's already nine that's the value that we're going to put there we don't need to guess what it is and you'll see in the next case is we're going to go through all the free digits and we're going to guess and make a guess but here we're just going to set it to that value which is going to be nine we're going to send back the solution cop cop and now we're in the 10's digit so that's a minus two so that's the case if it is a digit if it's not a digit just realizing I probably don't need to call this again because I think I just called it up here nothing's changed um I'm just going to get rid of this I don't need that there we'll come back to that if it broke it we'll find out here soon enough so if it's not already a digit then what I'm going to do is I'm going to go through this list of free digits that I already computed up here these are all the remaining dig digits are there and I'm going to guess that this next one that we're going to replace it with is whatever the first one is then the second one then the third one then the fourth one and this is where you can start to see this recursion adding up and so now we've completed one whole column what we've done we've completed um the 10 or the ones column so that index was minus one so now we're going to go back up through this one and we're going to look at the 10's column and then we're just going to repeat the same logic and so that logic is every single column um we're just going to work our way backwards until all the letters are replaced and then we're going to test our solution so if I run this now I think I got all the code in place and we should see that we are getting our test pass now one of the things you'll see is you know at the for example the bottom of the test there a lot of verbiage in here so see see all that recursion so we're seeing all the different things we're trying all right so if we scroll all the way to the end what we're going to see is this Max buffer size is reached which happens because um we're being very verbose here and that's why I have it so that I can flip this off so I'll set this to false I'll run the test again um now we get the five kind of example tests they all pass um there's not really anything being output here I think what I'm going to do is just so you can visually see maybe we'll just make these normal prints and you probably will see some of them being printed out right so this one and there was a member H we checked it twice so we do it down here and maybe there's some refactoring I can do here make this one common function okay so now we see them all so it's finding hey this is the solution this is the one that works and that matches up with you know what the solution that they expected was and then when you click attempt all um let's see if it does it with the print in there we do we pass up all our tests so it's an interesting solution there's a bit of complexity in there originally had just planned on videoing this from the beginning to the end but then I got caught into it was like an hour and a half video nobody wants to sit through an hour and a half I don't even know how long this one is this is long enough but anyways I just decided to carry on with it cuz I was interested in the solution but I just went back and just kind of videoed the solution itself so you can see uh my thought process a little bit more but we're going to submit this um the difficulty I guess it was a 3Q all right so that's it for this one we'll see you next time
15950	https://www.khanacademy.org/math/algebra/x2f8bb11595b61c86:foundation-algebra/x2f8bb11595b61c86:combine-like-terms/v/combining-like-terms	Khan Academy Skip to main content If you're seeing this message, it means we're having trouble loading external resources on our website. If you're behind a web filter, please make sure that the domains .kastatic.org and .kasandbox.org are unblocked. Explore Browse By Standards Explore Khanmigo Math: Pre-K - 8th grade Math: High school & college Math: Multiple grades Math: Illustrative Math-aligned Math: Eureka Math-aligned Math: Get ready courses Test prep Science Economics Reading & language arts Computing Life skills Social studies Partner courses Khan for educators Select a category to view its courses Search AI for Teachers FreeDonateLog inSign up Search for courses, skills, and videos Use of cookies Cookies are small files placed on your device that collect information when you use Khan Academy. Strictly necessary cookies are used to make our site work and are required. Other types of cookies are used to improve your experience, to analyze how Khan Academy is used, and to market our service. You can allow or disallow these other cookies by checking or unchecking the boxes below. You can learn more in our cookie policy Accept All Cookies Strictly Necessary Only Cookies Settings Privacy Preference Center When you visit any website, it may store or retrieve information on your browser, mostly in the form of cookies. This information might be about you, your preferences or your device and is mostly used to make the site work as you expect it to. The information does not usually directly identify you, but it can give you a more personalized web experience. Because we respect your right to privacy, you can choose not to allow some types of cookies. Click on the different category headings to find out more and change our default settings. However, blocking some types of cookies may impact your experience of the site and the services we are able to offer. More information Allow All Manage Consent Preferences Strictly Necessary Cookies Always Active Certain cookies and other technologies are essential in order to enable our Service to provide the features you have requested, such as making it possible for you to access our product and information related to your account. For example, each time you log into our Service, a Strictly Necessary Cookie authenticates that it is you logging in and allows you to use the Service without having to re-enter your password when you visit a new page or new unit during your browsing session. Functional Cookies [x] Functional Cookies These cookies provide you with a more tailored experience and allow you to make certain selections on our Service. For example, these cookies store information such as your preferred language and website preferences. Targeting Cookies [x] Targeting Cookies These cookies are used on a limited basis, only on pages directed to adults (teachers, donors, or parents). We use these cookies to inform our own digital marketing and help us connect with people who are interested in our Service and our mission. We do not use cookies to serve third party ads on our Service. Performance Cookies [x] Performance Cookies These cookies and other technologies allow us to understand how you interact with our Service (e.g., how often you use our Service, where you are accessing the Service from and the content that you’re interacting with). Analytic cookies enable us to support and improve how our Service operates. For example, we use Google Analytics cookies to help us measure traffic and usage trends for the Service, and to understand more about the demographics of our users. We also may use web beacons to gauge the effectiveness of certain communications and the effectiveness of our marketing campaigns via HTML emails. Cookie List Clear [x] checkbox label label Apply Cancel Consent Leg.Interest [x] checkbox label label [x] checkbox label label [x] checkbox label label Reject All Confirm My Choices
15951	https://pubmed.ncbi.nlm.nih.gov/24752448/	Contemporary cost-effectiveness analysis comparing sequential bacillus Calmette-Guerin and electromotive mitomycin versus bacillus Calmette-Guerin alone for patients with high-risk non-muscle-invasive bladder cancer - PubMed Clipboard, Search History, and several other advanced features are temporarily unavailable. Skip to main page content An official website of the United States government Here's how you know The .gov means it’s official. Federal government websites often end in .gov or .mil. Before sharing sensitive information, make sure you’re on a federal government site. The site is secure. The https:// ensures that you are connecting to the official website and that any information you provide is encrypted and transmitted securely. Log inShow account info Close Account Logged in as: username Dashboard Publications Account settings Log out Access keysNCBI HomepageMyNCBI HomepageMain ContentMain Navigation Search: Search AdvancedClipboard User Guide Save Email Send to Clipboard My Bibliography Collections Citation manager Display options Display options Format Save citation to file Format: Create file Cancel Email citation Email address has not been verified. Go to My NCBI account settings to confirm your email and then refresh this page. 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Contemporary cost-effectiveness analysis comparing sequential bacillus Calmette-Guerin and electromotive mitomycin versus bacillus Calmette-Guerin alone for patients with high-risk non-muscle-invasive bladder cancer Bassel G Bachir1,Alice Dragomir,Armen G Aprikian,Simon Tanguay,Adrian Fairey,Girish S Kulkarni,Rodney H Breau,Peter C Black,Wassim Kassouf Affiliations Expand Affiliation 1 Division of Urology, Department of Surgery, McGill University Health Center, Montreal, Quebec, Canada. PMID: 24752448 DOI: 10.1002/cncr.28731 Free article Item in Clipboard Randomized Controlled Trial Contemporary cost-effectiveness analysis comparing sequential bacillus Calmette-Guerin and electromotive mitomycin versus bacillus Calmette-Guerin alone for patients with high-risk non-muscle-invasive bladder cancer Bassel G Bachir et al. Cancer.2014. Free article Show details Display options Display options Format Cancer Actions Search in PubMed Search in NLM Catalog Add to Search . 2014 Aug 15;120(16):2424-31. doi: 10.1002/cncr.28731. Epub 2014 Apr 18. Authors Bassel G Bachir1,Alice Dragomir,Armen G Aprikian,Simon Tanguay,Adrian Fairey,Girish S Kulkarni,Rodney H Breau,Peter C Black,Wassim Kassouf Affiliation 1 Division of Urology, Department of Surgery, McGill University Health Center, Montreal, Quebec, Canada. PMID: 24752448 DOI: 10.1002/cncr.28731 Item in Clipboard Full text links Cite Display options Display options Format Abstract Background: Sequential bacillus Calmette-Guerin (BCG) and electromotive mitomycin (sequential therapy) have been shown in a randomized prospective trial to be superior to therapy with BCG alone in patients with high-risk non-muscle-invasive bladder cancer. The objective of the current study was to compare the costs and benefits of these 2 treatment strategies by performing a 5-year and 10-year cost-effectiveness study. Methods: A Markov model was developed to estimate the incremental cost-effectiveness ratio over a 5-year and 10-year period. Estimates of disease progression, death, and treatment efficacy were obtained from what to the authors' knowledge is the only randomized trial comparing the 2 therapies. Costs included: 1) medical costs (physician fees); 2) drug costs (preparation and instillation); and 3) hospital costs (procedure fees, admission fees, and tests and procedures done during surveillance). Patients were allowed a second course of induction therapy. Results: Sequential therapy was found to be associated with a higher initial material cost for induction and maintenance. The average effectiveness for the patients treated with therapy with BCG alone was 4.39 years with a mean cost of $9236 (95% confidence interval, $9118-$9345) per patient. The sequential group resulted in an average effectiveness of 4.65 years, with a mean cost of $16,468 (95% confidence interval, $16,371-$16,527). The 5-year incremental cost-effectiveness ratio of sequential versus BCG-alone therapy was $27,815 per life-year gained. The corresponding figure over a 10-year period was $8618 per life-year gained. Conclusions: The results of the current study suggest that sequential therapy is a cost-effective treatment for patients with high-risk non-muscle-invasive bladder cancer. Keywords: bacillus Calmette-Guerin (BCG); cost-effectiveness; electromotive mitomycin; non-muscle invasive bladder cancer; sequential therapy. © 2014 American Cancer Society. PubMed Disclaimer Comment in Re: Contemporary Cost-Effectiveness Analysis Comparing Sequential bacillus Calmette-Guerin and Electromotive Mitomycin versus bacillus Calmette-Guerin Alone for Patients with High-Risk Non-Muscle-Invasive Bladder Cancer.Chang SS.Chang SS.J Urol. 2015 Sep;194(3):666. doi: 10.1016/j.juro.2015.06.005. Epub 2015 Jun 9.J Urol. 2015.PMID: 26292855 No abstract available. Similar articles Re: Contemporary Cost-Effectiveness Analysis Comparing Sequential bacillus Calmette-Guerin and Electromotive Mitomycin versus bacillus Calmette-Guerin Alone for Patients with High-Risk Non-Muscle-Invasive Bladder Cancer.Chang SS.Chang SS.J Urol. 2015 Sep;194(3):666. doi: 10.1016/j.juro.2015.06.005. Epub 2015 Jun 9.J Urol. 2015.PMID: 26292855 No abstract available. Sequential combination of mitomycin C plus bacillus Calmette-Guérin (BCG) is more effective but more toxic than BCG alone in patients with non-muscle-invasive bladder cancer in intermediate- and high-risk patients: final outcome of CUETO 93009, a randomized prospective trial.Solsona E, Madero R, Chantada V, Fernandez JM, Zabala JA, Portillo JA, Alonso JM, Astobieta A, Unda M, Martinez-Piñeiro L, Rabadan M, Ojea A, Rodriguez-Molina J, Beardo P, Muntañola P, Gomez M, Montesinos M, Martinez Piñeiro JA; Members of Club Urológico Español de Tratamiento Oncológico.Solsona E, et al.Eur Urol. 2015 Mar;67(3):508-16. doi: 10.1016/j.eururo.2014.09.026. Epub 2014 Oct 6.Eur Urol. 2015.PMID: 25301758 Clinical Trial. Sequential bacillus Calmette-Guérin/Electromotive Drug Administration of Mitomycin C as the Standard Intravesical Regimen in High Risk Nonmuscle Invasive Bladder Cancer: 2-Year Outcomes.Gan C, Amery S, Chatterton K, Khan MS, Thomas K, O'Brien T.Gan C, et al.J Urol. 2016 Jun;195(6):1697-703. doi: 10.1016/j.juro.2016.01.103. Epub 2016 Feb 2.J Urol. 2016.PMID: 26845426 Treatment of non muscle invasive bladder tumor related to the problem of bacillus Calmette-Guerin availability. Consensus of a Spanish expert's panel. Spanish Association of Urology.Fernández-Gómez JM, Carballido-Rodríguez J, Cozar-Olmo JM, Palou-Redorta J, Solsona-Narbón E, Unda-Urzaiz JM; Spanish Association of Urology.Fernández-Gómez JM, et al.Actas Urol Esp. 2013 Jul-Aug;37(7):387-94. doi: 10.1016/j.acuro.2013.04.002. Epub 2013 Jun 14.Actas Urol Esp. 2013.PMID: 23773824 English, Spanish. Bacillus Calmette-Guérin immunotherapy for genitourinary cancer.Gandhi NM, Morales A, Lamm DL.Gandhi NM, et al.BJU Int. 2013 Aug;112(3):288-97. doi: 10.1111/j.1464-410X.2012.11754.x. Epub 2013 Mar 20.BJU Int. 2013.PMID: 23517232 Review. See all similar articles Cited by Cost-Effectiveness and Economic Impact of Bladder Cancer Management: An Updated Review of the Literature.Joyce DD, Sharma V, Williams SB.Joyce DD, et al.Pharmacoeconomics. 2023 Jul;41(7):751-769. doi: 10.1007/s40273-023-01273-8. Epub 2023 Apr 23.Pharmacoeconomics. 2023.PMID: 37088844 Review. Local Drug Delivery in Bladder Cancer: Advances of Nano/Micro/Macro-Scale Drug Delivery Systems.Marchenko IV, Trushina DB.Marchenko IV, et al.Pharmaceutics. 2023 Dec 3;15(12):2724. doi: 10.3390/pharmaceutics15122724.Pharmaceutics. 2023.PMID: 38140065 Free PMC article.Review. Biodegradable ring-shaped implantable device for intravesical therapy of bladder disorders.Kim H, Lee SH, Wentworth A, Babaee S, Wong K, Collins JE, Chu J, Ishida K, Kuosmanen J, Jenkins J, Hess K, Lopes A, Morimoto J, Wan Q, Potdar SV, McNally R, Tov C, Kim NY, Hayward A, Wollin D, Langer R, Traverso G.Kim H, et al.Biomaterials. 2022 Sep;288:121703. doi: 10.1016/j.biomaterials.2022.121703. Epub 2022 Aug 17.Biomaterials. 2022.PMID: 36030104 Free PMC article. Humanistic and Economic Burden of Non-Muscle Invasive Bladder Cancer: Results of Two Systematic Literature Reviews.Lee LJ, Kwon CS, Forsythe A, Mamolo CM, Masters ET, Jacobs IA.Lee LJ, et al.Clinicoecon Outcomes Res. 2020 Nov 23;12:693-709. doi: 10.2147/CEOR.S274951. eCollection 2020.Clinicoecon Outcomes Res. 2020.PMID: 33262624 Free PMC article.Review. Is frozen section analysis of ureteral margins at time of radical cystectomy useful?Satkunasivam R, Hu B, Daneshmand S.Satkunasivam R, et al.Curr Urol Rep. 2015 Jun;16(6):38. doi: 10.1007/s11934-015-0506-x.Curr Urol Rep. 2015.PMID: 25940187 Review. 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15952	http://dynamicmathematicslearning.com/diagonal-area-formula-quad.html	Area Formula for Quadrilateral in terms of its Diagonals Area Formula for Quadrilateral in terms of its Diagonals The area of any quadrilateral (convex, concave, crossed) is equal to half the product of the diagonals multiplied by the sine of the angle between them. As shown in the dynamic sketch below the area ABCD = ½ AC x BD x sin ∠DEA. It's surprising though that this simple formula appears to be completely absent from high school textbooks. The reader/viewer is requested to investigate the validity of the result by dragging any of the red vertices A, B, C or D. Point B Point A Segment Point D Segment Point C Segment Segment Segment Segment m A C = 8.30 cm m B D = 12.53 cm Line Inter section E Quadrilateral Area A B C D = 51.78 cm 2 Line Inter section E m ∠ D E A = 84.63 ° 1 2 ⋅ m A C ⋅ m B D ⋅ sin m ∠ D E A = 51.78 cm 2 These are the keyboard-only Toolplay instructions Challenge: 1) Can you explain why (prove that) the above result is true? 2) Drag the vertices so that ABCD becomes concave or crossed. Can you also explain why (prove) that the result remains true in these cases? Related Links Finding the Area of a Crossed Quadrilateral Crossed Quadrilateral Properties Bretschneider's Quadrilateral Area Formula & Brahmagupta's Formula International Mathematical Talent Search (IMTS) Problem Generalized A Geometric Paradox Explained (Another variation of an IMTS problem) Another parallelogram area ratio Area Parallelogram Partition Theorem: Another Example of the Discovery Function of Proof Area ratios of some polygons inscribed in quadrilaterals and triangles An Area Preserving Transformation: Shearing Sylvie's Theorem Some Parallelo-hexagon Area Ratios Cyclic Hexagon Alternate Angles Sum Theorem Free Download of Geometer's Sketchpad Copyright © 2016 KCP Technologies, a McGraw-Hill Education Company. All rights reserved. Release: 2017Q2beta, Semantic Version: 4.5.1-alpha, Build Number: 1021, Build Stamp: ip-10-149-70-76/20160706123640 Back to "Dynamic Geometry Sketches" Back to "Student Explorations" Michael de Villiers, created with WebSketchpad, 29 January 2020; updated 3 April 2024.
15953	https://math.indianapolis.iu.edu/~roederr/publications/Roeder25.pdf	AROUND THE BOUNDARY OF COMPLEX DYNAMICS. ROLAND K.W. ROEDER Abstract. We introduce the exciting ﬁeld of complex dynamics at an undergraduate level while reviewing, reinforcing, and extending the ideas learned in an typical ﬁrst course on complex analysis. Julia sets and the famous Mandelbrot set will be introduced and interesting prop-erties of their boundaries will be described. We will conclude with a discussion of problems at the boundary between complex dynamics and other areas, including a nice application of the material we have learned to a problem in astrophysics. Preface These notes were written for the 2015 Thematic Program on Boundaries and Dynamics held at Notre Dame University. They are intended for an advanced undergraduate student who is majoring in mathematics. In an ideal world, a student reading these notes will have already taken under-graduate level courses in complex variables, real analysis, and topology. As the world is far from ideal, we will also review the needed material. There are many fantastic places to learn complex dynamics, including the books by Beardon , Carleson-Gamelin , Devaney [11, 12], Milnor , and Steinmetz , as well as the Orsay Notes by Douady and Hubbard, the surveys by Blanchard and Lyubich [32, 31], and the invitation to transcendental dynamics by Shen and Rempe-Gillen . The books by Devaney and the article by Shen and Rempe-Gillen are especially accessible to undergraduates. We will take a complementary approach, following a somewhat diﬀerent path through some of the same material as presented in these sources. We will also present modern connections at the boundary between complex dynamics and other areas. None of the results presented here are new. In fact, I learned most of them from the aforementioned textbooks and from courses and informal discussions with John Hubbard and Mikhail Lyubich. Our approach is both informal and naive. We make no eﬀort to provide a comprehensive or historically complete introduction to the subject. Many important results will be omitted. Rather, we will simply have fun doing mathematics. 1 2 ROLAND K.W. ROEDER Dedicated to Emile and Eli. Acknowledgments I am grateful to Notre Dame University for their hos-pitality during the thematic program on boundaries and dynamics. Ivan Chio, Youkow Homma, Lyndon Ji, Scott Kaschner, Dmitry Khavinson, Seung-Yeop Lee, Rodrigo P´ erez, and Mitsuhiro Shishikura provided many helpful comments. All of the computer-drawn images of basins of attrac-tion, ﬁlled Julia sets, and the Mandelbrot set were created using the Frac-talstream software that was written by Matthew Noonan. This work was partially supported by NSF grant DMS-1348589. Lecture 1: “Warm up” Let us start at the very beginning: 1.1. Complex Numbers. Recall that a complex number has the form z = x + iy, where x, y ∈R and i satisﬁes i2 = −1. One adds, subtracts, multiplies, and divides complex numbers using the following rules: (a + bi) ± (c + di) = (a ± c) + (b ± d)i, (a + bi)(c + di) = ac + adi + bci + bdi2 = (ac −bd) + (ad + bc)i, and a + bi c + di = a + bi c + di c −di c −di = (ac + bd) + (bc −ad)i c2 + d2 . The set of complex numbers forms a ﬁeld C under the operations of addition and multiplication. The real part of z = x + iy is Re(z) = x and the imaginary part of z = x + iy is Im(z) = y. One typically depicts a complex number in the complex plane using the horizontal axis to measure the real part and the vertical axis to measure the imaginary part; See Figure 1. One can also take the real or imaginary part of more complicated expressions. For example, Re(z2) = x2 −y2 and Im(z2) = 2xy. The complex conjugate of z = x + iy is z = x −iy and the modulus of z is \|z\| = p x2 + y2 = √ zz. In the complex plane, z is obtained by reﬂecting z across the real axis and \|z\| is the distance from z to the origin 0 = 0 + 0i. The argument of z ̸= 0 is the angle counterclockwise from the positive real axis to z. A helpful tool is the: Triangle Inequality. For every z, w ∈C we have \|z\| −\|w\| ≤\|z + w\| ≤\|z\| + \|w\|. A complex polynomial p(z) of degree d is an expression of the form p(z) = adzd + ad−1zd−1 + · · · + a1z + a0 AROUND THE BOUNDARY OF COMPLEX DYNAMICS. 3 1 \|z\| arg(z) Im(z) z = x −iy z = x + iy x = Re(z) Re(z) y = Im(z) 0 i Figure 1. The complex plane. where ad, . . . , a0 are some given complex numbers with ad ̸= 0. Historically, complex numbers were introduced so that the following theorem holds: Fundamental Theorem of Algebra. A polynomial p(z) of degree d has d complex zeros z1, . . . , zd, counted with multiplicity. In other words, a complex polynomial p(z) can be factored over the complex numbers as p(z) = c(z −z1)(z −z2) · · · (z −zd), (1) where c ̸= 0 and some of the roots zj may be repeated. (The number of times zj is repeated in (1) is the multiplicity of zj as a root of p.) Multiplying and dividing complex numbers is often simpler in polar form. Euler’s Formula states eiθ = cos θ + i sin θ for any θ ∈R. We can therefore represent any complex number z = x + iy by z = reiθ where r = \|z\| and θ = arg(z). Suppose z = reiθ and w = seiφ and n ∈N. The simple formulae zw = rsei(θ+φ), zn = rneinθ, and z w = r sei(θ−φ). (2) follow from the rules of exponentiation. Multiplication and taking powers of complex numbers in polar form are depicted geometrically in Figure 2. 4 ROLAND K.W. ROEDER i z = reiθ z2 = r2ei2θ 0 Re(z) θ Im(z) 1 z3 = r3ei3θ θ θ θ i z4 = r4ei4θ 0 φ θ Re(z) θ zw = rsei(θ+φ) Im(z) z = reiθ 1 w = seiφ Figure 2. Multiplication and taking powers in polar form. 1.2. Iterating Linear Maps. A linear map L : C →C is a mapping of the form L(z) = az, where a ∈C \ {0}. Suppose we take some initial condition z0 ∈C and repeatedly apply L: z0 / L(z0) / L(L(z0)) / L(L(L(z0))) / · · · . (3) For any natural number n ≥1 let L◦n : C →C denote the composition of L with itself n times. We will often also use the notation zn := L◦n(z0). The sequence {zn}∞ n=0 ≡{L◦n(z0)}∞ n=0 is called the sequence of iterates of z0 under L. It is also called the orbit of z0 under L. Remark. The notion of linear used above is from your course on linear algebra: a linear map must satisfy L(z + w) = L(z) + L(w) for all z, w ∈C and L(cz) = cL(z) for all z, c ∈C. For this reason, mappings of the form z 7→az + b are not considered linear. Instead, they are called aﬃne. (See Exercise 1.) The number a is called a parameter of the system. We think of it as describing the overall state of the system (think, for example, temperature or barometric pressure) that is ﬁxed for all iterates n. One can change the parameter to see how it aﬀects the behavior of sequences of iterates (for example, if the temperature is higher, does the orbit move farther in each step?). Our rules for products and powers in polar form (2) allow us to under-stand the sequence of iterates (3). Suppose z0 = reiθ and a = seiθ with r, s > 0. Then, the behavior of the iterates depends on s = \|a\|, as shown in Figure 3. AROUND THE BOUNDARY OF COMPLEX DYNAMICS. 5 \|a\| > 1 implies 0 is unstable i φ φ z3 = az2 i Im(z) a = seiφ φ z0 = reiθ z3 = rs3ei(θ+3φ) z4 z5 φ φ φ 1 z2 = rs2ei(θ+2φ) z1 = rsei(θ+φ) φ φ \|a\| < 1 implies 0 is stable Re(z) Re(z) Im(z) z4 1 z5 z0 φ φ z1 = az0 z2 = az1 a = seiφ z7 z6 φ φ φ φ Figure 3. Iterating the linear map L(z) = az. Above: \|a\| < 1 implies orbits spiral into 0. Below: \|a\| > 0 implies s spiral away from 0. Not Shown: \|a\| = 1 implies orbits rotate around 0 at constant modulus. Remark. For a linear map L(z) = az with \|a\| ̸= 1 the orbits {zn} and {wn} for any two non-zero initial conditions z0 and w0 have the same dynamical behavior. If \|a\| < 1 then lim n→∞zn = 0 = lim n→∞wn and if \|a\| > 1 then lim n→∞zn = ∞= lim n→∞wn. 6 ROLAND K.W. ROEDER This is atypical for dynamical systems—the long term behavior of the orbit usually depends greatly on the initial condition. For example, we will soon see that when iterating the quadratic mapping p(z) = z2 + i 4 there are many initial conditions whose orbits remain bounded and many whose orbit escapes to ∞. There will also be many initial conditions whose orbits have completely diﬀerent behavior! Linear maps are just too simple to have interesting dynamical properties. Exercise 1. An aﬃne mapping A : C →C is a mapping given by A(z) = az + b, where a, b ∈C and a ̸= 0. Show that iteration of aﬃne mappings produces no dynamical behavior that was not seen when iterating linear mappings. 1.3. Iterating quadratic polynomials. Matters become far more inter-esting if one iterates quadratic mappings pc : C →C given by pc(z) = z2+c. Here, c is a parameter, which we sometimes include in the notation by means of a subscript, writing pc(z), and sometimes omit, writing simply p(z). Remark. Like in Exercise 1, one can show that quadratic mappings of the form pc(z) = z2 + c actually capture all of the types of dynamical behavior that can arise when iterating a more general quadratic mapping q(z) = az2 + bz + c. Applying the mapping pc can be understood geometrically in two steps: one ﬁrst squares z using the geometric interpretation provided in polar coordinates (2). One then translates (shifts) the result by c. This two-step process is illustrated in Figure 4. i θ θ pc Re(z) Im(z) 1 z = reiθ z2 = r2ei2θ c pc(z) = z2 + c Figure 4. Geometric interpretation of applying pc(z) = z2 + c. AROUND THE BOUNDARY OF COMPLEX DYNAMICS. 7 Remark. Solving the exercises in this subsection may require some of the basic complex analysis from the following subsection. They are presented here for better ﬂow of the material. Example 1. Exploring the dynamics of pc : C →C for c = i 4. In Fig-ure 5 we show the ﬁrst few iterates under p(z) = z2+ i 4 of two diﬀerent orbits: {zn} of initial condition z0 = i and {wn} of initial condition w0 = 1.1i. Note that orbit {zn} seems to converge to a point z ≈−0.05 + .228i while orbit {wn} seems to escape to ∞. w1 = −1.2 + 0.3i 1 Re Im z0 = i z1 = −1 + i 4 z3 = 0.8 −0.2i z6 = 0.1 + 0.3i z4 = 0.6 −0.1i z5 = 0.4 + 0.1i w0 = 1.1i z2 = 15 16 −i 4 w2 = 1.4 −0.4i z7 = −0.1 + 0.3i Figure 5. Orbits {zn} for initial condition z0 = i and {wn} for w0 = 1.1i under p(z) = z2 + i 4. Exercise 2. Use the quadratic formula to prove that there exists z• ∈C that is close to −0.05 + .228i and satisﬁes p(z•) = z•. Such a point is called a ﬁxed point for p(z) because if you use z• as initial condition the orbit is a constant sequence {z•, z•, z•, . . .}. Show that there is a second ﬁxed point z∗for p(z) with z∗≈1.05−.228i. Compute \|p′(z•)\| and \|p′(z∗)\|, where p′(z) = 2z is the derivative of p(z) = z2 + i 4. Use the behavior of linear maps, as shown in Figure 3, to 8 ROLAND K.W. ROEDER make a prediction about the behavior of orbits for p(z) near each of these ﬁxed points. Exercise 3. Let z• be the ﬁxed point for p(z) discovered in Exercise 2. Prove that for any point z0 suﬃciently close to z• the orbit {zn} under p(z) = z2 + i 4 converges to z•. (I.e., prove that there exists δ > 0 such that for any z0 satisfying \|z0 −z•\| < δ and any ϵ > 0 there exists N ∈N such that for all n ≥N we have \|zn −z•\| < ϵ.) Why does your proof fail if you replace the ﬁxed point z• with z∗? Now, prove that the orbit of z0 = i converges to z•. Exercise 4. Prove that there exists r > 0 such that for any initial condition z0 with \|z0\| > r the orbit {zn} of z0 under p(z) = z2 + i 4 escapes to inﬁnity. (I.e., prove that there exists r > 0 such that for any z0 satisfying \|z0\| > r and any R > 0 there exists N ∈N such that for all n ≥N we have \|zn\| > R.) Now prove that the orbit of w0 = 1.1i escapes to inﬁnity. Example 2. Exploring the dynamics of pc : C →C for c = −1. In Figure 6 we show the ﬁrst few iterates under p(z) = z2 −1 of two diﬀerent orbits: {zn} of initial condition z0 ≈0.08 + 0.66i and {wn} of initial condition w0 = √ 2 2 (1+i). Orbit {zn} seems to converge to a periodic behavior (‘periodic orbit’) while {wn} seems to escape to ∞. In fact, the periodic orbit that {zn} seems to converge to is easy to ﬁnd for this mapping. If we use initial condition u0 = 0 we have u1 = p−1(u0) = 02 −1 = −1. Then, u2 = p(u1) = p(−1) = (−1)2 −1 = 0 = u0. We conclude that the orbit of u0 = 0 is periodic with period two: 0 p−1 / −1 p−1 ^ (Subsequently, this periodic orbit will be denoted 0 ↔1.) The following two exercises are in the context of Example 2. Exercise 5. Make precise the statement that if z0 is an initial condition suﬃciently close to 0, then its orbit “converges to the periodic orbit 0 ↔1”. Prove the statement. Now, suppose z0 ≈0.08 + 0.66i and prove that its orbit converges to the periodic orbit 0 ↔1. Exercise 6. Find an initial condition z0 ∈C such that for any ϵ > 0 there are AROUND THE BOUNDARY OF COMPLEX DYNAMICS. 9 z4 ≈−1.39 + 0.04i w2 = −7 4 i z2 ≈1.03 −0.30i i 1 Im Re z0 ≈0.08 + 0.66i z5 ≈0.93 −0.11i z7 ≈−1.02 + 0.06i z8 ≈0.04 −0.12i z10 ≈0 z6 ≈−0.15 −0.20i z9 ≈−1 z1 ≈−1.43 + 0.11i z3 ≈−.03 −0.62i w0 = √ 2 2 + √ 2 2 i w1 = −1 + i Figure 6. Orbits {zn} of z0 ≈0.08 + 0.66i and {wn} of w0 = √ 2 2 (1 + i) under the quadratic polynomial p−1(z) = z2 −1. (1) inﬁnitely many initial conditions w0 with \|w0 −z0\| < ϵ having orbit {wn} under p−1 that remains bounded, and (2) inﬁnitely many initial conditions u0 with \|u0 −z0\| < ϵ having orbit {un} under p−1 that escapes to inﬁnity. Hint: work within R and consider the graph of p(x) = x2 −1. Example 3. Exploring the dynamics of pc : C →C for c = 1 2. As in the previous two examples, we will try a couple of arbitrary initial conditions. Figure 7 shows the orbits of initial conditions z0 = 0 and w0 ≈0.4 + 0.6i under p(z) = z2 + 1 2. Both orbits seem to escape to ∞. Exercise 7. Prove that for any real initial condition z0 ∈R the orbit {zn} under p(z) = z2 + 1 2 escapes to inﬁnity. Exercise 8. Determine whether there is any initial condition z0 for which the orbit under p1/2 remains bounded. 10 ROLAND K.W. ROEDER w4 ≈0.7 + 0.2i w6 ≈1.4 + 0.7i w3 ≈0.5 + 0.2i w2 ≈0.4 + 0.3i i Im Re z0 = 0 z1 = 1 2 z2 = 3 4 z3 = 17 16 1 w0 ≈0.4 + 0.6i w1 ≈0.3 + 0.5i w5 ≈1 + 0.3i Figure 7. Orbits {zn} of z0 = 0 and {wn} of w0 ≈0.4+0.6i under the quadratic polynomial p 1 2 (z) = z2 + 1 2. Exercise 9. Repeat the type of exploration done in Examples 1 - 3 for c = 0, c = −2, c = i, and c = −0.1 + 0.75i. Try other values of c. 1.4. Questions. During our explorations we’ve discovered several ques-tions. Some of them were answered in the exercises, but several of them are still open, including: (1) Does every quadratic map have some initial condition z0 whose orbit escapes to ∞? (2) Does every quadratic map have some periodic orbit z0 7→z1 7→z2 7→· · · 7→zn 7→z0 which attracts the orbits of nearby initial conditions? Perhaps we didn’t look hard enough for one when c = 1 2? (3) Can a map pc(z) have more than one such attracting periodic orbit? (4) For any m ≥1 does there exist a parameter c such that pc(z) has an attracting periodic orbit of period m? Exercise 10. Answer Question 1 by showing that for any c there is a radius R(c) such that for any initial condition z0 with \|z0\| > R(c) the orbit {zn} escapes to ∞. AROUND THE BOUNDARY OF COMPLEX DYNAMICS. 11 Generalize your result to prove that for any polynomial q(z) of degree at least 2 there is some R > 0 so that any initial condition z0 with \|z0\| > R has orbit {zn} that escapes to ∞. 1.5. Crash course in complex analysis. In order to answer the ques-tions posed in the previous subsection and explore the material more deeply, we will need some basic tools from complex analysis. We have slightly adapted the follows results from the textbook by Saﬀand Snider . We present at most sketches of the proofs and leave many of the details to the reader. This subsection is rather terse. The reader may want to initially skim over it and then move forward to see how the material is used in the later lectures. We begin with some topological properties of C. The open disc of radius r > 0 centered at z0 is D(z0, r) := {z ∈C : \|z −z0\| < r}. Deﬁnition 1. A set S ⊂C is open if for every z ∈S there exists r > 0 such that D(z, r) ⊂S. A set S ⊂C is closed if its complement C \ S is open. Exercise 11. Prove that for any z0 ∈C and any r > 0 the “open disc” D(z0, r) is actually open. Then prove that the set D(z0, r) := {z ∈C : \|z −z0\| ≤r} is closed. It is called the closed disc of radius r centered at z0. Deﬁnition 2. The boundary of S ⊂C is ∂S := {z ∈C : D(z, r) contains points in S and in C \ S for every r > 0}. Deﬁnition 3. A set S ⊂C is disconnected if there exist open sets U and V with (i) S ⊂U ∪V , (ii) S ∩U ̸= ∅and S ∩V ̸= ∅, and (iii) U ∩V = ∅. A set S ⊂C is connected if it is not disconnected. An open connected U ⊂C is called a domain. Any set denoted U in this subsection will be assumed to be a domain. If z0 ∈U, a neighborhood of z0 will be another domain V ⊂U with z0 ∈V . (A round disc D(z0, r) for some r > 0 suﬃciently small will always suﬃce.) Deﬁnition 4. A contour γ ⊂U is a piecewise smooth function γ : [0, 1] →U. (Here, the notation implicitly identiﬁes the function γ : [0, 1] →U with its image γ[0, 1] ≡γ ⊂U.) 12 ROLAND K.W. ROEDER A contour γ is closed if γ(0) = γ(1). A closed contour γ is simple if γ(s) ̸= γ(t) for t ̸= s unless t = 0 and s = 1 or vice-versa. (Informally, a simple closed contour as a loop that does not cross itself.) A simple closed contour is positively oriented if as you follow the contour, the region it encloses is on your left. (Informally, this means that it goes counterclockwise.) Remark. An open set S is connected if and only if for every two points z, w ∈S there is a contour γ ⊂S with γ(0) = z and γ(1) = w. Deﬁnition 5. A domain U ⊂C is simply connected if any closed con-tour γ ⊂U can be continuously deformed within U to some point z0 ∈U. We refer the reader to [43, Section 4.4, Deﬁnition 5] for the formal deﬁnition of continuously deformed. In these notes, we will only need that the disc D(z0, r) is simply connected. It follows from the fact that any closed contour γ ⊂D(z0, r) can be aﬃnely scaled within D(z0, r) down to the center z0. Remark. You have seen Deﬁnition 5 in your multivariable calculus class, where it was used in the statement of Green’s Theorem. Deﬁnition 6. A set K ⊂C is compact if for any collection {Wλ}λ∈Λ of open sets with K ⊂ [ λ∈Λ Wλ there are a ﬁnite number of sets Wλ1, . . . , Wλn so that K ⊂Wλ1 ∪· · · ∪Wλn. Heine-Borel Theorem. A set S ⊂C is compact if and only if it is closed and bounded. Exercise 12. Suppose K1 ⊃K2 ⊃K3 ⊃· · · is a nested sequence of non-empty connected compact sets in C. Prove that T n≥1 Kn is non-empty and connected. We are now ready to start doing complex calculus. The notion of limit is deﬁned in exactly the same as in calculus, except that modulus \| · \| takes the place of absolute value. Deﬁnition 7. Let z0 ∈U and let f : U \ {z0} →C be a function. We say that limz→z0 f(z) = L for some L ∈C if for every ϵ > 0 there is a δ > 0 such that 0 < \|z −z0\| < δ implies \|f(z) −L\| < ϵ. If we write f(z) = u(x, y) + iv(x, y) AROUND THE BOUNDARY OF COMPLEX DYNAMICS. 13 with u : R2 →R and v : R2 →R, then limz→z0 f(z) = L if and only if lim (x,y)→(x0,y0) u(x, y) = Re(L) and lim (x,y)→(x0,y0) v(x, y) = Im(L). (The limits on the right hand side are taken as in the sense of your multi-variable calculus class.) Deﬁnition 8. f : U →C is continuous if for every z0 ∈U we have limz→z0 f(z) = f(z0). Deﬁnition 9. f : U →C is diﬀerentiable at z0 ∈U if f′(z0) := lim h→0 f(z0 + h) −f(z0) h exists. Remark. The usual rules for diﬀerentiating sums, products, and quotients, as well as the chain rule hold for complex derivatives. They are proved in the same way as in your calculus class. Remark. It is crucial in Deﬁnition 9 that one allows h to approach 0 from any direction and that the resulting limit is independent of that direction. Now for the most important deﬁnition in this whole set of notes: Deﬁnition 10. f : U →C is analytic (or holomorphic) if it is diﬀer-entiable at every z0 ∈U. We will see that analytic functions have marvelous properties! It will be the reason why studying the iteration of analytic functions is so fruitful. Exercise 13. Show that f(z) = z is analytic on all of C and that g(z) = z is not analytic in a neighborhood of any point of C. (In fact, it is “anti-analytic”.) Exercise 14. Show that any complex polynomial p(z) = adzd + ad−1zd−1 + · · · + a1z + a0 gives an analytic function p : C →C. Deﬁnition 11. Suppose U and V are domains. A mapping f : U →V is called conformal if it is analytic and has an analytic inverse f−1 : V →U. Cauchy-Riemann Equations. Let f : U →C be given by f(z) = u(x, y) + iv(x, y) with ∂u ∂x, ∂u ∂y , ∂v ∂x, and ∂v ∂y continuous on U. Then f is analytic on U ⇔ ∂u ∂x = ∂v ∂y and ∂u ∂y = −∂v ∂x for all (x, y) ∈U. 14 ROLAND K.W. ROEDER Inverse Function Theorem. Suppose f : U →C is analytic and f′(z0) ̸= 0. Then, there is an open neighborhood V of f(z0) in C and an analytic func-tion g : V →U such that g(f(z0)) = z0 and for all w ∈V we have f(g(w)) = w and all z ∈g(V ) we have g(f(z)) = z. Moreover, g′(f(z0)) = 1 f′(z0). Exercise 15. Show that f(z) = z2−1 satisﬁes the hypotheses of the inverse function theorem for any z ̸= 0. Use the quadratic equation to explicitly ﬁnd the function g(z) whose existence is asserted by the Inverse Function Theorem. What goes wrong with g at −1 = f(0)? Exponential and Logarithm. According to Euler’s Formula, if z = x+iy with x, y ∈R then ez = ex+iy = ex (cos y + i sin y) , which can be veriﬁed to be analytic on all of C by using the Cauchy-Riemann Equations. It satisﬁes (ez)′ = ez, which is never 0. Let S := {z ∈C : −π < Im(z) < π} and C† := C \ (−∞, 0]. Then, the exponential function maps the strip S bijectively onto C†. Therefore, it has an inverse function Log(z) : C† →S, which is analytic by the inverse function theorem. (This function is called the Principal Branch of the Logarithm. One can deﬁne other branches that are analytic on domains other than C†; see [43, Section 3.3].) Deﬁnition 12. Suppose f : U →C is an analytic function. A point z ∈U with f′(z) = 0 is called a critical point of f. A point w ∈C with w = f(z) for some critical point z is called a critical value. The neighborhood V provided by the inverse function theorem could be very small. When combined with the Monodromy Theorem [1, p. 295-297], one can control the size of the domain, so long as it is simply connected: Simply-Connected Inverse Function Theorem. Suppose f : U →C is an analytic function and V ⊂f(U) is a simply connected domain that doesn’t contain any of the critical values of f. Given any w• ∈V and any z• ∈f−1(w•) there is a unique analytic function g : V →C with g(w•) = z•, f(g(w)) = w for all w ∈V , and g(f(z)) = z for all z ∈g(V ). Remark. Our name for the previous result is not standard. Use it with caution! AROUND THE BOUNDARY OF COMPLEX DYNAMICS. 15 If f : U →C is continuous and γ ⊂U is a contour, then the integral Z γ f(z)dz is deﬁned in terms of a suitable complex version of Riemann sums; see [43, Section 4.2]. For our purposes, we can take as deﬁnition Z γ f(z)dz := Z 1 0 f(γ(t))γ′(t)dt, which is stated as Theorem 4 from [43, Section 4.2]. Exercise 16. Let γ be the positively oriented unit circle in C. Show that Z γ dz z = 2πi, (4) which is perhaps “the most important contour integral”. Cauchy’s Theorem. If f : U →C is analytic and U is simply connected, then for any closed contour γ ⊂D we have Z γ f(z)dz = 0. Sketch of proof: The following is “cribbed” directly from [43, p. 192-193]. Write f(z) = u(x, y) + iv(x, y) and γ(t) = (x(t), y(t)). Then, Z γ f(z)dz = Z 1 0 f(γ(t))γ′(t)dt = Z 1 0 u(x(t), y(t)) + iv(x(t), y(t)) dx dt + idy dt dt = Z 1 0 u(x(t), y(t))dx dt −v(x(t), y(t))dy dt dt + i Z 1 0 v(x(t), y(t))dx dt + u(x(t), y(t))dy dt dt. The real and imaginary parts above are just the parameterized versions of the real contour integrals Z γ u(x, y)dx −v(x, y)dy and Z γ v(x, y)dx + u(x, y)dy 16 ROLAND K.W. ROEDER considered in your multivariable calculus class. Since U is simply connected, Green’s Theorem gives Z γ u(x, y)dx −v(x, y)dy = Z Z D −∂v ∂x −∂u ∂y dxdy and Z γ v(x, y)dx + u(x, y)dy = Z Z D ∂u ∂x −∂v ∂y dxdy. Since f is analytic, the Cauchy-Riemann Equations imply that both inte-grands are 0. Thus, R γ f(z)dz = 0. □ Remark. In the proof we have used the additional assumption that the partial derivatives ∂u ∂x, ∂u ∂y , ∂v ∂x, and ∂v ∂y are all continuous functions of (x, y). This was needed in order for us to apply Green’s Theorem. This hypothesis is not needed, but the general proof of Cauchy’s Theorem is more compli-cated; see, for example, [1, Section 4.4]. There is also an amazing “converse” to Cauchy’s Theorem Morera’s Theorem. If f : U →C is continuous and if Z γ f(z)dz = 0. for any closed contour γ ⊂U, then f is analytic in U. Cauchy Integral Formula. Let γ be a simple closed positively oriented contour. If f is analytic in some simply connected domain U containing γ and z0 is any point inside of γ, then f(z0) = 1 2πi Z γ f(z) z −z0 dz Sketch of proof: Refer to Figure 8 throughout the proof. For any ϵ > 0 we can apply Cauchy’s Theorem to the contour η proving that Z γ f(z) z −z0 dz = Z γ′ f(z) z −z0 dz, where γ′ is the positively oriented circle \|z −z0\| = ϵ. Since f(z) is ana-lytic it is continuous, implying that if we choose ϵ > 0 suﬃciently small, f(z) ≈f(z0) on γ′. Then, Z γ′ f(z) z −z0 dz ≈f(z0) Z γ′ 1 z −z0 dz = 2πif(z0), with the last equality coming from (4). □ AROUND THE BOUNDARY OF COMPLEX DYNAMICS. 17 γ′ z0 η γ Figure 8. Illustration of the proof of the Cauchy Integral Formula. Exercise 17. Use the fact that if \|f(z) −g(z)\| < ϵ for all z on a contour γ then Z γ f(z)dz − Z γ g(z)dz < ϵ length(γ) to make rigorous the estimates ≈in the proof of the Cauchy Integral For-mula. Let us write the Cauchy Integral Formula slightly diﬀerently: f(z) = 1 2πi Z γ f(ζ) ζ −z dζ, (5) where z is any point inside of γ. (This makes it more clear that we think of z as an independent variable.) By diﬀerentiating under the integral sign (after checking that it’s allowed) we obtain: Cauchy Integral Formula For Higher Derivative. Let γ be a simple closed positively oriented contour. If f is analytic in some simply connected domain U containing γ and z is any point inside of γ, then f(n)(z) = n! 2πi Z γ f(ζ) (ζ −z)n+1 dz. (6) In particular, an analytic function is inﬁnitely diﬀerentiable! Cauchy Estimates. Suppose f(z) is analytic on a domain containing the disc D(z0, r) and suppose \|f(z)\| < M on the boundary ∂D(z0, r) = {z ∈ C : \|z −z0\| = r}. Then, for any n ∈N we have f(n)(z0) ≤n!M rn . Exercise 18. Prove the Cauchy Estimates, supposing (6). 18 ROLAND K.W. ROEDER Suppose D(z0, r) ⊂U and f : U →C is analytic. If we parameterize ∂D(0, r) by γ(t) = z0 + reit, then the Cauchy Integral Formula becomes f(z0) = 1 2π Z 2π 0 f(z0 + reit)dt. From this, one sees that it is impossible to have \|f(z0)\| ≥\|f(z0 + reit)\| for all t ∈[0, 2π] without the inequality actually being an equality for all t. From this, it is straightforward to prove: Maximum Modulus Principle. Suppose f(z) is analytic in a domain U and \|f(z)\| achieves its maximum at a point z0 ∈U. Then f(z) is constant on U. If, moreover, U is compact and f extends continuously to U, then f achieves its maximum modulus on the boundary of U. Meanwhile, by using the geometric series to write 1 ζ −z = 1 ζ · 1 1 −z ζ = 1 ζ ∞ X n=0 z ζ n , for any z ζ < 1, the Cauchy Integral Formula (5) implies: Existence of Power Series. Let f be analytic on a domain U and suppose the disc D(z0, r) is contained in U. Then, we can write f(z) as a power series f(z) = ∞ X n=0 an(z −z0)n that converges on D(z0, r). The multiplicity of a zero z0 for an analytic function f(z) is deﬁned as the order of the smallest non-zero term in the power series expansion of f(z) around z0. Argument Principle. Suppose f : U →C is analytic and γ ⊂U is a positively oriented simple closed contour such that all points inside of γ are in U. Then, the number of zeros of f (counted with multiplicities) is equal to the change in arg(f(z)) as z traverses γ once in the counter-clockwise direction. Deﬁnition 13. Let fn : U →C be a sequence of functions and f : U →C be another function. Let K ⊂U be a compact set. The sequence {fn} converges to f uniformly on K if for every ϵ > 0 there is a δ > 0 such that for every z ∈K \|fn(z) −f(z)\| < ϵ. AROUND THE BOUNDARY OF COMPLEX DYNAMICS. 19 Note that the order of quantiﬁers in Deﬁnition 13 is crucial. If δ was allowed to depend on z, we would have the weaker notion of pointwise convergence. Uniform Limits Theorem. Suppose fn : U →C is a sequence of analytic functions and f : U →C is another (potentially non-analytic) function. If for any compact K ⊂U we have that {fn} converges uniformly to f on K, then f : U →C is also analytic. Moreover, for any k ≥1, the k-th derivatives f(k) n (z) converge uniformly to f(k)(z) on any compact K ⊂U. Sketch of the proof: By restricting to a smaller domain, we can suppose U is simply connected. For any contour γ ⊂U, Cauchy’s Theorem gives R γ fn(z)dz = 0. Since the convergence is uniform on the compact set γ ⊂U, we have Z γ f(z)dz = Z γ lim n→∞fn(z)dz = lim n→∞ Z γ fn(z)dz = 0. Thus, Morera’s Theorem gives that f(z) is analytic. Convergence of the derivatives follows from the Cauchy Integral Formula For Higher Derivatives. □ The following exercises illustrate the power of the Uniform Limits The-orem. Exercise 19. Suppose that for some R > 0 the power series ∞ X n=0 an(z −z0)n (7) converges for each z ∈D(z0, R). Prove that for any 0 < r < R the power series converges uniformly on the closed disc D(z0, r). Use Exercise 14 and the Uniform Limits Theorem to conclude that power series (7) deﬁnes an analytic function f : D(z0, R) →C. Exercise 20. Suppose we have a sequence of polynomials pn : [0, 1] →R and that pn(x) converges uniformly on [0, 1] to some function f : [0, 1] →R. Does f even have to be diﬀerentiable? We close this section with the following famous result: Schwarz Lemma. Let D := D(0, 1) be the unit disc and suppose f : D →D is analytic with f(0) = 0. Then (a) \|f′(0)\| ≤1, and (b) \|f′(0)\| = 1 if and only if f(z) = eiθz for some θ ∈R. 20 ROLAND K.W. ROEDER Sketch of the proof. By the Existence of Power Series Theorem we can write f as a power series converging on D: f(z) = a1z + a2z2 + a3z3 · · · , where the constant term is 0 because f(0) = 0. Therefore, F(z) := f(z) z = a1 + a2z + a3z3 · · · is also analytic on D, by Exercise 19. Applying the Maximum Modulus Principle to F(z) we see that for any 0 < r < 1 and any ζ satisfying \|ζ\| < r \|F(ζ)\| ≤max{\|z\|=r}\|f(z)\| r ≤1 r. Since this holds for any 0 ≤r ≤1, we ﬁnd that \|F(ζ)\| ≤1 for any ζ ∈D. Part (a) follows because F(0) = f′(0). If \|f′(0)\| = 1, then \|F(0)\| = 1, implying that F attains its maximum at a point of D. The Maximum Modulus Principle implies that F(z) is constant, i.e. F(z) = c for some c with \|c\| = 1. Any such c is of the form eiθ for some θ ∈R, so by the deﬁnition of F, we have f(z) = eiθz for all z ∈D. □ Remark. There was nothing special about radius 1. If f : D(0, r) →D(0, r) for some r > 0 and f(0) = 0, then (a) and (b) still hold. Lecture 2: “Mandelbrot set from the inside out” We will work our way to the famous Mandelbrot set from an unusual perspective. 2.1. Attracting periodic orbits. In Section 1.3 we saw that the qua-dratic maps pc(z) = z2 + c for c = i 4, −1, and −0.1 + 0.75i seemed to have attracting periodic orbits of periods 1, 2, and 3, respectively. In this subsection we will make that notion precise and prove two results about attracting periodic orbits. We will also see that the set of initial conditions whose orbits converge to an attracting periodic orbit can be phenomenally complicated. While we are primarily interested in iterating quadratic polynomials pc(z) = z2 + c, it will also be helpful to consider iteration of higher de-gree polynomials q(z). Deﬁnition 14. A sequence z0 q − →z1 q − →z2 q − →· · · q − →zm = z0 is called a periodic orbit of period m for q if zn ̸= z0 for each 1 ≤ n ≤m −1. The members of such a periodic orbit for q are called periodic AROUND THE BOUNDARY OF COMPLEX DYNAMICS. 21 points of period m for q. A periodic point of period 1 is called a ﬁxed point of q. If z0 is a periodic point of period m for q, then it is a ﬁxed point for the polynomial s(z) = q◦m(z). Meanwhile, if z0 is a ﬁxed point for s(z), then it is a periodic point of period j for q, where j divides m. Thus, we can often reduce the study of periodic points to that of ﬁxed points. Deﬁnition 15. A ﬁxed point z∗of q is called attracting if there is some r > 0 such that such q D(z∗, r) ⊂D(z∗, r) and for any initial condition z0 ∈D(z∗, r) the orbit {zn} under q satisﬁes lim zn = z∗. A periodic orbit z0 →z1 →· · · →zm = z0 is attracting if for each n = 0, . . . , m −1 the point zn is an attracting ﬁxed point for s(z) = q◦m(z). Deﬁnition 16. The multiplier of a periodic orbit z0 →z1 →· · · →zm = z0 is λ = q′(z0) · q′(z1) · · · q′(zm−1). Note that if s(z) = q◦m(z), then the chain rule gives that s′(zj) = q′(z0) · q′(z1) · · · q′(zm−1) = λ for each 0 ≤j ≤m −1. Thus the multiplier of the periodic orbit z0 →z1 →· · · →zm = z0 under q is the same as the multiplier of each point zj, when considered as a ﬁxed point of s(z). The next lemma tells us that the same criterion we had in Section 1.3 for 0 being attracting under a linear map applies to ﬁxed points of non-linear maps. Attracting Periodic Orbit Lemma. A periodic orbit z0 →z1 →· · · →zm = z0 of q is attracting if and only if its multiplier satisﬁes \|λ\| < 1. Proof. Replacing q by a suitable iterate we can suppose the periodic orbit is a ﬁxed point z∗of q. If z∗̸= 0 then we can consider the new polynomial q(z + z∗) −z∗for which 0 replaces z∗as the ﬁxed point of interest. (We call this a shift of coordinates .) Suppose 0 is an attracting ﬁxed point for q. Then, there exists r > 0 so that q D(0, r) ⊂D(0, r) and so that the orbit {zn} of any initial condition z0 ∈D(0, r) satisﬁes limn→∞zn = 0. Since q(0) = 0, the Schwarz Lemma implies that \|q′(0)\| ≤1. If \|q′(0)\| = 1, then the Schwarz Lemma implies that q is a rigid rotation z 7→eiθz. This would violate that the orbit of any initial condition z0 ∈D(0, r) converges to 0. Therefore, \|q′(0)\| < 1. Now, suppose 0 is a ﬁxed point for q with multiplier λ = q′(0) of modulus less than one. We will consider the case λ ̸= 0, leaving the case λ = 0 as 22 ROLAND K.W. ROEDER Exercise 23, below. We have q(z) = λz + a2z2 + · · · + adzd = λ 1 + a2 λ z + · · · + ad λ zd−1 z. Since limz→0 1 + a2 λ z + · · · + ad λ zd−1 = 1 and \|λ\| < 1 there exists ϵ > 0 so that if \|z\| < ϵ then 1 + a2 λ z + · · · + ad λ zd−1 < 1 + 1 −\|λ\| 2\|λ\| . Thus, for any \|z\| < ϵ we have \|q(z)\| = λ 1 + a2 λ z + · · · + ad λ zd−1 \|z\| ≤1 + \|λ\| 2 \|z\|. (8) In particular, q (D(0, r)) ⊂D(0, r) and (8) implies that for any z0 ∈D(0, r) the orbit satisﬁes \|zn\| ≤ 1+\|λ\| 2 n r →0. We conclude that 0 is an attracting ﬁxed point for q. □ Exercise 21. Use the Attracting Periodic Orbit Lemma to verify that (a) z∗= 1 2 − √1−i 2 is an attracting ﬁxed point for p(z) = z2 + i 4, (b) 0 ↔1 is an attracting periodic orbit of period 2 for p(z) = z2 −1, and (c) If c satisﬁes c3 + 2c2 + c + 1 = 0, then 0 →c →c2 + c →0 is an attracting periodic orbit of period 3. (One of the solutions for c is the parameter c ≈−0.12 + 0.75i studied in Exercise 9.) Exercise 22. Verify that there exists r > 0 such that for any initial con-dition z0 ∈R with \|z0\| < r the orbit under q(z) = z −z3 converges to 0. Why is 0 not attracting as a complex ﬁxed point? Exercise 23. Prove that if z∗is a ﬁxed point for a polynomial q having multiplier λ = 0, then z∗is attracting. Deﬁnition 17. Suppose O = z0 →z1 →· · · →zm = z0 is an attracting periodic orbit. The basin of attraction A(O) is A(O) := {z ∈C : s◦n(z) →zj as n →∞for some 0 ≤j ≤m −1}, where s(z) = q◦m(z). The immediate basin A0(O) is the union of the connected components of A(O) containing the points z0, . . . , zm−1. Computer generated images of the basins of attraction for the attract-ing periodic orbits discussed in Exercise 21 are shown in Figures 9 - 11. Notice the remarkable complexity of the boundaries of the basins of attrac-tion, something we would never have guessed during our experimentation in Section 1.3. AROUND THE BOUNDARY OF COMPLEX DYNAMICS. 23 Remark on computer graphics: We used Fractalstream to create Figures 9-12, 16-20, and 22-23. Other useful programs include Dynamics Explorer and the Boston University Java Applets . Im(z) 3 2 i 3 2 Re(z) Figure 9. Basin of attraction of the ﬁxed point z∗= 1 2 − √1−i 2 for p(z) = z2 + i 4. It is natural to ask how complicated the dynamics for iteration of q can be near an attracting ﬁxed point. The answer is provided by Kœnig’s Theorem and B¨ ottcher’s Theorem. Kœnig’s Theorem. Suppose z• is an attracting ﬁxed point for q with multiplier λ ̸= 0. Then, there exists a neighborhood U of z• and a conformal map φ : U →φ(U) ⊂C so that for any w ∈φ(U) we have φ ◦q ◦φ−1(w) = λw. (9) In other words, Theorem 2.1 gives that there is a neighborhood U of z• in which there is a coordinate system w = φ(z) in which the non-linear mapping q becomes linear! This explains why the the same geometric spirals shown on the top of Figure 3 for the linear map appear suﬃciently close 24 ROLAND K.W. ROEDER 2 i Re(z) Im(z) Figure 10. Basin of attraction of the period two cycle 0 ↔−1 for p(z) = z2 −1. to an attracting ﬁxed point z• for a non-linear map. This is illustrated in Figure 12. Proof. Shifting the coordinates if necessary, we can suppose z• = 0. The Attracting Periodic Orbit Lemma gives that the multiplier of 0 satisﬁes \|λ\| < 1. Therefore, as in the second half of the proof of the Attracting Periodic Orbit Lemma, we can ﬁnd some r > 0 and \|λ\| < a < 1 so that for any z ∈D(0, r) we have \|zn\| ≤anr, (10) where zn := q◦n(z). Since q(0) = 0 we have q(z) = λz + s(z), (11) where s(z) = a2z2 +a3z3 +· · · adzd. In particular, there exists b > 0 so that \|s(z)\| ≤b\|z\|2. (12) for all z ∈D(0, r). Let φn : D(0, r) →C be given by φn(z) := zn λn , which satisﬁes φn(0) = 0, since 0 is a ﬁxed point. Notice that φn(q(z)) = zn+1 λn = λ · zn+1 λn+1 = λφn+1(z). (13) Suppose we can prove that φn converges uniformly on D(0, r) to some func-tion φ : D(0, r) →C. Then, φ will be analytic by the Uniform Limits AROUND THE BOUNDARY OF COMPLEX DYNAMICS. 25 Im(z) 3 2 i 1 Re(z) Figure 11. Top: basin of attraction of the attracting pe-riod 3 cycle 0 →c →c2 + c for c ≈−0.12 + 0.75i. Bottom: zoomed-in view of the boxed region from the left. 26 ROLAND K.W. ROEDER Figure 12. An orbit converging to the attracting ﬁxed point for p(z) = z2 + i 4. Here, λ = 1 −√1 −2 i ≈0.8e1.9i. Theorem. Meanwhile, the left and right sides of (13) converge to φ(q(z)) and λφ(z), respectively, implying φ(q(z)) = λφ(z). (14) Since φn(0) = 0 for each φn we will also have φ(0) = 0. To see that the φn converge uniformly on D(0, r), let us rewrite it as φn(z) = zn λn = z0 · z1 λz0 · z2 λz1 · z3 λz2 · · · zn λzn−1 . By (11), the general term of the product becomes zk λzk−1 = q(zk−1) λzk−1 = λzk−1 + s(zk−1) λzk−1 = 1 + s(zk−1) λzk−1 . By the estimates (12) and (10) on \|zn\| we have s(zk−1) λzk−1 ≤b\|zk\| λ ≤bakr λ . (15) We will now make r smaller, if necessary, to ensure that the right hand side of (15) is less than 1 2. AROUND THE BOUNDARY OF COMPLEX DYNAMICS. 27 To show that the φn converge uniformly on D(0, r), it is suﬃcient to show that the inﬁnite product z1 λz0 · z2 λz1 · z3 λz2 · · · zn λzn−1 · · · does. Such a product converges if and only if logarithms of the ﬁnite partial products converge, i.e. if and only if the inﬁnite sum Logφ(z) = ∞ X k=1 Log 1 + s(zk−1) λzk−1 (16) converges. (We can take the logarithms on the right hand side of (16) because our bound of (15) by 1 2 implies that 1 + s(zk−1) λzk−1 ∈C \ (−∞, 0].) Using the estimate \|Log(1 + w)\| ≤2\|w\| for any \|w\| < 1 2 and (15) we see that the k-th term is geometrically small: Log 1 + s(zk−1) λzk−1 ≤2bakr λ . This proves convergence of (16) It remains to show that φ is conformal when restricted to a small enough neighborhood U ⊂D(0, r) of 0. By the chain rule, each φn satisﬁes φ′ n(0) = 1. Since the φn converge uniformly to φ in a neighborhood of 0, the Cauchy Integral Formula (6) implies that φ′ n(0) →φ′(0). Thus φ′(0) = 1. By the Inverse Function Theorem, there is a neighborhood V of 0 = φ(0) and an analytic function g : V →D(0, r) so that φ(g(w)) = w for every w ∈V . If we let U = g(V ), then φ : U →V is conformal. To obtain (9), precompose (14) with φ−1 = g on V . □ Extended Exercise 1. Adapt the proof of Kœnig’s Theorem to prove: B¨ ottcher’s Theorem. Suppose p(z) has a ﬁxed point z• of multiplier λ = 0 and thus is of the form p(z) = z• + ak(z −z•)k + ak+1(z −z•)k+1 + · · · + ad(z −z•)d for some 2 ≤k < d. Then, there exists a neighborhood U of z• and a conformal map φ : U →φ(U) ⊂C so that for any w ∈φ(U) we have φ ◦p ◦φ−1(w) = wk. 28 ROLAND K.W. ROEDER 2.2. First Exploration of the Parameter Space: The Set M0. Let us try to understand the space of parameters c ∈C for the quadratic poly-nomial maps pc(z) = z2 + c. Consider M0 := {c ∈C : pc(z) has an attracting periodic orbit}. We have already seen in Section 1.3 that c = i 4 and c = −1 are in M0 and that c = 1 2 is probably not in M0. We will now use the Attracting Periodic Orbit Lemma to ﬁnd some regions that are in M0. The ﬁxed points of pc(z) = z2 + c are z∗= 1 2 + √1 −4c 2 and z• = 1 2 − √1 −4c 2 and, since p′ c(z) = 2z, their multipliers are λ∗= 1 + √ 1 −4c and λ• = 1 − √ 1 −4c. If \|λ∗\| = 1, then 1 + √ 1 −4c = eiθ for some θ ∈R. Solving for c, we ﬁnd c = eiθ 2 −ei2θ 4 . The resulting curve C is a “Cardiod”, shown in Figure 13. Im(z) are repelling Re(z) pc has an attracting ﬁxed point both ﬁxed points of pc Figure 13. pc(z) = z2 + c has an attracting ﬁxed point if and only if c lies inside the Cardiod c = eiθ 2 −ei2θ 4 , where 0 ≤θ ≤2π, depicted here. AROUND THE BOUNDARY OF COMPLEX DYNAMICS. 29 In each of the two regions of C \ C we choose the points c = 0 and c = 1, which result in λ∗= 2 and λ∗= 1 + √ 3i, respectively. Thus, neither of the regions from C\C corresponds to parameters c for which z∗is an attracting ﬁxed point. Thus, we conclude that the smallest \|λ∗\| can be is 1, occurring exactly on the Cardiod C. Doing the same computations with the multiplier λ• of the second ﬁxed point z•, we also ﬁnd that \|λ•\| = 1 if and only if c is on the Cardiod C. However, at c = 0 we have λ• = 0 and at c = 1 we have λ• = 1 − √ 3i, which is of modulus greater than 1. Therefore, according to the Attracting Periodic Orbit Lemma, ﬁxed point z• is attracting if and only if c is inside of the Cardiod C. We summarize the past three paragraphs with: Lemma 1. pc(z) = z2 + c has an attracting ﬁxed point if and only if c lies inside the Cardiod curve C := n c = eiθ 2 −ei2θ 4 : 0 ≤θ ≤2π o . To ﬁnd periodic orbits of period two, we solve p◦2 c (z) = z2 + c 2 +c = z. In addition to the two ﬁxed points z∗and z•, we ﬁnd z0 = −1 2 + √−3 −4 c 2 and z1 = −1 2 − √−3 −4 c 2 . One can check that pc(z0) = z1 and pc(z1) = z0. These points are equal if c = −3 4, otherwise, they are indeed a periodic orbit of period 2. The multiplier of this periodic orbit is λ = −1 + √ −3 −4 c −1 − √ −3 −4 c = 4 + 4c, which has modulus 1 if and only if \|c + 1\| = 1 4. Since λ = 0 for c = −1 (inside the circle) and λ = 4 for c = 0 (outside the circle) we ﬁnd: Lemma 2. pc(z) = z2 + c has an periodic orbit of period 2 if and only if c lies inside the circle \|c + 1\| = 1 4. In Figure 14 we show the regions of M0 that we have discovered. Exercise 24. If possible, determine the region of parameters c for which pc(z) = z2 + c has an attracting periodic orbit of period 3. As n increases, this approach becomes impossible. We need a diﬀerent approach, which requires a deeper study of attracting periodic orbits. 2.3. Second Exploration of the Parameter Space: The Mandelbrot set M. Fatou-Julia Lemma. Let q be a polynomial of degree d ≥2. Then, the immediate basin of attraction for any attracting periodic orbit contains at least one critical point of q. In particular, since q has d −1 critical points (counted with multiplicity), q can have no more than d−1 distinct attracting periodic orbits. 30 ROLAND K.W. ROEDER ﬁxed point pc has an attracting period two orbit pc has an attracting Im(z) Re(z) Figure 14. The regions in the parameter plane where pc(z) = z2 + c has an attracting ﬁxed point and where pc has an attracting periodic orbit of period 2. Combined, they form a subset of M0. The following proof is illustrated in Figure 15. Proof. Replacing q by an iterate, we can suppose that the attracting pe-riodic orbit is a ﬁxed point z• of q. Performing a shift of coordinates, we suppose z• = 0. If 0 has multiplier λ = 0, then 0 is already a critical point in the imme-diate basin A0(0). We therefore suppose 0 has multiplier λ ̸= 0. By the Attracting Periodic Orbit Lemma, \|λ\| < 1. Suppose for contradiction that there is no critical point for q in A0(0). According to Exercise 10 there is some R > 0 so that any initial condition z0 with \|z0\| > R has orbit {zn} that escapes to ∞. In particular, A0(0) ⊂D(0, R). (17) We claim that q(A0(0)) = A0(0). Since A(0) is forward invariant, q(A0(0)) ⊂A(0). Because A0(0) is connected, so is q(A0(0)), which is therefore contained in one of the connected components of A(0). Since 0 = q(0) ∈q(A0(0)), we have q(A0(0)) ⊂A0(0). AROUND THE BOUNDARY OF COMPLEX DYNAMICS. 31 gn A0(0) 2r gn(D(0, 2r)) R r 0 Figure 15. Illustration of the proof of the Fatou-Julia Lemma. Conversely, suppose z∗∈A0(0). Let γ be a simple contour in A0(0) connecting z∗to 0 and avoiding any critical values of q. (By hypothesis, such critical values would be images of critical points that are not in A0(0).) Then, q−1(γ) is a union of several simple contours. Since q−1(0) = 0, one of them is a simple contour ending at 0. The other end is a point z#, which is therefore in A0(0). By construction q(z#) = z∗. To simplify notation, let f := q\|A0(0) : A0(0) →A0(0), which satisﬁes (1) f(A0(0)) = A0(0) and (2) f has no critical points. These properties persists under iteration, giving that fn(A0(0)) = A0(0) and fn has no critical points for every n ≥1. (The latter uses the Chain Rule.) Let r > 0 be chosen suﬃciently small so that D(0, 2r) ⊂A0(0). Since D(0, 2r) is simply connected, the Simply-Connected Inverse Function The-orem gives for each n ≥1 an analytic function gn : D(0, 2r) →A0(0) ⊂D(0, R) 32 ROLAND K.W. ROEDER with gn(0) = 0 and f◦n(gn(w)) = w for all w ∈(0, 2r). Its derivative satisﬁes g′ n(0) = 1 (fn)′(0) = 1 λn , (18) which can be made arbitrarily large by choosing n suﬃciently large, since \|λ\| < 1. Meanwhile, we can apply the Cauchy Estimates 1.5 to the closed disc D(0, r) ⊂D(0, 2r). They assert that \|g′ n(0)\| ≤R r , where R is the bound on the radius of A0(0) given in (17). This is a contradiction to (18). We conclude that the immediate basin A0(0) contains a critical point of q. □ Exercise 25. Use the Fatou-Julia Lemma and the result of Exercise 7 to (ﬁnally) prove that p(z) = z2 + 1 2 does not have any attracting periodic orbit. This answers Question 2 from Section 1.4 in the negative. Remark. The Fatou-Julia Lemma also answers our Question 3 from Sec-tion 1.4 by telling us that a quadratic polynomial can have at most one attracting periodic orbit. In Section 2.2 we were interested in the set M0 := {c ∈C : pc(z) has an attracting periodic orbit}. Using the Attracting Periodic Orbit Lemma to ﬁnd regions in the complex plane for which pc(z) = z2 + c has an attracting periodic point of period n became hopeless once n is large. The results for n = 1 and 2 are shown in Figure 14. If we decide to lose control over what period the attracting periodic point has, the Fatou-Julia Lemma gives us some very interesting information: Corollary. (Consequence of Fatou-Julia Lemma) If pc(z) has an at-tracting periodic orbit, then the orbit {p◦n c (0)} of the critical point 0 remains bounded. This motivates one to deﬁne another set: Deﬁnition 18. The Mandelbrot set is M := {c ∈C : p◦n c (0) remains bounded for all n ≥0}. (19) A computer image of the Mandelbrot set is depicted in Figure 16. One sees small “dots” at the left end and top and bottom of the ﬁgure. They are in M, but it is not at all clear if they are connected to the main cardiod and period two disc of M that are shown in Figure 14. If one looks closer, one AROUND THE BOUNDARY OF COMPLEX DYNAMICS. 33 sees many more such “dots”. In Section 2.3 we will use a smart coloring of C \ M to better understand this issue. We will then state a theorem of Douady and Hubbard, which clears up this mystery. The Mandelbrot set was initially discovered around 1980, but the histor-ical details are a bit controversial. We refer the reader to Appendix G from for an unbiased account. (The reader who seeks out controversy may enjoy .) Figure 16. The Mandelbrot set M, shown in black. The region shown is approximately −2.4 ≤Re(z) ≤1 and −1.6 ≤Im(z) ≤1.6. The corollary to the Fatou-Julia Lemma implies that M0 ⊂M. In other words, the Mandelbrot set is an “outer approximation” to our set M0. The reader should compare Figure 16 with Figure 14 for a better appreciation of much progress we’ve made! Exercise 26. Prove that M0 ̸= M by exhibiting a parameter c for which p◦n c (0) remains bounded but with pc having no attracting periodic orbit. Density of Hyperbolicity Conjecture. M0 = M. 34 ROLAND K.W. ROEDER Although this conjecture is currently unsolved, the corresponding result for real polynomials pc(x) = x2 + c with x, c ∈R was proved by Lyubich and by Graczyk-´ Swi¸ atek . Both proofs use complex techniques to solve the real problem. We have approached the deﬁnition of M from an unusual perspective, i.e. “from the inside out”. In the next section we will use the ﬁxed point at ∞for pc to study M again, but “from the outside in”. It is the more traditional way of introducing M. Lecture 3: “Complex Dynamics from the Outside In” Deﬁnition 19. The ﬁlled Julia set of pc(z) = z2 + c is Kc := {z ∈C : p◦n c (z) remains bounded for all n ≥0}. If pc has an attracting periodic orbit O, then the basin of attraction A(O) is contained in Kc. However, Kc is deﬁned for any c ∈C, even if pc has no attracting periodic orbit in C. There is a natural way to extend pc as a function pc : C ∪{∞} →C ∪{∞}. (More formally, the space C ∪{∞} is called the Riemann Sphere; see [43, Section 1.7].) This extension satisﬁes pc(∞) = ∞and, by your solution to Exercise 10, ∞always has a non-empty basin of attraction: A(∞) := {z ∈C : p◦n c (z) →∞} = C \ Kc. Thus, ∞is an attracting ﬁxed point of pc for any parameter c ∈C. In this way, the deﬁnition of Kc is always related to basin of attraction for an attracting ﬁxed point, even if pc has no attracting periodic point in C. A detailed study of A(∞) will help us to prove nice theorems later in this subsection. Remark. The Fatou-Julia Lemma still applies to the extended function pc : C ∪{∞} →C ∪{∞}. If you follow through the details of how this extension is done, you ﬁnd that ∞is a critical point of pc for every c. Deﬁnition 20. The Julia set of pc(z) = z2 +c is Jc := ∂Kc, the boundary of Kc. Exercise 27. Check that for any c ∈C the sets Kc and Jc are totally invariant meaning that z ∈Kc ⇔pc(z) ∈Kc and z ∈Jc ⇔pc(z) ∈Jc. Exercise 28. Use the Cauchy Estimates and invariance of Jc to prove that any repelling periodic point for pc is in Jc. Before drawing some computer images of Julia sets, it will be helpful to study A(∞) a bit more. AROUND THE BOUNDARY OF COMPLEX DYNAMICS. 35 Deﬁnition 21. A harmonic function h : C →R is a function with continuous second partial derivatives h(x + iy) ≡h(x, y) satisfying ∂2h ∂x2 + ∂2h ∂y2 = 0. One can use the Cauchy-Riemann Equations to verify that the real or imaginary part of an analytic function is harmonic and also that any har-monic function can be written (locally) as the real or imaginary part of some analytic function. Thus, there is a close parallel between the theory of analytic functions and of harmonic functions. We will only need two facts which follow directly from their analytic counterparts: Maximum Principle. Suppose h(z) is harmonic in a domain U and h(z) achieves its maximum or minimum at a point z0 ∈U. Then h(z) is constant on U. If, moreover, U is compact and h extends continuously to U, then h achieves its maximum and minimum on the boundary of U. Uniform Limits of Harmonic Functions. Suppose hk : U →R is a sequence of harmonic functions and h : U →R is some other function. If for any compact K ⊂U we have that {hk} converges uniformly to h on K, then h is harmonic on U. Moreover, any (repeated) partial derivative of hk converges uniformly to the corresponding partial derivative of h on any compact K ⊂U. Lemma 3. The following limit exists Gc(z) := lim n→∞ 1 2n log+ \|p◦n c (z)\| where log+(x) = max(log(x), 0) for any parameter c ∈C and any z ∈C. For each c the resulting function Gc : C →R is called the Green function associated to pc. It satisﬁes: (i) Gc is continuous on C and harmonic on A(∞), (ii) Gc(pc(z)) = 2Gc(z), (iii) G(z) ≈log \|z\| for \|z\| suﬃciently large, and (iv) G(z) = 0 iﬀz ∈Kc. The Green Function Gc is interpreted as describing the rate at which the orbit of initial condition z0 = z escapes to inﬁnity under iteration of pc. (The proof of Lemma 3 is quite similar to the proofs of Kœnig’s Theorem and B¨ ottcher’s Theorem, so we will omit it.) It is customary when drawing ﬁlled Julia sets on the computer to color A(∞) = C \ Kc according to the values of Gc(z). This is especially helpful for parameters c at which pc has no attracting periodic orbit. Using how the colors cycle one can “view” where Kc should be. In Figure 17 we show 36 ROLAND K.W. ROEDER the ﬁlled Julia sets for four diﬀerent values of c. (Among these is c = 1 2, from Example 3. We can now see where the bounded orbits are.) For the parameter values c = i 4, −1, and c ≈−0.12 + 0.75i, the ﬁlled Julia set is the closure of the basin of attracting periodic orbit. Thus, Figures 9-11 also depict the ﬁlled Julia sets for these parameter values. Remark. Like the ancient people who named the constellations, people doing complex dynamics also have active imaginations. They have named the ﬁlled Julia sets for c = −1 the “basilica” and the ﬁlled Julia set for c ≈−0.12 + 0.75i “Douady’s Rabbit”. The Green function also helps us to make better computer pictures of the Mandelbrot set. The value Gc(0) expresses the rate at which the critical point 0 of pc escapes to ∞under iteration of pc. Thus, points c with larger values of Gc(0) should be farther away from M. Therefore, it is customary to color C \ M according to the values of Gc(0), as in Figure 18. It is interesting to compare Figures 18 and 16. It now looks more plausible that the black “dots” in Figure 16 might be connected to the “main part” of M. The Green function is not only useful for making nice pictures. It also plays a key role in the proof of: Topological Characterization of the Mandelbrot Set. Kc is con-nected if and only if c ∈M. We illustrate this theorem with Figure 19. The reader may also enjoy comparing the parameter values shown in Figure 18 with their Filled Julia Sets shown in previous ﬁgures. According to the deﬁnition (19) of M, this statement is equivalent to Topological Characterization of the Mandelbrot Set’. Kc is con-nected if and only if the orbit {p◦n c (0)} of the critical point 0 of pc remains bounded. Although the Mandelbrot set was not deﬁned at the time of Fatou and Julia’s work (they lived from 1878-1929 and 1893-1978, respectively), the proof of the ’Topological Characterization of the Mandelbrot Set’ is due to them. Sketch of the proof: We will consistently identify C with R2 when taking partial derivatives and gradients of Gc : C →R. We claim that Gc(z) has a critical point at z0 ∈A(∞) if and only if p◦n(z0) = 0 for some n ≥0. Consider the ﬁnite approximates Gc,n(z) := 1 2n log+ \|p◦n c (z)\|, which one can check converge uniformly to Gc(z) on any compact subset of C. For points z ∈A(∞) we can drop the subscript + and use that log \|z\| AROUND THE BOUNDARY OF COMPLEX DYNAMICS. 37 c = −0.8 + 0.3i c ≈−0.92 + 0.25i c = i c = 1 2 Figure 17. Filled Julia sets for four values of c. The basin of attraction for ∞is colored according to the value of Gc(z). is diﬀerentiable on C \ {0}. Combined with the chain rule, we see that ∂ ∂xGc,n(z) = ∂ ∂yGc,n(z) = 0 if and only if (p◦n)′(z) = 0. This holds if and only if p◦m(z) = 0 for some 0 ≤m ≤n −1. The claim then follows from the Uniform Limits of Harmonic Functions Theorem. Suppose that the critical point 0 has bounded orbit under pc. Then, according to the previous paragraph, Gc has no critical points in A(∞). For any t > 0 let Lt := {z ∈C : Gc(z) ≤t}. 38 ROLAND K.W. ROEDER −0.8 + 0.3i −1 i i 4 1 2 −0.12 + 0.75i Figure 18. Mandelbrot set with the approximate locations of parameters c = i 4, −1, −0.12 + 0.75i, i, −0.8 + 0.3i, and 1 2 labeled. By deﬁnition, if t < s then Lt ⊂Ls. For each t > 0, Lt is closed and bounded since Gc : C →R is continuous and Gc(z) →∞as \|z\| →∞, respectively. Therefore, by the Heine-Borel Theorem, Lt is compact. Since Kc ̸= ∅and Gc(z) = 0 on Kc, each Lt is non-empty. Since z ∈Kc if and only if Gc(z) = 0, we can write Kc as a nested intersection of non-empty compact sets: Kc = \ n≥1 L1/n. If we can show that Lt is connected for each t > 0, then Exercise 12 will imply that Kc is connected. Since Gc(z) ≈log \|z\| for \|z\| suﬃciently large, there exists t0 > 0 suﬃ-ciently large so that Lt0 is connected (it is almost a closed disc of radius log t0). We will show that for any 0 < t1 < t0 the sets Lt1 and Lt0 are homeomorphic (i.e, there is a continuous bijection with continuous inverse from Lt1 to Lt0). Since Lt0 is connected, this will imply that Lt1 is also connected. AROUND THE BOUNDARY OF COMPLEX DYNAMICS. 39 Mandelbrot set Julia Sets Figure 19. Left: Zoomed-in view of the Mandelbrot set near the cusp at c = 1 4. Right: two ﬁlled Julia sets corre-sponding to points inside M and outside of M. The following is a standard construction from Morse Theory; see [36, Theorem 3.1]. Because Gc is harmonic and has no critical points on A(∞), −∇Gc is a non-vanishing smooth vector ﬁeld on A(∞). It is a relatively standard smoothing construction to deﬁne a new vector ﬁeld V : R2 →R2 that is smooth on all of C ≡R2 and equals −∇Gc ∥∇Gc∥2 for z ∈C \ Lt1/2. For any t ∈[0, ∞) let Φt : R2 →R2 denote the ﬂow obtained by inte-grating V . According to the existence and uniqueness theorem for ordinary diﬀerential equations (see, e.g., ), Φt : R2 →R2 is a homeomorphism for each t ∈[0, ∞). (We’re using that V “points inward” from ∞so that the solutions exist for all time.) For any z0 ∈C \ Lt1/2 and any 0 ≤t ≤Gc(z0) −t1/2 we have d dtGc(Φt(z0)) = ∇Gc(Φt(z0)) · d dtΦt(z0) = ∇Gc(Φt(z0)) · V (Φt(z0)) = ∇Gc(Φt(z0)) · −∇Gc(Φt(z0)) ∥∇Gc(Φt(z0))∥2 = −1. In particular, Φt0−t1(Lt0) = Lt1, implying that Lt0 is homeomorphic to Lt1. 40 ROLAND K.W. ROEDER Now suppose that 0 has unbounded orbit under pc. In this case, 0 and all of its iterated preimages under pc are critical points of Gc. Since pc has a simple critical point at 0, one can check that these critical points of Gc are all “simple” in that the Hessian matrix of second derivatives has non-zero determinant. Moreover, by the Maximum Principle, they cannot be local minima or local maxima. They are therefore saddle points. From the property Gc(p(z)) = 2Gc(z), the saddle point at z = 0 is the one with the largest value of Gc. There are two paths along which we can start at 0 and walk uphill in the steepest way possible—call them γ1 and γ2. Since 0 is the highest critical point, they lead all the way from 0 out to ∞. Together with 0, these two paths divide C into two domains U1 and U2. Meanwhile, there are two directions that one can walk downhill from a saddle point. Walking the fastest way downhill leads to two paths η1 and η2 which lead to points in U1 and in U2 along which Gc(z) < Gc(0). To make this idea rigorous, one considers the ﬂow associated to the vector ﬁeld −∇Gc. The saddle point 0 becomes a saddle type ﬁxed point for the ﬂow with the paths γ1 and γ2 being the stable manifold of this ﬁxed point. The paths η1 and η2 are the unstable manifolds of this ﬁxed point. (See again .) The union γ1 ∪γ2 ∪{0} divides the complex plane into two domains U1 and U2 with η1 ⊂U1 and η2 ⊂U2. We claim that both of these domains contain points of Kc. Suppose for contradiction that one of them (say U1) does not. Then, U1 ⊂A(∞) and hence Gc would be harmonic on U1. However, Gc(z) ∼log \|z\| for \|z\| large and Gc(z) > Gc(0) for points z ∈γ1 ∪γ2. Since Gc(z) < Gc(0) for points on η1, this would violate the Maximum Principle. □ Remark. A stronger statement actually holds: if Kc is disconnected, then it is a Cantor Set. (See for the deﬁnition of Cantor Set.) In particular, it is totally disconnected: for any z, w ∈Kc there exist open sets U, V ⊂C such that Kc ⊂U ∪V , z ∈U, w ∈V , and U ∩V = ∅. This follows from the fact that once Gc has the critical point 0 ∈A(∞) then it actually has inﬁnitely many critical points in A(∞). These additional critical points of Gc are the iterated preimages of 0 under pc. For a somewhat diﬀerent proof from the one presented above, including a proof of this stronger statement, see [11, 12]. Exercise 29. Prove that if c ̸= 0 then log \|z2 + c\| has a saddle-type critical point at z = 0. AROUND THE BOUNDARY OF COMPLEX DYNAMICS. 41 γ1 γ2 η1 0 η2 U1 U2 Figure 20. Stable and unstable trajectories of −∇Gc for the critical point 0. Hint: Write z = x + iy and c = a + bi and use that log \|z2 + c\| = 1 2 log z2 + c z2 + c . We will now state (without proofs) several interesting properties of the Mandelbrot set: Theorem. (Douady-Hubbard ) The Mandelbrot set M is connected. (Nessim Sibony gave an alternate proof around the same time.) This theo-rem clears up the mystery about the black “dots” in Figure 16. The following very challenging extended exercise leads the reader through a proof that M is connected that is related to the coloring of C\M according to the value of Gc(0). (It will be somewhat more convenient to consider Gc(c) = Gc(pc(0)) = 2Gc(0).) Extended Exercise 2. Let H : C →R be given by H(c) = Gc(c). Prove that (1) H is continuous, (2) H is harmonic on C \ M, (3) H is identically 0 on M, (4) lim\|c\|→∞H(c) = ∞, and (5) H has no critical points in C \ M. (Step 5 is the hardest part.) Use these facts to adapt the proof of the topo-logical characterization of the Mandelbrot Set to prove that M is connected. 42 ROLAND K.W. ROEDER HausdorﬀDimension extends the classical notion of dimension from lines and planes to more general metric spaces. As the formal deﬁnition is a bit complicated, we instead illustrate the notion with a few examples. A line has Hausdorﬀdimension equal to 1 and the plane has Hausdorﬀdimension equal to 2. A contour has Hausdorﬀdimension equal to 1 because, if you zoom in suﬃciently far near any of the smooth points, the contour appears more and more like a straight line. However, sets of a “fractal nature” can have non-integer HausdorﬀDimension. One example is the Koch Curve, which a simple closed curve in the plane that is obtained as the limit of the iterative process shown in Figure 21. No matter how far you zoom in, the Koch Curve looks the same as a larger copy of itself, and not like a line! This results in the Koch Curve having Hausdorﬀdimension equal to log(4)/ log(3) ≈1.26. We refer the reader to for a gentle introduction to HausdorﬀDimension. Figure 21. The Koch Curve. If S ⊂C contains an open subset of C, then it is easy to see that it has HausdorﬀDimension equal to 2. It is much harder to imagine a subset of C that contains no such open set having HausdorﬀDimension 2. Therefore, the following theorem shows that the boundary ∂M of the Mandelbrot set M has amazing complexity. It also shows that for many parameters c from ∂M the Julia set Jc has amazing complexity. Theorem. (Shishikura ) The boundary of the Mandelbrot set ∂M has Hausdorﬀdimension equal to 2. Moreover, for a dense set of parameters c from the boundary of M the Julia set Jc has HausdorﬀDimension equal to 2. Another interesting property of the Mandelbrot set is the appearance of “small copies” within itself. (Some of these were the “dots” from Figure 16.) Figure 22 shows a zoomed in view of M, where several small copies of M are visible. These copies are explained by the renormalization theory [15, 34]. It would be remiss to not include one of the most famous conjectures about the Mandelbrot set. We ﬁrst need AROUND THE BOUNDARY OF COMPLEX DYNAMICS. 43 Figure 22. Zoomed-in view of part of the Mandelbrot set showing two smaller copies. The approximate location where we have zoomed in is marked by the tip of the arrow in the inset ﬁgure. Deﬁnition 22. A topological space X is locally connected if for every point x ∈X and any open set V ⊂X that contains x there is another connected open set U with x ∈U ⊂V . MLC Conjecture. The Mandelbrot set M is locally connected. According to the Orsay Notes of Douady and Hubbard, if this were the case, then one could have a very nice combinatorial description of M. Given a proposed way that pc acts on the Julia set Jc (described by means of the so called Hubbard Tree) one can use this combinatorial description of M to ﬁnd the desired value of c. To better appreciate the diﬃculty in proving the MLC Conjecture, we include one more zoomed-in image of the Mandelbrot set in Figure 23. 44 ROLAND K.W. ROEDER Figure 23. Another zoomed-in view of part of the Mandel-brot set. Let us ﬁnish the section, and our discussion of iterating quadratic poly-nomials, by returning to mathematics that can be done by undergraduates. The reader is now ready to answer Question 4 from Section 1.4: Extended Exercise 3. Prove that for every m ≥1 there exists a pa-rameter c ∈C such that pc(z) has an attracting periodic orbit of period exactly m. Hint: prove that there is a parameter c such that p◦m c (0) = 0 and p◦j c (0) ̸= 0 for each 0 ≤j < m. Lecture 4: “Complex dynamics and astrophysics.” Most of the results discussed in Sections 1-3 of these notes are now quite classical. Let us ﬁnish our lectures with a beautiful and quite modern application of the Fatou-Julia Lemma to a problem in astrophysics [26, 24]. We also mention that there are connections between complex dynamics and the Ising model from statistical physics (see [7, 6] and the references therein) and the study of droplets in a Coulomb gas [30, 29]. AROUND THE BOUNDARY OF COMPLEX DYNAMICS. 45 4.1. Gravitational Lensing. Einstein’s Theory of General Relativity pre-dicts that if a point mass is placed directly between an observer and a light source, then the observer will see a ring of light, called an “Einstein Ring”. The Hubble Space Telescope has suﬃcient power to see these rings—one such image is shown in Figure 24. If the point mass is moved slightly, the observer will see two diﬀerent images of the same light source. With more complicated distributions of mass, like n point masses, the observer can see more complicated images, resulting from a single point light source. Such an image is shown in Figure 25. (Thanks to NASA for these images and their interpretations.) There are many excellent surveys on gravitational lensing that are writ-ten for the mathematically inclined reader, including [25, 39, 48], as well as the book . We will be far more brief, with the goal of this lecture being to explain how Rhie and Khavinson-Neumann answered the question: What is the maximum number of images that a single light source can have when lensed by n point masses? We will tell some of the history of how this problem was solved and then focus on the role played by the Fatou-Julia Lemma. Figure 24. An Einstein Ring. For more information, see 46 ROLAND K.W. ROEDER Figure 25. Five images of the same quasar (boxed) and three images of the same galaxy (circled). The middle image of the quasar (boxed) is behind the small galaxy that does the lensing. For more information, see image_feature_575.html Suppose that n point masses lie on a plane that is nearly perpendicular to the line of sight between the observer and the light source and that they lie relatively close to the line of sight. If we describe their positions relative to the line of sight to the light source by complex numbers zj and their normalized masses by σj > 0 for 1 ≤j ≤n, then the images of the light source seen by the observer are given by solutions z to the Lens Equation: (20) z = n X j=1 σj z −zj . The “mysterious” appearance of complex conjugates on the right hand side of this equation makes it diﬃcult to study. It will be explained in Sec-tion 4.3, where we derive (20). Exercise 30. Verify that (20) gives a full circle of solutions (Einstein Ring) when there is just one mass at z1 = 0. Then, verify that when z1 ̸= 0 there are two solutions. Can you ﬁnd a conﬁguration of two masses so that (20) has ﬁve solutions? AROUND THE BOUNDARY OF COMPLEX DYNAMICS. 47 Remark. Techniques from complex analysis extend nicely to lensing by mass distributions more complicated than ﬁnitely many points, including elliptical and spiral galaxies. The right hand side of (20) is of the form r(z), where r(z) is a rational function r(z) = p(z) q(z) of degree n. (The degree of a rational function is the maximum of the degrees of its numerator and denominator.) Thus, our physical question becomes the problem of bounding the number of solutions to an equation of the form z = r(z) (21) in terms of n = deg(r(z)). Sadly, the Fundamental Theorem of Algebra cannot be applied to zq(z) −p(z) = 0 (22) because the resulting equation is a polynomial in both z and z. If one writes z = x+iy with x, y ∈R, one can change (22) to a system of two real polynomial equations a(x, y) := Re z q(z) −p(z) = 0 and b(x, y) := Im z q(z) −p(z) = 0, each of which has degree n + 1. So long as there are no curves of common zeros for a(x, y) and b(x, y), Bezout’s Theorem (see, e.g., ) gives a bound on the number of solutions by (n + 1)2. In 1997, Mao, Petters, and Witt exhibited conﬁgurations of n point masses at the vertices of a regular polygon in such a way that 3n+1 solutions were found. They conjectured a linear bound for the number of solutions to (20). For large n this would be signiﬁcantly better than the bound given by Bezout’s Theorem. In 2003, Rhie showed that if one takes the conﬁguration of masses considered by Mao, Petters, and Witt and places a suﬃciently small mass centered at the origin, then one ﬁnds 5n−5 solutions to (20). (We refer the reader also to [8, Section 5] for an another exposition on Rhie’s examples.) In order to address a problem on harmonic mappings C →C posed by Wilmshurst in , in 2003 Khavinson and ´ Swi¸ atek studied the number of solutions to z = p(z) where p(z) is a complex polynomial. They proved Theorem. (Khavinson-´ Swi¸ atek ) Let p(z) be a complex polynomial of degree n ≥2. Then, z = p(z) has at most 3n −2 solutions. Khavinson and Neumann later adapted the techniques from to prove 48 ROLAND K.W. ROEDER Theorem. (Khavinson-Neumann ) Let r(z) be a rational function of degree n ≥2. Then, z = r(z) has at most 5n −5 solutions. Apparently, Khavinson and Neumann solved this problem because of its mathematical interest and only later were informed that they had actually completed the solution to our main question of this lecture: When lensed by n point masses, a single light source can have at most 5n−5 images. Remark. Geyer used a powerful theorem of Thurston to show that for every n ≥2 there is a polynomial p(z) for which z = p(z) has 3n −2 solutions, thus showing that Theorem 4.1 is sharp. It would be interesting to see an “elementary” proof. 4.2. Sketching the proof of the 5n−5 bound. We provide a brief sketch of the proof of the Khavinson-Neumann upper bound in the special case that r(z) = n X j=1 σj z −zj , (23) with each σj > 0. It is the case arising in the Lens Equation (20). The locations {z1, . . . , zn} of the masses are called poles of r(z). They satisfy limz→zj \|r(z)\| = ∞for any 1 ≤j ≤n. The function f : C \ {z1, . . . , zn} →C given by f(z) = z −r(z) (24) is an example of a harmonic mapping with poles since its real and imaginary parts are harmonic. It is orientation preserving near a point z• with \|r′(z•)\| < 1 and orientation reversing (like a reﬂection z 7→z) near points with \|r′(z•)\| > 1. A zero z• of f is simple if \|r′(z•)\| ̸= 1 and a simple zero is called sense preserving if \|r′(z•)\| < 1 and sense-reversing if \|r′(z•)\| > 1. Step 1: Reduction to Simple Zeros. Suppose r(z) is of the form (23) and f(z) = z−r(z) has k zeros, some of which are not simple. Then, one can show that there is an arbitrarily small perturbation of the locations of the masses so that the resulting rational function s(z) produces g(z) = z −s(z) having at least as many zeros as f(z) all of which are simple. Therefore, it suﬃces to consider rational functions r(z) of the form (23) such that each zero of f(z) = z −r(z) is simple. AROUND THE BOUNDARY OF COMPLEX DYNAMICS. 49 Step 2: Argument Principle for Harmonic Mappings. Suﬀridge and Thompson adapted the Argument Principle to harmonic mappings with poles f : C \ {z1, . . . , zk} →C. Since r(z) has the form (23), limz→∞\|r(z)\| = 0. Therefore, we can choose R > 0 suﬃciently large so that all of the poles of r(z) lie in D(0, R) and the change of argument for f(z) while traversing γ = ∂D(0, R) counter clockwise is 1. This variant of the argument principle then gives (m+ −m−) + n = 1 (25) where m+ is the number of sense preserving zeros, m−is the number of sense reversing zeros, and n is the number of poles. (We are using that all of the zeros and poles are simple so that they do not need to be counted with multiplicities.) Step 3: Fatou-Julia Lemma Bound on m+: Zeros of f(z) correspond to ﬁxed points for the anti-analytic mapping z 7→r(z). Moreover, sense preserving zeros correspond to attracting ﬁxed points (those with \|r′(z•)\| < 1). Since the coeﬃcients of r(z) are real, taking the second iterate yields Q(z) = r r(z) = r(r(z)), which is an analytic rational mapping of degree n2. Such a mapping has 2n2 −2 critical points and an adaptation of the Fatou-Julia Lemma implies that each attracting ﬁxed point of Q attracts a critical point. However, the chain rule gives that critical points of Q(z) = r(r(z)) are the critical points of r(z) and their inverse images under r(z). Since a generic point has n inverse images under r, this can be used to show that each attracting ﬁxed point of Q(z) actually attracts n + 1 critical points of Q. Therefore, Q(z) has at most 2n −2 attracting ﬁxed points. Since any sense preserving zero for f(z) is an attracting ﬁxed point for Q, we conclude that m+ ≤2n −2. Step 4: Completing the proof: Since m+ ≤2n −2, Equation (25) implies m−≤3n −3. Therefore, the total number of zeros is m+ + m−≤5n −5. The reader is encouraged to see for the full details, including how to prove the bound for general rational functions r(z). 50 ROLAND K.W. ROEDER PL S I DLS DL DS θ β O ˜ α α PS L ξ Figure 26. S is the light source, I is an image, O is the observer, L is a point mass, PL is the lens plane, PS is the source plane. 4.3. Derivation of the Lens Equation. This derivation is a synthesis of ideas from and [41, Section 3.1] that was written jointly with Ble-her, Homma, and Ji when preparing . Since it was not included in the published version of , we present it here. We will ﬁrst derive the Lens Equation for one point mass using Figure 26, and then adapt it to N point masses. Suppose the observer is located at point O, the light source at a point S, and a mass M at point L. Also, suppose PL is the plane perpendicular to OL that contains L, and PS is the plane perpendicular to OL that contains S. Due to the point mass, an image, I, will be created at angle α with respect to S. Einstein derived using General Relativity that the bending angle is ˜ α = 4GM c2ξ , (26) where G is the universal gravitational constant and c is the speed of light, see . The observer O describes the location S of the light source using an angle β and the perceived location I using another angle θ (see Figure 26). By a small angle approximation, ξ = DLθ, which we substitute into (26) AROUND THE BOUNDARY OF COMPLEX DYNAMICS. 51 obtaining ˜ α(θ) = 4GM c2DLθ. (27) A small angle approximation also gives that DSI = DLS ˜ α(θ) = DSα(θ). Substituting this into β = θ −α(θ) gives (28) β = θ −DLS DSDL · 4GM c2θ . For β ̸= 0, exactly two images are produced. When β = 0, the system is rotationally symmetric about OL, thereby producing an Einstein Ring, whose angular radius is given by Equation (28). In order to describe systems of two or more masses, we need to describe locations in the source plane PS and the lens plane PL using two-dimensional vectors of angles (polar and azimuthal angles) as observed from O. Complex numbers will be a good way to do this: α = α(1) + iα(2), ˜ α = ˜ α(1) + i˜ α(2), β = β(1) + iβ(2), and θ = θ(1) + iθ(2). When there is only one mass, the whole conﬁguration must still lie in one plane, as in Figure 26. In particular, all four complex numbers have the same argument, forcing us to replace the θ on the right hand side of (27) with θ: ˜ α(θ) = 4GM c2DLθ and hence β = θ −DLS DSDL · 4GM c2θ . (29) This is why the complex conjugate arises in the Lens Equation (20). We now generalize to n point masses. Let L be the center of mass of the n masses, and redeﬁne SL as the plane that is perpendicular to OL and contains L. We assume that the distance between L and the individual point masses is extremely small with respect to the pairwise distances between O, PL, and PS. Now consider the projection of the n point masses onto SL. We continue to let β ∈C describe the location of the center of mass and we describe the location of the jth point mass by ϵj = ϵ(1) j + iϵ(2) j ∈C. It has mass Mj. In general, the bending angle is expressed as an integral expressed linearly in terms of the mass distribution, see [41, Equation 3.57]. In particular, with point masses, the bending angle decomposes to a sum of bending angles, one for each point mass. Each is computed as in the one mass system: ˜ αj = 4GMj c2DLθj where θj = θ −ϵj. 52 ROLAND K.W. ROEDER We obtain β = θ − n X j=1 αj = θ −DLS DSDL n X j=1 4GMj c2 θ −ϵj . Letting w = β, z = θ, zj = ϵj, and σj = DLS DSDL · 4GMj c2 gives (30) w = z − n X j=1 σj z −zj . Equation (30) requires the assumption that the center of mass is the origin, i.e. P σjzj = 0. A translation by w allows us to ﬁx the position of the light source at the origin and vary the location of the center of mass. This simpliﬁes Equation (30) to Equation (20). 4.4. Wilmshurst’s Conjecture. Let us ﬁnish our notes with an open problem that can be explored by undergraduates. In , Wilmshurst con-sidered equations of the form p(z) = q(z) (31) where p(z) and q(z) are polynomials of degree n and m, respectively. By conjugating the equation, if necessary, one may suppose n ≥m. If m = n, then one can have inﬁnitely many solutions (e.g. p(z) = zn = q(z)), but once n > m Wilmshurst showed that there are ﬁnitely many solutions. He conjectured that the number of solutions to (31) is at most 3n−2+m(m−1). Unfortunately, this conjecture is false! Counterexamples were found when m = n −3 by Lee, Lerario, and Lundberg . They propose: Conjecture. (Lee, Lerario, Lundberg) If deg(p(z)) = n, deg(q(z)) = m, and n > m, then the number of solutions to p(z) = q(z) is bounded by 2m(n −1) + n. Note that this conjectured bound is not intended to be sharp. For example, Wilmshurst proved his conjecture in the case that m = n −1, providing a stronger bound in that case . This problem was further studied using certiﬁed numerics by Hauenstein, Lerario, Lundberg, and Mehta . Their work provides further evidence for this conjecture. Question. Can techniques from complex dynamics be used to prove this conjecture? AROUND THE BOUNDARY OF COMPLEX DYNAMICS. 53 References Lars V. Ahlfors. Complex analysis. McGraw-Hill Book Co., New York, third edition, 1978. An introduction to the theory of analytic functions of one complex variable, International Series in Pure and Applied Mathematics. H. Asada. Perturbation theory of n point mass gravitational lens. Monthly Notices of the Royal Astronomical Society, 394(2):818–830. Alan F. Beardon. Iteration of rational functions, volume 132 of Graduate Texts in Mathematics. Springer-Verlag, New York, 1991. Complex analytic dynamical sys-tems. Steven R. Bell, Brett Ernst, Sean Fancher, Charles R. Keeton, Abi Komanduru, and Erik Lundberg. Spiral galaxy lensing: a model with twist. Math. Phys. Anal. Geom., 17(3-4):305–322, 2014. Paul Blanchard. Complex analytic dynamics on the Riemann sphere. Bull. Amer. Math. Soc. (N.S.), 11(1):85–141, 1984. Pavel Bleher, Mikhail Lyubich, and Roland Roeder. Lee-Yang-Fisher zeros for DHL and 2D Rational Dynamics, II. Global Pluripotential Interpretation. Preprint availi-ble at: Pavel Bleher, Mikhail Lyubich, and Roland Roeder. Lee-Yang zeros for DHL and 2D Rational Dynamics, I. Foliation of the Physical Cylinder. Preprint availible at: Pavel M. Bleher, Youkow Homma, Lyndon L. Ji, and Roland K. W. Roeder. Counting zeros of harmonic rational functions and its application to gravitational lensing. Int. Math. Res. Not. IMRN, (8):2245–2264, 2014. Boston University Java Applets for Dynamical Systems. DYSYS/applets/ Lennart Carleson and Theodore W. Gamelin. Complex dynamics. Universitext: Tracts in Mathematics. Springer-Verlag, New York, 1993. Robert L. Devaney. A ﬁrst course in chaotic dynamical systems. Addison-Wesley Studies in Nonlinearity. Addison-Wesley Publishing Company, Advanced Book Pro-gram, Reading, MA, 1992. Theory and experiment, With a separately available computer disk. Robert L. Devaney. An introduction to chaotic dynamical systems. Studies in Non-linearity. Westview Press, Boulder, CO, 2003. Reprint of the second (1989) edition. A. Douady and J. H. Hubbard. ´ Etude dynamique des polynˆ omes complexes. Partie I, volume 84 of Publications Math´ ematiques d’Orsay [Mathematical Publications of Orsay]. Universit´ e de Paris-Sud, D´ epartement de Math´ ematiques, Orsay, 1984. Adrien Douady and John Hamal Hubbard. It´ eration des polynˆ omes quadratiques complexes. C. R. Acad. Sci. Paris S´ er. I Math., 294(3):123–126, 1982. Adrien Douady and John Hamal Hubbard. On the dynamics of polynomial-like map-pings. Ann. Sci. ´ Ecole Norm. Sup. (4), 18(2):287–343, 1985. Dynamics Explorer software. Written by Brian and Suzanne Boyd. http:// sourceforge.net/projects/detool/ Gerald A. Edgar. Measure, topology, and fractal geometry. Undergraduate Texts in Mathematics. Springer-Verlag, New York, 1990. C. D. Fassnacht, C. R. Keeton, and D. Khavinson. Gravitational lensing by elliptical galaxies, and the Schwarz function. In Analysis and mathematical physics, Trends Math., pages 115–129. Birkh¨ auser, Basel, 2009. Fractalstream dynamical systems software. Written by Matthew Noonan. https: //code.google.com/p/fractalstream/downloads/list. 54 ROLAND K.W. ROEDER Lukas Geyer. Sharp bounds for the valence of certain harmonic polynomials. Proc. Amer. Math. Soc., 136(2):549–555, 2008. Jacek Graczyk and Grzegorz ´ Swiatek. Generic hyperbolicity in the logistic family. Ann. of Math. (2), 146(1):1–52, 1997. Jonathan Hauenstein, Antonio Lerario, Erik Lundberg, and Dhagash Mehta. Exper-iments on the zeros of harmonic polynomials using certiﬁed counting. To appear in Experimental Mathematics. See also John Horgan. Mandelbrot set-to: Did the father of fractals “discover” his namesake set? Scientiﬁc American, 262:30–34. Dmitry Khavinson and Genevra Neumann. On the number of zeros of certain rational harmonic functions. Proc. Amer. Math. Soc., 134(4):1077–1085 (electronic), 2006. Dmitry Khavinson and Genevra Neumann. From the fundamental theorem of algebra to astrophysics: a “harmonious” path. Notices Amer. Math. Soc., 55(6):666–675, 2008. Dmitry Khavinson and Grzegorz ´ Swi¸ atek. On the number of zeros of certain har-monic polynomials. Proc. Amer. Math. Soc., 131(2):409–414, 2003. Frances Kirwan. Complex algebraic curves, volume 23 of London Mathematical So-ciety Student Texts. Cambridge University Press, Cambridge, 1992. Seung-Yeop Lee, Antonio Lerario, and Lundberg Erik. Remarks on wilmshurst’s theorem. To appear in the Indiana University Math Journal. See also org/pdf/1308.6474.pdf. Seung-Yeop Lee and Nikolai Makarov. Sharpness of connectivity bounds for quadra-ture domains. Preprint: Seung-Yeop Lee and Nikolai Makarov. Topology of quadrature domains. To appear in the Journal of the AMS; see also Mikhail Lyubich. Six Lectures on Real and Complex Dynamics (based on European Lectures given in Barcelona, Copenhagen and St Petersburg in May-June 1999). See Mikhail Lyubich. Dynamics of rational transformations: topological picture. Uspekhi Mat. Nauk, 41(4(250)):35–95, 239, 1986. Mikhail Lyubich. Dynamics of quadratic polynomials. I, II. Acta Math., 178(2):185– 247, 247–297, 1997. Mikhail Lyubich. Baby Mandelbrot sets, renormalization and MLC. Gaz. Math., (113):45–50, 2007. S. Mao, A. O. Petters, and H. J. Witt. Properties of point mass lenses on a regular polygon and the problem of maximum number of images. In The Eighth Marcel Grossmann Meeting, Part A, B (Jerusalem, 1997), pages 1494–1496. World Sci. Publ., River Edge, NJ, 1999. J. Milnor. Morse theory. Based on lecture notes by M. Spivak and R. Wells. Annals of Mathematics Studies, No. 51. Princeton University Press, Princeton, N.J., 1963. John Milnor. Dynamics in one complex variable, volume 160 of Annals of Mathe-matics Studies. Princeton University Press, Princeton, NJ, third edition, 2006. Lawrence Perko. Diﬀerential equations and dynamical systems, volume 7. Springer Science & Business Media, 2001. A. O. Petters. Gravity’s action on light. Notices Amer. Math. Soc., 57(11):1392–1409, 2010. Arlie O. Petters, Harold Levine, and Joachim Wambsganss. Singularity theory and gravitational lensing, volume 21 of Progress in Mathematical Physics. Birkh¨ auser Boston, Inc., Boston, MA, 2001. With a foreword by David Spergel. AROUND THE BOUNDARY OF COMPLEX DYNAMICS. 55 Arlie O. Petters, Harold Levine, and Joachim Wambsganss. Singularity theory and gravitational lensing, volume 21 of Progress in Mathematical Physics. Birkh¨ auser Boston, Inc., Boston, MA, 2001. With a foreword by David Spergel. S. H. Rhie. n-point gravitational lenses with 5(n −1) images. Preprint. arXiv: astro-ph/0305166. E.B. Saﬀand A.D. Snider. Fundamentals of Complex Analysis with Applications to Engineering and Science, Third Edition. Prentice Hall, 2003. Zhaiming Shen and Lasse Rempe-Gillen. The exponential map is chaotic: An invi-tation to transcendental dynamics. To appear in American Mathematical Monthly. See also Mitsuhiro Shishikura. The Hausdorﬀdimension of the boundary of the Mandelbrot set and Julia sets. Ann. of Math. (2), 147(2):225–267, 1998. Norbert Steinmetz. Rational iteration, volume 16 of de Gruyter Studies in Mathe-matics. Walter de Gruyter & Co., Berlin, 1993. Complex analytic dynamical systems. James Stewart. Calculus, 7th Edition. Cengage Learning, 2012. Norbert Straumann. Complex Formulation of Lensing Theory and Applications. Hel-vetica Physica Acta., 70(6):894-908, 1997. T. J. Suﬀridge and J. W. Thompson. Local behavior of harmonic mappings. Complex Variables Theory Appl., 41(1):63–80, 2000. A. S. Wilmshurst. The valence of harmonic polynomials. Proc. Amer. Math. Soc., 126(7):2077–2081, 1998. Index analytic, 13–20, 24, 27, 31, 35, 49 anti-analytic, 13, 49 Argument Principle, 18, 49 Attracting Periodic Orbit Lemma, 21, 22, 24, 28–30, 32 B¨ ottcher’s Theorem, 23, 27 Bezout’s Theorem, 47 boundary, 11, 17, 18, 34, 35, 42 Cauchy Estimates, 17, 32, 34 Cauchy Integral Formula, 16–18, 27 Cauchy Integral Formula For Higher Derivatives, 17, 19 Cauchy’s Theorem, 15, 16, 19 Cauchy-Riemann Equations, 13, 14, 16, 35 closed disc, 11, 19, 32, 38 compact, 12, 18, 19, 35, 36, 38 complex polynomial, 2, 3, 13, 47 computer software, 2, 23 conformal, 13, 23, 27 connected, 11, 12, 22, 30, 32, 36, 38, 41, 43 locally connected, 43 continuous, 13, 15, 16, 35, 38 contour, 11, 12, 15–19, 31, 42 closed, 12 continuously deformed, 12 positively oriented, 12 simple, 12 critical point, 14, 29–32, 36, 37, 39–41, 49 critical value, 14, 31 Density of Hyperbolicity Conjecture, 33 diﬀerentiable, 13, 17, 19, 37 disconnected, 11, 40 domain, 11–14, 17–19, 35, 40 simply connected, 12, 14–17, 19, 31 Douady, 1, 33, 36, 41, 43 Fatou-Julia Lemma, 29, 31, 32, 44, 45, 49 ﬁlled Julia set, 34–36, 39 ﬁxed point, 7, 8, 21–24, 26–30, 34, 40, 49 Fundamental Theorem of Algebra, 3, 47 Graczyk, 34 Green function, 35, 36 harmonic function, 35 harmonic mapping, 47–49 Hauenstein, 52 HausdorﬀDimension, 42 Heine-Borel Theorem, 12, 38 holomorphic, 13 Hubbard, 1, 33, 41, 43 initial condition, 4–11, 20–22, 30 Inverse Function Theorem, 14, 27 Julia set, 34, 42, 43 Kœnig’s Theorem, 23 Khavinson, 45, 47, 48 Lee, 52 Lens Equation, 46, 48, 50, 51 Lerario, 52 Lundberg, 52 Lyubich, 1, 34 Mandelbrot set, 20, 29, 32, 33, 36, 38, 39, 41–44 Maximum Modulus Principle, 18, 20 Maximum Principle, 35, 40 Mehta, 52 MLC Conjecture, 43 neighborhood, 11, 13, 14, 23, 27 Neumann, 45, 47, 48 orbit, 4–11, 20–22, 26, 29 Orsay Notes, 1, 43 parameter, 4, 6, 10, 22, 28–30, 33, 35, 36, 38, 42, 44 periodic orbit, 20, 21 attracting, 10, 20–26, 28–30, 32–36, 44, 49 basin of attraction, 22 immediate basin, 22 multiplier, 21–24, 27–30 periodic points, 21, 29 Rhie, 45, 47 ´ Swi¸ atek, 34, 47 56 AROUND THE BOUNDARY OF COMPLEX DYNAMICS. 57 Schwarz Lemma, 19, 21 shift of coordinates, 21, 24, 30 Shishikura, 42 Simply-Connected Inverse Function Theorem, 14, 31 Topological Characterization of the Mandelbrot Set, 36 Uniform Limits of Harmonic Functions, 35, 37 Uniform Limits Theorem, 19, 26 Wilmshurst’s Conjecture, 52
15954	https://www.bosterbio.com/blog/post/a-guide-to-10-blotting-types?srsltid=AfmBOoo9EOYp4dPnhMz8LK3ftiD6e7onolJmOwlqefN-F4STevbh0g9y	Blotting Techniques Guide: Top 10 Methods Explained \| Boster Bio This website uses cookies to ensure you get the best experience on our website. Got it! 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These techniques enable scientists to uncover intricate details about genetic material, gene expression, and protein interactions. From the foundational Southern blotting, which revolutionized DNA analysis, to the versatile Western blotting, a staple in protein research, blotting methods have become indispensable tools in the laboratory. This blog will delve into 10 blotting types, each with unique applications and methodologies. We'll explore the nuances of each technique, highlighting their roles in studying DNA, RNA, and proteins. If you’re looking to learn about the different types of blotting techniques for your research, this guide is for you! 1. Southern blotting Southern blotting, the original blotting technique, is named after its inventor Edwin Southern who developed the method in 1975 for transferring DNA from a gel to a membrane, enabling the identification of specific DNA sequences. The naming convention for subsequent blotting techniques was influenced by this original name. To this day, Southern blotting remains a common technique for DNA analysis and identification. The process involves digesting DNA with restriction enzymes, separating the fragments by gel electrophoresis, and transferring them onto a membrane. Labeled DNA or RNA probes then hybridize to the target sequences. Popular in genetics and molecular biology, Southern blotting is used for the detection of specific DNA sequences, study of DNA methylation patterns, RFLP (Restriction Fragment Length Polymorphism) analysis for genetic fingerprinting, identification of gene mutations and polymorphisms, and gene mapping and cloning. 2. Northern blotting Northern blotting was named by James Alwine, David Kemp, and George Stark in a 1977 paper as a playful pun on Southern blotting, indicating its use for RNA rather than DNA. This technique is similar to Southern blotting, but focuses on RNA analysis. RNA samples are separated by gel electrophoresis, transferred to a membrane, and hybridized with labeled DNA or RNA probes to detect and quantify specific RNA transcripts. Northern blotting is commonly used to analyze gene expression patterns, determine mRNA size and abundance, study RNA processing and degradation, and detect alternative splicing events. To quantify RNA expression levels beyond blotting techniques, a solid qpcr service provides a highly sensitive and accurate alternative for transcript analysis. 3. Western blotting Western blotting, also called immunoblotting, is widely used for protein detection and analysis, making it the most popular blotting technique. For streamlined and reproducible results, many researchers turn to a dedicated Western Blotting Service to handle the full workflow with optimized protocols. Proper western blot sample preparation is essential to ensure reliable and reproducible protein detection throughout the process. Understanding the western blot principle is key to effectively applying this method and interpreting its results accurately. The method was first described by Towbin et al. in 1979 and the term "western" was coined by W. Neal Burnette in 1981 as a tongue-in-cheek reference to the direction naming theme established by Southern and Northern blotting, this time focusing on proteins instead of nucleic acids. This technique detects specific proteins by separating them via gel electrophoresis (typically SDS-PAGE), transferring them to a membrane (usually PVDF or nitrocellulose: AR0135-02, AR0135-04), and using primary and secondary antibodies for detection. The secondary antibody is usually conjugated to an enzyme or a fluorescent tag that produces a detectable signal when exposed to a substrate or under specific conditions. For your western blot experiment, you can explore Boster’s catalog to browse primary antibodies and secondary antibodies validated for western blot as well as find western blot reagents you will need. In protein research and diagnostics, Western blot is used for the detection and quantification of specific proteins in a sample, analysis of protein expression levels across different conditions or treatments, detection of post-translational modifications (e.g., phosphorylation, glycosylation), study of protein-protein interactions, and confirmation of protein identity and purity in recombinant protein production. For researchers interested in analyzing protein expression directly within intact cells, consider our in-cell western blot service, which offers a high-throughput, quantitative alternative to traditional blotting techniques. To learn more about western blotting, download our Western Blot eBook, which discusses the principle, protocol, troubleshooting tips, and FAQs for western blot. 4. Eastern blotting Following the directional theme set by Southern and Western blotting, Eastern blotting is used to analyze post-translational modifications (PTMs) of proteins, such as glycosylation or phosphorylation. Many scientists deem Eastern blotting as a variation of Western blotting. The Eastern blot technique involves transferring proteins separated by gel electrophoresis onto a membrane, followed by detection using specialized probes or antibodies targeting the PTM of interest. Though less frequently performed than other blotting techniques, Eastern blotting is useful for analyzing PTMs (e.g., glycosylation), studying protein modifications (e.g., phosphorylation, lipidation), and detecting and characterizing glycoproteins and other modified proteins. 5. Far-Western Blotting Far-Western blotting, derived from Western blotting, focuses on protein-protein interactions. It involves transferring proteins separated by gel electrophoresis onto a membrane and identifies interactions using labeled proteins or peptides to probe for binding with the immobilized target proteins. Growing in popularity, Far-Western blotting is valuable for identifying and studying protein-protein interactions, mapping interaction domains, and screening potential binding partners. 6. Southwestern blotting Southwestern blotting combines features of Southern and Western blotting techniques, focusing on DNA-binding protein detection. This technique involves transferring denatured proteins from a gel onto a membrane, followed by incubation with labeled DNA probes to identify proteins that bind to specific DNA sequences. Though less common than Southern and Western blotting, Southwestern blotting is useful for detecting DNA-binding proteins, studying protein-DNA interactions, and identifying transcription factors and other regulatory proteins. 7. Reverse Northern blotting Reverse Northern blotting indicates the reversal of the Northern blotting process. Instead of transferring RNA and probing with DNA, Reverse Northern blotting transfers DNA and probes with labeled RNA. This technique involves immobilizing DNA on a membrane and hybridizing it with labeled RNA or cDNA probes to detect DNA sequences. Although less common than standard Northern blotting, it is used for gene expression analysis using cDNA or genomic DNA arrays, screening differentially expressed genes in various conditions, and studying changes in transcript levels in response to treatments. 8. Colony blotting Colony blotting is named for its application in screening microbial colonies, such as bacterial or yeast colonies. It involves transferring entire colonies from a culture plate onto a membrane, where specific nucleic acids or proteins are detected through DNA hybridization analysis. Primarily used in microbiology and cloning, colony blotting helps screen colonies for target DNA or RNA sequences, identify recombinant clones containing specific genetic inserts, and rapidly detect plasmid-containing colonies. Dot blotting Dot blotting is frequently used as a quick and simple screening method for the presence or absence of nucleic acids and proteins. Named for the dot-like application of samples on the membrane, it is a simplified version of Western blotting, where samples are directly spotted onto a membrane for detection without prior gel electrophoresis. Popular for rapid screening, dot blotting is used to screen specific nucleic acids or proteins, relatively quantify target molecules, and analyze large numbers of samples simultaneously. It should be noted that dot blots do not provide information about molecular weight, so false positive signals or the presence of modified proteins are difficult to identify. Slot blotting Slot blotting is similar to dot blotting, but less commonly used. Named for the slot-like application of samples, slot blotting involves applying samples in rectangular slots on a membrane, allowing more uniform application and the quantification of target molecules in a sample without the need for gel electrophoresis. This technique is helpful for analyzing nucleic acids or proteins without electrophoresis, comparing the relative abundance of target molecules in different samples, and screening multiple samples in high-throughput formats. As with dot blots, slot blots are also unable to provide information about the size of the target protein. Table of Blotting Types We have provided a table below that highlights the features of each blotting type. \| Blotting Type \| Target Molecule \| Detection Method \| Technique Description \| Applications \| --- --- \| Southern Blotting \| DNA \| Labeled DNA/RNA probes \| DNA fragments are separated by electrophoresis, transferred to a membrane, and probed. \| Detection of DNA sequences, genetic fingerprinting, gene mapping \| \| Northern Blotting \| RNA \| Labeled DNA/RNA probes \| RNA is separated by electrophoresis, transferred to a membrane, and hybridized with probes. \| Analysis of gene expression, RNA processing, alternative splicing \| \| Western Blotting \| Proteins \| Primary/secondary antibodies \| Proteins are separated by SDS-PAGE, transferred to a membrane, and detected with antibodies. \| Protein detection, expression analysis, post-translational modifications \| \| Eastern Blotting \| Modified proteins \| Specific probes/antibodies \| Proteins are transferred to a membrane and probed for post-translational modifications. \| Analysis of glycosylation and other post-translational modifications \| \| Far-Western Blotting \| Proteins (interactions) \| Labeled proteins/peptides \| Proteins are transferred to a membrane, probed with labeled proteins to detect interactions. \| Study of protein-protein interactions, mapping interaction domains \| \| Southwestern Blotting \| DNA-binding proteins \| Labeled DNA probes \| Proteins are transferred to a membrane and probed with labeled DNA to detect binding. \| Detection of DNA-binding proteins, study of protein-DNA interactions \| \| Reverse Northern Blotting \| DNA \| Labeled RNA/cDNA probes \| DNA is immobilized on a membrane and hybridized with labeled RNA/cDNA probes. \| Gene expression profiling, screening for differentially expressed genes \| \| Colony Blotting \| DNA/RNA in colonies \| Labeled probes \| Microbial colonies are transferred to a membrane and probed for specific sequences. \| Screening for specific sequences, identifying recombinant clones \| \| Dot Blotting \| Nucleic acids/proteins \| Labeled probes/antibodies \| Samples are spotted directly onto a membrane for rapid detection. \| Rapid screening, quantification of target molecules \| \| Slot Blotting \| Nucleic acids/proteins \| Labeled probes/antibodies \| Samples are applied in slots on a membrane for quantitative analysis. \| Quantitative analysis, comparing relative abundance of target molecules \| Blotting techniques are important tools in molecular biology and biochemistry. Each blotting method caters to different experimental needs, enabling detailed analysis of DNA, RNA, proteins, and their interactions. By utilizing each technique’s unique advantages and applications, researchers can deepen our understanding of molecular interactions and processes. Boster Bio supports these workflows with advanced solutions like our Recombinant Antibody Production Service, ideal for creating specialized reagents for your blotting experiments. References Alwine, J. C., Kemp, D. J., & Stark, G. R. (1977). Method for detecting specific RNAs in agarose gels by transfer to diazobenzyloxymethyl-paper and hybridization with DNA probes. Proceedings of the National Academy of Sciences, 74(12), 5350-5354. Bowen, B., Steinberg, J., Laemmli, U. K., & Weintraub, H. (1979). The detection of DNA-binding proteins by protein blotting. Nucleic Acids Research, 8(1), 1-20. Brown T. (2001). Dot and slot blotting of DNA. Current protocols in molecular biology, Chapter 2, Unit2.9B. Burnette, W. N. (1981). "Western blotting": Electrophoretic transfer of proteins from sodium dodecyl sulfate–polyacrylamide gels to unmodified nitrocellulose and radiographic detection with antibody and radioiodinated protein A. Analytical Biochemistry, 112(2), 195-203. Grunstein, M., & Hogness, D. S. (1975). Colony hybridization: a method for the isolation of cloned DNAs that contain a specific gene. Proceedings of the National Academy of Sciences, 72(10), 3961-3965. Irwin, M. H., & Pinkert, C. A. (2014). Transgenic Animal Technology. In M. H. Irwin & C. A. Pinkert (Eds.), Transgenic Animal Technology (3rd ed., pp. 543-564). Elsevier Inc. Chapter 20. Ishikawa, D., & Taki, T. (2000). Thin-layer chromatography blotting using polyvinylidene difluoride membrane (Far-Eastern blotting) and its applications. In T. Taki (Ed.), Sphingolipid metabolism and cell signaling, Part B. Methods in Enzymology (Vol. 312, pp. 145–157). Academic Press. Kafatos, F. C., Jones, C. W., & Efstratiadis, A. (1979). Determination of nucleic acid sequence homologies and relative concentrations by a dot hybridization procedure. Nucleic Acids Research, 7(6), 1541-1552. Leca‐Bouvier, B., & Blum, L. J. (2005). Biosensors for Protein Detection: A Review. Analytical Letters, 38(10), 1491–1517. Machida, K., & Mayer, B. J. (2009). Detection of protein-protein interactions by far-western blotting. Methods in Molecular Biology (Clifton, N.J.), 536, 313–329. Mahmood, T., & Yang, P. C. (2012). Western blot: technique, theory, and trouble shooting. North American Journal of Medical Sciences, 4(9), 429–434. Primrose, S. B., & Twyman, R. (2009). Principles of Genome Analysis and Genomics (3rd ed.). John Wiley & Sons. Southern, E. M. (1975). Detection of specific sequences among DNA fragments separated by gel electrophoresis. Journal of Molecular Biology, 98(3), 503-517. Towbin, H., Staehelin, T., & Gordon, J. (1979). 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15955	https://www.geeksforgeeks.org/engineering-mathematics/understanding-chebyshevs-inequality-with-an-example/	Understanding Chebyshev's Inequality with an example Last Updated : 20 Aug, 2024 Suggest changes 1 Like In this article, we will discuss the overview of Chebyshev's inequality algorithm, and will cover the Understanding Chebyshev's inequality with an example. Pre-requisite is to go to given below link to understand Markov's theorem to get more deep mathematical insights behind the Chebyshev's inequality, and it's proof. Let's discuss it one by one. Chebyshev's inequality : It is based on the concept of variance. It says that given a random variable R, then ∀ x > 0, The probability that the random variable R deviates from its expected value in either side by at least x is given as follows. P(∣R−Ex(R)∣>=X)<=Var(R)/X∗XP(\|R - Ex(R)\|>=X) < = Var(R)/XXP(∣R−Ex(R)∣>=X)<=Var(R)/X∗XP(∣R−Ex(R)∣>=X)<=Var(R)/X∗XP(\|R - Ex(R)\|>=X) < = Var(R)/XXP(∣R−Ex(R)∣>=X)<=Var(R)/X∗XP(∣R−Ex(R)∣>=X)<=Var(R)/X∗XP(\|R - Ex(R)\|>=X) < = Var(R)/XXP(∣R−Ex(R)∣>=X)<=Var(R)/X∗XP(∣R− P(∣ R − Ex(R)∣>= E x(R) ∣ >= X)<= X) <= Var(R)/X∗ Va r(R)/ X ∗ X X //equation -1 Where it represents the following values as follows. Var(R) - It denotes variance of Random Variable R.Ex(R) - It denotes the Expected value of Random Variable R. Proof : We know Markov's inequality in Probability as follows. P(R>=X)<=Ex(R)/XP(R>=X) <= Ex(R)/XP(R>=X)<=Ex(R)/XP(R>=X)<=Ex(R)/XP(R>=X) <= Ex(R)/XP(R>=X)<=Ex(R)/XP(R>=X)<=Ex(R)/XP(R>=X) <= Ex(R)/XP(R>=X)<=Ex(R)/XP(R>= P(R >= X)<= X) <= Ex(R)/X E x(R)/ X //equation -2 Put : R - Ex(R) in place of R in this and square this and then apply Markov's inequality, we get the following expression as follows. P(∣R−Ex(R)∣>=X)=P(∣R−Ex(R)∣∗∣R−Ex(R)∣>=X∗X)P(\|R-Ex(R)\|>=X) = P(\|R-Ex(R)\| \|R-Ex(R) \|> = XX)P(∣R−Ex(R)∣>=X)=P(∣R−Ex(R)∣∗∣R−Ex(R)∣>=X∗X)P(∣R−Ex(R)∣>=X)=P(∣R−Ex(R)∣∗∣R−Ex(R)∣>=X∗X)P(\|R-Ex(R)\|>=X) = P(\|R-Ex(R)\| \|R-Ex(R) \|> = XX)P(∣R−Ex(R)∣>=X)=P(∣R−Ex(R)∣∗∣R−Ex(R)∣>=X∗X)P(∣R−Ex(R)∣>=X)=P(∣R−Ex(R)∣∗∣R−Ex(R)∣>=X∗X)P(\|R-Ex(R)\|>=X) = P(\|R-Ex(R)\| \|R-Ex(R) \|> = XX)P(∣R−Ex(R)∣>=X)=P(∣R−Ex(R)∣∗∣R−Ex(R)∣>=X∗X)P(∣R− P(∣ R − Ex(R)∣>= E x(R) ∣ >= X)= X) = P(∣R− P(∣ R − Ex(R)∣∗ E x(R) ∣ ∗ ∣R− ∣ R − Ex(R)∣>= E x(R) ∣ >= X∗ X ∗ X) X) //equation -3 P(∣R−Ex(R)∣∗∣R−Ex(R)∣>=X∗X)<=Ex(R−Ex(R)∗R−Ex(R))/X∗XP(\|R-Ex(R)\| \|R-Ex(R)\| > = XX) <= Ex(R-Ex(R) R-Ex(R)) / XXP(∣R−Ex(R)∣∗∣R−Ex(R)∣>=X∗X)<=Ex(R−Ex(R)∗R−Ex(R))/X∗XP(∣R−Ex(R)∣∗∣R−Ex(R)∣>=X∗X)<=Ex(R−Ex(R)∗R−Ex(R))/X∗XP(\|R-Ex(R)\| \|R-Ex(R)\| > = XX) <= Ex(R-Ex(R) R-Ex(R)) / XXP(∣R−Ex(R)∣∗∣R−Ex(R)∣>=X∗X)<=Ex(R−Ex(R)∗R−Ex(R))/X∗XP(∣R−Ex(R)∣∗∣R−Ex(R)∣>=X∗X)<=Ex(R−Ex(R)∗R−Ex(R))/X∗XP(\|R-Ex(R)\| \|R-Ex(R)\| > = XX) <= Ex(R-Ex(R) R-Ex(R)) / XXP(∣R−Ex(R)∣∗∣R−Ex(R)∣>=X∗X)<=Ex(R−Ex(R)∗R−Ex(R))/X∗XP(∣R− P(∣ R − Ex(R)∣∗ E x(R) ∣ ∗ ∣R− ∣ R − Ex(R)∣>= E x(R) ∣ >= X∗ X ∗ X)<= X) <= Ex(R− E x(R − Ex(R)∗ E x(R) ∗ R− R − Ex(R))/X∗ E x(R))/ X ∗ X X //equation -4 We also know that the following expression and with the help of this we can evaluate. Var(R)=Ex((R−Ex(R))∗∣(R−Ex(R)))Var(R) = Ex((R-Ex(R)) \|(R-Ex(R)))Var(R)=Ex((R−Ex(R))∗∣(R−Ex(R)))Var(R)=Ex((R−Ex(R))∗∣(R−Ex(R)))Var(R) = Ex((R-Ex(R)) \|(R-Ex(R)))Var(R)=Ex((R−Ex(R))∗∣(R−Ex(R)))Var(R)=Ex((R−Ex(R))∗∣(R−Ex(R)))Var(R) = Ex((R-Ex(R)) \|(R-Ex(R)))Var(R)=Ex((R−Ex(R))∗∣(R−Ex(R)))Var(R)= Va r(R) = Ex((R− E x((R − Ex(R))∗ E x(R)) ∗ ∣(R− ∣(R − Ex(R))) E x(R))) Now, put this in the fourth equation and replace LHS of the fourth equation with LHS of the third equation, we get Chebychev's inequality as follows. Result : P(∣R−Ex(R)∣>=X)<=Var(R)/X∗XP(\|R - Ex(R)\| >=X) <= Var(R)/XXP(∣R−Ex(R)∣>=X)<=Var(R)/X∗XP(∣R−Ex(R)∣>=X)<=Var(R)/X∗XP(\|R - Ex(R)\| >=X) <= Var(R)/XXP(∣R−Ex(R)∣>=X)<=Var(R)/X∗XP(∣R−Ex(R)∣>=X)<=Var(R)/X∗XP(\|R - Ex(R)\| >=X) <= Var(R)/XXP(∣R−Ex(R)∣>=X)<=Var(R)/X∗XP(∣R− P(∣ R − Ex(R)∣>= E x(R) ∣ >= X)<= X) <= Var(R)/X∗ Va r(R)/ X ∗ X X Corollary of Chebyshev's inequality : If we replace X with cVar(R), where c > 0 ,then we get the below equation which can be proven very easily as follows. P(∣R−Ex(R)∣>=c∗Var(R))<=1/c∗cP(\|R -Ex(R)\| >= cVar(R))<= 1/ccP(∣R−Ex(R)∣>=c∗Var(R))<=1/c∗cP(∣R−Ex(R)∣>=c∗Var(R))<=1/c∗cP(\|R -Ex(R)\| >= cVar(R))<= 1/ccP(∣R−Ex(R)∣>=c∗Var(R))<=1/c∗cP(∣R−Ex(R)∣>=c∗Var(R))<=1/c∗cP(\|R -Ex(R)\| >= cVar(R))<= 1/ccP(∣R−Ex(R)∣>=c∗Var(R))<=1/c∗cP(∣R− P(∣ R − Ex(R)∣>= E x(R) ∣ >= c∗ c ∗ Var(R))<= Va r(R)) <= 1/c∗ 1/ c ∗ c c Example of Chebyshev's inequality : Let's understand the concept with the help of an example for better understanding as follows. Example-1 : Let us say that Random Variable R = IQ of a random person. And average IQ of a person is 100, i.e, Ex(R) = 100. And Variance in R is 15. (Assuming R >0). Then what is the probability that if we pick a random person, his/her IQ is at least 250 ? Solution - To solve this, we will use corollary of the Chebyshev's inequality as follows. P(R>=250) = P( R-100 >=150 ), Comparing with the Corollary, we can say that the following result as follows. since 150 = 10 Varianceso, c = 10. Therefore, answer is upper bounded by 1/100 which is ≤1 %. Example-2 : If we solve the same problem using Markov's theorem without using the variance, we get the upper bound as follows. P ( R >= 250 ) < = Ex(R) / 250 = 100/250 = 2/5 = 40%. So, the Same problem is upper bounded by 40 % by Markov's inequality and by 1% by Chebyshev's inequality. Thus, we can say that Chebyshev's inequality gives a better bound on probability compared to Markov's if we know the variance of random variable R. Solved Examples Example 1: Suppose the mean height of a group of students is 170 cm with a standard deviation of 5 cm. Use Chebyshev's inequality to find the probability that a student's height is between 160 cm and 180 cm. Here, 𝜇 = 170cm and σ=5 cm. The interval 160 cm to 180 cm is μ±2σ. Chebyshev's inequality states that P(∣X−μ∣≥kσ)≤ (1/k2) For k=2, P(∣X−170∣≥10)≤ 1/4 =0.25. So, P(160≤X≤180)≥1−0.25=0.75. Thus, the probability that a student's height is between 160 cm and 180 cm is at least 75%. . Example 2: A dataset has a mean of 50 and a standard deviation of 8. What is the minimum probability that a randomly selected data point lies within 34 and 66? Here, 𝜇 = 50, 𝜎 = 8, and the interval is 34 to 66. The distance from the mean is 𝑘 = (66 − 50)/8 = 2 Using Chebyshev's inequality: P(∣X−50∣≥16)≤ (1/4) =0.25. Thus, 𝑃(34≤𝑋≤66)≥0.75 Practice Problems A class’s test scores have a mean of 80 and a standard deviation of 10. Use Chebyshev’s inequality to determine the probability that a student’s score is between 60 and 100. The mean score of a group of students in a test is 75 with a standard deviation of 8. Find the minimum probability that a randomly selected student's score is between 67 and 83. A factory produces items with an average weight of 50 grams and a standard deviation of 5 grams. Estimate the probability that a randomly chosen item weighs between 40 grams and 60 grams. For a set of data with a mean of 45 and a standard deviation of 12, what is the minimum probability that a randomly selected value is within 24 units of the mean? The average height of trees in a forest is 12 meters with a standard deviation of 2 meters. Use Chebyshev's inequality to find the probability that a tree’s height is between 8 meters and 16 meters. The average rainfall in a region is 100 mm with a standard deviation of 20 mm. Find the probability that the rainfall in a randomly selected year is between 60 mm and 140 mm. In a survey, the mean response time is 30 seconds with a standard deviation of 6 seconds. Estimate the probability that a response time is between 18 seconds and 42 seconds. A company reports an average processing time of 200 minutes with a standard deviation of 25 minutes. Determine the probability that the processing time is between 150 and 250 minutes. For a population with a mean of 100 and a standard deviation of 15, find the probability that a randomly selected value lies between 70 and 130. Given a random variable 𝑋 with a mean of 20 and a standard deviation of 3, use Chebyshev’s inequality to estimate the probability that 𝑋 is within 9 units of the mean. 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15956	https://lematematiche.dmi.unict.it/index.php/lematematiche/article/download/14/13/68	SOME REMARKS ON THE CANTOR PAIRING FUNCTION 55 LE MATEMATICHE Vol. LXII (2007) - Fasc. I, pp. 55-65 SOME REMARKS ON THE CANTOR PAIRING FUNCTION MERI LISI In this paper, some results and generalizations about the Cantor pairing function are given. In particular, it is investigated a very compact expression for the n-degree generalized Cantor pairing function (g.C.p.f., for short), that permits to obtain n−tupling functions which have the characteristics to be n-degree polynomials with rational coefficients. A recursive formula for the n-degree g.C.p.f. is also provided. Introduction. In mathematics a pairing function is a process to uniquely encode two natural numbers into a single natural number. Any pairing function can be used in set theory to prove that integers and rational numbers have the same cardinality as natural numbers. In theoretical computer science, pairing functions are used to encode a function defined on a vector of natural numbers f : N k → N into a new function, , , , . Definition 1.1. The function 〈· , ·〉 : N2 → N, such that (1) 〈x1, x2〉 = (x1 + x2)2 + 3x1 + x2 2 , ∀ (x1, x2) ∈ N2, is called the Cantor “pairing” function. Entrato in redazione il 28 Novembre 2006. 56 MERI LISI The fundamental property of the Cantor pairing function is given by the following theorem. Theorem 1.1. The Cantor pairing function is a bijection from N 2 onto N .Proof. In order to prove the theorem, consider the straight lines x1 + x2 = k, with k ∈ N. It is clear that the “point” ( ¯x1, ¯x2) belongs to x1 +x2 = ¯ x1 + ¯ x2 , or, more precisely, to the intersection of x1 +x2 = ¯ x1 + ¯ x2with the first quadrant of the euclidean plane. Hence, for any ( ¯x1 + ¯ x2) fixed, moving along the line x1 + x2 = k towards increasing ordinates ( k = 0, . . . , ¯x1 + ¯ x2 ), let us list all the couples of natural numbers we “meet” and associate to each one of these an increasing natural number, starting from zero. Thanks to a simple (pascal) algorithm, we have a bijective corre-spondence between N2 and N. The variable out put represents the natural number associated to (x1, x2):program pairing var i, j, k, x1, x2, out put : 0 .. maxint; begin out put := 0for k = 0 to (x1 + x2) do begin for (i = 0 to k) and ( j = k to 0 ) do begin if (x1 = i) and (x2 = j) then write (out put );else begin i := i + 1; j := j − 1; out put := out put + 1; end; end; k := k + 1; end. end. By using the previous algorithm, we can build Table 1, where we list the couples of natural numbers (x1, x2). We stop at (3, 0).SOME REMARKS ON THE CANTOR PAIRING FUNCTION 57 Remark 1.1. In what follows we shall use the following notation: N n (k) indicates the number of n-tuples satisfying condition n ∑ i=1 x i = k. For the pairing function, we have: (2) N2(k) = k + 1. (x1,x2)(0,0) (0,1) (1,0) (0,2) (1,1) (2,0) (0,3) (1,2) (2,1) (3,0) output 0123456789 k0123 N2(k)1234 Table 1 The Cantor pairing function is a second degree polynomial, with rational coef fi cients, , , , . Rudolph Fueter proposed in 1923 four conjectures about the set of polynomial pairing functions, . By composition, we can obtain “tripling ” functions, “quadrupling ” functions and so on. By using (1), a possible tripling function is given by 〈· , ·, ·〉 : N3 → N, such that: (3) 〈x1, x2, x3〉 = 〈 x1, 〈x2, x3〉〉 = { x1 + [(x2 + x3)2 + 3x2 + x3 ] 2 }2 3x1 + [(x2 + x3)2 + 3x2 + x3 ] 2 2 . However, note that, listing the terns of natural numbers (x1, x2, x3) in a way similar to that used for the pairing function (in this case, we consider the planes x1 + x2 + x3 = k, with k ∈ N), we are not able to identify a generalized rule of order. In fact, by means of an analogous algorithm, we can build Table 2 (we stop at (2, 0, 0)): (x1,x2,x3) (0,0,0) (0,0,1) (1,0,0) (0,1,0) (1,0,1) (2,0,0) (0,0,2) (1,1,0) (2,0,1) (3,0,0) output 0123456789 Table 2 Note also that the tripling function (3), obtained by means of a composition, is a fourth degree polynomial. Spontaneous questions arise: 58 MERI LISI is it possible to have tripling functions which are third degree polynomials, quadrupling functions which are fourth degree polynomials and so on? And moreover: does a generalized formula exist, that permits to obtain these polynomials in a simple way? The n-degree generalized Cantor pairing function. Theorem 2.1. For any n ∈ N, the so-called “n-tupling ” function , 〈· , ..., ·〉 : Nn → N, such that (4) 〈x1, ..., x n 〉 = n ∑ h=1 { 1 h! h−1 ∏ j=0 [( h∑ i=1 x i ) j ]} , ∀(x1, ..., x n ) ∈ Nn , is an n-degree polynomial, with rational coef fi cients. Moreover it is a one-to-one correspondence from Nn onto N.Proof. We can prove the theorem by induction on the degree n. In order to do this, before studying the inductive case, we analyze some particular cases. Case n = 1From de fi nition (4), we have 〈x1〉 = x1 .In this case, 〈·〉 is trivially a bijection from N to N.Case n = 2From de fi nition (4), we have: 〈x1, x2〉 = 2 ∑ h=1 { 1 h! h−1 ∏ j=0 [( h∑ i=1 x i ) j ]} = x1 + (x1 + x2)( x1 + x2 + 1) 2 = (x1 + x2)2 + 3x1 + x2 2 . This is the Cantor pairing function, we have just proved in Theorem 1.1 to be a bijection from N2 to N .Case n = 3Let us build a bijection from N3 to N, by de fi ning a way of ordering terns similar to that followed for couples of the pairing function, at the SOME REMARKS ON THE CANTOR PAIRING FUNCTION 59 previous step: fi rst organize terns of natural numbers (x1, x2, x3) according to their plane of belongings ∑3 i=1 x i = k, with increasing k ≥ 0 ( k ∈ N). Then, for any plane, put the “points ” in order as follows: (x1, x2, x3) < ( y1, y2, y3) if x3 > y3 or if x3 = y3and < x1, x2 > is smaller than < y1, y2 >, where the way of organizing < x1, x2 > is that de fi ned for the pairing function < y1, y2 >. This permits to build Table 3. We stop at (2, 0, 0). (x1,x2,x3) (0,0,0) (0,0,1) (0,1,0) (1,0,0) (0,0,2) (0,1,1) (1,0,1) (0,2,0) (1,1,0) (2,0,0) 0 1 2 3 4 5 6 7 8 9 k012 N3(k)136 Table 3 Note that (5) N3(k) = k ∑ h=0 (h + 1) = 1 2! 2∏ j=1 (k + j) indicates the number of terns satisfying 3∑ i=1 x i = k. In particular, from (2), we have: (6) N3(k) = k ∑ h=0 N2(h). Hence, to identify a given z3 = ( ¯x1, ¯x2, ¯x3), with 3∑ i=1 ¯x i = m ( ¯x i , m ∈ N), we have to take into account all those points “coming before ” it. First, we have to consider all those “points ” such that 3∑ i=1 x i = g, for any x i , g ∈ N, with g < m. They are: 60 MERI LISI m−1 ∑ l=0 N3(l) = m−1 ∑ l=0 l ∑ h=0 N2(h) = m−1 ∑ l=0 l ∑ h=0 (h + 1) = m−1 ∑ l=0 1 2! 2∏ j=1 (l + j) = 1 3! 2∏ j=0 (m + j), where we used (5), (6) and relation (10) proved in Appendix. Then, we have to consider all those terns satisfying 3∑ i=1 ¯x i = m but “smaller ” than z3 . Since the rule of ordering the fi rst two terms of (x1, x2, x3) is that used for the Cantor pairing, we have that, starting from 0, the natural number associated to ( ¯x1, ¯x2, ¯x3) is: 1 3! 2∏ j=0 [( 3∑ i=1 ¯x i ) j ] 1 2! 1∏ j=0 [( 2∑ i=1 ¯x i ) j ] ¯ x1 = ( ¯x1 + ¯ x2 + ¯ x3)( ¯x1 + ¯ x2 + ¯ x3 + 1)( ¯x1 + ¯ x2 + ¯ x3 + 2) 3! ++( ¯x1 + ¯ x2)( ¯x1 + ¯ x2 + 1) 2! + ¯ x1 = ( ¯x1 + ¯ x2 + ¯ x3)3 + 3( ¯x1 + ¯ x2 + ¯ x3)2 + 2( ¯x1 + ¯ x2 + ¯ x3) 6 ++( ¯x1 + ¯ x2)2 + ( ¯x1 + ¯ x2) 2 + ¯ x1. Since (7) 〈x1, x2, x3〉 = 3 ∑ h=1 { 1 h! h−1 ∏ j=0 [( h∑ i=1 x i ) j ]} , we proved that the tripling function given by relation (7) represents a third degree polynomial with rational coef fi cients and it turns out to be a bijection from N3 to N.Consider the inductive case: assume < x1, . . . , x n > to be a bijection SOME REMARKS ON THE CANTOR PAIRING FUNCTION 61 from Nn to N, for any n ≥ 1 and that the n-tuples of natural numbers satisfying n ∑ i=1 x i = k are: (8) N n (k) = k ∑ h=0 N n−1(h) = k ∑ h=0 1 (n − 2)! n−2 ∏ j=1 (h + j) = 1 (n − 1)! n−1 ∏ j=1 (k + j). Now, we want to prove that 〈x1, . . . , x n+1〉 = n+1 ∑ h=1 { 1 h! h−1 ∏ j=0 [( h∑ i=1 x i ) j ]} is a polynomial of degree (n +1), with rational coef fi cients and a bijection from Nn+1 to N.By using a rule of ordering the (n + 1)−tuples similar to that followed for the n−tupling function, we build a bijection from N n+1 to N. In particular, ordinate the (n + 1)-tuples of natural numbers according to relation n+1 ∑ i=1 x i = k, with increasing k ≥ 0, k ∈ N. Organize the “points ” satisfying n+1 ∑ i=1 x i = k as follows: (x1, ..., x n+1) < ( y1, ..., yn+1) if x n+1 > yn+1 or if x n+1 = yn+1 and < x1, ..., x n > is smaller than < y1, . . . , yn >. Hence: (9) N n+1(k) = k ∑ h=0 N n (h) = k ∑ h=0 1 (n − 1)! n−1 ∏ j=1 (h + j) = 1 n! n−1 ∏ j=0 (k + j + 1) = 1 n! n ∏ j=1 (k + j), where we used the inductive assumption (8) and relation (10) of Appendix. To “identify ” z n+1 = ( ¯x1, . . . , ¯x n+1), with n+1 ∑ i=1 ¯x i = m, we have to 62 MERI LISI count all those (n +1)−tuples of natural numbers, satisfying the condition n+1 ∑ i=1 x i = k, with k = 0, . . . , m − 1: m−1 ∑ l=0 N n+1(l) = m−1 ∑ l=0 l ∑ h=0 N n (h) = m−1 ∑ l=0 1 n! n ∏ j=1 (l + j) = 1 (n + 1)! n ∏ j=0 (m + j), where we used (9) and relation (10) proved in Appendix. Then, we have to add all those (n + 1)−tuples satisfying n+1 ∑ i=1 x i = k,but “smaller ” than z n+1 . However, since the rule of ordering the fi rst n terms of (x1, . . . x n+1) is that followed for the n-tupling function, we have that, starting to count from 0, the natural number associated to ( ¯x1, . . . , ¯x n+1), ¯x i ∈ N, is given by: 1 (n + 1)! n ∏ j=0 [( n+1∑ i=1 ¯x i ) j ] + n ∑ h=1 { 1 h! h−1 ∏ j=0 [( h∑ i=1 ¯x i ) j ]} . Since 〈x1, ..., x n+1〉 = n+1 ∑ h=1 { 1 h! h−1 ∏ j=0 [( h∑ i=1 x i ) j ]} , we have that it represents a polynomial of degree (n + 1), with rational coef fi cients and which is a one-to-one correspondence from Nn+1 onto N . The theorem is so proved. Remark 2.1. The Cantor pairing function is based on the idea of counting anti-diagonals x + y = k and then of counting within a given diagonal by increasing ordinates. This geometrical device has been generalized to “Cantor n-tupling function ” which is a bijection from N n onto N .At fi rst, we used the level k of the hyperplane H k of the equation x1 + x2 + . . . + x n = k and then the level h in the hyperplane H k , having in turn for equation x1 + . . . + x n−1 = h, and so and up to obtaining the line x1 + x2 = constant. Hence, Cantor n-tupling function can also be expressed via binomial coef fi cients as follows, : 〈x1, ..., x n 〉 = ( x1 + ... + x n + n − 1 n ) + ( x1 + ... + x n−1 + n − 2 n − 1 )SOME REMARKS ON THE CANTOR PAIRING FUNCTION 63 +... + ( x1 + x2 + 12 ) + ( x11 ) . Finally, the Cantor pairing function can be considered a “primitive recursive ” function, , , . For the n-degree generalized Cantor pairing function (4) (indicated as 〈· , ·, . . . , ·〉 n ) a possible recursive formula could be given by the following: 〈x1, x2, . . . , x n 〉n =  0, if n∑ i=1 x i = 0 〈 n∑ i=1 x i − 1, 0, ..., 0〉n + 〈 x1, x2, ..., x n−1〉n−1 + 1, otherwise . Appendix. We can prove by induction on the index m, that, for any n ≥ 2, m ≥ 1: (10 ) m−1 ∑ k=0 1 (n − 1)! n−1 ∏ j=1 (k + j) = 1 n! n−1 ∏ j=0 (m + j). In fact, if m = 1, relation (10) is veri fi ed because, if n ≥ 2: 1 (n − 1)! n−1 ∏ j=1 j = 1 · 2 · . . . · (n − 1) (n − 1)! = 1 · 2 · . . . · n n! = 1 n! n−1 ∏ j=0 ( j + 1). Assume condition (10) to be satis fi ed for m ≥ 1, for any n ≥ 2, i.e.: (11 ) m−1 ∑ k=0 1 (n − 1)! n−1 ∏ j=1 (k + j) = 1 n! n−1 ∏ j=0 (m + j). Then, it is veri fi ed also for the index m + 1. In fact, if n ≥ 2: m ∑ k=0 1 (n − 1)! n−1 ∏ j=1 (k + j) = m−1 ∑ k=0 1 (n − 1)! n−1 ∏ j=1 (k + j) + 1 (n − 1)! n−1 ∏ j=1 (m + j). By using the inductive assumption (11), we get: 1 n! n−1 ∏ j=0 (m + j) + 1 (n − 1)! n−1 ∏ j=1 (m + j) = (m + n) n! n−1 ∏ j=1 (m + j), 64 MERI LISI and so, for any n ≥ 2: m ∑ k=0 1 (n − 1)! n−1 ∏ j=1 (k + j) = 1 n! n−1 ∏ j=0 (m + j + 1). Acknowledgments. The author would like to express her gratitude to Prof. A. Sorbi for his suggestions and to the referee who contributed to make the paper more readable. This work was partially supported by Par 2004 - Research Project of the University of Siena funds as well as by G.M.F.M. and M.U.R.S.T. research funds. REFERENCES G. Cantor, Contributions to the Founding of the Theory of Trans fi nite Numbers ,Dover, New York, 1955 G. Cantor, “Ein beitrag zur mannigfaltigkeitslehre ”, Journal f ¨ur die reine und angewandte Mathematik, 84, (1878), pp. 242 –258 P. Cegielski - D. Richard, “Decidability of the theory of the natural integers with the Cantor pairing function and the successor ”, Theor. Comput. Sci., 257, (2001), pp. 51 –77 P. Cegielski - D. Richard, “On arithmetical fi rst-order theories allowing encoding and decoding of lists ”, Theor. Comput. Sci., 222, (1999) pp. 55 –75 N. J. Cutland, Computability: An Introduction to Recursive Function Theory ,Cambridge University Press, Cambridge, 1986 M. D. Davis - R. Sigal - E. J. Wyuker, Computability, Complexity and Languages , Academic Press, Sudkamp, T. A., 1994 R. Dedekind, “The nature and meaning of numbers ”, in: R. Dedekind, Essays in the Theory of Numbers, Dover, New York, 1954 J. E. Hopcroft - R. Motwani - J. D. Ullman, Introduction to Automata Theory, Languages, and Computation , Addison-Wesley, Boston, 2001 N. D. Jones, Computability and Complexity from a Programming Perspective ,Foundations of Computing, MIT Press, Boston, 1997 H. Rogers, Theory of Recursive Functions and Effective Computability ,Cambridge, MA, MIT Press, 1987 C. Smory ´nski, Logical Number Theory I - An Introduction Springer –Verlag, Berlin, 1991 SOME REMARKS ON THE CANTOR PAIRING FUNCTION 65 T. A. Sudkamp, Languages and Machines: An Introduction to the Theory of Computer Science , Addison-Wesley, Longman, 2006 S. Wolfram, A New Kind of Science , Wolfram Media, Chamaign, IL, 2002 Dipartimento di Scienze Matematiche ed Informatiche “Roberto Magari ”, Pian dei Mantellini 44, 53100 - Siena, Italy, email: lisi7@unisi.it
15957	https://artofproblemsolving.com/wiki/index.php/Binomial_Theorem?srsltid=AfmBOorquXS4WcEhZBD_sHZGxFO_3y4xpsaL6GssApsG2HS8o3x61_Ib	Art of Problem Solving Binomial Theorem - 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 Binomial Theorem Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search Binomial Theorem The Binomial Theorem states that for real or complex, , and non-negativeinteger, where is a binomial coefficient. In other words, the coefficients when is expanded and like terms are collected are the same as the entries in the th row of Pascal's Triangle. For example, , with coefficients , , , etc. Contents 1 Proof 1.1 Proof via Induction 1.2 Proof using calculus 2 Generalizations 2.1 Proof 3 Usage 4 See also Proof There are a number of different ways to prove the Binomial Theorem, for example by a straightforward application of mathematical induction. The Binomial Theorem also has a nice combinatorial proof: We can write . Repeatedly using the distributive property, we see that for a term , we must choose of the terms to contribute an to the term, and then each of the other terms of the product must contribute a . Thus, the coefficient of is the number of ways to choose objects from a set of size , or . Extending this to all possible values of from to , we see that , as claimed. Similarly, the coefficients of will be the entries of the row of Pascal's Triangle. This is explained further in the Counting and Probability textbook [AoPS]. Proof via Induction Given the constants are all natural numbers, it's clear to see that . Assuming that , Therefore, if the theorem holds under , it must be valid. (Note that for ) Proof using calculus The Taylor series for is for all . Since , and power series for the same function are termwise equal, the series at is the convolution of the series at and . Examining the degree- term of each, which simplifies to for all natural numbers. Generalizations The Binomial Theorem was generalized by Isaac Newton, who used an infiniteseries to allow for complex exponents: For any real or complex, , and , . Proof Consider the function for constants . It is easy to see that . Then, we have . So, the Taylor series for centered at is Usage Many factorizations involve complicated polynomials with binomial coefficients. For example, if a contest problem involved the polynomial , one could factor it as such: . It is a good idea to be familiar with binomial expansions, including knowing the first few binomial coefficients. See also Combinatorics Multinomial Theorem Retrieved from " Categories: Theorems Combinatorics Algebra 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.
15958	https://www.finra.org/rules-guidance/notices/13-07	FINRA Utility Menu For the Public FINRA DATA FINRA Data provides non-commercial use of data, specifically the ability to save data views and create and manage a Bond Watchlist. For Industry Professionals FINPRO Registered representatives can fulfill Continuing Education requirements, view their industry CRD record and perform other compliance tasks. For Member Firms FINRA GATEWAY Firm compliance professionals can access filings and requests, run reports and submit support tickets. For Case Participants DR PORTAL Arbitration and mediation case participants and FINRA neutrals can view case information and submit documents through this Dispute Resolution Portal. Need Help? \| Check System Status Log In to other FINRA systems FINRA Utility Menu For the Public FINRA DATA FINRA Data provides non-commercial use of data, specifically the ability to save data views and create and manage a Bond Watchlist. 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Need Help? \| Check System Status Log In to other FINRA systems FINRA Main Navigation FINRA Requests Comment on Proposed FINRA Rules Governing Markups, Commissions and Fees Markups, Commissions and Fees \| \| \| --- \| \| Regulatory Notice \| \| \| Notice Type Request for Comment Consolidated FINRA Rulebook \| Suggested Routing Compliance Legal Operations Senior Management \| \| Key Topics 5% Policy Commissions Government Securities Markups and Markdowns Rulebook Consolidation Service Charges and Fees \| Referenced Rules & Notices FINRA Rule 0150 FINRA Rule 2010 FINRA Rule 2111 NASD IM-2310-3 NASD IM-2440-1 NASD IM-2440-2 NASD Rule 2430 NASD Rule 2440 NYSE Rule 375 and Interpretation 375/01 Regulatory Notice 11-08 SEA Section 3 SEA Section 19 \| Executive Summary As part of the process to develop a new, consolidated rulebook (the Consolidated FINRA Rulebook),1 FINRA is requesting comment on proposed FINRA rules governing markups, markdowns, commissions and fees. FINRA initially sought comment on the proposed rules in Regulatory Notice 11-08. In response to the comments received, FINRA is proposing several changes to the proposed rules. These changes include, among other things, amendments to: (1) retain the 5% markup policy in NASD IM-2440-1 (Mark-Up Policy); (2) revise certain of the relevant factors used to determine the reasonableness of markups and commissions; (3) eliminate the requirement to provide commission schedules for equity securities transactions to retail customers; and (4) extend the proposed markup rules to transactions in certain government securities. This Notice requests comment on the revised proposal. The text of the proposed rules can be found at www.finra.org/notices/13-07. Questions concerning this Notice should be directed to: Action Requested FINRA encourages all interested parties to comment on the proposal. Comments must be received by April 1, 2013. Comments must be submitted through one of the following methods: To help FINRA process comments more efficiently, persons should use only one method to comment on the proposal. Important Notes: All comments received in response to this Notice will be made available to the public on the FINRA website. In general, FINRA will post comments as they are received.2 Before becoming effective, a proposed rule change must be authorized for filing with the Securities and Exchange Commission (SEC) by the FINRA Board of Governors, and then must be filed with the SEC pursuant to Section 19(b) of the Securities Exchange Act of 1934 (SEA).3 Background & Discussion In Regulatory Notice 11-08, FINRA sought comment on an initial proposal regarding proposed FINRA Rules 2121 (Fair Prices and Markups, Markdowns and Commissions) and 2122 (Markups and Markdowns for Transactions in Debt Securities, Except Municipal Securities) governing markups, markdowns and commissions (the proposed markup rules), and proposed FINRA Rule 2123 (Charges and Fees for Services Performed) governing fees. The proposed FINRA rules are derived from NASD Rule 2440 (Fair Prices and Commissions), NASD IM-2440-1 (Mark-Up Policy), NASD IM-2440-2 (Additional Mark-Up Policy for Transactions in Debt Securities, Except Municipal Securities), NASD Rule 2430 (Charges for Services Performed), and Incorporated NYSE Rule 375 (Missing the Market).4 FINRA received 25 comment letters in response to Regulatory Notice 11-08.5 FINRA now seeks comments on a revised proposal. Differences Between the Initial and Revised Proposals The significant differences between the initial proposal and the revised proposal are set forth below; however, interested parties should carefully read the proposed rule text for a complete and detailed understanding of the revised proposal. Revised Proposal Market Makers FINRA notes that proposed FINRA Rules 2121 and 2122 (like the current markup rules) do not address a market maker's allowance, subject to the limitations in regulation, to capture the trading spread between the bid and the ask prices and nothing in proposed FINRA Rules 2121 and 2122 affects that body of law and regulation. 1 The current FINRA rulebook consists of: (1) FINRA Rules; (2) NASD Rules; and (3) rules incorporated from NYSE (Incorporated NYSE Rules) (together, the NASD Rules and Incorporated NYSE Rules are referred to as the Transitional Rulebook). While the NASD Rules generally apply to all FINRA member firms, the Incorporated NYSE Rules apply only to those members of FINRA that are also members of the NYSE (Dual Members). The FINRA Rules apply to all FINRA member firms, unless such rules have a more limited application by their terms. For more information about the rulebook consolidation process, see Information Notice 03/12/08 Consolidation Process). For convenience, the Incorporated NYSE Rules are referred to as the NYSE Rules. 2 FINRA will not edit personal identifying information, such as names or email addresses, from submissions. Persons should submit only information that they wish to make publicly available. See Notice to Members 03-73 November 2003) (NASD Announces Online Availability of Comments) for more information. 3 See SEA Section 19 and rules thereunder. After a proposed rule change is filed with the SEC, the proposed rule change generally is published for public comment in the Federal Register. Certain imited types of proposed rule changes, however, take effect upon filing with the SEC. See SEA Section 19(b)(3) and SEA Rule 19b-4. 4 NASD Rule 2440, NASD IM-2440-1, and NASD IM-2440-2 govern markups, markdowns and commissions in transactions with customers. Fees or charges that are not transaction-related (e.g., charges for safekeeping or collecting dividends or interest for a customer) are governed by NASD Rule 2430 (Charges for Services Performed). NYSE Rule 375 (Missing the Market) addresses instances where, by reason of neglect to execute the order or otherwise, a member firm takes or supplies for its own account the securities named in the order. The rules are summarized in Regulator Notice 11-08. 5 The comments received in response to Regulatory Notice 11-08 are available on FINRA's website at www.finra.org/notices/11-08. 6 FINRA Rule 2111 took effect on July 9, 2012, and superseded NASD Rule 2310 (Recommendations to Customers (Suitability)), NASD IM-2310-1 (Possible Application of SEC Rules 15g-1 through 15g-9), NASD IM-2310-2 (Fair Dealing with Customers), and NASD IM-2310-3 (Suitability Obligations to Institutional Customers). 7 NASD Rule 2440, NASD IM-2440-1 and NYSE Rule 375 would be deleted with the adoption of proposed FINRA Rule 2121. 8 As discussed in more detail in Regulatory Notice 11-08, FINRA would delete NYSE Rule Interpretation 375/01, which provides that a member firm that has "missed the market" should contact the customer, inform the customer of the circumstances and permit the customer to choose one of two ways that the member firm then will use to fill the order. 9 This Notice does not address certain amendments discussed in Regulatory Notice 11-08 that are not changing under the revised proposal. For example, consistent with the initial proposal, FINRA proposes several conforming changes to FINRA Rule 2121 to add the term "reasonable" when referring to markups, markdowns and commissions that must be "fair" to incorporate the more widely used phrase "fair and reasonable." 10 The results are based on the 82 percent of the membership that responded to the survey. 11 See, e.g., Law Office of Scott T. Beall (Beall), Law Office of Steve A. Buchwalter, P.C. (Buchwalter), Churchill Financial, LLC (Churchill), Compliance-by-Proxy (CBP), Cornell Securities Law Clinic (Cornell), Barry D. Estell (Estell), William Gladden (Gladden), Ledbetter & Associates P.A. (Ledbetter), North American Securities Administrators Association (NASAA), Public Investors Arbitration Bar Association (PIABA), Jeffrey R. Sonn, Esq. (Sonn), St. John's School of Law Clinic (St. John's), and Wells Fargo Advisors (WFA). Six commenters favored retiring the 5% policy. See letters from Securities Industry and Financial Markets Association (SIFMA), Financial Services Institute (FSI), Cambridge Investment Research (Cambridge), JW Korth, Moloney Securities, Inc. (Moloney), and National Planning Holdings, Inc. (NPH). However, three of the commenters, SIFMA, FSI and Cambridge, stated that the 5% policy should not be withdrawn unless FINRA provided to the membership, before or at the same time, the "Markup Threshold Guidance" or similar guidance. 12 In light of the proposal to retain the 5% policy, FINRA does not intend at this time to provide "Markup Threshold Guidance" in a separate Regulatory Notice. 13 FINRA Rule 2121(b)(1) through (b)(4) as initially proposed would be renumbered, respectively, as proposed FINRA Rule 2121(b)(2) through (b)(5). 14 See letters from Beall, Buchwalter, Cornell, Estell, Gladden, Ledbetter, NASAA, and PIABA opposing the deletion of the proceeds provision. See letters from SIFMA and Roberts & Ryan Investments, Inc. (R&R) in favor of deleting the provision. 15 For example, it is not always clear when two transactions occurring close in time are related (the two transactions may represent unrelated investment decisions) or how close in time transactions must be to be considered "proceeds" transactions. In addition, the proceeds provision may not be applied when a customer decides to sell a position at one member firm and purchase a position at another member firm. 16 See, e.g., letters from Cambridge, CBP, Churchill, FSI, Moloney, NPH, Regal Bay Investment Groups, R& R, SIFMA, and WFA opposing the requirement to provide equity commission schedules to retail customers; letter from Juanita D. Hanley noting certain limitations of the proposed requirement; and letters from Cornell, St John's, NASAA, and Sonn supporting the proposed requirement. 17 Proposed FINRA Rule 2121(e) would be a new requirement for former NASD-only members. As discussed in the initial proposal, FINRA proposes to transfer these requirements because there are no similar requirements in the NASD markup rules regarding whether, and under what circumstances, a member firm may charge a commission if a member "misses the market." 18 NASD IM-2440-2 would be deleted with the adoption of FINRA Rule 2122. 19 See 17 CFR § 230.144A(a)(1). 20 For the purpose of the rule, the proposal would adopt the definition of "non-investment grade debt security" in NASD IM-2440-2(b)(9) with no change. 21 See supra note 6. 22 NASD Rule 2430 would be deleted with the adoption of FINRA Rule 2123. See Regulator Notice 11-08. See also Notice to Members 92-11 (Fees and Charges for Services). 23 This is largely for historical reasons. The Government Securities Act Amendments of 1993 (GSAA) eliminated the statutory limitations on NASD's authority to apply sales practice rules to transactions in exempted securities, except municipal securities. NASD undertook to review the specific application of certain of its rules, including the NASD markup rules then in effect (Rules 2440 and IM-2440-1), to the government securities market. See Notice to Members 96-66 (October 1996). NASD IM-2440-2 (the debt markup interpretation)-approved in 2007- had not been adopted at the time FINRA Rule 0150 (then NASD Rule 0116) went into effect. See Securities Exchange Act Release No. 44631 (July 31, 2001), 66 FR 41283 (August 7, 2001) (Order Approving File No. SR-NASD-2000-038 (NASD Rule 0116)); see also Securities Exchange Act Release No. 55638 (April 16, 2007), 72 FR 20150 (April 23, 2007) (Order Approving File No. SR-NASD-2003-141 (NASD IM-2440-2)). 24 In Notice to Members 96-66, FINRA noted that actions for conduct generally encompassed by the NASD markup rules, among others, in the government securities market may be brought under FINRA Rule 2010. See also Securities Exchange Act Release No. 37588 (August 20, 1996), 61 FR 44100 (August 27, 1996) (Order Approving File No. SR-NASD-95-39) (1996 Approval Order). 25 "U.S. Treasury security" is defined in FINRA Rule 6710(p) to mean a security issued by the U.S. Department of the Treasury to fund the operations of the federal government or to retire such outstanding securities. 26 See supra note 24, the 1996 Approval Order at 61 FR 44104. \| Date \| Commenter \| --- \| \| 2013-03-01 \| Roberto A. Eder Sr. J.D. Comments on Regulatory Notice 13-07 \| \| 2013-03-15 \| SIFMA Comments on Regulatory Notice 13-07 \| \| 2013-03-26 \| PIABA Comments on Regulatory Notice 13-07 \| \| 2013-03-28 \| Cornell Securities Law Clinic Comments on Regulatory Notice 13-07 \| \| 2013-04-01 \| First Asset Financial Inc. Comments on Regulatory Notice 13-07 \| \| 2013-04-01 \| Financial Services Institute Comments on Regulatory Notice 13-07 \| \| 2013-04-02 \| Bond Dealers of America Comments on Regulatory Notice 13-07 \| \| 2013-04-18 \| Heritage Bank of Nevada Comments on Regulatory Notice 13-07 \| Notice Attachment Referenced Rules & Notices FINRA Utility Menu FINRA Main Navigation General Inquiries 301-590-6500 Securities Helpline for Seniors® 844-574-3577 (Mon-Fri 9am-5pm ET) File a Regulatory Tip To report on abuse or fraud in the industry Arbitration & Mediation FINRA operates the largest securities dispute resolution forum in the United States File an Investor Complaint File a complaint about fraud or unfair practices. Small Firm Help Line 833-26-FINRA (Mon-Fri 9am-6pm ET) Office of the Ombuds Report a concern about FINRA at 888-700-0028 Footer Legal Links © 2025 FINRA. All Rights Reserved. FINRA is a Registered Trademark of the Financial Industry Regulatory Authority, Inc.
15959	https://artofproblemsolving.com/community?srsltid=AfmBOoqmvR1DhYDErNjfcJokFlqGbk8v4q3oqHzkCKufExS4_w2ady2l	AoPS Community Math Message Boards & Community Forums \| AoPS 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 1-12. 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Community Middle School Math Grades 5-8, Ages 10-13, MATHCOUNTS, AMC 8 Grades 5-8, Ages 10-13, MATHCOUNTS, AMC 8 3 V New Topic k Locked Middle School Math Grades 5-8, Ages 10-13, MATHCOUNTS, AMC 8 Grades 5-8, Ages 10-13, MATHCOUNTS, AMC 8 3 V New Topic k Locked t 51,944 topics w 1 user TOPICS G Topic First Poster Last Poster H k a September Highlights and 2025 AoPS Online Class Information jwelsh 0 It’s back to school time! A new academic year is a new opportunity to train for the upcoming competition season. Our Worldwide Online Olympiad Training (WOOT) classes begin the first week of September, so be sure to take advantage of the only program where you learn the skills and strategies necessary to succeed at the Olympiad Level. This year alone, Art of Problem Solving students from around the globe racked up over 100 total medals at the International Mathematics Olympiad, including 25 gold medals in ‘25! "As a parent, I'm deeply grateful to AoPS. Tiger has taken very few math courses outside of AoPS, except for a local Math Circle that doesn't focus on Olympiad math. AoPS has been one of the most important resources in his journey. Without AoPS, Tiger wouldn't be where he is today — especially considering he's grown up in a family with no STEM background at all." — Doreen Dai, parent of IMO US 2025 Team Member Tiger Zhang Be sure to read the article by our CEO, Ben Kornell, about his experience at the IMO in 2025! Below is a list of the different WOOT programs and the competitions they cover. [list][]MathWOOT Level 1: Designed for AIME qualifiers ready to make the jump to Olympiads. []MathWOOT Level 2: Designed for Olympiad qualifiers ready to increase their scores in national and international Olympiads. []CodeWOOT: USACO, IOI []PhysicsWOOT: USAPhO, SIN, IPhO, F=ma exam []ChemWOOT: USNCO, IChO For those not quite ready for CodeWOOT and are interested in coding competitions, check out our USACO Bronze course! To be eligible, students should already be comfortable enough with C++, Java, or Python to write simple programs using basic concepts like arrays, maps/sets, if statements, and for loops. Note that either Java or Python is sufficient for the USACO Bronze and Silver levels (AoPS offers Python courses), but USACO Gold and above (and most programming contests) essentially require C++, and the IOI only supports C++. Keep in mind some other important dates coming up fast! [list][]AMC 10/12 competitions are right around the corner! Take advantage of our accelerated AMC 10 and AMC 12 Problem Series courses which run from October until right before the competitions begin! In these courses, you will learn test taking strategies and have the opportunity to take a practice exam - efficiently train to succeed in this year’s competitions! []The AMC 10A/12A will be held on November 5th []The AMC 10B/12B will be held on November 13th. []USAMTS (United States of America Mathematical Talent Search) will release their first round of problems soon after Labor day! This contest is another method for qualifying for AIME. As an extra bonus, students can use the comments they receive on their USAMTS solutions to hone their skills for Olympiad level contests! The Round 1 problems will be available at Students should submit solutions to the Round 1 problems by mid-October. []Admissions for MIT Primes opens on October 1st! MIT PRIMES is a free, year-long program, in which high school students work on individual and group research projects and participate in reading groups under the guidance of academic mentors. MIT PRIMES includes three sections: mathematics, computer science, and computational and physical biology. Our full course list for upcoming classes is below: All classes run 7:30pm-8:45pm ET/4:30pm - 5:45pm PT unless otherwise noted. Introductory: Grades 5-10 Prealgebra 1 Self-Paced Prealgebra 1 Friday, Sep 5 - Jan 16 Monday, Sep 8 - Jan 12 Tuesday, Sep 16 - Jan 20 (4:30 - 5:45 pm ET/1:30 - 2:45 pm PT) Sunday, Sep 21 - Jan 25 Thursday, Sep 25 - Jan 29 Wednesday, Oct 22 - Feb 25 Tuesday, Nov 4 - Mar 10 Friday, Dec 12 - Apr 10 Prealgebra 2 Self-Paced Prealgebra 2 Tuesday, Sep 9 - Jan 13 Thursday, Sep 25 - Jan 29 Sunday, Oct 19 - Feb 22 Monday, Oct 27 - Mar 2 Wednesday, Nov 12 - Mar 18 Introduction to Algebra A Self-Paced Introduction to Algebra A Friday, Sep 5 - Jan 16 Thursday, Sep 11 - Jan 15 Sunday, Sep 28 - Feb 1 Monday, Oct 6 - Feb 9 Tuesday, Oct 21 - Feb 24 Sunday, Nov 9 - Mar 15 Friday, Dec 5 - Apr 3 Introduction to Counting & Probability Self-Paced Introduction to Counting & Probability Wednesday, Sep 3 - Nov 19 Sunday, Sep 21 - Dec 14 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT) Friday, Oct 3 - Jan 16 Sunday, Oct 19 - Jan 25 Tuesday, Nov 4 - Feb 10 Sunday, Dec 7 - Mar 8 Introduction to Number Theory Friday, Sep 12 - Dec 12 Sunday, Oct 26 - Feb 1 Monday, Dec 1 - Mar 2 Introduction to Algebra B Self-Paced Introduction to Algebra B Sunday, Sep 7 - Jan 11 Thursday, Sep 11 - Jan 15 Wednesday, Sep 24 - Jan 28 Sunday, Oct 26 - Mar 1 Tuesday, Nov 4 - Mar 10 Monday, Dec 1 - Mar 30 Introduction to Geometry Sunday, Sep 7 - Mar 8 Thursday, Sep 11 - Mar 12 Wednesday, Sep 24 - Mar 25 Sunday, Oct 26 - Apr 26 Monday, Nov 3 - May 4 Friday, Dec 5 - May 29 Paradoxes and Infinity Sat & Sun, Sep 13 - Sep 14 (1:00 - 4:00 PM PT/4:00 - 7:00 PM ET) Intermediate: Grades 8-12 Intermediate Algebra Sunday, Sep 28 - Mar 29 Wednesday, Oct 8 - Mar 8 Sunday, Nov 16 - May 17 Thursday, Dec 11 - Jun 4 Intermediate Counting & Probability Sunday, Sep 28 - Feb 15 Tuesday, Nov 4 - Mar 24 Intermediate Number Theory Wednesday, Sep 24 - Dec 17 Precalculus Tuesday, Sep 9 - Feb 24 Sunday, Sep 21 - Mar 8 Monday, Oct 20 - Apr 6 Sunday, Dec 14 - May 31 Advanced: Grades 9-12 Calculus Sunday, Sep 7 - Mar 15 Wednesday, Sep 24 - Apr 1 Friday, Nov 14 - May 22 Contest Preparation: Grades 6-12 MATHCOUNTS/AMC 8 Basics Wednesday, Sep 3 - Nov 19 Tuesday, Sep 16 - Dec 9 Sunday, Sep 21 - Dec 14 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT) Monday, Oct 6 - Jan 12 Thursday, Oct 16 - Jan 22 Tues, Thurs & Sun, Dec 9 - Jan 18 (meets three times a week!) MATHCOUNTS/AMC 8 Advanced Thursday, Sep 4 - Nov 20 Friday, Sep 12 - Dec 12 Monday, Sep 15 - Dec 8 Sunday, Oct 5 - Jan 11 Tues, Thurs & Sun, Dec 2 - Jan 11 (meets three times a week!) Mon, Wed & Fri, Dec 8 - Jan 16 (meets three times a week!) AMC 10 Problem Series Mon & Wed, Sep 15 - Oct 22 (meets twice a week!) Mon, Wed & Fri, Oct 6 - Nov 3 (meets three times a week!) Tue, Thurs & Sun, Oct 7 - Nov 2 (meets three times a week!) AMC 10 Final Fives Sunday, Sep 7 - Sep 28 Tuesday, Sep 9 - Sep 30 Monday, Sep 22 - Oct 13 Sunday, Sep 28 - Oct 19 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT) Wednesday, Oct 8 - Oct 29 Thursday, Oct 9 - Oct 30 AMC 12 Problem Series Mon & Wed, Sep 15 - Oct 22 (meets twice a week!) Tues, Thurs & Sun, Oct 7 - Nov 2 (meets three times a week!) AMC 12 Final Fives Thursday, Sep 4 - Sep 25 Sunday, Sep 28 - Oct 19 Tuesday, Oct 7 - Oct 28 AIME Problem Series A Thursday, Oct 23 - Jan 29 AIME Problem Series B Tuesday, Sep 2 - Nov 18 F=ma Problem Series Tuesday, Sep 16 - Dec 9 Friday, Oct 17 - Jan 30 WOOT Programs Visit the pages linked for full schedule details for each of these programs! MathWOOT Level 1 MathWOOT Level 2 ChemWOOT CodeWOOT PhysicsWOOT Programming Introduction to Programming with Python Sunday, Sep 7 - Nov 23 Tuesday, Dec 2 - Mar 3 Intermediate Programming with Python Friday, Oct 3 - Jan 16 USACO Bronze Problem Series Wednesday, Sep 3 - Dec 3 Thursday, Oct 30 - Feb 5 Tuesday, Dec 2 - Mar 3 Physics Introduction to Physics Tuesday, Sep 2 - Nov 18 Sunday, Oct 5 - Jan 11 Wednesday, Dec 10 - Mar 11 Physics 1: Mechanics Sunday, Sep 21 - Mar 22 Sunday, Oct 26 - Apr 26 0 replies jwelsh Sep 2, 2025 0 replies J H A weird congruent case of triangles RopuToran 0 Given two acute triangles and satisfying ,,. Prove that 0 replies RopuToran an hour ago 0 replies J H 9 Pythagorean Triples ZMB038 146 Please put some of the ones you know, and try not to troll/start flame wars! Thank you :D 146 replies ZMB038 May 19, 2025 melloncandy Today at 3:21 AM J H Percent & Saline Mix Kushagra2012 2 How many liters of a % saline solution must be added to liters of a % saline solution to produce a % saline solution? %? %. Any pattern? 2 replies Kushagra2012 Saturday at 9:32 PM evt917 Today at 2:35 AM J H Two Liar Problems matharama 11 1 Frederick never tells the truth on Tuesdays, Thursdays, and Saturdays. He always tells the truth on Sundays, Mondays, Wednesdays, and Fridays. One day Matthew had this conversation with Frederick: "What day is it today?" asked Matthew. "Saturday," Frederick replied. "And what day will it be tomorrow?" "Wednesday." On which day did this conversation take place? First to answer correctly goes on the leaderboard! Leaderboard 2 The Liar Paradox Important Note! There is no correct solution. Therefore, there is no leaderboard! A man say that he is lying. Is what he says true or false? 11 replies matharama Sep 26, 2025 Wax_Wax Today at 2:03 AM J High School Math Grades 9-12, Ages 13-18 Grades 9-12, Ages 13-18 3 V New Topic k Locked High School Math Grades 9-12, Ages 13-18 Grades 9-12, Ages 13-18 3 V New Topic k Locked t 122,700 topics w 1 user TOPICS G Topic First Poster Last Poster H k a September Highlights and 2025 AoPS Online Class Information jwelsh 0 It’s back to school time! A new academic year is a new opportunity to train for the upcoming competition season. Our Worldwide Online Olympiad Training (WOOT) classes begin the first week of September, so be sure to take advantage of the only program where you learn the skills and strategies necessary to succeed at the Olympiad Level. This year alone, Art of Problem Solving students from around the globe racked up over 100 total medals at the International Mathematics Olympiad, including 25 gold medals in ‘25! "As a parent, I'm deeply grateful to AoPS. Tiger has taken very few math courses outside of AoPS, except for a local Math Circle that doesn't focus on Olympiad math. AoPS has been one of the most important resources in his journey. Without AoPS, Tiger wouldn't be where he is today — especially considering he's grown up in a family with no STEM background at all." — Doreen Dai, parent of IMO US 2025 Team Member Tiger Zhang Be sure to read the article by our CEO, Ben Kornell, about his experience at the IMO in 2025! Below is a list of the different WOOT programs and the competitions they cover. [list][]MathWOOT Level 1: Designed for AIME qualifiers ready to make the jump to Olympiads. []MathWOOT Level 2: Designed for Olympiad qualifiers ready to increase their scores in national and international Olympiads. []CodeWOOT: USACO, IOI []PhysicsWOOT: USAPhO, SIN, IPhO, F=ma exam []ChemWOOT: USNCO, IChO For those not quite ready for CodeWOOT and are interested in coding competitions, check out our USACO Bronze course! To be eligible, students should already be comfortable enough with C++, Java, or Python to write simple programs using basic concepts like arrays, maps/sets, if statements, and for loops. Note that either Java or Python is sufficient for the USACO Bronze and Silver levels (AoPS offers Python courses), but USACO Gold and above (and most programming contests) essentially require C++, and the IOI only supports C++. Keep in mind some other important dates coming up fast! [list][]AMC 10/12 competitions are right around the corner! Take advantage of our accelerated AMC 10 and AMC 12 Problem Series courses which run from October until right before the competitions begin! In these courses, you will learn test taking strategies and have the opportunity to take a practice exam - efficiently train to succeed in this year’s competitions! []The AMC 10A/12A will be held on November 5th []The AMC 10B/12B will be held on November 13th. []USAMTS (United States of America Mathematical Talent Search) will release their first round of problems soon after Labor day! This contest is another method for qualifying for AIME. As an extra bonus, students can use the comments they receive on their USAMTS solutions to hone their skills for Olympiad level contests! The Round 1 problems will be available at Students should submit solutions to the Round 1 problems by mid-October. []Admissions for MIT Primes opens on October 1st! MIT PRIMES is a free, year-long program, in which high school students work on individual and group research projects and participate in reading groups under the guidance of academic mentors. MIT PRIMES includes three sections: mathematics, computer science, and computational and physical biology. Our full course list for upcoming classes is below: All classes run 7:30pm-8:45pm ET/4:30pm - 5:45pm PT unless otherwise noted. Introductory: Grades 5-10 Prealgebra 1 Self-Paced Prealgebra 1 Friday, Sep 5 - Jan 16 Monday, Sep 8 - Jan 12 Tuesday, Sep 16 - Jan 20 (4:30 - 5:45 pm ET/1:30 - 2:45 pm PT) Sunday, Sep 21 - Jan 25 Thursday, Sep 25 - Jan 29 Wednesday, Oct 22 - Feb 25 Tuesday, Nov 4 - Mar 10 Friday, Dec 12 - Apr 10 Prealgebra 2 Self-Paced Prealgebra 2 Tuesday, Sep 9 - Jan 13 Thursday, Sep 25 - Jan 29 Sunday, Oct 19 - Feb 22 Monday, Oct 27 - Mar 2 Wednesday, Nov 12 - Mar 18 Introduction to Algebra A Self-Paced Introduction to Algebra A Friday, Sep 5 - Jan 16 Thursday, Sep 11 - Jan 15 Sunday, Sep 28 - Feb 1 Monday, Oct 6 - Feb 9 Tuesday, Oct 21 - Feb 24 Sunday, Nov 9 - Mar 15 Friday, Dec 5 - Apr 3 Introduction to Counting & Probability Self-Paced Introduction to Counting & Probability Wednesday, Sep 3 - Nov 19 Sunday, Sep 21 - Dec 14 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT) Friday, Oct 3 - Jan 16 Sunday, Oct 19 - Jan 25 Tuesday, Nov 4 - Feb 10 Sunday, Dec 7 - Mar 8 Introduction to Number Theory Friday, Sep 12 - Dec 12 Sunday, Oct 26 - Feb 1 Monday, Dec 1 - Mar 2 Introduction to Algebra B Self-Paced Introduction to Algebra B Sunday, Sep 7 - Jan 11 Thursday, Sep 11 - Jan 15 Wednesday, Sep 24 - Jan 28 Sunday, Oct 26 - Mar 1 Tuesday, Nov 4 - Mar 10 Monday, Dec 1 - Mar 30 Introduction to Geometry Sunday, Sep 7 - Mar 8 Thursday, Sep 11 - Mar 12 Wednesday, Sep 24 - Mar 25 Sunday, Oct 26 - Apr 26 Monday, Nov 3 - May 4 Friday, Dec 5 - May 29 Paradoxes and Infinity Sat & Sun, Sep 13 - Sep 14 (1:00 - 4:00 PM PT/4:00 - 7:00 PM ET) Intermediate: Grades 8-12 Intermediate Algebra Sunday, Sep 28 - Mar 29 Wednesday, Oct 8 - Mar 8 Sunday, Nov 16 - May 17 Thursday, Dec 11 - Jun 4 Intermediate Counting & Probability Sunday, Sep 28 - Feb 15 Tuesday, Nov 4 - Mar 24 Intermediate Number Theory Wednesday, Sep 24 - Dec 17 Precalculus Tuesday, Sep 9 - Feb 24 Sunday, Sep 21 - Mar 8 Monday, Oct 20 - Apr 6 Sunday, Dec 14 - May 31 Advanced: Grades 9-12 Calculus Sunday, Sep 7 - Mar 15 Wednesday, Sep 24 - Apr 1 Friday, Nov 14 - May 22 Contest Preparation: Grades 6-12 MATHCOUNTS/AMC 8 Basics Wednesday, Sep 3 - Nov 19 Tuesday, Sep 16 - Dec 9 Sunday, Sep 21 - Dec 14 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT) Monday, Oct 6 - Jan 12 Thursday, Oct 16 - Jan 22 Tues, Thurs & Sun, Dec 9 - Jan 18 (meets three times a week!) MATHCOUNTS/AMC 8 Advanced Thursday, Sep 4 - Nov 20 Friday, Sep 12 - Dec 12 Monday, Sep 15 - Dec 8 Sunday, Oct 5 - Jan 11 Tues, Thurs & Sun, Dec 2 - Jan 11 (meets three times a week!) Mon, Wed & Fri, Dec 8 - Jan 16 (meets three times a week!) AMC 10 Problem Series Mon & Wed, Sep 15 - Oct 22 (meets twice a week!) Mon, Wed & Fri, Oct 6 - Nov 3 (meets three times a week!) Tue, Thurs & Sun, Oct 7 - Nov 2 (meets three times a week!) AMC 10 Final Fives Sunday, Sep 7 - Sep 28 Tuesday, Sep 9 - Sep 30 Monday, Sep 22 - Oct 13 Sunday, Sep 28 - Oct 19 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT) Wednesday, Oct 8 - Oct 29 Thursday, Oct 9 - Oct 30 AMC 12 Problem Series Mon & Wed, Sep 15 - Oct 22 (meets twice a week!) Tues, Thurs & Sun, Oct 7 - Nov 2 (meets three times a week!) AMC 12 Final Fives Thursday, Sep 4 - Sep 25 Sunday, Sep 28 - Oct 19 Tuesday, Oct 7 - Oct 28 AIME Problem Series A Thursday, Oct 23 - Jan 29 AIME Problem Series B Tuesday, Sep 2 - Nov 18 F=ma Problem Series Tuesday, Sep 16 - Dec 9 Friday, Oct 17 - Jan 30 WOOT Programs Visit the pages linked for full schedule details for each of these programs! MathWOOT Level 1 MathWOOT Level 2 ChemWOOT CodeWOOT PhysicsWOOT Programming Introduction to Programming with Python Sunday, Sep 7 - Nov 23 Tuesday, Dec 2 - Mar 3 Intermediate Programming with Python Friday, Oct 3 - Jan 16 USACO Bronze Problem Series Wednesday, Sep 3 - Dec 3 Thursday, Oct 30 - Feb 5 Tuesday, Dec 2 - Mar 3 Physics Introduction to Physics Tuesday, Sep 2 - Nov 18 Sunday, Oct 5 - Jan 11 Wednesday, Dec 10 - Mar 11 Physics 1: Mechanics Sunday, Sep 21 - Mar 22 Sunday, Oct 26 - Apr 26 0 replies jwelsh Sep 2, 2025 0 replies J H IOQM 2025 Region 04 and 05 P17 Lunatic_Lunar7986 5 If where are positive integers, find . 5 replies Lunatic_Lunar7986 Today at 3:34 AM Lunatic_Lunar7986 28 minutes ago J H IOQM 2025 Region 04 and 05 P25 Lunatic_Lunar7986 3 How many natural numbere are there such that ? 3 replies Lunatic_Lunar7986 Today at 3:14 AM Royrik123456 30 minutes ago J H Divisibility Mathelets 3 Determine the smallest positive integer with the property that is divisible by both and . 3 replies Mathelets 2 hours ago HAL9000sk 36 minutes ago J H IOQM 2025 Region 04 and 05 P16 Lunatic_Lunar7986 3 Find the number of ordered pairs , where and are positive integers such that and is a perfect square. 3 replies Lunatic_Lunar7986 Today at 2:55 AM Lunatic_Lunar7986 44 minutes ago J Contests & Programs AMC and other contests, summer programs, etc. AMC and other contests, summer programs, etc. 3 V New Topic k Locked Contests & Programs AMC and other contests, summer programs, etc. AMC and other contests, summer programs, etc. 3 V New Topic k Locked t 35,030 topics w 3 users TOPICS G Topic First Poster Last Poster H k a September Highlights and 2025 AoPS Online Class Information jwelsh 0 It’s back to school time! A new academic year is a new opportunity to train for the upcoming competition season. Our Worldwide Online Olympiad Training (WOOT) classes begin the first week of September, so be sure to take advantage of the only program where you learn the skills and strategies necessary to succeed at the Olympiad Level. This year alone, Art of Problem Solving students from around the globe racked up over 100 total medals at the International Mathematics Olympiad, including 25 gold medals in ‘25! "As a parent, I'm deeply grateful to AoPS. Tiger has taken very few math courses outside of AoPS, except for a local Math Circle that doesn't focus on Olympiad math. AoPS has been one of the most important resources in his journey. Without AoPS, Tiger wouldn't be where he is today — especially considering he's grown up in a family with no STEM background at all." — Doreen Dai, parent of IMO US 2025 Team Member Tiger Zhang Be sure to read the article by our CEO, Ben Kornell, about his experience at the IMO in 2025! Below is a list of the different WOOT programs and the competitions they cover. [list][]MathWOOT Level 1: Designed for AIME qualifiers ready to make the jump to Olympiads. []MathWOOT Level 2: Designed for Olympiad qualifiers ready to increase their scores in national and international Olympiads. []CodeWOOT: USACO, IOI []PhysicsWOOT: USAPhO, SIN, IPhO, F=ma exam []ChemWOOT: USNCO, IChO For those not quite ready for CodeWOOT and are interested in coding competitions, check out our USACO Bronze course! To be eligible, students should already be comfortable enough with C++, Java, or Python to write simple programs using basic concepts like arrays, maps/sets, if statements, and for loops. Note that either Java or Python is sufficient for the USACO Bronze and Silver levels (AoPS offers Python courses), but USACO Gold and above (and most programming contests) essentially require C++, and the IOI only supports C++. Keep in mind some other important dates coming up fast! [list][]AMC 10/12 competitions are right around the corner! Take advantage of our accelerated AMC 10 and AMC 12 Problem Series courses which run from October until right before the competitions begin! In these courses, you will learn test taking strategies and have the opportunity to take a practice exam - efficiently train to succeed in this year’s competitions! []The AMC 10A/12A will be held on November 5th []The AMC 10B/12B will be held on November 13th. []USAMTS (United States of America Mathematical Talent Search) will release their first round of problems soon after Labor day! This contest is another method for qualifying for AIME. As an extra bonus, students can use the comments they receive on their USAMTS solutions to hone their skills for Olympiad level contests! The Round 1 problems will be available at Students should submit solutions to the Round 1 problems by mid-October. []Admissions for MIT Primes opens on October 1st! MIT PRIMES is a free, year-long program, in which high school students work on individual and group research projects and participate in reading groups under the guidance of academic mentors. MIT PRIMES includes three sections: mathematics, computer science, and computational and physical biology. Our full course list for upcoming classes is below: All classes run 7:30pm-8:45pm ET/4:30pm - 5:45pm PT unless otherwise noted. Introductory: Grades 5-10 Prealgebra 1 Self-Paced Prealgebra 1 Friday, Sep 5 - Jan 16 Monday, Sep 8 - Jan 12 Tuesday, Sep 16 - Jan 20 (4:30 - 5:45 pm ET/1:30 - 2:45 pm PT) Sunday, Sep 21 - Jan 25 Thursday, Sep 25 - Jan 29 Wednesday, Oct 22 - Feb 25 Tuesday, Nov 4 - Mar 10 Friday, Dec 12 - Apr 10 Prealgebra 2 Self-Paced Prealgebra 2 Tuesday, Sep 9 - Jan 13 Thursday, Sep 25 - Jan 29 Sunday, Oct 19 - Feb 22 Monday, Oct 27 - Mar 2 Wednesday, Nov 12 - Mar 18 Introduction to Algebra A Self-Paced Introduction to Algebra A Friday, Sep 5 - Jan 16 Thursday, Sep 11 - Jan 15 Sunday, Sep 28 - Feb 1 Monday, Oct 6 - Feb 9 Tuesday, Oct 21 - Feb 24 Sunday, Nov 9 - Mar 15 Friday, Dec 5 - Apr 3 Introduction to Counting & Probability Self-Paced Introduction to Counting & Probability Wednesday, Sep 3 - Nov 19 Sunday, Sep 21 - Dec 14 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT) Friday, Oct 3 - Jan 16 Sunday, Oct 19 - Jan 25 Tuesday, Nov 4 - Feb 10 Sunday, Dec 7 - Mar 8 Introduction to Number Theory Friday, Sep 12 - Dec 12 Sunday, Oct 26 - Feb 1 Monday, Dec 1 - Mar 2 Introduction to Algebra B Self-Paced Introduction to Algebra B Sunday, Sep 7 - Jan 11 Thursday, Sep 11 - Jan 15 Wednesday, Sep 24 - Jan 28 Sunday, Oct 26 - Mar 1 Tuesday, Nov 4 - Mar 10 Monday, Dec 1 - Mar 30 Introduction to Geometry Sunday, Sep 7 - Mar 8 Thursday, Sep 11 - Mar 12 Wednesday, Sep 24 - Mar 25 Sunday, Oct 26 - Apr 26 Monday, Nov 3 - May 4 Friday, Dec 5 - May 29 Paradoxes and Infinity Sat & Sun, Sep 13 - Sep 14 (1:00 - 4:00 PM PT/4:00 - 7:00 PM ET) Intermediate: Grades 8-12 Intermediate Algebra Sunday, Sep 28 - Mar 29 Wednesday, Oct 8 - Mar 8 Sunday, Nov 16 - May 17 Thursday, Dec 11 - Jun 4 Intermediate Counting & Probability Sunday, Sep 28 - Feb 15 Tuesday, Nov 4 - Mar 24 Intermediate Number Theory Wednesday, Sep 24 - Dec 17 Precalculus Tuesday, Sep 9 - Feb 24 Sunday, Sep 21 - Mar 8 Monday, Oct 20 - Apr 6 Sunday, Dec 14 - May 31 Advanced: Grades 9-12 Calculus Sunday, Sep 7 - Mar 15 Wednesday, Sep 24 - Apr 1 Friday, Nov 14 - May 22 Contest Preparation: Grades 6-12 MATHCOUNTS/AMC 8 Basics Wednesday, Sep 3 - Nov 19 Tuesday, Sep 16 - Dec 9 Sunday, Sep 21 - Dec 14 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT) Monday, Oct 6 - Jan 12 Thursday, Oct 16 - Jan 22 Tues, Thurs & Sun, Dec 9 - Jan 18 (meets three times a week!) MATHCOUNTS/AMC 8 Advanced Thursday, Sep 4 - Nov 20 Friday, Sep 12 - Dec 12 Monday, Sep 15 - Dec 8 Sunday, Oct 5 - Jan 11 Tues, Thurs & Sun, Dec 2 - Jan 11 (meets three times a week!) Mon, Wed & Fri, Dec 8 - Jan 16 (meets three times a week!) AMC 10 Problem Series Mon & Wed, Sep 15 - Oct 22 (meets twice a week!) Mon, Wed & Fri, Oct 6 - Nov 3 (meets three times a week!) Tue, Thurs & Sun, Oct 7 - Nov 2 (meets three times a week!) AMC 10 Final Fives Sunday, Sep 7 - Sep 28 Tuesday, Sep 9 - Sep 30 Monday, Sep 22 - Oct 13 Sunday, Sep 28 - Oct 19 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT) Wednesday, Oct 8 - Oct 29 Thursday, Oct 9 - Oct 30 AMC 12 Problem Series Mon & Wed, Sep 15 - Oct 22 (meets twice a week!) Tues, Thurs & Sun, Oct 7 - Nov 2 (meets three times a week!) AMC 12 Final Fives Thursday, Sep 4 - Sep 25 Sunday, Sep 28 - Oct 19 Tuesday, Oct 7 - Oct 28 AIME Problem Series A Thursday, Oct 23 - Jan 29 AIME Problem Series B Tuesday, Sep 2 - Nov 18 F=ma Problem Series Tuesday, Sep 16 - Dec 9 Friday, Oct 17 - Jan 30 WOOT Programs Visit the pages linked for full schedule details for each of these programs! MathWOOT Level 1 MathWOOT Level 2 ChemWOOT CodeWOOT PhysicsWOOT Programming Introduction to Programming with Python Sunday, Sep 7 - Nov 23 Tuesday, Dec 2 - Mar 3 Intermediate Programming with Python Friday, Oct 3 - Jan 16 USACO Bronze Problem Series Wednesday, Sep 3 - Dec 3 Thursday, Oct 30 - Feb 5 Tuesday, Dec 2 - Mar 3 Physics Introduction to Physics Tuesday, Sep 2 - Nov 18 Sunday, Oct 5 - Jan 11 Wednesday, Dec 10 - Mar 11 Physics 1: Mechanics Sunday, Sep 21 - Mar 22 Sunday, Oct 26 - Apr 26 0 replies jwelsh Sep 2, 2025 0 replies J H force overlay inversion vibes v4913 68 Source: USAMO 2023/6 Let be a triangle with incenter and excenters ,, and opposite ,, and , respectively. Let be an arbitrary point on the circumcircle of that does not lie on any of the lines ,, or . Suppose the circumcircles of and intersect at two distinct points and . If is the intersection of lines and , prove that . Proposed by Zach Chroman 68 replies v4913 Mar 23, 2023 bin_sherlo 2 hours ago J H States Question xHypotenuse 3 Hello everyone, I have a bit of confusion regarding states. Here is the question: when should I recursively define states, and when should I "forwardly" define states? For example, a "forwardly" defined states problem would be 2017 AMC 12A Problem 22. In this case, the states would be "forwardly" defined because we calculate the states based on how to go to the "next" state, not how we arrived upon it. An example of a "recursively" defined state would be 2021 aime ii, problem 8. This is because our states depend on how we arrived upon it, not how we go to the next one. From my understanding, we use recursively defined states if there is a set time/#steps involved but we use "forwardly" defined states in any expected value or "eventually reach" problems (any states problem with an absorbing state). I'm sorry if this question is unclear; I really have no idea how to phrase it. That being said, if anyone can understand/help, that would be really appreciate it! Thanks :) 3 replies xHypotenuse Sep 26, 2025 nsking_1209 4 hours ago J H 9 What do you guys most strongest in, for Amc 10 littleduckysteve 0 Just for fun, what are you guys good at for amc 10? 0 replies littleduckysteve 5 hours ago 0 replies J H 9 What do you guys find difficult about the amc 10. littleduckysteve 0 Just for fun, let's see what people find difficult on the Amc 10. I will make another poll for strengths too. 0 replies littleduckysteve 5 hours ago 0 replies J High School Olympiads Regional, national, and international math olympiads Regional, national, and international math olympiads 3 V New Topic k Locked High School Olympiads Regional, national, and international math olympiads Regional, national, and international math olympiads 3 V New Topic k Locked t 317,096 topics w 10 users TOPICS G Topic First Poster Last Poster H k a September Highlights and 2025 AoPS Online Class Information jwelsh 0 It’s back to school time! A new academic year is a new opportunity to train for the upcoming competition season. Our Worldwide Online Olympiad Training (WOOT) classes begin the first week of September, so be sure to take advantage of the only program where you learn the skills and strategies necessary to succeed at the Olympiad Level. This year alone, Art of Problem Solving students from around the globe racked up over 100 total medals at the International Mathematics Olympiad, including 25 gold medals in ‘25! "As a parent, I'm deeply grateful to AoPS. Tiger has taken very few math courses outside of AoPS, except for a local Math Circle that doesn't focus on Olympiad math. AoPS has been one of the most important resources in his journey. Without AoPS, Tiger wouldn't be where he is today — especially considering he's grown up in a family with no STEM background at all." — Doreen Dai, parent of IMO US 2025 Team Member Tiger Zhang Be sure to read the article by our CEO, Ben Kornell, about his experience at the IMO in 2025! Below is a list of the different WOOT programs and the competitions they cover. [list][]MathWOOT Level 1: Designed for AIME qualifiers ready to make the jump to Olympiads. []MathWOOT Level 2: Designed for Olympiad qualifiers ready to increase their scores in national and international Olympiads. []CodeWOOT: USACO, IOI []PhysicsWOOT: USAPhO, SIN, IPhO, F=ma exam []ChemWOOT: USNCO, IChO For those not quite ready for CodeWOOT and are interested in coding competitions, check out our USACO Bronze course! To be eligible, students should already be comfortable enough with C++, Java, or Python to write simple programs using basic concepts like arrays, maps/sets, if statements, and for loops. Note that either Java or Python is sufficient for the USACO Bronze and Silver levels (AoPS offers Python courses), but USACO Gold and above (and most programming contests) essentially require C++, and the IOI only supports C++. Keep in mind some other important dates coming up fast! [list][]AMC 10/12 competitions are right around the corner! Take advantage of our accelerated AMC 10 and AMC 12 Problem Series courses which run from October until right before the competitions begin! In these courses, you will learn test taking strategies and have the opportunity to take a practice exam - efficiently train to succeed in this year’s competitions! []The AMC 10A/12A will be held on November 5th []The AMC 10B/12B will be held on November 13th. []USAMTS (United States of America Mathematical Talent Search) will release their first round of problems soon after Labor day! This contest is another method for qualifying for AIME. As an extra bonus, students can use the comments they receive on their USAMTS solutions to hone their skills for Olympiad level contests! The Round 1 problems will be available at Students should submit solutions to the Round 1 problems by mid-October. []Admissions for MIT Primes opens on October 1st! MIT PRIMES is a free, year-long program, in which high school students work on individual and group research projects and participate in reading groups under the guidance of academic mentors. MIT PRIMES includes three sections: mathematics, computer science, and computational and physical biology. Our full course list for upcoming classes is below: All classes run 7:30pm-8:45pm ET/4:30pm - 5:45pm PT unless otherwise noted. Introductory: Grades 5-10 Prealgebra 1 Self-Paced Prealgebra 1 Friday, Sep 5 - Jan 16 Monday, Sep 8 - Jan 12 Tuesday, Sep 16 - Jan 20 (4:30 - 5:45 pm ET/1:30 - 2:45 pm PT) Sunday, Sep 21 - Jan 25 Thursday, Sep 25 - Jan 29 Wednesday, Oct 22 - Feb 25 Tuesday, Nov 4 - Mar 10 Friday, Dec 12 - Apr 10 Prealgebra 2 Self-Paced Prealgebra 2 Tuesday, Sep 9 - Jan 13 Thursday, Sep 25 - Jan 29 Sunday, Oct 19 - Feb 22 Monday, Oct 27 - Mar 2 Wednesday, Nov 12 - Mar 18 Introduction to Algebra A Self-Paced Introduction to Algebra A Friday, Sep 5 - Jan 16 Thursday, Sep 11 - Jan 15 Sunday, Sep 28 - Feb 1 Monday, Oct 6 - Feb 9 Tuesday, Oct 21 - Feb 24 Sunday, Nov 9 - Mar 15 Friday, Dec 5 - Apr 3 Introduction to Counting & Probability Self-Paced Introduction to Counting & Probability Wednesday, Sep 3 - Nov 19 Sunday, Sep 21 - Dec 14 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT) Friday, Oct 3 - Jan 16 Sunday, Oct 19 - Jan 25 Tuesday, Nov 4 - Feb 10 Sunday, Dec 7 - Mar 8 Introduction to Number Theory Friday, Sep 12 - Dec 12 Sunday, Oct 26 - Feb 1 Monday, Dec 1 - Mar 2 Introduction to Algebra B Self-Paced Introduction to Algebra B Sunday, Sep 7 - Jan 11 Thursday, Sep 11 - Jan 15 Wednesday, Sep 24 - Jan 28 Sunday, Oct 26 - Mar 1 Tuesday, Nov 4 - Mar 10 Monday, Dec 1 - Mar 30 Introduction to Geometry Sunday, Sep 7 - Mar 8 Thursday, Sep 11 - Mar 12 Wednesday, Sep 24 - Mar 25 Sunday, Oct 26 - Apr 26 Monday, Nov 3 - May 4 Friday, Dec 5 - May 29 Paradoxes and Infinity Sat & Sun, Sep 13 - Sep 14 (1:00 - 4:00 PM PT/4:00 - 7:00 PM ET) Intermediate: Grades 8-12 Intermediate Algebra Sunday, Sep 28 - Mar 29 Wednesday, Oct 8 - Mar 8 Sunday, Nov 16 - May 17 Thursday, Dec 11 - Jun 4 Intermediate Counting & Probability Sunday, Sep 28 - Feb 15 Tuesday, Nov 4 - Mar 24 Intermediate Number Theory Wednesday, Sep 24 - Dec 17 Precalculus Tuesday, Sep 9 - Feb 24 Sunday, Sep 21 - Mar 8 Monday, Oct 20 - Apr 6 Sunday, Dec 14 - May 31 Advanced: Grades 9-12 Calculus Sunday, Sep 7 - Mar 15 Wednesday, Sep 24 - Apr 1 Friday, Nov 14 - May 22 Contest Preparation: Grades 6-12 MATHCOUNTS/AMC 8 Basics Wednesday, Sep 3 - Nov 19 Tuesday, Sep 16 - Dec 9 Sunday, Sep 21 - Dec 14 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT) Monday, Oct 6 - Jan 12 Thursday, Oct 16 - Jan 22 Tues, Thurs & Sun, Dec 9 - Jan 18 (meets three times a week!) MATHCOUNTS/AMC 8 Advanced Thursday, Sep 4 - Nov 20 Friday, Sep 12 - Dec 12 Monday, Sep 15 - Dec 8 Sunday, Oct 5 - Jan 11 Tues, Thurs & Sun, Dec 2 - Jan 11 (meets three times a week!) Mon, Wed & Fri, Dec 8 - Jan 16 (meets three times a week!) AMC 10 Problem Series Mon & Wed, Sep 15 - Oct 22 (meets twice a week!) Mon, Wed & Fri, Oct 6 - Nov 3 (meets three times a week!) Tue, Thurs & Sun, Oct 7 - Nov 2 (meets three times a week!) AMC 10 Final Fives Sunday, Sep 7 - Sep 28 Tuesday, Sep 9 - Sep 30 Monday, Sep 22 - Oct 13 Sunday, Sep 28 - Oct 19 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT) Wednesday, Oct 8 - Oct 29 Thursday, Oct 9 - Oct 30 AMC 12 Problem Series Mon & Wed, Sep 15 - Oct 22 (meets twice a week!) Tues, Thurs & Sun, Oct 7 - Nov 2 (meets three times a week!) AMC 12 Final Fives Thursday, Sep 4 - Sep 25 Sunday, Sep 28 - Oct 19 Tuesday, Oct 7 - Oct 28 AIME Problem Series A Thursday, Oct 23 - Jan 29 AIME Problem Series B Tuesday, Sep 2 - Nov 18 F=ma Problem Series Tuesday, Sep 16 - Dec 9 Friday, Oct 17 - Jan 30 WOOT Programs Visit the pages linked for full schedule details for each of these programs! MathWOOT Level 1 MathWOOT Level 2 ChemWOOT CodeWOOT PhysicsWOOT Programming Introduction to Programming with Python Sunday, Sep 7 - Nov 23 Tuesday, Dec 2 - Mar 3 Intermediate Programming with Python Friday, Oct 3 - Jan 16 USACO Bronze Problem Series Wednesday, Sep 3 - Dec 3 Thursday, Oct 30 - Feb 5 Tuesday, Dec 2 - Mar 3 Physics Introduction to Physics Tuesday, Sep 2 - Nov 18 Sunday, Oct 5 - Jan 11 Wednesday, Dec 10 - Mar 11 Physics 1: Mechanics Sunday, Sep 21 - Mar 22 Sunday, Oct 26 - Apr 26 0 replies jwelsh Sep 2, 2025 0 replies J H Even number of irreducible polynomials Quantum-Phantom 11 Source: Canada MO 2024/3 Let be the number of positive integers with digits in base (where for all and ) such that the polynomial is irreducible in . Prove that is even. (A polynomial is irreducible in if it cannot be factored into two non-constant polynomials with rational coefficients.) 11 replies Quantum-Phantom Mar 8, 2024 ray66 18 minutes ago J H radical axis m4thbl3nd3r 0 Let be the incenter of the triangle and be the midpoint of . Denote as the incircle touchpoint on . Prove that the m idline of triangle is the radical axis of and . 0 replies m4thbl3nd3r 20 minutes ago 0 replies J H Find the largest real constant Jackson0423 6 Find the largest real constant such that for all real numbers , 6 replies Jackson0423 Aug 31, 2025 Penguuin1 28 minutes ago J H Nice Symmedian property Ahiles 51 Source: BMO 2009 Problem 2 Let be a line parallel to the side of a triangle , with on the side and on the side . The lines and meet at point . The circumcircles of triangles and meet at two distinct points and . Prove that . Liubomir Chiriac, Moldova 51 replies Ahiles Apr 30, 2009 Lunatic_Lunar7986 an hour ago J College Math Topics in undergraduate and graduate studies Topics in undergraduate and graduate studies 3 V New Topic k Locked College Math Topics in undergraduate and graduate studies Topics in undergraduate and graduate studies 3 V New Topic k Locked t 114,929 topics w 3 users TOPICS G Topic First Poster Last Poster H k a September Highlights and 2025 AoPS Online Class Information jwelsh 0 It’s back to school time! A new academic year is a new opportunity to train for the upcoming competition season. Our Worldwide Online Olympiad Training (WOOT) classes begin the first week of September, so be sure to take advantage of the only program where you learn the skills and strategies necessary to succeed at the Olympiad Level. This year alone, Art of Problem Solving students from around the globe racked up over 100 total medals at the International Mathematics Olympiad, including 25 gold medals in ‘25! "As a parent, I'm deeply grateful to AoPS. Tiger has taken very few math courses outside of AoPS, except for a local Math Circle that doesn't focus on Olympiad math. AoPS has been one of the most important resources in his journey. Without AoPS, Tiger wouldn't be where he is today — especially considering he's grown up in a family with no STEM background at all." — Doreen Dai, parent of IMO US 2025 Team Member Tiger Zhang Be sure to read the article by our CEO, Ben Kornell, about his experience at the IMO in 2025! Below is a list of the different WOOT programs and the competitions they cover. [list][]MathWOOT Level 1: Designed for AIME qualifiers ready to make the jump to Olympiads. []MathWOOT Level 2: Designed for Olympiad qualifiers ready to increase their scores in national and international Olympiads. []CodeWOOT: USACO, IOI []PhysicsWOOT: USAPhO, SIN, IPhO, F=ma exam []ChemWOOT: USNCO, IChO For those not quite ready for CodeWOOT and are interested in coding competitions, check out our USACO Bronze course! To be eligible, students should already be comfortable enough with C++, Java, or Python to write simple programs using basic concepts like arrays, maps/sets, if statements, and for loops. Note that either Java or Python is sufficient for the USACO Bronze and Silver levels (AoPS offers Python courses), but USACO Gold and above (and most programming contests) essentially require C++, and the IOI only supports C++. Keep in mind some other important dates coming up fast! [list][]AMC 10/12 competitions are right around the corner! Take advantage of our accelerated AMC 10 and AMC 12 Problem Series courses which run from October until right before the competitions begin! In these courses, you will learn test taking strategies and have the opportunity to take a practice exam - efficiently train to succeed in this year’s competitions! []The AMC 10A/12A will be held on November 5th []The AMC 10B/12B will be held on November 13th. []USAMTS (United States of America Mathematical Talent Search) will release their first round of problems soon after Labor day! This contest is another method for qualifying for AIME. As an extra bonus, students can use the comments they receive on their USAMTS solutions to hone their skills for Olympiad level contests! The Round 1 problems will be available at Students should submit solutions to the Round 1 problems by mid-October. []Admissions for MIT Primes opens on October 1st! MIT PRIMES is a free, year-long program, in which high school students work on individual and group research projects and participate in reading groups under the guidance of academic mentors. MIT PRIMES includes three sections: mathematics, computer science, and computational and physical biology. Our full course list for upcoming classes is below: All classes run 7:30pm-8:45pm ET/4:30pm - 5:45pm PT unless otherwise noted. Introductory: Grades 5-10 Prealgebra 1 Self-Paced Prealgebra 1 Friday, Sep 5 - Jan 16 Monday, Sep 8 - Jan 12 Tuesday, Sep 16 - Jan 20 (4:30 - 5:45 pm ET/1:30 - 2:45 pm PT) Sunday, Sep 21 - Jan 25 Thursday, Sep 25 - Jan 29 Wednesday, Oct 22 - Feb 25 Tuesday, Nov 4 - Mar 10 Friday, Dec 12 - Apr 10 Prealgebra 2 Self-Paced Prealgebra 2 Tuesday, Sep 9 - Jan 13 Thursday, Sep 25 - Jan 29 Sunday, Oct 19 - Feb 22 Monday, Oct 27 - Mar 2 Wednesday, Nov 12 - Mar 18 Introduction to Algebra A Self-Paced Introduction to Algebra A Friday, Sep 5 - Jan 16 Thursday, Sep 11 - Jan 15 Sunday, Sep 28 - Feb 1 Monday, Oct 6 - Feb 9 Tuesday, Oct 21 - Feb 24 Sunday, Nov 9 - Mar 15 Friday, Dec 5 - Apr 3 Introduction to Counting & Probability Self-Paced Introduction to Counting & Probability Wednesday, Sep 3 - Nov 19 Sunday, Sep 21 - Dec 14 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT) Friday, Oct 3 - Jan 16 Sunday, Oct 19 - Jan 25 Tuesday, Nov 4 - Feb 10 Sunday, Dec 7 - Mar 8 Introduction to Number Theory Friday, Sep 12 - Dec 12 Sunday, Oct 26 - Feb 1 Monday, Dec 1 - Mar 2 Introduction to Algebra B Self-Paced Introduction to Algebra B Sunday, Sep 7 - Jan 11 Thursday, Sep 11 - Jan 15 Wednesday, Sep 24 - Jan 28 Sunday, Oct 26 - Mar 1 Tuesday, Nov 4 - Mar 10 Monday, Dec 1 - Mar 30 Introduction to Geometry Sunday, Sep 7 - Mar 8 Thursday, Sep 11 - Mar 12 Wednesday, Sep 24 - Mar 25 Sunday, Oct 26 - Apr 26 Monday, Nov 3 - May 4 Friday, Dec 5 - May 29 Paradoxes and Infinity Sat & Sun, Sep 13 - Sep 14 (1:00 - 4:00 PM PT/4:00 - 7:00 PM ET) Intermediate: Grades 8-12 Intermediate Algebra Sunday, Sep 28 - Mar 29 Wednesday, Oct 8 - Mar 8 Sunday, Nov 16 - May 17 Thursday, Dec 11 - Jun 4 Intermediate Counting & Probability Sunday, Sep 28 - Feb 15 Tuesday, Nov 4 - Mar 24 Intermediate Number Theory Wednesday, Sep 24 - Dec 17 Precalculus Tuesday, Sep 9 - Feb 24 Sunday, Sep 21 - Mar 8 Monday, Oct 20 - Apr 6 Sunday, Dec 14 - May 31 Advanced: Grades 9-12 Calculus Sunday, Sep 7 - Mar 15 Wednesday, Sep 24 - Apr 1 Friday, Nov 14 - May 22 Contest Preparation: Grades 6-12 MATHCOUNTS/AMC 8 Basics Wednesday, Sep 3 - Nov 19 Tuesday, Sep 16 - Dec 9 Sunday, Sep 21 - Dec 14 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT) Monday, Oct 6 - Jan 12 Thursday, Oct 16 - Jan 22 Tues, Thurs & Sun, Dec 9 - Jan 18 (meets three times a week!) MATHCOUNTS/AMC 8 Advanced Thursday, Sep 4 - Nov 20 Friday, Sep 12 - Dec 12 Monday, Sep 15 - Dec 8 Sunday, Oct 5 - Jan 11 Tues, Thurs & Sun, Dec 2 - Jan 11 (meets three times a week!) Mon, Wed & Fri, Dec 8 - Jan 16 (meets three times a week!) AMC 10 Problem Series Mon & Wed, Sep 15 - Oct 22 (meets twice a week!) Mon, Wed & Fri, Oct 6 - Nov 3 (meets three times a week!) Tue, Thurs & Sun, Oct 7 - Nov 2 (meets three times a week!) AMC 10 Final Fives Sunday, Sep 7 - Sep 28 Tuesday, Sep 9 - Sep 30 Monday, Sep 22 - Oct 13 Sunday, Sep 28 - Oct 19 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT) Wednesday, Oct 8 - Oct 29 Thursday, Oct 9 - Oct 30 AMC 12 Problem Series Mon & Wed, Sep 15 - Oct 22 (meets twice a week!) Tues, Thurs & Sun, Oct 7 - Nov 2 (meets three times a week!) AMC 12 Final Fives Thursday, Sep 4 - Sep 25 Sunday, Sep 28 - Oct 19 Tuesday, Oct 7 - Oct 28 AIME Problem Series A Thursday, Oct 23 - Jan 29 AIME Problem Series B Tuesday, Sep 2 - Nov 18 F=ma Problem Series Tuesday, Sep 16 - Dec 9 Friday, Oct 17 - Jan 30 WOOT Programs Visit the pages linked for full schedule details for each of these programs! MathWOOT Level 1 MathWOOT Level 2 ChemWOOT CodeWOOT PhysicsWOOT Programming Introduction to Programming with Python Sunday, Sep 7 - Nov 23 Tuesday, Dec 2 - Mar 3 Intermediate Programming with Python Friday, Oct 3 - Jan 16 USACO Bronze Problem Series Wednesday, Sep 3 - Dec 3 Thursday, Oct 30 - Feb 5 Tuesday, Dec 2 - Mar 3 Physics Introduction to Physics Tuesday, Sep 2 - Nov 18 Sunday, Oct 5 - Jan 11 Wednesday, Dec 10 - Mar 11 Physics 1: Mechanics Sunday, Sep 21 - Mar 22 Sunday, Oct 26 - Apr 26 0 replies jwelsh Sep 2, 2025 0 replies J H $n-a_n$ exists? SatisfiedMagma 4 Source: MU(theta) Let and . Show that exists and find its value. 4 replies +1 w SatisfiedMagma Saturday at 10:39 PM Alphaamss 26 minutes ago J H Disjoint segments ZETA_in_olympiad 2 Source: 2022 Miklós Schweitzer Problem 5 Is it possible to select a non-degenerate segment from each line of the plane such that any two selected segments are disjoint? 2 replies ZETA_in_olympiad Nov 12, 2022 alinazarboland 2 hours ago J H Question on Goldbach Conjecture Verification MDB001 6 Hello, I was looking at the AoPS page on the Goldbach Conjecture, and something came to mind. When people say the conjecture has been verified up to 10^{18}, what exactly is the verification method? Is it done by generating all primes up to that bound and checking their sums to cover all even numbers in the interval? Or is it done the other way around, starting with an even number and directly decomposing it into two primes? If the method is the first, then the second seems more intriguing. Because if one could start from an even number, especially a large one, say 100 digits or more, and directly retrieve its two primes, what would that imply for the problem itself? Thank you for clarifying. I would be very curious about your perspectives. 6 replies MDB001 Yesterday at 9:23 AM MDB001 3 hours ago J H six seven integrals aidan0626 1 These problems were inspired by Six SevenTM. Six SevenTM is an incredible piece of art, and it is only fitting that those two elegant numbers were honored with problems. Six SevenTM is love. Six SevenTM is life. Six SevenTM is me. I am Six SevenTM. Six SevenTM is inside of me. 1. 2. 3. 4. 5. 6-7. where 1 reply aidan0626 4 hours ago UniqueScorpion89 4 hours ago J Site Support Tech support and questions about AoPS classes and materials Tech support and questions about AoPS classes and materials 3 V New Topic k Locked Site Support Tech support and questions about AoPS classes and materials Tech support and questions about AoPS classes and materials 3 V New Topic k Locked t 18,906 topics w 1 user TOPICS G Topic First Poster Last Poster H k a September Highlights and 2025 AoPS Online Class Information jwelsh 0 It’s back to school time! A new academic year is a new opportunity to train for the upcoming competition season. Our Worldwide Online Olympiad Training (WOOT) classes begin the first week of September, so be sure to take advantage of the only program where you learn the skills and strategies necessary to succeed at the Olympiad Level. This year alone, Art of Problem Solving students from around the globe racked up over 100 total medals at the International Mathematics Olympiad, including 25 gold medals in ‘25! "As a parent, I'm deeply grateful to AoPS. Tiger has taken very few math courses outside of AoPS, except for a local Math Circle that doesn't focus on Olympiad math. AoPS has been one of the most important resources in his journey. Without AoPS, Tiger wouldn't be where he is today — especially considering he's grown up in a family with no STEM background at all." — Doreen Dai, parent of IMO US 2025 Team Member Tiger Zhang Be sure to read the article by our CEO, Ben Kornell, about his experience at the IMO in 2025! Below is a list of the different WOOT programs and the competitions they cover. [list][]MathWOOT Level 1: Designed for AIME qualifiers ready to make the jump to Olympiads. []MathWOOT Level 2: Designed for Olympiad qualifiers ready to increase their scores in national and international Olympiads. []CodeWOOT: USACO, IOI []PhysicsWOOT: USAPhO, SIN, IPhO, F=ma exam []ChemWOOT: USNCO, IChO For those not quite ready for CodeWOOT and are interested in coding competitions, check out our USACO Bronze course! To be eligible, students should already be comfortable enough with C++, Java, or Python to write simple programs using basic concepts like arrays, maps/sets, if statements, and for loops. Note that either Java or Python is sufficient for the USACO Bronze and Silver levels (AoPS offers Python courses), but USACO Gold and above (and most programming contests) essentially require C++, and the IOI only supports C++. Keep in mind some other important dates coming up fast! [list][]AMC 10/12 competitions are right around the corner! Take advantage of our accelerated AMC 10 and AMC 12 Problem Series courses which run from October until right before the competitions begin! In these courses, you will learn test taking strategies and have the opportunity to take a practice exam - efficiently train to succeed in this year’s competitions! []The AMC 10A/12A will be held on November 5th []The AMC 10B/12B will be held on November 13th. []USAMTS (United States of America Mathematical Talent Search) will release their first round of problems soon after Labor day! This contest is another method for qualifying for AIME. As an extra bonus, students can use the comments they receive on their USAMTS solutions to hone their skills for Olympiad level contests! The Round 1 problems will be available at Students should submit solutions to the Round 1 problems by mid-October. []Admissions for MIT Primes opens on October 1st! MIT PRIMES is a free, year-long program, in which high school students work on individual and group research projects and participate in reading groups under the guidance of academic mentors. MIT PRIMES includes three sections: mathematics, computer science, and computational and physical biology. Our full course list for upcoming classes is below: All classes run 7:30pm-8:45pm ET/4:30pm - 5:45pm PT unless otherwise noted. 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Refreshing fixes it, but it is still reproduceable. 12 replies RollingPanda4616 Sep 20, 2025 Staragon Today at 3:50 AM J H Is this a Glitch? ZuzabKoit 5 While I was looking at a topic it suddenly started scrolling. I was doing nothing, but I had to close the tab or hit the return to top button for it to stop. Even when I tried scrolling, it overrided my mouse and kept scrolling. Is my account getting hacked or is this a glitch? 5 replies ZuzabKoit Sep 26, 2025 KangarooPrecise Today at 1:37 AM J H [RESOLVED] 8char limit wrongfully imposed jkim0656 12 Hi SS, I noticed that if you write a message and surround it in brackets [], it says that the message is shorter than eight characters even when it is definitely longer. Chrome Browser HP laptop I did notice that if you write at least 8 characters after the text in [], it lets the message through. 12 replies jkim0656 Yesterday at 1:05 AM SpeedCuber7 Yesterday at 1:47 PM J Other Forums and Collections ? 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15960	https://www.collinsdictionary.com/us/dictionary/english/anything	English French German Italian Spanish Portuguese Hindi Chinese Korean Japanese More English Italiano Português 한국어 简体中文 Deutsch Español हिंदी 日本語 English French German Italian Spanish Portuguese Hindi Chinese Korean Japanese Definitions Summary Synonyms Sentences Pronunciation Collocations Conjugations Grammar Credits × Definition of 'anything' COBUILD frequency band anything (ɛnɪθɪŋ ) 1. indefinite pronoun You use anything in statements with negative meaning to indicate in a general way that nothing is present or that an action or event does not or cannot happen. We can't do anything. She couldn't see or hear anything at all. Synonyms: a thing, NEhng [text messaging] More Synonyms of anything 2. indefinite pronoun You use anything in questions and conditional clauses to ask or talk about whether something is present or happening. What happened, is anything wrong? Did you find anything? 3. indefinite pronoun [PRON cl/group] You can use anything before words that indicate the kind of thing you are talking about. More than anything else, he wanted to become a teacher. Anything that's cheap this year will be even cheaper next year. 4. indefinite pronoun You use anything to emphasize a possible thing, event, or situation, when you are saying that it could be any one of a very large number of things. [emphasis] He is young, fresh, and ready for anything. 5. indefinite pronoun [PRON prep] You use anything in expressions such as anything near, anything close to and anything like to emphasize a statement that you are making. [emphasis] Doctors have decided the only way he can live anything near a normal life is to give him an operation. 6. indefinite pronoun [PRON from n to n, PRON between n and n] When you do not want to be exact, you use anything to talk about a particular range of things or quantities. ...Chinese herbs that have cured anything from colds to broken bones. 7. See anything but 8. See would not do sth for anything/would not be sth for anything 9. See if anything 10. See or anything Collins COBUILD Advanced Learner’s Dictionary. Copyright © HarperCollins Publishers English Easy Learning GrammarIndefinite pronounsWhat type of pronoun are the words 'someone', 'everybody', 'anything', 'nothing', etc.? What are the different types of indefinite pronoun? What determiner ...Read more American English pronunciation ! It seems that your browser is blocking this video content. To access it, add this site to the exceptions or modify your security settings, then refresh this page. British English pronunciation ! It seems that your browser is blocking this video content. To access it, add this site to the exceptions or modify your security settings, then refresh this page. You may also like English Quiz ConfusablesSynonyms of 'anything'Language Lover's BlogFrench Translation of 'anything'Translate your textPronunciation PlaylistsWord of the day: 'hwyl'Spanish Translation of 'anything'English GrammarCollins AppsEnglish Quiz ConfusablesSynonyms of 'anything'Language Lover's BlogFrench Translation of 'anything'Translate your textPronunciation PlaylistsWord of the day: 'hwyl'Spanish Translation of 'anything'English GrammarCollins AppsEnglish Quiz ConfusablesSynonyms of 'anything'Language Lover's Blog COBUILD frequency band anything in American English (ˈəniˌθɪŋ ) pronoun 1. any object, event, fact, etc. do you know anything about it? noun 2. a thing, no matter of what kind do anything you want adverb 3. in any way; at all is he anything like his father? Idioms: anything but Webster’s New World College Dictionary, 5th Digital Edition. Copyright © 2025 HarperCollins Publishers. COBUILD frequency band anything in American English (ˈeniˌθɪŋ) pronoun 1. any thing whatever; something, no matter what Do you have anything for a toothache? noun 2. a thing of any kind 3. See anything goes adverb 4. in any degree; to any extent; in any way; at all Does it taste anything like chocolate? 5. See anything but Most material © 2005, 1997, 1991 by Penguin Random House LLC. Modified entries © 2019 by Penguin Random House LLC and HarperCollins Publishers Ltd Word origin [bef. 900; ME ani thing, eni thing, OE ǣnig thing. See any, thing1] COBUILD frequency band anything in British English (ˈɛnɪˌθɪŋ ) pronoun 1. any object, event, action, etc, whatever anything might happen noun 2. a thing of any kind have you anything to declare? adverb 3. in any way she wasn't anything like her parents 4. See anything but 5. See like anything Collins English Dictionary. Copyright © HarperCollins Publishers Examples of 'anything' in a sentence anything These examples have been automatically selected and may contain sensitive content that does not reflect the opinions or policies of Collins, or its parent company HarperCollins. We welcome feedback: report an example sentence to the Collins team. Read more… This is not about paying a donor back or anything like that. Wall Street Journal (2023) Was anything too dangerous to do? The Guardian (2015) Anything else is not the right way. The Guardian (2016) They are as good as anything else. The Guardian (2019) She has been forced to surrender any hope of doing anything else in politics. The Guardian (2019) No one who knows anything about elite sport will be surprised at this. Times, Sunday Times (2017) Nobody talks about anything other than the next game but there is a bigger picture. Times, Sunday Times (2011) They were unavailable for comment yesterday but have talked previously about the agony of not hearing anything. Times, Sunday Times (2009) Yet bargaining is not a process in which anything goes. A Conceptual View of Human Resource Management: Strategic Objectives, Environments, Functions If anything happens it will be a tragedy for all concerned. Times, Sunday Times (2013) Trends of anything View usage over: Source: Google Books Ngram Viewer In other languages anything British English: anything /ˈɛnɪˌθɪŋ/ PRONOUN You use anything to talk about a thing, when it does not matter which one. I can't see anything. American English: anything /ˈɛnɪθɪŋ/ Arabic: أَيُّ شَيء Brazilian Portuguese: qualquer coisa Chinese: 任何事 Croatian: išta Czech: něco Danish: noget Dutch: iets European Spanish: algo interrogación Finnish: mitään French: n’importe quoi German: irgendetwas Greek: τίποτα Italian: niente Japanese: 何でも Korean: 아무것도 Norwegian: noe Polish: cokolwiek European Portuguese: qualquer coisa Romanian: nimic Russian: что-нибудь Spanish: algo Swedish: något Thai: สิ่งใด, ใดๆ, อะไร Turkish: herhangi bir şey Ukrainian: що-небудь Vietnamese: bất cứ cái gì Translate your text for free Browse alphabetically anything anyone's guess/anybody's guess anyplace anyroad anything anything but anything goes All ENGLISH words that begin with 'A' Related terms of anything as anything if anything or anything anything but anything goes View more related words ## Wordle Helper ## Scrabble Tools Quick word challenge Quiz Review Question: 1 - Score: 0 / 5 SPORTS What is this an image of? shootingpooltable tennisskiing SPORTS What is this an image of? wrestlingpoolcanoeingsnooker SPORTS What is this an image of? squashice skatingkaratetable tennis SPORTS Drag the correct answer into the box. archery weightlifting shooting netball SPORTS Drag the correct answer into the box. shotput pilates squash ice skating Your score: New collocations added to dictionary Collocations are words that are often used together and are brilliant at providing natural sounding language for your speech and writing. 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15961	https://www.goodreads.com/book/show/545365.The_Developing_Human	The Developing Human: Clinically Oriented Embryology by Keith L. Moore \| Goodreads Home My Books Browse ▾ Recommendations Choice Awards Genres Giveaways New Releases Lists Explore News & Interviews Genres Art Biography Business Children's Christian Classics Comics Cookbooks Ebooks Fantasy Fiction Graphic Novels Historical Fiction History Horror Memoir Music Mystery Nonfiction Poetry Psychology Romance Science Science Fiction Self Help Sports Thriller Travel Young Adult More Genres Community ▾ Groups Quotes Ask the Author People Sign in Join Jump to ratings and reviews Want to Read Kindle $57.59 Rate this book The Developing Human: Clinically Oriented Embryology Keith L. Moore, T.V.N. Persaud 3.87 212 ratings 12 reviews Want to Read Kindle $57.59 Rate this book This bestselling resource comprehensively covers human embryology and teratology, presenting all of the complex clinical and scientific concepts in an engaging, lucid, and practical way. Completely revised and updated, the New Edition emphasises the clinical aspects throughout by using a wealth of case studies, clinical correlations, and hundreds of outstanding illustrations. GenresMedicineMedicalNonfictionTextbooksReferenceSchool 544 pages, Hardcover First published January 1, 1973 Book details & editions 100 people are currently reading 810 people want to read About the author Keith L. Moore 99 books 22 followers Follow Follow Keith L. Moore (MSc PhD FIAC FRSM FAAA) is a physician specializing in embryology and human development. Readers also enjoyed Around the World in Eighty Days Jules Verne 3.95 278k The Kiss Quotient Helen Hoang 3.87 490k All similar books Ratings & Reviews What do you think? 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Community Reviews 3.87 212 ratings 12 reviews 5 stars 81 (38%) 4 stars 61 (28%) 3 stars 47 (22%) 2 stars 8 (3%) 1 star 15 (7%) Search review text Filters Displaying 1 - 12 of 12 reviews Al-anoud Al-jarbou 50 reviews 235 followers Follow Follow August 8, 2009 الكتاب رائع من جد دلالة على عظمة خلق الله قصة إسلامه موجودة في كتاب الذين هدى الله للدكتور زغلول النجار نص القصة : كيث مور وهو من أشهر العلماء في علم الأجنة ويعرفه تقريباً كل أطباء العالم,فهو له كتاب يدرس في معظم كليات الطب في العالم وقد ترجم هذا الكتاب لأكثر من 25 لغة فهو صاحب الكتاب الشهير (The Developing Human). الرجل في وقف هذا الرجل وسط جمع في احد المؤتمرات قائلاً: "إن التعبيرات القرآنية عن مراحل تكون الجنين في الإنسان لتبلغ من الدقة والشمول ما لم يبلغه العلم الحديث, وهذا إن دل علي شيء فإنما يدل علي أن هذا القرآن لا يمكن أن يكون إلا كلام الله, وأن محمداً رسول الله" .فقيل له: هل أنت مسلم؟!؟.قال: لا ولكني أشهد أن القرآن كلام الله وأن محمداً مرسل من عند الله.فقيل له:إذاً فأنت مسلم,قال: أنا تحت ضغوط اجتماعية تحول دون إعلان إسلامي الآن ولكن لا تتعجبوا إذا سمعتم يوماً أن كيث مور قد دخل الإسلام. ولقد وصلنا في العام الماضي أنه قد أعلن إسلامه فعلاً فلله الحمد والمنة. وفي مؤتمر الإعجاز العلمي الأول للقرآن الكريم والسنة المطهرة والذي عقد في القاهرة عام 1986 وقف الأستاذ الدكتور، كيث مور محاضرته قائلاً : (إنني أشهد بإعجاز الله في خلق كل طور من أطوار القرآن الكريم، ولست أعتقد أن محمداً صلى الله عليه وسلم أو أي شخص آخر يستطيع معرفة ما يحدث في تطور الجنين لأن هذه التطورات لم تكتشف إلا في الجزء الأخير من القرن العشرين، وأريد أن أؤكد على أن كل شيء قرأته في القرآن الكريم عن نشأة الجنين وتطوره في داخل الرحم ينطبق على كل ما أعرفه كعالم من علماء الأجنة البارزين). علماً أن مراحل خلق الإنسان (بني آدم ) التي ذكرها القرآن هي سبع مراحل. قال تعالى: ((وَلَقَدْ خَلَقْنَا الإِنسَانَ مِنْ سُلالَةٍ مِنْ طِينٍ،ثُمَّ جَعَلْنَاهُ نُطْفَةً فِي قَرَارٍ مَكِينٍ، ثُمَّ خَلَقْنَا النُّطْفَةَ عَلَقَةً فَخَلَقْنَا الْعَلَقَةَ مُضْغَةً فَخَلَقْنَا الْمُضْغَةَ عِظَامًا فَكَسَوْنَا الْعِظَامَ لَحْمًا ثُمَّ أَنشَأْنَاهُ خَلْقًا آخَرَ فَتَبَارَكَ اللَّهُ أَحْسَنُ الْخَالِقِينَ )) [المؤمنون 12ـ14:] وقد أثبت علم الأجنة هذه المراحل وصحتها وتطابقها مع المراحل المذكورة في القرآن. وهذه المراحل هي:1- أصل الإنسان (سلالة من طين) 2- النطفة 3- العلقة 4- المضغة 5-العظام 6- الإكساء باللحم 7- النشأة. وقد اعتبر المؤتمر الخامس للإعجاز العلمي في القرآن والسنة والذي عقد في موسكو (أيلول 1995) هذا التقسيم القرآني لمراحل خلق الجنين وتطوره صحيحاً ودقيقاً وأوصى في مقرراته على اعتماده كتصنيف علمي للتدريــس علماً أن الأستاذ الدكتور كيث مور Keith Mooreوهو من أشهر علماء التشريح وعلم الأجنة في العالم ورئيس هذا القسم في جامعة تورنتو بكندا (والذي كان أحد الباحثين المشاركين في المؤتمر المذكور ) ، ألف كتاباً يعد من أهم المراجع الطبية في هذا الاختصاص ( مراحل خلق الإنسان _ علم الأجنة السريري ) وضمنه ذكر هذه المراحل المذكورة في القرآن، وربط في كل فصل من فصول الكتاب التي تتكلم عن تطور خلق الجنين وبين الحقائق العلمية والآيات والأحاديث المتعلقة بها وشرحها وعلق عليها بالتعاون مع الشيخ الزنداني وزملائه Show more 9 likes 1 comment Like Comment Juna 34 reviews Follow Follow January 11, 2009 keith moore was the teacher of my graduate embryology professor dr. fabian eluma whos an utterly brilliant obgyn, genetics counselor, and was global advisor on hiv/aids to the former president. now im biased of course because the subject fascinates me but this was one of the only textbooks i ever had that was a pageturner! 40 weeks of pregnancy that reads like a story. its really dope everybody else was selling theirs back i kept mine.. Show more 2 likes Like Comment fioo ! ♡ ∗ ˚ ˖࣪ ∗ ‎˖ ݁ . ° · ˚ ₊ 160 reviews 36 followers Follow Follow Read October 22, 2021 NSYNC's BYE, BYE, BYE plays in the background 1 like Like Comment Marbook_gr 175 reviews Follow Follow August 19, 2023 Καλο 1 like Like Comment Elizabeth 29 reviews 5 followers Follow Follow August 5, 2013 I used this book as reference throughout the first three years of med school. I haven't read it through, but have read a fair bit of it. It was good for outlining the detailed steps in various processes - e.g. development of the heart and gut - that I had a bit of trouble visualizing. It was especially helpful during my clinical rotation in pediatric surgery, as I could read more about how the normal processes of development occur and contract those with how/when/why things go wrong and what results from that. It uses a lot of pictures to teach concepts, though I still found some sections confusing and wished there had been more (I used youtube to fill in the gaps for visualizing the processes). I enjoyed reading this book more than most textbooks - the information is presented in a way that increased my interest and curiosity in the subject. Overall, it's a good book for students learning embryology and development, and I think would be a helpful resource for those in e.g. pediatric surgery, though they would probably eventually need a more in-depth resource. Show more med-books 1 like Like Comment Abdullah F 4 reviews 4 followers Follow Follow Read November 23, 2009 in my opinion best reference in embryology 1 like Like Comment Brian 57 reviews 6 followers Follow Follow July 25, 2011 My parents are doctors of internal medicine, but had to learn about this subject. We therefore grew up with this book. 1 like Like Comment aby l.s 160 reviews Follow Follow February 28, 2025 Moore embrio 🥰💕💅🥹❤️‍🩹>>>>> Moore anatomía 😡☠️🫠🫨🤮 Like Comment Gala 25 reviews 6 followers Follow Follow April 26, 2015 Textbook for my Embryology class. Reads like a dictionary- heavy on definitions, light on interconnections. Some of the images are helpful illustrations/ photographs. Used it to become acquainted with the vocabulary- relied on youtube videos to understand the processes. Like Comment Einer 28 reviews 1 follower Follow Follow December 15, 2012 It needs to be organized. It's information is perfect but being disorganized that book can be bad. Like Comment Rheema 6 reviews 1 follower Follow Follow November 18, 2013 Great layout over-all, but I still find trouble visualizing some concepts..... medical Like Comment Arantxa Author 1 book Follow Follow November 7, 2014 Libro muy completo para el estudio de la embriología. médicos Like Comment Displaying 1 - 12 of 12 reviews Join the discussion Add a quoteStart a discussionAsk a question Can't find what you're looking for? Get help and learn more about the design. Help center Company About us Careers Terms Privacy Interest Based Ads Ad Preferences Help Work With Us Authors Advertise Authors & ads blog Connect © 2025 Goodreads, Inc.
15962	https://www.overleaf.com/articles/griffith-quantum-mechanics-time-dependent-perturbation-theory-cheatsheet-ucb-137b-final/jwynrzctvqgp	Griffith Quantum Mechanics Time Dependent Perturbation theory CheatSheet (UCB 137B final) - Overleaf, Online LaTeX Editor Skip to content menu Product Features AI Solutions For business For universities For government For publishers Customer stories Templates Pricing Help & resources Documentation Help guides Why LaTeX? Blog Contact us Sign up Log in arrow_left_alt Back to all community articles Griffith Quantum Mechanics Time Dependent Perturbation theory CheatSheet (UCB 137B final) Open as TemplateView SourceView PDF Author: Liuke LYU Last Updated: 6 years ago License: Creative Commons CC BY 4.0 Abstract: This is a cheatsheet made for the final exam of Berkeley Physics 137B. It mainly covers chapters from Variational Method to the end (except for Adiabetic Approximation). Tags: HandoutMathPhysicsCheat sheet \begin{now} Discover why over 20 million people worldwide trust Overleaf with their work. Sign up for freeExplore all plans Source ``` %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % writeLaTeX Example: A quick guide to LaTeX % % Source: Dave Richeson (divisbyzero.com), Dickinson College % % A one-size-fits-all LaTeX cheat sheet. Kept to two pages, so it % can be printed (double-sided) on one piece of paper % % Feel free to distribute this example, but please keep the referral % to divisbyzero.com % %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % How to use writeLaTeX: % % You edit the source code here on the left, and the preview on the % right shows you the result within a few seconds. % % Bookmark this page and share the URL with your co-authors. They can % edit at the same time! % % You can upload figures, bibliographies, custom classes and % styles using the files menu. % % If you're new to LaTeX, the wikibook is a great place to start: % % %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \documentclass[a4paper]{article} \usepackage{amssymb,amsmath,amsthm,amsfonts} \usepackage{multicol,multirow} \usepackage{calc} \usepackage{ifthen} \usepackage[landscape]{geometry} \usepackage[colorlinks=true,citecolor=blue,linkcolor=blue]{hyperref} \ifthenelse{\lengthtest { \paperwidth = 11in}} { \geometry{top=.5in,left=.5in,right=.5in,bottom=.5in} } {\ifthenelse{ \lengthtest{ \paperwidth = 297mm}} {\geometry{top=1cm,left=1cm,right=1cm,bottom=1cm} } {\geometry{top=1cm,left=1cm,right=1cm,bottom=1cm} } } \pagestyle{empty} \makeatletter \renewcommand{\section}{\@startsection{section}{1}{0mm}% {-1ex plus -.5ex minus -.2ex}% {0.5ex plus .2ex}%x {\normalfont\large\bfseries}} \renewcommand{\subsection}{\@startsection{subsection}{2}{0mm}% {-1explus -.5ex minus -.2ex}% {0.5ex plus .2ex}% {\normalfont\normalsize\bfseries}} \renewcommand{\subsubsection}{\@startsection{subsubsection}{3}{0mm}% {-1ex plus -.5ex minus -.2ex}% {1ex plus .2ex}% {\normalfont\small\bfseries}} \makeatother \setcounter{secnumdepth}{0} \setlength{\parindent}{0pt} \setlength{\parskip}{0pt plus 0.5ex} % ----------------------------------------------------------------------- \title{Griffith QM Time Dependent Perturbation Theory CheatSheet} \begin{document} \raggedright \footnotesize \begin{center} \Large{\textbf{Griffith QM Time Dependent Perturbation Theory CheatSheet (UCB 137B)}} \ \end{center} \begin{multicols}{3} \setlength{\premulticols}{1pt} \setlength{\postmulticols}{1pt} \setlength{\multicolsep}{1pt} \setlength{\columnsep}{2pt} \section{TIPT} \begin{align} H = H_0 + H' \ E_n = E_n^0 + E_n^1 \ \|\psi_n\rangle = \|\psi_n^0\rangle + \|\psi_n^1\rangle \ E_n^1 = \langle \psi_n^0 \| H' \| \psi_n^0 \rangle \ \|\psi_n^1\rangle = \sum_{m\neq n} \frac{\langle \psi_m^0 \| H' \| \psi_n^0 \rangle}{E_n^0-E_m^0} \| \psi_m^0\rangle \end{align} \subsection{Degenerate Case} \begin{align} \text{Degenerate space: } {\|i\rangle} \to E\ \quad W_{ab} = \langle a \| H' \| b \rangle \text{ Non-Diagonal} \ \text{Eigenvalue and Eivenvectors} \to E_n^1, \|\hat{i}\rangle \end{align} \section{Variational Method} \begin{align} \langle H \rangle(\lambda) &= \frac{\langle \psi(x,\lambda)\|H\|\psi(x,\lambda) \rangle}{\langle \psi(x,\lambda)\|\psi(x,\lambda) \rangle} \ \langle H \rangle(\lambda) &\geq E_{.g.s} \ \frac{d}{d\lambda} \langle H \rangle(\lambda_0) = 0 &\Rightarrow \langle H \rangle(\lambda_0) \approx E_{.g.s} \end{align} \section{WKB Method} \begin{align} \frac{d^2 \psi(x)}{dx^2} &= -k^2(x) \psi(x)& \ k(x)&=\frac1\hbar \sqrt{2m(E-V(x)} \ \phi(x) &= \int^x k(x) dx \ \psi(x) &= \frac1{\sqrt{k(x)}} (C_+ e^{i \phi(x)} + C_- e^{-i \phi(x)})\ &= \frac1{\sqrt{k(x)}} (C_1 \sin{\phi(x)} + C_2 \cos{\phi(x)}) \end{align} \subsection{Energy Level} \begin{align} \int_{R_{classical}} k(x) dx = n\pi \ \text{one $\infty$ wall} \quad n \to n-1/4 \ \text{No $\infty$ wall} \quad n \to n-1/2 \end{align} \subsection{Tunneling} \begin{align} T = e^{-2\gamma} \ \gamma = \int_{R_{forbidden}} k(x) dx \end{align} \section{TDPT} \begin{align} H = H_0 + V(t) \ \text{Eigenstate of $H_0$: } \|n\rangle, E_n\ \text{transition: } \|i\rangle \to \| f\rangle \ V_{f i}(t) = \langle f \| V(t) \| i \rangle \ \omega_{f i} = (E_f-E_i)/\hbar \ c_f(T) = \frac{-i}{\hbar} \int_0^T V_{f i}(t) e^{-i \omega_{f i} t} dt \end{align} \subsection{Constant Perturbation} \begin{align} V(t) = \begin{cases} V, & 0 \leq t \leq T \ 0, & \text{otherwise} \end{cases} \ V_{fi}(t)= constant \ P_{i\to f}(t)= \|c_f(t)\|^2 = 4 \frac{\|V_{fi}\|^2}{\hbar^2} \frac{\sin^2(\omega_{fi})t/2}{\omega_{fi}^2} \ \omega_{fi} \to 0 \quad \text{(degenerate states):} \ \|c_f(t)\|^2 = \frac{\|V_{fi}\|^2}{\hbar^2} t^2 \end{align} \subsection{Absorption} \begin{align} V(t) = V \sin(\omega t) \ P_{i\to f}(t) = \frac{\|V_{fi}\|^2}{\hbar^2} \frac{\sin^2((\omega_{fi}-\omega)t/2)}{(\omega_{fi}-\omega)^2} \end{align} \subsection{Simulated Emission} \begin{align} E_i > E_f,\quad \omega_{fi}<0 \ P_{i\to f}(t) = \frac{\|V_{fi}\|^2}{\hbar^2} \frac{\sin^2((\omega_{fi}+\omega)t/2)}{(\omega_{fi}+\omega)^2} \end{align} \subsection{Fermi Golden Rule} \begin{align} E_i \to E_f\text{ (continuous states)} \ P_{i\to f} = \frac{2\pi}{\hbar} \|\langle f \| V\| i\rangle \|^2 \rho(E_f) t \ \end{align} \subsection{Selection Rule} For spherical symmetric potential: \begin{align} \langle n',l',m'\| \vec{r} \| n,l,m \rangle &\neq 0 \text{ when:} \ \Delta l &= \pm 1 \text{ and:}\ \Delta l &= \pm 1 \text{ or } 0 \end{align} \section{Scattering} \begin{align} \psi(r,\theta) &= e^{ikz} + f(\theta) \frac{e^{ikr}}{r}, \text{ for large r} \ k &= \frac{\sqrt{2mE}}\hbar \ \frac{d \sigma}{d \Omega} &= \|f(\theta)\|^2 \ \sigma &= \int d\omega \frac{d \sigma}{d \Omega} \end{align} \subsection{Born Approximation} \begin{align} f(\theta) = -\frac{m}{2\pi \hbar^2} \int V(\vec{r}) e^{i(\vec{k}'-\vec{k})\cdot \vec{r}} d^3\vec{r} \intertext{Low Energy:} f(\theta) = -\frac{m}{2\pi \hbar^2} \int V(\vec{r}) d^3\vec{r} \intertext{Spherical symmetric:} f(\theta) = -\frac{2m}{\hbar^2 \kappa} \int_0^\infty r V(r) \sin(\kappa r) dr \ \kappa = 2k\sin(\theta/2) \end{align} \subsubsection{Yukawa Potential} \begin{align} V(r) = V_0 \frac{e^{-r/R}}{r} \ f(\theta) = -\frac{2mV_0 R^2}{\hbar^2} \frac{1}{1+4k^2R^2\sin^2(\theta/2)} \ \sigma = (\frac{2mV_0R^2}{\hbar^2})^2 \frac{4\pi}{1+4k^2R^2} \end{align} \subsubsection{Rutherford Scattering} Let $V_0 = q_1q_2/4\pi \epsilon_0$, $R=\infty$: \begin{align} f(\theta) = -\frac{2mq_1q_2}{4\pi\epsilon_0\hbar^2\kappa^2} \end{align} \subsection{Partial Waves} \begin{align} f(\theta) &= \frac1k \sum_{i=0}^\infty (2l+1)e^{i \delta_l} \sin(\delta_l) P_l(\cos(\theta)) \ \sigma &= \frac{4\pi}{k^2}\sum_{l=0}^\infty (2l+1) \sin^2(\delta_l) \end{align} \subsubsection{Optical Theorem} \begin{equation} Im[f(0)] = \frac{k \sigma}{4\pi} \end{equation} \subsubsection{Hard Ball} \begin{align} \delta_l = \tan^{-1}(\frac{j_l(ka)}{\eta_l(ka)}) \ ka << 1 \to \sigma = 4\pi a^2 \end{align} \section{Useful Models} \subsection{Density of States} \begin{align} E &= \hbar^2 k^2 /2m \ dN &= \frac{L^3}{(2\pi)^3} d^3k = \frac{L^3}{(2\pi)^3} d\Omega dk \ dN &= \frac{L^3}{(2\pi)^3} 4\pi \frac{m}{\hbar^2 k} dE \ \rho(E) &= \frac{dN}{dE} = \frac{L^3}{2\pi^2} \frac{mk}{\hbar^2} \end{align} \subsection{infinite square well} \begin{align} H(x) = \frac{p^2}{2m} + \begin{cases} 0, & 0\leq x\leq a \ \infty, & \text{otherwise} \end{cases} \ E_n = \frac1{2m} (\frac{n \pi \hbar}{a})^2 \ \psi_n = \sqrt{\frac{2}{a}} \sin(\frac{n\pi x}{a}) e^{-iE_nt/\hbar} \end{align} \subsection{Harmonic Oscillator} \begin{align} H(x) &= \frac{p^2}{2m} + \frac12 m\omega^2x^2 \ E_n &= (n+1/2)\hbar \omega \ \psi_n(x) &= \frac1{\sqrt{2^n n!}} (\frac{m\omega}{\pi \hbar})^{1/4} e^{-\zeta^2/2} H_n(\zeta) \ \zeta &= \sqrt{\frac{m\omega}{\hbar}x} \end{align} \subsection{Virial Theorem} \begin{align} 2 \langle T \rangle = \langle \vec{r} \cdot \nabla V \rangle \quad \text{(3D)}\ 2 \langle T \rangle = \langle x \frac{dV}{dx} \rangle \quad \text{(1D)}\ 2 \langle T \rangle = n \langle V \rangle \quad (V\propto r^n) \ \langle T\rangle = -E_n, \quad \langle V \rangle = 2 E_n \quad \text{(hydrogen)}\ \langle T\rangle = \langle V \rangle = E_n/2 \quad \text{(harmonic oscillator)}\ \end{align} \section{Math} \subsection{Legendre Polynomials} Domain: $(-1,1)$ \ Even, Odd, Even, Odd ... \begin{align} P_0(x) &= 1 \ P_1(x) &= x \ P_2(x) &= \frac12 (3x^2-1) \ P_3(x) &= \frac12 (5x^3 - 3x) \end{align} \subsection{Hankel Functions} Solution to Radial Shrodinger Equation: \begin{align} -\frac{\hbar^2}{2m} \frac{1}{r^2} \frac{\partial}{\partial r} (r^2 R_{El}) + [V(r) + \frac{\hbar^2 l (l+1)}{2m r^2}]R_{El} = E R_{El} \ V = 0 \to R_{El} = j_l(kr) \ V \neq 0 \to R_{El} = j_l(kr+\delta_l) \ r \to \infty \Rightarrow R_{El} = \frac{\sin(kr-l\pi/2+\delta_l(E))}{kr} \end{align} When $kr >> 1$ \begin{align} j_l(kr) &\to \frac{\sin{kr-l\pi/2}}{kr} \ \eta_l(kr) &\to \frac{-\cos{kr-l\pi/2}}{kr} \ h_l(kr) &\to \frac{e^{i(kr-l\pi/2)}}{ikr} \ h_l^(kr) &\to \frac{e^{-i(kr-l\pi/2)}}{-ikr} \ j_l(kr) &= \frac{1}{2} (h_l(kr)+h^(kr)) \end{align} \subsection{Hermite Polynomials} Domain: $(-\infty,\infty)$ \ Even, Odd, Even, Odd ... \begin{align} H_0(x) &= 1 \ H_1(x) &= 2x \ H_2(x) &= 4x^2-2 \ H_3(x) &= 8x^3-12x \end{align} \subsection{Spherical Harmonics} \begin{align} \|l,m \rangle &= Y_l^m(\theta, \phi) \ Y_0^0(\theta,\phi) &= \frac12 \frac1{\sqrt{\pi}} \ Y_1^0(\theta,\phi) &= \frac12 \sqrt{\frac3\pi} \cos{\theta} \ Y_1^{-1}(\theta,\phi) &= \frac12 \sqrt{\frac{3}{2\pi}} \sin{\theta} e^{-i \phi} \ Y_1^{-1}(\theta,\phi) &= -\frac12 \sqrt{\frac{3}{2\pi}} \sin{\theta} e^{i \phi} \end{align} \subsection{Green's Function} For a Linear Operator $\hat{D}_x$ \begin{align} \text{Homogeneous solution: }\hat{D}x \psi_0(x) = 0 \ \text{Hard Problem: }\hat{D}_x \psi(x) = f(x) \ \text{Simple Problem: }\hat{D}_x G(x,x') = \delta(x-x') \ \psi(x) = \psi_0(x) + \int{\text{f Domain}} G(x,x') f(x') dx' \end{align} \subsection{Some Integrals} \begin{align} \Gamma(n+1) &= n! \ \Gamma(z+1) &= z \Gamma(z) \ \int_0^\infty x^n e^{-ax} dx &= \frac{n!}{a^{n+1}} \ \int_0^\infty e^{-ax^b} dx &= a^{-1/b} \Gamma(1/b+1) \ \int_0^\infty e^{-ax} \sin{bx} dx &= \frac{b}{a^2+b^2} \ \int_0^\infty e^{-ax} \cos{bx} dx &= \frac{a}{a^2+b^2} \ \int_{-\infty}^\infty e^{-ax^2+bx} dx &= \sqrt{\frac{\pi}{a}} e^{\frac{b^2}{4a}} \ \int_0^\infty e^{-ax^2}x^n dx &= I_n(a)\ I_0=\frac12 \sqrt{\frac{\pi}{a}}, I_1&=\frac{1}{2a}, I_2=\frac1{4a} \sqrt{\frac{\pi}{a}}, I_3=\frac{1}{2a^2} \end{align} \end{multicols} \end{document} ``` Close About About us Careers Blog Solutions For business For universities For government For publishers Customer stories Learn 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15963	https://www.ncbi.nlm.nih.gov/books/NBK558944/	Poliomyelitis - StatPearls - NCBI Bookshelf An official website of the United States government Here's how you know The .gov means it's official. Federal government websites often end in .gov or .mil. Before sharing sensitive information, make sure you're on a federal government site. The site is secure. The https:// ensures that you are connecting to the official website and that any information you provide is encrypted and transmitted securely. Log inShow account info Close Account Logged in as: username Dashboard Publications Account settings Log out Access keysNCBI HomepageMyNCBI HomepageMain ContentMain Navigation Bookshelf Search database Search term Search Browse Titles Advanced Help Disclaimer NCBI Bookshelf. A service of the National Library of Medicine, National Institutes of Health. StatPearls [Internet]. Treasure Island (FL): StatPearls Publishing; 2025 Jan-. StatPearls [Internet]. Show details Treasure Island (FL): StatPearls Publishing; 2025 Jan-. Search term Poliomyelitis Jonathan G. Wolbert; Michael Rajnik; Helena M. Swinkels; Karla Higginbotham. Author Information and Affiliations Authors Jonathan G. Wolbert 1; Michael Rajnik 2; Helena M. Swinkels 3; Karla Higginbotham 4. Affiliations 1 PBCGME 2 Uniformed Service University of the Health Sciences 3 University of British Columbia 4 St. Lucie Medical Last Update: October 6, 2024. Go to: Continuing Education Activity Poliomyelitis is a highly transmissible infection nearing global eradication. However, recent outbreaks and vaccine-derived cases highlight the ongoing risk. Most infections are asymptomatic, with 70% to 95% presenting as a self-limiting flu-like illness. However, up to 1 in 200 cases involve rapid-onset flaccid paralysis, with the potential for lifelong disability or death. Polio persists in areas with poor water, sanitation, and hygiene infrastructure, with sporadic cases reported globally. Clinicians must remain vigilant for polio, especially in countries with low vaccination rates. Vaccine-derived strains, which cause paralytic disease indistinguishable from wild-type polio, complicate eradication efforts. Early recognition of symptoms, timely reporting to public health authorities, and reinforcing vaccination remain critical for patient outcomes and global eradication efforts. This activity provides healthcare professionals with a comprehensive overview of polio identification, management, and control. Emphasizing the role of the interprofessional healthcare team and highlighting effective communication and collaboration strategies, it also discusses clinicians' ability to recognize polio and postpolio syndrome, report polio to public health authorities, and implement polio prevention measures, thereby improving patient outcomes and contributing to the global effort to eradicate polio. Objectives: Compare clinical manifestations and apply diagnostic testing to differentiate poliomyelitis from similar infections as possible causes of acute flaccid paralysis. Improve polio vaccination rates as a primary prevention strategy against poliomyelitis infection. Identify early signs of post-polio syndrome in patients with a history of poliomyelitis. Implement effective communication as an interprofessional healthcare team member to ensure comprehensive care for patients with poliomyelitis and prevent future infections. Access free multiple choice questions on this topic. Go to: Introduction Poliomyelitis, or polio, is an infection transmitted via the fecal-oral and oral-oral routes by the poliovirus. Today, polio primarily affects children younger than 5 in countries with poor water, sanitation, and hygiene infrastructure. While extremely rare outside these areas,polio presents a risk to populations with low vaccination rates, including in industrialized nations. The 2014 declaration deeming the global spread of polioviruses a public health emergency of international concern under the International Health Regulations remains in place.[WHO. Vaccine Position] Previously one of the most feared diseases in the world, polio caused widespread morbidity and mortality in children during multiple epidemics between the 1900s and 1960s. The incidence of polio began to decline following the development of the injectable polio vaccine (IPV) and oral polio vaccine (OPV). Intensified worldwide vaccination efforts starting in the 1980s by the World Health Organization (WHO) and partners under the Global Polio Eradication Initiative (GPEI) have almost completely eradicated the disease.Vaccination protocols worldwide now recommend administering at least 3 doses of IPV (IPV-only schedule) or at least 3 doses of bivalent OPV plus at least 2 doses of IPV (combined OPV-IPV schedule). The latter is recommended in countries where poliovirus incidence rates remain unchanged or are at risk of increasing.See StatPearls' companion article "Polio Vaccine" for more information on polio vaccinations. Healthcare professionals must maintain a high index of suspicion to diagnose acute poliomyelitis in patients who present with new-onset paralysis following a viral prodrome and are living in an endemic region or under-vaccinated population. The presentation of polio is highly variable, ranging from asymptomatic to a transient flu-like viral illness to paralysis, quadriplegia, and even respiratory failure and death. Many polio survivors experience a poor quality of life. Alternate diagnoses must be considered as polio is rare, and other more common diseases and conditions can present similarly. In particular, acute flaccid myelitis (AFM) presents with a polio-like syndrome. This condition is caused by epidemic enteroviral infections, most commonly enterovirus D68. As no antiviral treatment for acute poliomyelitis exists, prevention is critical. Paralytic deficits are often permanent, resulting in chronic pain, deformities, weakness, and eventual osteoporosis and fragility fracture. Between 25% and 40% of the estimated 12 to 20 million polio survivors worldwide will also develop postpolio syndrome (PPS).A diagnosis of PPS requires new-onset, progressive muscle weakness or fatiguability, mental fatigue, or pain in a patient diagnosed with poliomyelitis up to 40 years prior.PPS is not contagious and is rarely life-threatening, but it often affects patients' independence and quality of life. While the near eradication of polio represents a tremendous achievement for global public health, multiple challenges and delays remain. Vaccine-derived polioviruses (VDPVs) revert or recombine to wild-type neurovirulence and transmissibility, causing outbreaks and distrust in vaccination programs. Poliovirus typically escapes early detection; most infections are asymptomatic, and the characteristic paralysis occurs in a small minority of those it infects.[CDC. Polio_HCP]Delays in detecting circulating wild poliovirus (WPV) and VDPVs pose a significant risk to eradication efforts. Political unrest or indecision and declining or competing interests threaten efforts to eradicate the disease. Recent polio cases and the discovery of poliovirus in wastewater in industrialized nations emphasize that, until completely eradicated, polio remains a global risk.[GPEI. Polio Eradication] All suspected cases of poliomyelitis in the United States should be immediately reported to the United States Centers for Disease Control and Prevention (CDC). Clinicians should suspect polio in patients with acute-onset flaccid paralysis,decreased or absent tendon reflexes, and without sensory or cognitive deficits.Contact the CDC Emergency Operations Center within 4 hours of the suspected diagnosisto ensure appropriate diagnostic testing, surveillance, and public health follow-up at770-488-7100.[CDC.Polio_Clinical] Go to: Etiology Poliovirus, enterovirus subtype C, is a small,single positive-stranded ribonucleic acid virus that is acid-resistant, non-enveloped, and encapsulated.This causative agent of acute polio and postpolio syndrome belongs to the family Picornaviridae and reproduces only in humans and some great apes.Poliovirus is extremely contagious, with seroconversion occurring in more than 90% of household contacts.Wild poliovirus (WPV) has 3 distinct serotypes (1, 2, and 3), each with a slightly different capsid protein (WPV1, WPV2, and WPV3).Surviving an infection confers complete and lifelong immunity, but only to the specific infecting serotype.[GPEI. Polio Eradication] Vaccine-associated paralytic polio(VAPP) and circulating vaccine-derived polioviruses (cVDPVs) cause illness that is clinically indistinguishable from poliomyelitis caused by WPV.[WHO. Vaccine Position]VAPP results from the proliferation and mutation of polioviruses in OPV recipients or their contacts. Replicating viruses in the intestine revert to neurovirulent variants demonstrating close genetic similarity to the vaccine strain. VAPP is an adverse event triggered by immunization and is reportable under the Vaccine Adverse Event Reporting System. VDPVs result from the naturally selective proliferation of OPV viruses in populations with low herd immunity (cVDPVs) or prolonged carriage in immunocompromised individuals (iVDVPs). Recombination with other enteroviruses or reversion of OPV strains allows the resulting cVDPVs to regain the neurovirulence and transmissibility of WPV.[WHO. Vaccine Position]Genetic divergence of cVDPVs from the attenuated OPV virus indicates prolonged replication and circulation. cVDPVs derived from OPV type 2 (cVDPV2)are the most common, although cVDVP types 1 and 3 also occur. Go to: Epidemiology Poliovirus primarily spreads via the fecal-oral route and less commonly via respiratory droplets or oral-oral contact with nasopharyngeal secretions.[CDC. Polio_HCP]In temperate climates, infections occur more commonly during the late summer and fall.Children younger than 5 face the highest infection risk, although unvaccinated individuals of any age can become infected.The infection is asymptomatic in 75% to 90% of patients.[WHO. About_Polio][CDC. About Polio]Central nervous system invasion resulting in paralytic disease occurs in up to 1 in every 200 patients with a poliovirus infection. The mortality rate of paralytic polio can be as high as 20%,although most references cite a rate closer to 10%.[CDC. Polio_HCP]The risk and severity of paralysis increase with patient age and immune status.[WHO. Vaccine Position]The incubation period is usually 7 to 10 days but may range from 4 to 35 days.[WHO. About_Polio] Polio has afflicted humans since ancient times, first exposing infants at an early age when maternal antibodies are present and then repeatedly over the lifetime, resulting in persistently high antibody levels. As such, paralytic polio was likely uncommon. With the advent of modern sanitation, medical anthropologists believe early exposure to poliovirus became less common, and maternal antibody levels decreased, leading to massive outbreaks with high rates of paralysis in the early 20th century.[GPEI. Polio History]By the mid-20th century, the virus killed or paralyzed about 500,000 people worldwide yearly. A United States (US) outbreak in 1916 killed more than 8000 people;another in 1952 killed 3000 and resulted in 20,000 cases of paralysis.Case rates in the US dropped by 99% only 6 years after the first poliovirus vaccine became available. The most recent case of WPV acquired in the US was in 1979. The most recent case in all of the Americas occurred in 1991.[CDC. Yellow_Book]WPV1 has been the primary cause of most of the world's paralytic polio cases throughout history and is the only circulating subtype; WPV1 remains endemic in Pakistan and Afghanistan. In 1988, 350,000 cases of endemic poliomyelitis occurred across 125 countries. Based on success in the Americas,the World Health Assembly adopted a resolution to eradicate WPV worldwide by 2000 under the Global Polio Eradication Initiative.[GPEI. Polio Eradication][CDC. VPD Manual_Polio]While eradication remains elusive, establishing the polio vaccine in routine childhood immunization schedules has led to a greater than 99.9% worldwide reduction in WPV cases since 1988.[GPEI. Polio Eradication]The declaration of WPV2 eradication occurred in 2015, 6 years after the last case was reported in India. The WHO declared the eradication of WPV3 in October 2019, 7 years after the last reported case.The WHO declared Africa polio-free in 2020. A worldwide low of 6 cases of WPV1 was reported for 2021. Sporadic outbreaks continue to occur, including in countries previously declared polio-free.From late 2021 to early 2022, WPV1 caused paralytic polio in 9 patients in Malawi and Mozambique; this outbreak was traced to a transmission in Pakistan.The total number of reported WPV1 cases was 22 in 2022 and 12 in 2023. The WHO reported 23 cases between 2023 and 2024, for which recent statistics are available.[GPEI. Polio Eradication]When polio is reintroduced into a country, transmission must be traced and halted within 12 months for that country to maintain polio-free status. VAPP is extremely rare, occurring an estimated 0.42 times per million doses of OPV [CDC. VPD Manual_Polio]or 3.8 times (range 2.9 to 4.7) per million births in countries that use trivalent OPV.[WHO. Vaccine Position]In high-resource countries, VAPP occurs most commonly after the first dose of OPV. In immunocompromised individuals, the risk of VAPP is approximately 3000 times higher than in healthy patients. In low-resource countries, VAPP primarily affects children aged 1 to 4 receiving second or subsequent doses.[WHO. Vaccine Position]Providing IPV before OPV reduces or eliminates VAPP risk.[WHO. Vaccine Position] First detected in an outbreak on the island of Hispaniola in 2001, cVDPVs are responsible for most cases of poliomyelitis today.cVDPV-associated paralysis occurs primarily in sub-Saharan Africa and Asia.A total of 345 people reported an onset of paralysis in the 12 months leading up to July 2024. Forty-nine cases of cVDPV type 1 were reported in Mozambique, DR Congo, and Madagascar, and 296 cases of cVDPV type 2 were reported in 20 other countries.[GPEI. Polio Eradication] This is a reduction from 881 total cases in 2022 and 524 in 2023. While the number of cases in 2023 was reduced compared to 2022, cVDPV cases were more widespread geographically in 2023. In 2022, an unvaccinated young adult from New York presented with acute flaccid paralysis; stool testing revealed VDPV2. Environmental testing of samples taken before and after the patient's illness onset subsequently found VDPV2 in wastewater in surrounding counties. The virus was likely imported from a country that administers OPV but where the population remains under-vaccinated. In addition to this case, a case of polio in a young boy in Gaza, sporadic cases of WPV polio in Africa,and the discovery of cVDPV in wastewater across multiple countries, including the United Kingdom and Israel, emphasize the importance of maintaining high vaccination rates across all populations until polio is eradicated globally. For further information on vaccination rates in the United States and globally, see StatPearls' companion article, "Polio Vaccine." Go to: Pathophysiology Following exposure, poliovirus replicates in oropharynx and gastrointestinal tract tissues. Maternal antibodies or those acquired during a previous infection or oral vaccination often confine the virus to these tissues.In the absence of antibodies, a typically transient and asymptomatic viremia occurs.However, in a small minority of viremic cases, the virus enters the central nervous system (CNS) by crossing the blood-brain barrier or via peripheral nerve retrograde axonal transport. Depending on the virus type and location of invasion, CNS invasion causes non-paralytic aseptic meningitis in up to 5% of symptomatic infections.[CDC. Polio_HCP]CNS invasion can also result in infection and destruction of alpha motor neurons in the anterior horns of the spinal cord, the cranial nerve nuclei of the medulla oblongata, and the pons and midbrain. Inflammatory infiltration of these motor neurons causes cell destruction and eventual glial cell scarring. For patients who recover from acute polio, regeneration of reversibly damaged neurons, collateral axonal sprouting, and the development of other reparative and compensatory processes can result in the slow improvement of paralysis. Human axons may be able to reinnervate as much as 5 times their original muscle fiber territory.In chronic cases, de-innervated muscles atrophy and cause functional disorders of the limbs, trunk, and respiratory muscles, including the diaphragm. In children, deformities progressively develop due to the influence of muscular imbalances or dysfunction on the growing skeleton. This can result in joint contractures, kyphosis, scoliosis, foot deformities, or leg-length discrepancies. Chronic overloading results in joint instabilities, degenerative changes, and pain.Following infection, immunocompetent individuals develop complete immunity to poliovirus, which involves various mucosal and humoral immune responses. Mucosal immunity reduces viral replication and excretion, thereby reducing transmission risk.[WHO. Vaccine Position]Therefore, patients with B-cell immunodeficiency are at an increased risk of paralytic disease and prolonged shedding. Go to: Histopathology Enlarged motor neurons are characteristic but not pathognomonic of infection with poliovirus. In acute poliomyelitis, poliovirus destroys the original innervating neurons in the muscle fibers. Neighboring neurons attempt to reinnervate the denervated tissue; this leaves the surrounding motor units innervating more muscle fibers than before. This extra workload enlarges the affected neuron and is suspected to be part of the fatigue associated with post-polio syndrome later in life. Go to: History and Physical Until it is eradicated globally, clinicians must consider polio in the differential diagnosis of patients presenting with acute flaccid paralysis.The rapid onset of pure motor deficits without sensory or cognitive involvement following a flu-like prodrome is suggestive of paralytic polio. However, not all patients with acute polio report a prodrome. Various conditions can cause acute onset flaccid paralysis, and the clinician must differentiate between poliovirus infection, other viral infections, and noninfectious causes. Disease caused by cVDPVs is clinically indistinguishable from that caused by WPVs.More information can be found in the Differential Diagnosis section. Obtaining a thorough medical history and performing a comprehensive neurological exam are crucial. In addition to a complete review of systems, the history must include the patient's vaccination status, recent travel, ill contacts, recent attendance at large gatherings, socioeconomic status, and living situation.Clinicians should consider polio in patients with sudden-onset limb weakness and a history of fever or gastrointestinal symptoms. This suspicion should be heightened further for patients who are also incompletely vaccinated or unvaccinated and residing in or have recently traveled to regions with known polio transmission or in contact with someone who has traveled to such an area. The cranial nerves should be examined, and reflexes, tone, and strength in all limbs should be documented.Adequate hand hygiene and contact precautions must be observed during the physical examination of any patient with suspected polio. The acute, prodromal, or pre-paralytic stage of poliovirus infection is similar to many other viral illnesses;fever, malaise, headache, myalgia, fatigue, nausea, vomiting, abdominal pain, and sore throat are commonly observed.[WHO. About_Polio]In almost all cases, the infection resolves in 2 to 10 days without further sequelae.[WHO. About_Polio] Meningitis develops within a few days after the prodrome resolves. While usually asymptomatic, recurrent fever, neck stiffness, back pain, and muscle spasms may be observed.These symptoms resolve within 2 to 10 days, typically resulting in a rapid and complete recovery.[WHO. Vaccine Position] The return of the fever with severe muscle pain, fasciculations, spasms, and hyperreflexia occurring 1 to 3 days after apparent recovery heralds paralytic polio.[WHO. Vaccine Position]Weakness or flaccid paralysis ensues in an asymmetric distribution, with proximal to distal progression. Maximum paralysis is reached within 2 to 4 days, rarely progressing further after fever resolution.[CDC. Polio_HCP][CDC. Polio_HCP] The clinical manifestations depend on the site of viral invasion and the extent of viral replication, resulting in paralysis with a variable distribution and severity, affecting whole or parts of muscles or entire muscle groups.In most cases, spinal paralysis results in paralysis of the muscles of the lower limbs and, less commonly, of the arms, trunk, or diaphragm.[WHO. About_Polio]Autonomic dysfunction occurs infrequently, resulting in bladder emptying problems, constipation, or sweat secretion abnormalities.Poliovirus can invade the bulbar aspect of the brain, affecting the control of breathing and other vital functions, as well as the muscles for facial and eye movements and swallowing. Bulbar invasion can lead to respiratory insufficiency,vegetative crises, tachycardia, hypertension, and collapse. Fortunately, invasion of this severity rarely occurs. Go to: Evaluation Rapid investigation of acute flaccid paralysis is critical to identifying a possible polio infection and implementing control measures. Poliovirus infection is assessed through viral cultures or detection of viral ribonucleic acid in stool or throat irrigation or swabs.[CDC. Polio_HCP]The CDC recommends 2 specimens taken at least 24 hours apart during the first 14 days after the onset of paralysis.Specimens must be maintained at -20 ºC and shipped frozen.[CDC. Polio_HCP] Clinical confirmation of polio requires 1 of the following: Poliovirus detected by sequencing of the capsid region of the genome by the CDC Poliovirus Laboratory Poliovirus identified in an appropriate clinical specimen (eg, stool [preferred], cerebrospinal fluid, oropharyngeal secretions) using a validated assay, AND specimen is unavailable for sequencing by the CDC Poliovirus Laboratory.[CDC. Polio_HCP] Confirmed cases of nonparalytic poliovirus infections are those that meet laboratory but not clinical criteria. Maximum viral excretion in the stool and nasopharyngeal fluids begins 2 to 3 days before (and lasts for a week after) symptom onset.However, the virus may be present in the nasopharyngeal secretions for up to 2 weeks and 3 to 6 weeks in the stool, even in asymptomatic individuals.[CDC. Polio][CDC. Yellow_Book]Finding poliovirus in cerebral spinal fluid is uncommon. A cerebral spinal fluid test result negative for poliovirus does not exclude polio. Polioviruses undergo reverse transcriptase polymerase chain reaction testing following culture to differentiate between WPV and VDPVs. Serology is not helpful due to the high seroprevalence of anti-polio antibodies due to vaccination or previous asymptomatic infection. Researchers continue to seek direct molecular methods to diagnose polio, replacing culture-dependent methods.[WHO. Vaccine Position] Magnetic resonance imaging of the brain and spinal cord may identify typical patterns of poliomyelitis and rule out other pathologies, such as spinal cord infarction.Characteristic electromyographic findings develop later in the course of illness. Go to: Treatment / Management Supportive care is the mainstay of treatment for acute poliomyelitis, as no effective antiviral therapy exists.Therapy may include providing medication for fever and irritation, preventing respiratory tract infections, and managing respiratory paralysis via mechanical ventilation if required. Physiotherapists use splints to relieve pain and spasms; splints also play an imperative role in preventing the development of deformities. Following the acute stage, rehabilitation, exercise, counseling, and education help patients achieve their best functional status.Orthopedic surgery may be warranted to help patients walk and correct factors potentially leading to deformity as the patient grows and ages. Surgery may include joint contracture release, correction of muscle imbalances around the joint to prevent deformities, or correction of deformities.Following phase 1 trials, the capsid inhibitor pocapavir is available for compassionate use in patients who cannot clear the virus due to primary immunodeficiency disorder.[WHO. Vaccine Position] Go to: Differential Diagnosis Globally, acute flaccid paralysis occurs at a rate of about 1 of every 100,000 children younger than 15 per year. These are due to a wide variety of infectious and noninfectious causes. Other enteroviruses cause acute flaccid paralysis with greater frequency than poliovirus, including enteroviruses A71 and D68, coxsackievirus A, and rarely enteroviruses such as C105 and C109.Like polio, enterovirus infections are frequently asymptomatic or present with a mild viral prodrome. While paralytic polio predominantly affects the lower limbs, most other enterovirus-related acute flaccid paralysis have an upper extremity predilection. Other infections that may present similarly include adeno-associated virus, West Nile virus, varicella-zoster virus, Japanese encephalitis virus, rabies virus, and C lostridium botulinum.A careful history and physical exam usually differentiate these infections from polio.Noninfectious considerations include spinal cord infarctions, transverse myelitis, acquired axonal neuropathies, myasthenia gravis, and Lambert-Eaton myasthenic syndrome. Guillain-Barré syndrome is a common cause of acute flaccid paralysis, clinically differentiated from polio by ascending, symmetrical onset, sensory defects, and the absence of preceding meningeal symptoms.See StatPearls' companion article "Acute Flaccid Myelitis"for more information on acute flaccid paralysis associated with other viruses. Go to: Prognosis Most poliovirus infections are asymptomatic or self-limiting.For those with poliomyelitis (ie, symptomatic poliovirus infection), 10% to 15% will die due to bulbar involvement, with respiratory and cardiovascular collapse. For those who survive, recovery and residual paralysis stages follow. Convalescence or recovery begins when the acute, flu-like illness resolves and no further paralysis occurs. Neighboring nerves or axonal regeneration reinnervates paralyzed muscles. Recovery can last up to 2 years, with maximal improvement occurring in the first 6 months.Approximately 60% of polio survivors have permanent deficits. The more severe the acute phase of the disease, the greater the likelihood of residual deficits and the development of postpolio syndrome in the future. Go to: Complications Morbidity following paralytic polio commonly results from biomechanical alterations due to muscle weakness, imbalances of muscle power, deformities, and complications of polio-related surgeries. Pain,easy fatiguability, limited function,and reduced quality of life are common. Osteoporosis develops due to lack of function and other factors, with frequent fractures.Approximately 25% to 40% of patients with previous paralytic polio also develop PPS over their lifetime.[CDC. Polio_HCP]The diagnosis of post-polio syndrome (PPS) requires at least 1 year of new and progressive muscle weakness or fatiguability following decades of stable paralysis after the acute infection. Other symptoms include generalized fatigue, muscle or joint pain, worsening respiratory function, and dysphagia.The true prevalence of PPS is not known, partly due to a lack of established biomarkers to diagnose postpolio syndrome. While the prevalence of PPS is decreasing in industrialized nations due to the elimination of the virus, it remains highly prevalent in developing nations. The pathogenic mechanisms resulting in PPS are incompletely understood. The most common theory involves degeneration of the enlarged motor units formed following the initial infection.Longitudinal research results demonstrate a parallel between a decreasing size and number of active motor units and a gradual strength decline in PPS. The precise etiology of this decline is unclear; it likely results from the overuse of remaining motor units, stress-induced aging of motor units, deteriorating neuromuscular junctions, weight gain, deconditioning, and other factors.Although some authors suggest subgroups of patients may have a predominantly neuroinflammatory pathogenesis of their PPS, recent evidence does not support the role of humoral or cellular pro-inflammatory processes.Normal age-related sarcopenia contributes significantly and independently to decreasing muscle mass and strength; however, PPS may accelerate the decline in some patients, with an annual loss of up to 8% of muscle strength. PPS is a clinical diagnosis using Halstead, March of Dimes, or other criteria. Neither electromyography nor muscle biopsy reliably differentiates patients with or without PPS. Clinicians must carefully exclude other treatable medical and orthopedic conditions that may cause or contribute to the condition.Depending on the presenting symptomatology, the differential includes a wide variety of disorders, including the following: Metabolic: Iron deficiency, hypothyroidism, type 2 diabetes Sleep: Apnea, restless leg syndrome Inflammatory: Rheumatoid arthritis, polymyalgia rheumatica Chronic pain: Fibromyalgia The severity and speed of progression are variable and are influenced by factors such as the severity of acute paralysis, age of onset, and socioeconomic status.Complications include respiratory failure, fractures, decreased function, and worsening quality of life. Tailored muscle strengthening, aquatic conditioning, and pain control are the mainstays of treatment. Go to: Postoperative and Rehabilitation Care Rehabilitation, exercise, counseling, and education are crucial for patients with polio or PPS to maintain function, prevent further morbidity, and achieve the best possible health.Of patients previously diagnosed with polio,60% to 80% reported falling at least once in the past year.One-third of falls result in a fragility fracture of the polio-affected limb. Based on experience and data extrapolated from study results in other conditions, physiatrists recommend energy conservation techniques, pacing strategies, and lifestyle and activity modification to prevent premature fatigue, weight gain, and other factors that can worsen symptoms. Assistive equipment (eg, grab bars, raised toilet seats, modified utensils) can maximize functioning and decrease energy use. Patients often require mobility aids such as canes, walkers, wheelchairs, or scooters. Fitted ankle-foot or knee-ankle foot orthoses can mitigate gait disturbances. Orthoses made with lightweight carbon fiber are about 30% lighter and can increase comfort and function but are more expensive than those made with thermoplastics and metal. An interdisciplinary approach to fitting orthotics is vital due to the difficulty of assessing and mitigating individual gait patterns and parameters. Research results have shown that non-fatiguing strengthening exercises increase muscle strength and prevent further decline in patients with moderate weakness due to PPS. Muscle training lessens muscle pain and fatigue and improves functional activity levels and quality of life. Physiatrists recommend intermittent (eg, alternate days) submaximal exercise be interspersed with short rest periods to avoid fatigue and overuse. Study results suggest individualized exercise programs within a comprehensive rehabilitation program provide long-term positive effects on physical, functional, and psychological outcomes. Studies of aerobic exercise in patients with PPS have yielded mixed results. In some patients, muscles may already be maximally adapted for daily life activities, limiting the potential for improvement. However,whole-body aerobic exercise, such as non-swimming aquatic exercises in heated water,recruits more muscles and benefits patient well-being, cardiovascular conditioning, and pain relief. Medications and supplements, such as creatine and intravenous immunoglobulin, may improve strength and lessen pain and fatigue in some patients with polio or PPS. However, more research is needed to determine which subgroups of patients may respond to which medication. Go to: Deterrence and Patient Education Infection Prevention and Control Hand hygiene and contact precautions are imperative for clinicians or household members interacting with patients. As household members may be exposed to poliovirus before the patient becomes ill, household members must watch for symptoms and maintain excellent hygiene to avoid potentially infecting others. Chlorine, heat, ultraviolet light, and formaldehyde rapidly inactivate poliovirus.The virus resists inactivation by lipid solvents and common detergents.[WHO. Vaccine Position] Depending on conditions, poliovirus can remain stable outside,with sunlight and repeated freeze-thaw cycles decreasing its activity.The virus retains infectivity for months at 4 ºC but only days at 30 ºC.[WHO. Vaccine Position] Routine Childhood Polio Vaccinations Polio vaccination programs worldwide use IPV with or without live-attenuated OPV.The first polio vaccine, licensed by Jonas Salk in 1955, consisted of inactivated (killed) viruses of all 3 WPV types. This first IPV protected children against severe polio but could not prevent human-to-human transmission due to its inability to induce a robust mucosal immune response in the gut. The original Salk IPV was replaced globally in 1961 by a trivalent OPV developed by Albert Sabin. Consisting of live, attenuated viruses of all 3 WPV types (OPV1, OPV2, and OPV3), this vaccine was inexpensive, easy to administer, and could stop transmission by inducing mucosal antibodies that prevented infection. While immensely successful in eliminating the wild-type poliovirus, the high virus mutability of OPV viruses also leads to VAPP and the emergence of cVDPVs. Healthcare professionals can administer the polio vaccine via the intramuscular or intradermal route. The vaccines are highly effective, providing greater than 99%protection after receiving all recommended doses.[GPEI. Polio Eradication]Their safety profile is excellent, with no systemic severe adverse reactions reported. No risk of VAPP exists with IPV use. OPV is similar to IPV in effectiveness. The WHO recommends OPV in endemic countries and those at higher risk for WPV reintroduction primarily because of cost, ease of administration, and ability to stop transmission. Countries at higher risk of reintroduction include those with poor water, sanitation, and hygiene infrastructure and connections with endemic countries. As of 2022, the WHO recommends one of the following primary immunization regimes for all children worldwide:at least 3 doses of IPV or at least 3 doses of bivalent OPV plus 2 doses of IPV. A booster dose is given with a minimum interval of 6 months following the initial dose. The United States was declared WPV-free in 1994 and switched solely to IPV in 2000.[CDC. About Polio]The CDC recommends children receive 4 doses of IPV at 2, 4, and 6 to 18 months, followed by a booster between 4 and 6 years of age. This vaccine schedule confers total immunity in more than 95% of patients. Adults at Increased Risk of Exposure The CDC recommends a single lifetime IPV booster for adults who have fully completed the childhood series but are at an increased risk for exposure. All travelers to countries that are endemic for polio or have an actively circulating virus are at risk, as are travelers working in high-risk settings (eg,humanitarian or healthcare) in countries adjacent to and with strong social connections to countries with an actively circulating virus. Others at increased risk include workers with potential occupational exposure to poliovirus, such as poliovirus laboratory workers and healthcare professionals in close contact with patients during a polio outbreak.Other situations may also increase risk, such as an outbreak or a low vaccination community with strong social ties to a country with a circulating virus. See StatPearls' companion article"Polio Vaccine" for further information on polio vaccine use. Go to: Enhancing Healthcare Team Outcomes With effective vaccines, poliomyelitis is an often-forgotten disease. However, continued polio endemicity in Pakistan and Afghanistan and cVDPVs elsewhere pose an ongoing global threat, made worse by political unrest,growing opposition to vaccination in developed countries, and other factors.The interprofessional healthcare team plays a vital role in polio eradication efforts across multiple levels of intervention. Polio Eradication Strategy 2022-2026 The GPEI, a public-private partnership under the WHO of national governments, the CDC, Rotary International, the United Nations Children's Fund, and the Bill & Melinda Gates Foundation, coordinates and leads global polio eradication efforts. The GPEI's Polio Eradication Strategy 2022 to 2026 has only 2 high-level goals: Permanently interrupt all poliovirus transmission in endemic countries. Stop cVDPV transmission and prevent outbreaks in non-endemic countries.[GPEI. Polio Eradication] As expected, polio eradication is proving more challenging than smallpox eradication. First, the poliovirus does not have optimal characteristics for eradication.Unlike smallpox,poliovirus has a high rate of asymptomatic illness, facilitating undetected viral spread. Many other diseases have clinically similar presentations to poliomyelitis, resulting in diagnostic delays. The natural genetic instability of the poliovirus results in cVDPVs, requiring the eventual coordinated stoppage of OPV. Multiple sociopolitical challenges also exist in the polio endgame.[GPEI.Polio Eradication]These include the following: Poor access to healthcare in remote areas and for transient population Difficulty delivering vaccinations in countries with poorly resourced healthcare systems Competing governmental priorities, such as the COVID-19 pandemic Inadequate development and maintenance of water, sanitation, and hygiene infrastructure Ongoing political conflict and uncertainty Political indecision, reduced financial resources, and decreased public awareness as polio cases dwindle The GPEI works with governments at all levels, civil society partners, and local healthcare professionals to build, maintain, and improve systems in areas with or at high risk for polio transmission to increase polio vaccination, detect cases of acute flaccid paralysis (AFP), and ensure adequate reporting and surveillance systems for rapid detection of poliovirus transmission. These depend highly on solid relationships with communities and seasonally migrant populations for trust building, capacity building, increasing local government ownership and control, using a gender lens, and ensuring accountability.[GPEI. Polio Eradication]Supporting national governments to have maximum control over polio prevention, surveillance, and outbreak responses is a key strategy to improve global polio eradication efforts.[GPEI. Polio Eradication] Improving Access to Quality Healthcare A comprehensive, coordinated interprofessional approach to polio diagnosis, management, and vaccination is essential to ensure quality care for individuals and populations. Clinicians must consider polio infection in the differential of acute flaccid paralysis and be familiar with diagnosing and managing poliovirus infections and PPS. Reporting suspected acute flaccid paralysis cases to public health authorities ensures appropriate testing, surveillance, and public health follow-up. Physiatrists, neurologists, orthotists, and occupational and physiotherapists are critical to polio and PPS rehabilitation. Clinicians, nurses, pharmacists, community workers, and public health teams working collaboratively to develop community partnerships and deliver consistent,culturally appropriate messages to patients can increase trust and improve vaccination rates. In countries with or at risk of WPV or cVDPVs, interprofessional teams can further stress the importance of hygiene and reinforce the best practices for treating the disease when it occurs. Surveillance and Reporting Delays in the detection and reporting of circulating WPV1 and VDPVs pose one of the most significant risks to polio eradication efforts.While global surveillance systems perform well overall, barriers to adequate surveillance continue, particularly in hard-to-reach areas and people with inadequate connections to the formal healthcare system.[WHO. Polio Surv]Surveillance activities declined in the first months of the COVID-19 pandemic, followed in 2020 by increased cases due to pandemic-related setbacks in surveillance and control activities. The GPEI uses the following 3 types of surveillance for poliovirus detection: Reporting of acute flaccid paralysis cases Environmental surveillance Monitoring of immunodeficiency-associated vaccine-derived poliovirus[WHO. Polio Surv] The WHO requires all countries to urgently report (within 24 hours) all cases of acute flaccid paralysis in patients younger than 15 and anyone with suspected polio. All other poliovirus detection from environmental samples, contacts of patients with polio, or surveillance in patients with primary immune disorders must also be reported to the WHO, although timelines are less stringent.[WHO. Polio Surv] Epidemiologists assess AFP surveillance-system performance by the number of nonpolio AFP cases (target: ≥2 per 100,000 children younger than 15) and the percent of adequately collected stool specimens from AFP patients (target: ≥80%with≥2 samples collected ≥24 hours apart within 14 days of paralysis onset, with the cool chain maintained throughout transit).[WHO. Polio Surv]Of the more than 100,000 samples tested by the Global Polio Laboratory Network each year, few are positive, indicating other diagnoses.[GPEI. Polio Eradication] Environmental surveillance significantly increases the poliovirus surveillance system sensitivity.This surveillance form is increasingly used to monitor the virus and implement public health measures before cases of AFP arise, such as community awareness campaigns, targeted increases in vaccination coverage, and active surveillance for AFP.Environmental surveillance is focused on priority countries (ie,patients with medium-high to very high risk of poliovirus transmission and with persistent surveillance gaps).[WHO. Polio Surv] As the world reaches polio eradication and cases decrease, the polio surveillance systems must continue to become increasingly sensitive.Detection of poliovirus excretion among people with primary immunodeficiency disorders is a new addition to the standard surveillance methods.[WHO. Polio Surv]While no iVDPV outbreaks have yet been detected, patients with primary immunodeficiency disorders have the risk of developing paralysis or seeding a cVDPV outbreak. Several high-risk countries are piloting iVDPV surveillance, with plans to expand and continue through the post-eradication period. All patients identified with VDPV are provided treatment. Research and Development Research increases the effectiveness of eradication activities, documents and verifies achievements, and identifies policy needs for the post-eradication era. Research priorities identified by the GPEI include the following: Basic immunology: To better understand patient responses to polio vaccines, we need an improved understanding of humoral and mucosal immunity, duration of immunity, and the effects of malnutrition and other variables on vaccination response. Antiviral therapy development: With early evidence of pocapavir resistance developing in polioviruses, researchers are investigating combinations and other therapies,such as polio monoclonal antibodies, for patients with prolonged excretion. Monitoring and evaluation: Policy developers and program providers require optimal surveillance methodologies and VAPP and VDPV seroprevalence studies to develop local strategies and improve program delivery. Genetics: A better understanding of virus mutation and phylogenetic links can help understand the evolution of VDPVs.[GPEI. Polio Eradication] For further information on vaccine research, optimal vaccination schedules, and improved vaccination program delivery, see StatPearls' companion article, "Polio Vaccines." Go to: Review Questions Access free multiple choice questions on this topic. Click here for a simplified version. Comment on this article. Go to: References 1. Greene SA, Ahmed J, Datta SD, Burns CC, Quddus A, Vertefeuille JF, Wassilak SGF. Progress Toward Polio Eradication - Worldwide, January 2017-March 2019. MMWR Morb Mortal Wkly Rep. 2019 May 24;68(20):458-462. [PMC free article: PMC6532951] [PubMed: 31120868] 2. Arita M. Poliovirus Studies during the Endgame of the Polio Eradication Program. Jpn J Infect Dis. 2017 Jan 24;70(1):1-6. [PubMed: 27795480] 3. Geiger K, Stehling-Ariza T, Bigouette JP, Bennett SD, Burns CC, Quddus A, Wassilak SGF, Bolu O. Progress Toward Poliomyelitis Eradication - Worldwide, January 2022-December 2023. MMWR Morb Mortal Wkly Rep. 2024 May 16;73(19):441-446. [PMC free article: PMC11115430] [PubMed: 38753550] 4. Helfferich J, Knoester M, Van Leer-Buter CC, Neuteboom RF, Meiners LC, Niesters HG, Brouwer OF. Acute flaccid myelitis and enterovirus D68: lessons from the past and present. Eur J Pediatr. 2019 Sep;178(9):1305-1315. [PMC free article: PMC6694036] [PubMed: 31338675] 5. Murphy OC, Messacar K, Benson L, Bove R, Carpenter JL, Crawford T, Dean J, DeBiasi R, Desai J, Elrick MJ, Farias-Moeller R, Gombolay GY, Greenberg B, Harmelink M, Hong S, Hopkins SE, Oleszek J, Otten C, Sadowsky CL, Schreiner TL, Thakur KT, Van Haren K, Carballo CM, Chong PF, Fall A, Gowda VK, Helfferich J, Kira R, Lim M, Lopez EL, Wells EM, Yeh EA, Pardo CA., AFM working group. Acute flaccid myelitis: cause, diagnosis, and management. Lancet. 2021 Jan 23;397(10271):334-346. [PMC free article: PMC7909727] [PubMed: 33357469] 6. Lo JK, Robinson LR. Post-polio syndrome and the late effects of poliomyelitis: Part 2. treatment, management, and prognosis. Muscle Nerve. 2018 Dec;58(6):760-769. [PubMed: 29752826] 7. Klapsa D, Wilton T, Zealand A, Bujaki E, Saxentoff E, Troman C, Shaw AG, Tedcastle A, Majumdar M, Mate R, Akello JO, Huseynov S, Zeb A, Zambon M, Bell A, Hagan J, Wade MJ, Ramsay M, Grassly NC, Saliba V, Martin J. Sustained detection of type 2 poliovirus in London sewage between February and July, 2022, by enhanced environmental surveillance. Lancet. 2022 Oct 29;400(10362):1531-1538. [PMC free article: PMC9627700] [PubMed: 36243024] 8. Irwin-Weyant ME, Boggs SR. Polio: Recognition of a Reemerging Infection. Pediatr Rev. 2023 Jul 01;44(7):419-421. [PubMed: 37391633] 9. Devaux CA, Pontarotti P, Levasseur A, Colson P, Raoult D. Is it time to switch to a formulation other than the live attenuated poliovirus vaccine to prevent poliomyelitis? Front Public Health. 2023;11:1284337. [PMC free article: PMC10801389] [PubMed: 38259741] 10. Berner M, Pany-Kucera D, Doneus N, Sladek V, Gamble M, Eggers S. Challenging definitions and diagnostic approaches for ancient rare diseases: The case of poliomyelitis. Int J Paleopathol. 2021 Jun;33:113-127. [PubMed: 33894575] 11. Mazzarello P, Varotto E, Galassi FM. A depiction of poliomyelitis in a 17th -century Piedmontese fresco? Neurol Sci. 2024 Jul;45(7):3517-3519. [PubMed: 38662105] 12. Al Awaidy ST, Khamis F. Wild Poliovirus Type 1 in Oman: A re-emerging threat that requires urgent, targeted and strategic preparedness. Sultan Qaboos Univ Med J. 2020 Feb;20(1):e1-e4. [PMC free article: PMC7065690] [PubMed: 32190363] 13. Davlantes E, Greene SA, Tobolowsky FA, Biya O, Wiesen E, Abebe F, Weldetsadik MB, Eboh VA, Chisema MN, da Conceição Mário B, Tinuga F, Bobo PM, Chigodo CK, Sethy G, Hellström JM, Goundara AM, Burny ME, Mwale JC, Jorba J, Makua KS, Howard W, Seakamela L, Okiror S, Thompson A, Ali A, Samba D, Agbo C, Kabamba L, Kazoka A, Zomahoun DL, Manneh F, Abdelrahim K, Kamugisha C, Umar AS. Update on Wild Poliovirus Type 1 Outbreak - Southeastern Africa, 2021-2022. MMWR Morb Mortal Wkly Rep. 2023 Apr 14;72(15):391-397. [PMC free article: PMC10121257] [PubMed: 37053125] 14. Platt LR, Estívariz CF, Sutter RW. Vaccine-associated paralytic poliomyelitis: a review of the epidemiology and estimation of the global burden. J Infect Dis. 2014 Nov 01;210 Suppl 1(Suppl 1):S380-9. [PMC free article: PMC10424844] [PubMed: 25316859] 15. Lo JK, Robinson LR. Postpolio syndrome and the late effects of poliomyelitis. Part 1. pathogenesis, biomechanical considerations, diagnosis, and investigations. Muscle Nerve. 2018 Dec;58(6):751-759. [PubMed: 29752819] 16. Wiechers DO. Acute and latent effect of poliomyelitis on the motor unit as revealed by electromyography. Orthopedics. 1985 Jul;8(7):870-2. [PubMed: 3006005] 17. Mehndiratta MM, Mehndiratta P, Pande R. Poliomyelitis: historical facts, epidemiology, and current challenges in eradication. Neurohospitalist. 2014 Oct;4(4):223-9. [PMC free article: PMC4212416] [PubMed: 25360208] 18. Chu ECP, Lam KKW. Post-poliomyelitis syndrome. Int Med Case Rep J. 2019;12:261-264. [PMC free article: PMC6690913] [PubMed: 31496835] 19. Howard RS. Poliomyelitis and the postpolio syndrome. BMJ. 2005 Jun 04;330(7503):1314-8. [PMC free article: PMC558211] [PubMed: 15933355] 20. Kawajiri S, Tani M, Noda K, Fujishima K, Hattori N, Okuma Y. Segmental zoster paresis of limbs: report of three cases and review of literature. Neurologist. 2007 Sep;13(5):313-7. [PubMed: 17848871] 21. Chopra JS, Banerjee AK, Murthy JM, Pal SR. Paralytic rabies: a clinico-pathological study. Brain. 1980 Dec;103(4):789-802. [PubMed: 7437890] 22. Ragonese P, Fierro B, Salemi G, Randisi G, Buffa D, D'Amelio M, Aloisio A, Savettieri G. Prevalence and risk factors of post-polio syndrome in a cohort of polio survivors. J Neurol Sci. 2005 Sep 15;236(1-2):31-5. [PubMed: 16014307] 23. Rekand T, Albrektsen G, Langeland N, Aarli JA. Risk of symptoms related to late effects of poliomyelitis. Acta Neurol Scand. 2000 Mar;101(3):153-8. [PubMed: 10705936] 24. Bertolasi L, Danese A, Monaco S, Turri M, Borg K, Werhagen L. Polio Patients in Northern Italy, a 50 Year Follow-up. Open Neurol J. 2016;10:77-82. [PMC free article: PMC5012079] [PubMed: 27651845] 25. Laffont I, Duflos C, Hirtz C, Bakhti K, Gelis A, Palayer C, Macioce V, Soler M, Pradalier F, Galtier F, Jentzer A, Lozano C, Vincent T, Morales RJ. Post-polio syndrome is not a dysimmune condition. Eur J Phys Rehabil Med. 2024 Apr;60(2):270-279. [PMC free article: PMC11112507] [PubMed: 38252127] 26. Bang H, Suh JH, Lee SY, Kim K, Yang EJ, Jung SH, Jang SN, Han SJ, Kim WH, Oh MG, Kim JH, Lee SG, Lim JY. Post-polio syndrome and risk factors in korean polio survivors: a baseline survey by telephone interview. Ann Rehabil Med. 2014 Oct;38(5):637-47. [PMC free article: PMC4221392] [PubMed: 25379493] 27. Kidd S, Clark T, Routh J, Cineas S, Bahta L, Brooks O. Use of Inactivated Polio Vaccine Among U.S. Adults: Updated Recommendations of the Advisory Committee on Immunization Practices - United States, 2023. MMWR Morb Mortal Wkly Rep. 2023 Dec 08;72(49):1327-1330. [PMC free article: PMC10715822] [PubMed: 38060431] 28. O'Reilly KM, Lamoureux C, Molodecky NA, Lyons H, Grassly NC, Tallis G. An assessment of the geographical risks of wild and vaccine-derived poliomyelitis outbreaks in Africa and Asia. BMC Infect Dis. 2017 May 26;17(1):367. [PMC free article: PMC5446690] [PubMed: 28549485] 29. Razum O, Sridhar D, Jahn A, Zaidi S, Ooms G, Müller O. Polio: from eradication to systematic, sustained control. BMJ Glob Health. 2019;4(4):e001633. [PMC free article: PMC6730569] [PubMed: 31544903] Disclosure:Jonathan Wolbert declares no relevant financial relationships with ineligible companies. Disclosure:Michael Rajnik declares no relevant financial relationships with ineligible companies. Disclosure:Helena Swinkels declares no relevant financial relationships with ineligible companies. Disclosure:Karla Higginbotham declares no relevant financial relationships with ineligible companies. Continuing Education Activity Introduction Etiology Epidemiology Pathophysiology Histopathology History and Physical Evaluation Treatment / Management Differential Diagnosis Prognosis Complications Postoperative and Rehabilitation Care Deterrence and Patient Education Enhancing Healthcare Team Outcomes Review Questions References Copyright © 2025, StatPearls Publishing LLC. This book is distributed under the terms of the Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International (CC BY-NC-ND 4.0) ( which permits others to distribute the work, provided that the article is not altered or used commercially. You are not required to obtain permission to distribute this article, provided that you credit the author and journal. 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15964	https://www.wiley.com/en-us/Engineering+Mechanics%3A+Dynamics%2C+9th+Edition-p-x001053661	Engineering Mechanics: Dynamics, 9th Edition James L. Meriam, L. G. Kraige, J. N. Bolton ISBN: 978-1-119-39098-5 March 2018 624 pages PREFER DIGITAL VERSIONS OF YOUR TEXTBOOKS? Get instant access to your Wiley eBook. Buy or rent eBooks for a period of up to 150 days. Request Digital Evaluation Copy To Purchase this product, please visit ## Engineering Mechanics: Dynamics, 9th Edition James L. Meriam,L. G. Kraige,J. N. Bolton E-Book Rental (150 Days) 978-1-119-39098-5R150 March 2018 $51.00 E-Book 978-1-119-39098-5 March 2018 $121.95 Loose-leaf 978-1-119-72417-9 July 2020 $142.95 Single Term Access to WileyPLUS 978EEGRP38332 $76.95 Single Term Access to WileyPLUS + Permanent Copy of eTextbook 978-1-119-66389-8 $97.95 Single Term Access to WileyPLUS + Textbook Rental (130 Days) 978-1-119-57059-2 $125.95 Single Term Access to WileyPLUS + Loose-Leaf Textbook 978-1-119-72465-0 $141.95 Description Engineering Mechanics: Dynamics provides a solid foundation of mechanics principles and helps students develop their problem-solving skills with an extensive variety of engaging problems related to engineering design. More than 50% of the homework problems are new, and there are also a number of new sample problems. To help students build necessary visualization and problem-solving skills, this product strongly emphasizes drawing free–body diagrams, the most important skill needed to solve mechanics problems. Related Resources Instructor View Instructor Companion Site Contact your Rep for all inquiries About the Author Dr. James L. Meriam has contributed to the field of engineering mechanics as one of the premier engineering educators during the second half of the twentieth century. Dr. Meriam earned his B.E., M. Eng., and Ph.D. degrees from Yale University. He had early industrial experience with Pratt and Whitney Aircraft and the General Electric Company. During the Second World War, he served in the U.S. Coast Guard. He was a member of the faculty of the University of California-Berkeley, Dean of Engineering at Duke University, a faculty member at the California Polytechnic State University, and visiting professor at the University of California-Santa Barbara. He retired in 1990. Professor Meriam always placed great emphasis on teaching, and this trait was recognized by his students wherever he taught. At Berkeley in 1963, he was the first recipient of the Outstanding Faculty Award of Tau Beta Pi, given primarily for excellence in teaching. In 1978, he received the Distinguished Educator Award for Outstanding Service to Engineering Mechanics Education from the American Society for Engineering Education, and in 1992 was the Society's recipient of the Benjamin Garver Lamme Award, which is ASEE's highest annual national award. Dr. L. G. Kraige, coauthor of the Engineering Mechanics series since the early 1980s, has also made significant contributions to mechanics education. Dr. Kraige earned his B.S., M.S., and Ph.D. degrees at the University of Virginia, principally in aerospace engineering, and he currently serves as Professor of Engineering Science and Mechanics at Virginia Polytechnic Institute and State University. In addition to his widely recognized research and publications in the field of spacecraft dynamics. Professor Kraige has devoted his attention to the teaching of mechanics at both introductory and advanced levels. His outstanding teaching has been widely recognized and has earned him teaching awards at the departmental, college, university, state, regional, and national levels. New to Edition Orion Foundations Module: (for pre-requisite topics only). New Video Solutions: Approximately 170 new video solutions walk students through a detailed step-by-step problem-solving process. Algorithmic Automatically Graded Problems: All of the end of section problems in the course are available. Student Practice Modules: Student practice modules allow students to practice on their own with: Approximately 100 Check Your Understanding questions. Approximately 100 automatically-graded problems and the ability to check their work with official step-by-step solutions (text or video format). QuickStart Assignments: QuickStart Assignments for each course section make course set-up a snap. High Quality Problems: The course employs a wide variety of high quality problems that are known for their accuracy, realism, applications, and variety. Students benefit from realistic applications that motivate their desire to learn and develop their problem solving skills. Sample Problems: Sample problems with worked solutions appear throughout the course, providing examples and reinforcing important concepts and ideas in engineering mechanics. Free Body Diagrams: The course offers comprehensive coverage of how to draw free body diagrams. Through content discussion and assignable homework problems, students learn that drawing free body diagrams is one of the most important skills needed to learn how to solve mechanics problems. Statics teaches students the appropriate techniques and then applies them consistently in solutions of mechanics problems. Additional Support Features: Rich pedagogical features support ease of use. Key Concepts are highlighted within the theory presentations and course section reviews include itemized summaries of the material covered. To Purchase this product, please visit
15965	https://curriculum.illustrativemathematics.org/MS/teachers/3/4/index.html	Illustrative Mathematics Grade 8, Unit 4 - Teachers \| IM Demo Skip to main content Professional LearningContact Us For full sampling or purchase, contact an IM Certified Partner:Imagine LearningKendall HuntKiddom Grade 8 Middle SchoolGrade 6Grade 7Grade 8 Unit 4 Grade 8Unit 1Unit 2Unit 3Unit 4Unit 5Unit 6Unit 7Unit 8Unit 9 8.4 Linear Equations and Linear Systems In this unit, students write and solve linear equations in one variable. These include equations in which the variable occurs on both sides of the equal sign, and equations with no solutions, exactly one solution, and infinitely many solutions. They learn that any one such equation is false, true for one value of the variable, or true for all values of the variable. They interpret solutions in the contexts from which the equations arose. Students write and solve systems of linear equations in two variables and interpret the solutions in the contexts from which the equations arose. They learn what is meant by a solution for a system of equations, namely that a solution of the system is a solution for each equation in the system. Students use the understanding that each pair of values that make an equation true are coordinates of a point on the graph of the equation and conversely that the coordinates of each point on the graph of an equation make the equation true. Thus, a pair of values that satisfies a system of equations are coordinates of a point that lies on the graphs of all the equations in the system, and, conversely, a point that lies on the graphs of all the equations in the system has coordinates that satisfy all the equations in the system. Students learn to understand and use the terms “system of equations,” “solution for the system of equations,” “zero solutions,” “no solution,” “one solution,” and “infinitely many solutions.” Read More Lessons Lessons Puzzle Problems 1 Number Puzzles Linear Equations in One Variable 2 Keeping the Equation Balanced 3 Balanced Moves 4 More Balanced Moves 5 Solving Any Linear Equation 6 Strategic Solving 7 All, Some, or No Solutions 8 How Many Solutions? 9 When Are They the Same? Systems of Linear Equations 10 On or Off the Line? 11 On Both of the Lines 12 Systems of Equations 13 Solving Systems of Equations 14 Solving More Systems 15 Writing Systems of Equations Let's Put It to Work 16 Solving Problems with Systems of Equations About IM In the News Curriculum Grades K-5 Grades 6-8 Grades 9-12 Professional Learning Standards and Tasks Jobs Privacy Policy Facebook Twitter IM Blog Contact Us 855-741-6284 What is IM Certified™? IM 6–8 Math was originally developed by Open Up Resources and authored by Illustrative Mathematics®, and is copyright 2017-2019 by Open Up Resources. It is licensed under the Creative Commons Attribution 4.0 International License (CC BY 4.0). OUR's 6–8 Math Curriculum is available at Adaptations and updates to IM 6–8 Math are copyright 2019 by Illustrative Mathematics, and are licensed under the Creative Commons Attribution 4.0 International License (CC BY 4.0). Adaptations to add additional English language learner supports are copyright 2019 by Open Up Resources, and are licensed under the Creative Commons Attribution 4.0 International License (CC BY 4.0). The second set of English assessments (marked as set "B") are copyright 2019 by Open Up Resources, and are licensed under the Creative Commons Attribution 4.0 International License (CC BY 4.0). Spanish translation of the "B" assessments are copyright 2020 byIllustrative Mathematics, and are licensed under the Creative Commons Attribution 4.0 International License (CC BY 4.0). The Illustrative Mathematics name and logo are not subject to the Creative Commons license and may not be used without the prior and express written consent of Illustrative Mathematics. This site includes public domain images or openly licensed images that are copyrighted by their respective owners. Openly licensed images remain under the terms of their respective licenses. See the image attribution section for more information.
15966	https://www.blacksacademy.net/texts/VectorEquationOfTheStraightLine.pdf	© blacksacademy.net 1 The Vector Equation of the Straight Line Prerequisites You should understand what a displacement vector is and that displacement vectors can be added and subtracted by adding and subtracting their components. Example (1) Given    2 4 r i k s j 2k Find 2 r s Solution           2 2 4 2 2 2 8 r s i k j 2k i j k Vector equations We can rearrange equations containing vectors in the same way that we can rearrange regular equations. Example (2) Make a the subject of the equation    1 6 3 2 a b c d Solution           1 6 3 2 1 3 2 a d b c a d b c = 1 = 6 This means that equations involving vectors can be solved by the usual algebraic manipulations. Example (3) Solve the vector equation                                     6 4 1 1 2 3 0 1 3 1 p © blacksacademy.net 2 Solution                                     6 4 1 1 2 3 0 1 3 1 p Subtracting            6 1 1 from both sides of this equation                                                                  4 1 6 4 3 6 5 2 3 0 1 2 0 1 1 3 1 1 3 3 1 1 p Vector equation of the straight line Vectors are introduced so that we can describe objects that exist in two and three-dimensional space. One such object is a straight line. We can use vector algebra to find the vector equation of the straight line joining two points A and B. Suppose     OA a is the vector joining the origin O to the point A and    OB b the vector joining O to B. = -AB b a A O B a b Then the vector   AB that joins A to B is such that     AB a b . Rearranging this equation to solve for   AB     AB b a Thus     AB b a is the vector from A to B. A straight line is fixed in space by reference to the origin O. So in this equation the vectors a and b must have fixed points of application, which is the origin. This means that they are position vectors. Example (4) The position vectors of the points A and B are given by       6 2 3 2 a i j k b i j k © blacksacademy.net 3 (a) Find the vector   AB from A to B. (b) Find the vector   BA . (c) Find the vector       2 BA p b . (d) Make a sketch showing the vectors   AB ,   BA and       2 BA p b , and determine for each whether it is a displacement or position vector. (e) Describe in terms of geometry the vector       s BA r b where s is any real number. Solution (a) It is easier to add vectors in column form.                         6 3 2 2 1 1 a b                                           3 6 9 2 2 0 9 2 1 1 2 AB b a i k (b)           9 2 9 2 BA AB i k i k (c)                                            9 15 3 2 2 2 0 2 15 2 3 1 2 3 BA p b i j k (d) For the sketch we make no attempt to represent any of these vectors in three dimensions – that is we do not draw three axes and plot the points A, B or P represented by      2 BA b . A (6, 2, 1)  O B ( 3, 2, 1)  a b p b = + 2BA P (15, 2, 3)  AB = b a  BA © blacksacademy.net 4 (This diagram shows the plane that is determined by the points O, A and B. The point P also lies in this plane.) We are given that     OA a and    OB b are position vectors, meaning that they fix the position of the points A and B relative to the origin O. Likewise the vector       2 BA p b must also be a position vector. It fixes the position of the point P given by    15 2 3 p i j k relative to the origin. But the vector   BA is a displacement vector. It is the displacement from B to A and is not fixed relative to B. For example we could write       2 BA q a If   BA were fixed at B this would not make any sense because we would then not be able to interpret adding the vector fixed at B to a vector at the point A fixed by     OA a . The vector   BA here represents any vector that has the same direction and magnitude as   BA . Likewise   AB is a displacement vector. The equation       2 BA p b says    OP p is the position vector of the point P found by adding the position vector of the point B given by    OB b to twice the displacement vector   BA that is equal to the displacement from B to A. However, we remark that it is not necessary at this level to distinguish systematically between position and displacement vectors as the vector algebra simply “works”. (e) The equation       s BA r b fixes the position of a point r on the line joining B to A. As s varies so this position varies. Therefore, it is the vector equation of the straight line joining B to A. Since s can take negative values, by substituting a negative value of s we get the vector equation of the line joining A to B. Geometrically, these are the same line, but travelled in opposite directions. So the vector equation of the straight line provides information (a) about the position of the line in space – i.e. it is the line joining A to B or, what is the same thing, the line joining B to A; and (b) the optional information about the direction in which a particle might be travelling along that line. Thus the equation       t AB r b for positive values of the parameter t shows a particle travelling along a line in the direction from A to B. The equation © blacksacademy.net 5       s BA r b where s is a positive parameter shows a particle travelling alone the same line but in the opposite direction from B to A. By using different parameters t or s we have also the possibility that the particles are travelling at a different speed. All of this may be of particular relevance if you are also studying mechanics, which a natural application of vector algebra. The last example shows that the vector equation of the straight line passing through the points A and B with position vectors     OA a and    OB b respectively is given by       t AB r b where t is a real valued parameter. O a b r a = + t AB AB = b a  AB A B The same line is described by any of the following       s BA r b      t AB r a       s BA r a Thus the most general form of the vector equation of a straight line is  t r p q where p is the position vector of any point on the line, q is a displacement vector of any size having the same direction, forwards or backwards, as the line, and t is a real-valued parameter. © blacksacademy.net 6 O p A B q P q A line is determined by (1) fixing any point P lying on the line and by (2) any displacement vector having the same direction, forwards or backwards, as the line. Example (5) The position vectors of the points A and B are given by                       3 4 1 3 2 5 a b (a) Find the vector equation of the line AB. (b) Find the position vector of the point C lying on AB such that 2 AC BC . Solution (a)                                      1 4 3 3 1 2 5 2 3 AB b a The vector equation of the line AB is                            1 4 3 2 5 3 t AB t r a r Note, the equation                            1 3 1 2 2 3 tAB t r b r describes the same line. © blacksacademy.net 7 (b) Since 2 AC BC then 2 3 AC AB O a b r a = + t AB AB = b a  A B C c AC AB = + a 2 3 Hence                                 2 3 1 4 2 14 13 3 2 7 3 3 3 5 3 AC AB a i j k Points of Intersection In two dimensions any two lines that are not parallel, and not collinear, must intersect at a unique point. Hence, given two vector equations of lines in two dimensions, it is possible to find the position vector of their point of intersection. Example (6)                                 1 2 Let be given by 2 1 is a real number 3 -1 and be given by 4 2 is a real number -1 1 Find their point of intersection. l l r s © blacksacademy.net 8 Solution                                                          2 1 Equating the two lines: 2 1 4 2 3 1 1 1 2 4 2 3 1 On adding these two equations: 2 10 5 3 3 and 3 3 2 10 Substituting into or into , both give 3 3 4 2 3 1 l l p              1 3 1 3 5 2 1 In two dimensions, equating the equations of two lines gives two equations in two unknowns. Provided one equation is not a multiple of another, this gives a unique solution. In three dimensions there is no necessity that two lines intersect. If two lines do intersect, then equating them gives three equations in two unknowns that do have a unique and consistent solution. If the two lines do not intersect, solving the system of three equations in pairs gives different values for the point of intersection and inconsistent values for the parameters describing the point on the line. Example (7)                                   1 2 Show that given by 3 2 1 1 1 2 and given by 0 3 2 0 1 3 do not intersect. l t l s r r © blacksacademy.net 9 Solution                                         Suppose they do intersect, then 3 2 0 3 1 1 = 2 0 1 2 1 3 3 2 3 (1) 1 2 (2) 1 2 1 3 (3) t s t s t t s             From (2) 3 Substituting in 1 3 6 3 3 Substituting in 3 1 6 1 3 2 The values 2 and 3 t s s s s s s The values   2 and 3 s s are inconsistent, hence there cannot be a point of intersection. This is a proof by contradiction. In this method of proof, you assume something to be true and deduce from it a contradiction. When you arrive at a contradiction, you conclude that the thing you assumed must be false. Here we prove that there cannot be a point of intersection by, firstly, assuming that there is a point of intersection, and then showing that this leads to a contradiction. Example (8) The position vectors of the points A and B are given by       2 3 , 5 4 a i j k b i j k (a) Find the vector equation of the line AB. (b) The vector equation of the line L is          7 4 3 5 5 p r i j k i j k where p is a constant. Given that AB and L intersect, find p and the point of intersection. © blacksacademy.net 10 Solution                                          5 2 7 ( ) 4 1 3 1 3 4 a AB b a The vector equation of the line AB is                                        2 7 1 3 2 3 7 3 4 3 4 s a b a i j k i j k (b) The two lines intersect. Therefore                                                               2 7 7 3 1 3 5 3 4 4 5 2 7 7 3 1 1 3 5 2 3 4 4 5 3 p p We first solve equations (1) and (3) simultaneously to find   and .                           3 7 5 1 5 4 7 3 15 35 25 15 12 21 23 46 2 3 Then from equation (2)     1 6 15 20 p p The point of intersection is found by substituting either     2 or 3 into                                             2 7 7 3 1 3 20 5 3 4 4 5 This gives                                              2 7 7 3 1 2 3 20 3 5 16 11 3 4 4 5 i 5j k © blacksacademy.net 11 Parametric and Cartesian equations of the straight line We have seen that the general form of the vector equation of the straight line is  t r a p where a is the position vector of any point on the line, p is a displacement vector of any size having the same direction, forwards or backwards, as the line, and t is a real-valued parameter. Suppose that we have          x y z a b c p q r r i j k a i j k p i j k Here, t, x, y, z are variables: a, b, c, p, q, r are constants. Then the vector form can be written                                       x a p y b t q z c r Uncoupling this equation – that is, separating the three components – gives       x a tp y b tq z c tr This is called the parametric form of the equation of the straight line. The parameter is t, and the equations show how the coordinates vary as t varies. If we solve these equations for t we obtain       x a y b z c t p q r This is called the Cartesian equation of the straight line. Example (9) Determine the point of intersection between the following pair of lines given in Cartesian form           4 1 3 1 4 5 and 3 3 4 1 1 2 x y z x y z Solution Putting these into parametric form                                    4 3 4 1 3 1 3 3 3 4 3 4 x x y z y t z                                   1 1 1 4 5 4 1 1 1 2 5 2 x x y z y s z © blacksacademy.net 12 At intersection                                     4 3 1 1 1 3 4 1 3 4 5 2 t s                                 4 3 1 1 1 3 4 2 3 4 5 2 3 1 2 5 5 2 2 2 6 8 2 4 4 3 5 2 3 2 2 1 Sub in 2 6 t s t s t s t s t t t s The point of intersection is                                                 4 3 7 1 3 2 3 4 7 x y z
15967	https://www.physicsclassroom.com/curriculum/force2D/forces5.pdf	Forces in Two Dimensions Name: © The Physics Classroom, 2009 Page 1 Inclined Plane Analysis Read from Lesson 3 of the Vectors and Motion in Two-Dimensions chapter at The Physics Classroom: MOP Connection: Forces in Two Dimensions: sublevels 5 and 6 Review: 1. A normal force is a force that is always directed __. a. upwards b. sideways c. perpendicular to the surface the object is touching 2. An object is upon a surface. The normal force is equal to the force of gravity _. a. in all situations b. only when the object is at rest c. only when the object is accelerating d. only when there is no vertical acceleration e. only when there is no vertical acceleration AND Fnorm and Fgrav are the only vertical forces Getting the Forces Right: 3. The object at the right has been placed on a tilted surface or inclined plane. If there is enough tilt, it will accelerate from rest and begin its motion down the incline. Draw a free-body diagram for the object sliding down the rough incline. Label the three forces according to type (Fgrav, Fnorm, Ffrict, Fair, Ftens, Fapp, etc.). Physics Tip: When you encounter a situation involving a force directed at angles to all other forces, immediately convert the uncooperative force(s) into two perpendicular components. Use SOH CAH TOA to resolve any uncooperative force into components directed at right angles to each other. One component should be in the direction of the acceleration; the other should be perpendicular to it. In the case of inclined planes, resolve the uncooperative force into components parallel and perpendicular to the inclined plane. 4. The force of gravity (or weight vector) is the uncooperative force. It is typically resolved into two components - one parallel to the plane and the other perpendicular to the plane. Given the diagram at the right with the two components of gravity represented as and , use trigonometric functions to write equations relating these components to the force of gravity. 5. For the three situations described below, use <, > , or = symbols to complete the statements. Object at rest. _ Ffrict __ Fnorm Object moves at constant speed. __ Ffrict __ Fnorm Object accelerates down incline. __ Ffrict ______ Fnorm Forces in Two Dimensions © The Physics Classroom, 2009 Page 2 Use equations for calculating the components of gravity (#4) and Newton's laws to fill in the blanks. 6. A 4.50-kg object is accelerating down an inclined plane inclined at 36.0° (with the horizontal) and having a coefficient of friction of 0.548. 7. A 65.0-kg crate remains at rest on an inclined plane that is inclined at 23.0° (with the horizontal). 8. A 41.3-kg box slides down an inclined plane (inclined at 29.1 degrees) at a constant speed of 2.1 m/s. The Tilted Head Trick Inclined plane problems can be easy. Resolve gravity into its components. Then, ignore the force of gravity. Finally, tilt the paper or your head and the problem becomes a simple Fnet= m•a problem.
15968	https://www.statisticshowto.com/absolute-value-function/	Skip to content Absolute Value Function: Definition Types of Functions > Contents: What is absolute value? The absolute value function: Domain & Range Absolutely Integrable (Summable) What is Absolute Value? The absolute value represents a number’s positive distance from zero on the number line. For example, the absolute value of -3 is 3, because it is three units from zero on the number line. As it shows a distance, absolute value is always non-negative; It is either positive or zero. The absolute value also refers to the magnitude of a number. You can find the magnitude of any number by removing the negative sign. For example, the absolute value of -10 is 10 and -1000 is 1000. Absolute value of a number is sometimes called the “modulus” of a number and is denoted by vertical bars on each side of the number. For example: \|-10\| = the abs. value of 10 = 10 \|-3\| = 3 \|-22\| = 22 Absolute Difference Absolute difference, \|x-y,\| is the distance of two numbers on the number line. For example the absolute difference of -5 and 4 is 9: \|-5 – 4\| = 9 \|-22 – 1\| = 23 \|4 – -2\| = 6 (If you ever wondered why “two negatives equal a positive” in algebra, this last example, 4 – -2 helps to explain why!). When do we use absolute value in statistics? In general. we use absolute values when we don’t care about the sign of the number — we just care about it’s magnitude or distance from zero. For example, in a hypothesis test, the absolute value is taken of a test statistic because it defines extremes (tail areas) at both ends of a probability distribution. When deciding to reject or support the null hypothesis, we care if a value falls into one of these tail areas — it doesn’t matter which one. For example, the sign (±) of a t-test tells us the direction of the difference in sample means. If the absolute value of the t-statistic is greater than or equal to the absolute value of the critical value, we reject the null hypothesis. If you are running a two-tailed test (as in the image below, which shows rejection regions shaded in black), you would never say “less than” with two tails shaded. The t-distribution is symmetrical, so the absolute value simplifies running hypothesis tests by only requiring you look at one rejection area — the positive area to the right. Absolute values are also used in calculating the mean deviation. A deviation in statistics is the difference between a data point and the mean. The absolute value of the deviation removes any minus signs that occur. If we didn’t use the absolute value, the pluses and minuses would cancel each other out — which can result in a mean deviation of zero! History According to Math Boys, the word absolute comes from a variant of the word absolve and has a meaning close to free from conditions or restriction. The term, as it relates to mathematics, was first found in 1950 in the elements of analytical geometry; comprehending the doctrine of the conic sections, and the general theory of curves and surfaces of the second order by John Radford Young (1799-1885). “we have AF the positive value of x equal to BA – BF, and for the negative value, BF must exceed BA, that is, F must be on the other side of A, as at F’, hence making AF’ equal to the absolute value of the negative root of the equation”. What is an Absolute Value Function? An absolute value function has an expression within absolute value symbols. For example, all of the following are absolute value functions: y = \|x\| y = \|x + 5\| y = \|x – 10\| + 9 Some authors take the term “absolute value function” to mean just the first function (y = \|x\|). Others use it to mean all functions that include an absolute value expression. The differing terminology may stem from the fact that other functions (e.g. items 2 & 3) are just transformations (shifts and stretches) of the original function y = \|x\|. Therefore, y = \|x\| is sometimes called the “absolute value parent function.” Check with your particular author (or your instructor) to make sure you know which terminology they are using. Absolute Value Parent Function An absolute value function has a unique “V” shape when plotted on a graph. This is due to the fact that the absolute value of a negative number makes that number positive. The absolute value parent function is written as: f(x) = │x│ where: f(x) = x if x > 0 0 if x = 0 -x if x < 0 As the definition has three pieces, this is also a type of piecewise function. It’s only true that the absolute value function will hit (0,0) for this very specific case. Many functions you will come across in calculus will differ in where on the coordinate plane they appear. For example: Example of Graphing an Absolute Value Function Graph the following function using a table of values: y = │x + 2│ Before choosing values for x, remember to include negative values for x. \| \| \| --- \| \| x \| y =│x + 2│ \| \| -5 \| -5 + 2 = 3 \| \| -3 \| -3 + 2 = 1 \| \| -2 \| -2 + 2 = 0 \| \| 0 \| 0 + 2 = 2 \| \| 2 \| 2 + 2 = 4 \| \| 5 \| 5 + 2 = 7 \| If you were to plot the points above you would see that we end up with a “V” shaped plotted line identical to the absolute value parent function above, albeit with the vertex located at (-2, 0). Notice also that you could have used any values for x, provided you included negative values; You would have ended up with the same result. Absolutely Integrable An absolutely integrable function (also called a summable function) has an integrable absolute value. “Integrable” means that you can find an integral. The “Absolutely” portion of the term refers to the fact that it’s the absolute value of the function that must be integrable on the real line (Feeman, 2015). In notation, we can write that as: On the other hand, a function that isn’t absolutely integrable has an infinite value for the integral. It’s written in notation as: Riemann Integral vs. Absolute A Riemann integral is the “classic” integral you’re introduced to in introductory calculus classes. Normally shortened to just “integral”, it implies that you can find the limit of an infinite number of tiny rectangles below a curve (a.k.a. Riemann sums). Absolute integration has a more rigid requirement: in addition to being able to find an integral, you must also be able to find the integral for the absolute values of the function. Absolutely Integrable Functions in Signal Processing Absolutely integrable functions have special importance in signal processing, because absolutely integrable continuous time or discrete time signals are stable and have Fourier transforms. Any signal generated in a lab is going to have these properties. If a signal isn’t absolutely integrable, then you can’t perform Fourier analysis. The “severe restriction” of absolute integrability means that many useful periodic functions like sin t and cos t are not absolutely integrable and do not have Fourier transforms (Dettman, 2012). However, they can be defined on an interval of one time period (e.g. -T/2, T/2) to enable expression by a Fourier series. An alternative is to analyze the signals or functions with Laplace transform (which converts a real-valued function (like time) to a complex-valued function). Absolutely Integrable Function Defined on an Interval Many times, you’ll deal with functions that aren’t absolutely integrable on their entire domain. However, the condition can be defined in terms of intervals. For example, take some function f, defined on the open interval (a,b). If the function is locally integrable (i.e. the function can be integrated on the defined interval)and \|f\| is improperly integrable on the same interval, then the function is absolutely integrable. An integrable function, defined on a closed interval [a. b], has an indefinite integral defined by (Mendoza, 2017): . This indefinite integral is continuous and differentiable almost everywhere. References Dettman, J. Applied Complex Variables. Dover Publications. 2012. Feeman, T. The Mathematics of Medical Imaging, Springer Undergraduate Texts in Mathematics and Technology, DOI 10.1007/978-3-319-22665-1 Hazelwinkel, M. Encyclopaedia of Mathematics (set). 1994. Kluwer. Mendoza, J. Which Integrable Functions Fail to be Absolutely Integrable? Real Analysis Exchange. Vol. 43 (1), 2017. pp. 243-248. NIST. 1.3.6.7.2. Critical Values of the Student’s t Distribution. Purdue University. Signal Properties. Comments? Need to post a correction? Please Contact Us. Leave a Comment You must be logged in to post a comment.
15969	https://cnshuochi.com/en/ennews/enhnews/2024-06-14/46.html	东营市硕驰化工有限公司东营市硕驰化工有限公司东营市硕驰化工有限公司 CN EN NEWS CENTER 新闻中心 About About Us PRODUCT Product NEWS CENTER News CONTACT US Contact Us LANGUAGE 语言 Detailed explanation of the preparation method for dichloroethane [ TIME：2024-06-11 HITS： ]
15970	https://flexbooks.ck12.org/cbook/ck-12-cbse-maths-class-7/section/8.2/primary/lesson/percentage-12566584/	Skip to content Elementary Math Grade 1 Grade 2 Grade 3 Grade 4 Grade 5 Math 6 Math 7 Math 8 Algebra I Geometry Algebra II Conventional Math 6 Math 7 Math 8 Algebra I Geometry Algebra II Probability & Statistics Trigonometry Math Analysis Precalculus Calculus What's the difference? 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Learn. Interact. eXplore. CCSS Math Concepts and FlexBooks aligned to Common Core NGSS Concepts aligned to Next Generation Science Standards Certified Educator Stand out as an educator. Become CK-12 Certified. Webinars Live and archived sessions to learn about CK-12 Other Resources CK-12 Resources Concept Map Testimonials CK-12 Mission Meet the Team CK-12 Helpdesk FlexLets Know the essentials. Pick a Subject Donate Sign Up 8.2 Percentage Written by:Neha Khandelwal Fact-checked by:The CK-12 Editorial Team Last Modified: Aug 01, 2025 Percentage Definition The term ‘per cent’ means ‘out of a hundred’. In mathematics, percentages are used like fractions and decimals, as ways to describe parts of a whole. When we are using percentages, the whole is considered to be made up of a hundred equal parts. The symbol % is used to show that a number is a percentage. You will see percentages almost everywhere. They appear in discounts in shops, bank interest rates, rates of inflation, in news, even on our devices. Percentages are important for understanding the financial aspects of everyday life. Knowledge of percentages also helps in understanding how to run a business. It helps in analysing and find an optimal solution to what price to charge for the product, how to increase the profits without increasing the price, etc. Percent as a Fraction A percentage is a fraction out of a hundred, and any fraction can be converted to a percentage. When converting fractions to percentages, it is easiest to convert the quantity to an equivalent fraction out of a hundred and then write the number as a percentage. Percentages can be represented using models. Although using 100 squares makes representing and identifying percentages much simpler, percentages can also be represented using squares other than 100. 1) Write down the percentage represented by the shaded section of each grid. i. '25 out of a hundred' are coloured. We can write 25100=25%, we say that '25 percent' of the grid is shaded. ii. '3 out of twenty' are coloured. We can write 320=3×520×5=15100=15%, we say that '15 percent' of the grid is shaded. Percentage of a Number When dealing with percents, there are always three things to focus on: the part, the whole, and the percent. If you have two of these three values, you can always find the third. The relationship percent=partwhole can be used to solve for a percent using the part and the whole. By multiplying the whole to the other side you get the following equation:percent⋅whole=part 2) Find 36% of 30. Part = ? Whole = 30. Percent = 36%→36100. Percent⋅Whole=Part36100⋅30=Part10.8=Part∴ 36% of 30 is 10.8. CK-12 Interactive: Percents and Fractions Use the interactive below to explore how percents are related to fractions. Conversion between Fraction and Percent To convert a fraction into a percent, we multiply the fraction by 100 and put the percent sign %. To convert a percent into a fraction, we divide it by 100 and remove the percent sign %. Percent as a Ratio Any percent can be converted to a fraction by dividing it by 100 and then this fraction can be expressed as a ratio. Conversion between Ratio and Percent To convert a ratio into a percent, we express the given ratio as a fraction and then multiply the fraction by 100 and put the percent sign %. To convert a percent into a ratio, we divide it by 100, remove the percent sign % and express the fraction as a ratio. CK-12 Interactive: Your Weight on Other Planets! In the next interactive, use ratios to convert how much you weigh on Earth to how much you would weigh on different planets. If your weight on Earth represents 100%, determine what percentage is your weight on other planets using the questions that follow. Percent as a Decimal Any percent can be converted to a fraction by dividing it by 100 and then this fraction can be expressed as a decimal (terminating or non-terminating but recurring). Conversion between Decimal and Percent To convert a decimal into a percent, we multiply the decimal by 100 and put the percent sign %. To convert a percent into a decimal, we divide it by 100 and remove the percent sign %. CK-12 Interactive: Converting between Fraction, Decimals and Percents Use the interactive below to practice converting decimals, fractions and percents. Percentages Greater than 100% Since percents are a number or ratio expressed as a fraction of 100, you can express any number as a percent. You can also use a percent to express a number greater than 1. Since 100% represents 1 whole, we can say 200% can be used to represent 2 wholes because 100% + 100% = 200%. Since 50% represents half of a whole, 250% can be used to represent two and a half wholes because 100% + 100% + 50% = 250%. Use the interactive to estimate the location of percents greater than 100% on a number line. Percentage Change Often we are interested in finding by what percentage a quantity changes, which is a comparison of a change in value to the original value. The new price is expressed as a percentage of the old price. For example, if the new price is 50% of the old price, this means that the new price is half the old price. If the new price is 200% of the old price, this means that the new price is double the old price. Percentage Change Formula The percentage change is calculated using the following formula:Percentage increase/decrease=New amount - Original amountOriginal amount×100 CK-12 Interactive - Percentage Change: Percent of House Price Use the interactive below to explore this idea. CK-12 Interactive - - Percentage Change: Car Depreciation Use the interactive below to explore percent decreases further. 3) The price of tea leaves increased by 20%. If the price of tea leaves was ₹360 per kg, what is the increased price of tea leaves per kg? Original price of tea leaves= ₹360. Let the increased price be ₹x. Then the amount of increase in the price is ₹x−₹360. ∴Percentage increase=New amount - Original amountOriginal amount×10020=x−360360×10020×360100=x−36072=x−360x=72+360=432∴ The increased price of sugar is ₹432 per kg. Percentage - Examples Example 1 Convert the following into percentages. i. 2 : 5 ii. 825222 iii. 0.006 i. To convert a ratio into a percentage, we first convert the ratio into a fraction and then the fraction is converted into a percentage (by multiplying it with 100)2:5=25222=25×100222=40%ii. To convert a fraction into a percentage, we multiply the fraction by 100. 825=825×100222=32%iii. To convert a decimal into a percentage, we multiply the decimal by 100. 0.006=0.006×100=0.6% Example 2 Out of her total salary, Shreya spends 40% on household expenditures, invests 20% salary and pays 18% salary as loan instalment. If her remaining salary is ₹10560, then find her total salary. Let Shreya's total salary be ₹x. The percentage of salary spent on household expenses, investment and loan=(40+20+18)%=78%. The percentage of salary left=(100−78)%=22%. According to the question, 22% of x=₹1056022100×x=10560222x=10560×10022222x=48000∴ Total salary of Shreya is ₹48000. Example 3 Tina and Karan contest an election. Tina gets 47% of the valid votes and is defeated by 1320 votes. Find the total number of valid votes cast in the election. Tina gets 47% of the valid votes Karan gets (100−47)%=53% of the valid votes. ∴ Percentage difference between the votes=53%−47%=6%. According to the question, 6% of valid votes=13206100× valid votes=1320valid votes=1320×1006=22000Hence, the number of valid votes cast in the election were 22000. Example 4 In a district, 40% of the population comprises children below 16 years of age. If the ratio of the number of men to women is 8 : 7, then find the number of women in that district, given that the total population of the district is 7500. Total population of the district =7500. Population of children=40100×7500 =3000 Rest of the population=7500−3000=4500. According to the question, Men : Women =8:7. Sum of the ratio=8+7=15. Population of women=715×4500 =2100 Hence, the number of women in the district =2100. Example 5 1% of the people of a village died due to an epidemic. A panic set in during which 15% of the remaining people left the village. If the population is then reduced to 1683, what was it originally? Let the original population of the village be x. Since 1% of people died due to an epidemic, the percentage of people who survived=(100−1)%=99%. Further, as 15% of the remaining people left the village, people left in the village=(1−15100) of 99% of x=85100 of 99100xAccording to the question, 85100×99100x=1683x=1683×100×10085×99=2000Hence, the original population of the village was 2000. Summary Percent means 'out of a hundred'. To convert a fraction into a percent, we multiply the fraction by 100 and put the percent sign %. To convert a percent into a fraction, we divide it by 100 and remove the percent sign %. To convert a decimal into a percent, we multiply the decimal by 100 and put the percent sign %. To convert a percent into a decimal, we divide it by 100 and remove the percent sign %. To convert a ratio into a percent, we express the given ratio as a fraction and then multiply the fraction by 100 and put the percent sign %. To convert a percent into a ratio, we divide it by 100, remove the percent sign % and express the fraction as a ratio. Percentage increase/ decrease=New amount - Original amountOriginal amount×100 x% of a given quantity=x100×Given quantity To express x as a percentage of y, percentage=(xy×100)%. Both quantities must be of the same kind (same units). If x% of a given quantity is y, then quantity =yx×100%. Percentage - Review Questions Write the following as a percent. 1550222 310222 1425222 20 out of 25 4 out of 10 Express each of the following as a fraction 40% 39% 50% 72% 66% 4% Write the following as a percent. 0.64 0.09 0.9 0.246 1.79 Express each of the following as a decimal. 60% 2% 89.4% 40% 36.3% Write the following as a percent. 4 : 5 2 : 3 12 : 125 3 : 8 9 : 4 Express each of the following as a ratio. 28% 45% 22.5% 43% 14.5% Find the percent of the number. 19% of 516 45% of 120 11% of 630 23% of 482 18% of 500 24% of 394 Find the value of x. 20% of x is 5 15% of x is 6 35% of x is 7 6% of x is 12 40% of x is 8 4% of x is 10 Out of 400 students at a school, 240 came on time, 100 were late and 60 were absent. Write down the percentage of students who came on time were late were absent At an RTO last week, 135 people took a driving test. The results are shown below. \| \| \| \| --- \| \| Number tested \| Number passed \| \| Male \| 75 \| 39 \| \| Female \| 60 \| 42 \| Find the percentage of males who passed females who passed males who failed females who failed people who passed people who failed In a box of 60 assorted cards, 18 were birthday cards, 15 were anniversary cards, 9 were get-well cards, and the rest were all-occasion cards. What percent of the box of cards was all-occasion cards? Rahul completed 35% of an 1850-km car trip on the first day. How far did he drive that day? If 16.5% of a 60-minute television show is commercials, how many minutes of commercials are there during this hour? Of the students at school, 35% walk or ride a bike to school, 1118 ride a bus to school, and 15% ride in a car to school. Which way of getting to school does the fewest number of students use? Radhika set a goal of practising the keyboard for 350 hours over her summer vacation. She has already practised for 225 hours. What percentage of her goal has she completed? In a biology experiment, Shreya found that the mass of a yeast culture increased by 8%. If the original mass of the culture was 2.5 grams, what was it at the end of the experiment? \| Image \| Reference \| Attributions \| --- \| \| \| Credit: CK-12 Source: CK-12 License: CK-12 Curriculum Materials License \| \| \| \| Credit: CK-12 Source: CK-12 License: CK-12 Curriculum Materials License \| Student Sign Up Are you a teacher? Having issues? Click here By signing up, I confirm that I have read and agree to the Terms of use and Privacy Policy Already have an account? Save this section to your Library in order to add a Practice or Quiz to it. (Edit Title)10/ 100 This lesson has been added to your library. \|Searching in: \| \| \| Looks like this FlexBook 2.0 has changed since you visited it last time. 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15971	https://my.clevelandclinic.org/health/diseases/rsv-respiratory-syncytial-virus	Abu Dhabi\|Canada\|Florida\|London\|Nevada\|Ohio\| Home/ Health Library/ Diseases & Conditions/ RSV (Respiratory Syncytial Virus) AdvertisementAdvertisement RSV (Respiratory Syncytial Virus) RSV is a respiratory infection caused by a virus. Babies, kids and adults can get it. It can cause mild symptoms, but sometimes — especially in babies and older adults — it can lead to severe difficulty breathing. Babies can get a monoclonal antibody treatment to help protect them. There’s a vaccine available for adults over 60 and pregnant women. Advertisement Cleveland Clinic is a non-profit academic medical center. Advertising on our site helps support our mission. We do not endorse non-Cleveland Clinic products or services. Policy Care at Cleveland Clinic Find a Primary Care Provider Schedule an Appointment Advertisement Advertisement Advertisement Advertisement ContentsWhat Is RSV?Symptoms and CausesDiagnosis and TestsManagement and TreatmentOutlook / PrognosisPrevention What Is RSV? RSV is a respiratory illness that can cause cold-like symptoms or, sometimes, serious illness. RSV stands for respiratory syncytial (sin-SISH-uhl) virus, the germ that makes you sick with RSV. Almost everyone gets it for the first time before the age of 2. But the protection (immunity) you get from being infected doesn’t last, so you can get it more than once. Advertisement Cleveland Clinic is a non-profit academic medical center. Advertising on our site helps support our mission. We do not endorse non-Cleveland Clinic products or services. Policy RSV is more common than you might think. We most often hear about babies getting severely ill from RSV, but adults and kids of all ages get it. Infants are more likely to get very sick from RSV because their immune systems are still developing. Their smaller airways are also more likely to get inflamed. You can also get RSV many times throughout your life — sometimes even twice in one year. For adults, it usually causes mild, cold-like symptoms. But adults over 65 and people with compromised immune systems are at a higher risk for serious illness. Symptoms and Causes RSV symptoms Cough Trouble breathing Pauses in breathing in young infants Runny or stuffy nose Sneezing Sore throat Headache Fatigue Fever Not feeling hungry Symptoms in babies or young kids might look a little different. They might seem fussy or irritable, or they might not want to play as they usually would. Contact your healthcare provider if your child is younger than a year old and has RSV symptoms. Go to an emergency room if you or your child has severe symptoms. These include: Shortness of breath or trouble breathing Wheezing or noisy breathing Bluish or grayish skin, lips or nails Skin pulling in between your child’s ribs when they breathe (retractions) Nostrils spreading out (flaring) when breathing Short, shallow or fast breathing, or pauses in breathing Advertisement RSV causes A virus — respiratory syncytial virus — causes RSV. It’s not caused by bacteria. RSV spreads through: Close contact with someone who has it Coughing and sneezing (respiratory droplets) Contaminated objects or surfaces (think toys, countertops and phones) Is RSV contagious? Yes, RSV is contagious while you have symptoms — usually three to eight days. RSV spreads easily from person to person. It can live on hard surfaces, like tables, for several hours. You may be able to spread RSV a day or two before you develop symptoms. Babies and people with compromised immune systems may be contagious even after symptoms go away, for up to four weeks. Complications of RSV RSV can cause complications that make it hard to breathe. They include: Bronchitis Pneumonia Bronchiolitis Worsening of existing conditions like asthma, congestive heart failure or COPD (chronic obstructive pulmonary disease) Respiratory failure Hypoxia or low oxygen levels Dehydration Ear infections Risk factors Babies born early (preterm) and kids under 2 years old have smaller airways and are more likely to get severely sick from RSV. You’re also at a higher risk if you: Are over age 65 Have a compromised immune system Have heart disease, including congenital heart disease Have chronic lung conditions, like asthma Diagnosis and Tests How doctors diagnose RSV A healthcare provider swabs your nose with a soft-tipped stick to test for RSV. They may get a chest X-ray to look for pneumonia and other complications. Management and Treatment How is RSV treated? There’s no specific treatment for RSV. If you or your child has severe RSV, you may need to stay in the hospital to recover. There, providers might help you get better with: Oxygen therapy to get more oxygen into your body Fluids through an IV to prevent dehydration Mechanical ventilation (a ventilator) if you can’t breathe on your own Antibiotics don’t treat RSV since it’s a virus. When should I see my healthcare provider? Call a pediatrician if your child has RSV symptoms and is under 12 months old. You should also call if your child has any of the following: A temperature above 100.4 degrees Fahrenheit or 38 degrees Celsius (if your baby is under 3 months old) A fever above 104 degrees F and 40 degrees C (at any age) Symptoms that don’t improve or get worse after a week Ear drainage or tugging at their ears (possible signs of an ear infection) Difficulty breathing or bad coughing fits Wheezing or noisy breathing Adults can often manage RSV symptoms on their own. Call a healthcare provider if you have symptoms and at least one of the following applies to you: You’re over age 65 You have a compromised immune system You have a heart or lung condition Advertisement Your provider will let you know if you should come in for a checkup and what severe symptoms to look out for. When should I go to the ER? Go to an emergency room if you or your child has severe RSV symptoms. These include signs of difficulty breathing — like wheezing, flaring nostrils, chest retractions, or bluish or grayish skin color. Care at Cleveland Clinic Find a Primary Care Provider Schedule an Appointment Outlook / Prognosis How long does RSV last? RSV can last a week or two. You might have a lingering cough for a while. Severe cases of RSV may last longer. Is there anything I can do to feel better? If you have mild symptoms, you can take care of yourself at home with: A cool-mist humidifier to help with breathing Nasal saline spray to help relieve cough and congestion Suctioning your child’s nose to remove mucus Plenty of fluids to avoid dehydration Over-the-counter (OTC) medications (like acetaminophen or ibuprofen) Always check with your provider or your child’s pediatrician before using any medications or giving them to kids. Prevention Can RSV be prevented? There are some things you can do to help protect yourself from RSV. Babies up to 8 months old and some toddlers at high risk can get a monoclonal antibody immunization. This gives them antibodies that can help them fight off an RSV infection. Pregnant women and adults 75 and older (or 60 and older with certain health conditions) are eligible for RSV vaccines. Talk to your healthcare provider about what’s best for you and your family. Advertisement Other steps you can take include: Wash your hands. Always wash your hands after using the bathroom and before eating or preparing food. Limit exposure to germs. Avoid taking your baby out in large crowds, especially during cold and flu season. Avoid play dates or other close contact with kids who are sick. Clean commonly touched surfaces. Disinfect toys, tables, doorknobs and other surfaces that people in your house touch often. A note from Cleveland Clinic RSV can be as mild as a cold. Or it can turn into something more serious. The important thing is to keep an eye out for signs of serious illness. If you feel something isn’t right — even if you think you or your child just has a cold — don’t hesitate to call a provider or seek emergency care. Advertisement Care at Cleveland Clinic Cleveland Clinic’s primary care providers offer lifelong medical care. From sinus infections and high blood pressure to preventive screening, we’re here for you. Find a Primary Care Provider Schedule an Appointment Medically Reviewed Last reviewed on 05/07/2025. Learn more about the Health Library and our editorial process. AdvertisementAdvertisement Ad Questions 216.444.2538 Appointments & Locations Request an Appointment Find a Primary Care Provider Rendered: Tue Sep 16 2025 05:48:29 GMT+0000 (Coordinated Universal Time)
15972	https://saylordotorg.github.io/text_general-chemistry-principles-patterns-and-applications-v1.0/s22-06-spontaneity-and-equilibrium.html	Spontaneity and Equilibrium Previous Section Table of Contents Next Section 18.6 Spontaneity and Equilibrium Learning Objective To know the relationship between free energy and the equilibrium constant. We have identified three criteria for whether a given reaction will occur spontaneously: Δ S univ> 0, Δ G sys< 0, and the relative magnitude of the reaction quotient Q versus the equilibrium constant K. (For more information on the reaction quotient and the equilibrium constant, see Chapter 15 "Chemical Equilibrium".) Recall that if Q<K, then the reaction proceeds spontaneously to the right as written, resulting in the net conversion of reactants to products. Conversely, if Q>K, then the reaction proceeds spontaneously to the left as written, resulting in the net conversion of products to reactants. If Q = K, then the system is at equilibrium, and no net reaction occurs. Table 18.3 "Criteria for the Spontaneity of a Process as Written" summarizes these criteria and their relative values for spontaneous, nonspontaneous, and equilibrium processes. Because all three criteria are assessing the same thing—the spontaneity of the process—it would be most surprising indeed if they were not related. The relationship between Δ S univ and Δ G sys was described in Section 18.5 "Free Energy". In this section, we explore the relationship between the standard free energy of reaction (Δ G°) and the equilibrium constant (K). Table 18.3 Criteria for the Spontaneity of a Process as Written \| Spontaneous \| Equilibrium \| Nonspontaneous \| :---: \| Δ S univ> 0 \| Δ S univ = 0 \| Δ S univ< 0 \| \| Δ G sys< 0 \| Δ G sys = 0 \| Δ G sys> 0 \| \| Q<K \| Q = K \| Q>K \| \| Spontaneous in the reverse direction. \| Free Energy and the Equilibrium Constant Because Δ H° and Δ S° determine the magnitude of Δ G° (Equation 18.26), and because K is a measure of the ratio of the concentrations of products to the concentrations of reactants, we should be able to express K in terms of Δ G° and vice versa. As you learned in Section 18.5 "Free Energy", Δ G is equal to the maximum amount of work a system can perform on its surroundings while undergoing a spontaneous change. For a reversible process that does not involve external work, we can express the change in free energy in terms of volume, pressure, entropy, and temperature, thereby eliminating Δ H from the equation for Δ G. Using higher math, the general relationship can be shown as follows: Equation 18.29 Δ G = V Δ P − S Δ T If a reaction is carried out at constant temperature (Δ T = 0), then Equation 18.29 simplifies to Equation 18.30 Δ G = V Δ P Under normal conditions, the pressure dependence of free energy is not important for solids and liquids because of their small molar volumes. For reactions that involve gases, however, the effect of pressure on free energy is very important. Assuming ideal gas behavior, we can replace the V in Equation 18.30 by nRT/P (where n is the number of moles of gas and R is the ideal gas constant) and express Δ G in terms of the initial and final pressures (P i and P f, respectively) as in Equation 18.20: Equation 18.31 Δ G=(n R T P)Δ P=n R T Δ P P=n R T ln(P f P i) If the initial state is the standard state with P i = 1 atm, then the change in free energy of a substance when going from the standard state to any other state with a pressure P can be written as follows: G − G° = nRT ln P This can be rearranged as follows: Equation 18.32 G = G° + nRT ln P As you will soon discover, Equation 18.32 allows us to relate Δ G° and K p. Any relationship that is true for K p must also be true for K because K p and K are simply different ways of expressing the equilibrium constant using different units. Let’s consider the following hypothetical reaction, in which all the reactants and the products are ideal gases and the lowercase letters correspond to the stoichiometric coefficients for the various species: Equation 18.33 a A+b B⇌c C+d D Because the free-energy change for a reaction is the difference between the sum of the free energies of the products and the reactants, we can write the following expression for Δ G: Equation 18.34 Δ G=∑m G products−∑n G reactants=(c G C+d G D)−(a G A+b G B) Substituting Equation 18.32 for each term into Equation 18.34, Δ G=[(c G°C+c R T ln P C)+(d G°D+d R T ln P D)]−[(a G°A+a R T ln P A)+(b G°B+b R T ln P B)] Combining terms gives the following relationship between Δ G and the reaction quotient Q: Equation 18.35 Δ G=Δ G°+R T ln(P c C P d D P a A P b B)=Δ G°+R T ln Q where Δ G° indicates that all reactants and products are in their standard states. In Chapter 15 "Chemical Equilibrium", you learned that for gases Q = K p at equilibrium, and as you’ve learned in this chapter, Δ G = 0 for a system at equilibrium. Therefore, we can describe the relationship between Δ G° and K p for gases as follows: Equation 18.36 0=Δ G°+R T ln K p Δ G°=−R T ln K p If the products and reactants are in their standard states and Δ G° < 0, then K p> 1, and products are favored over reactants. Conversely, if Δ G° > 0, then K p< 1, and reactants are favored over products. If Δ G° = 0, then K p = 1, and neither reactants nor products are favored: the system is at equilibrium. Note the Pattern For a spontaneous process under standard conditions, K eq and K p are greater than 1. Example 12 In Example 10, we calculated that Δ G° = −32.7 kJ/mol of N 2 for the reaction N 2(g)+3 H 2(g)⇌2 NH 3(g). This calculation was for the reaction under standard conditions—that is, with all gases present at a partial pressure of 1 atm and a temperature of 25°C. Calculate Δ G for the same reaction under the following nonstandard conditions: P N 2 = 2.00 atm, P H 2 = 7.00 atm, P NH 3 = 0.021 atm, and T = 100°C. Does the reaction favor products or reactants? Given:balanced chemical equation, partial pressure of each species, temperature, and Δ G° Asked for:whether products or reactants are favored Strategy: A Using the values given and Equation 18.35, calculate Q. B Substitute the values of Δ G° and Q into Equation 18.35 to obtain Δ G for the reaction under nonstandard conditions. Solution: A The relationship between Δ G° and Δ G under nonstandard conditions is given in Equation 18.35. Substituting the partial pressures given, we can calculate Q: Q=P 2 NH 3 P N 2 P 3 H 2=(0.021)2(2.00)(7.00)3=6.4×10−7 B Substituting the values of Δ G° and Q into Equation 18.35, Δ G=Δ G°+R T ln Q=−32.7 kJ+[(8.314 J/K)(373 K)(1 kJ 1000 J)ln(6.4×10−7)]=−32.7 kJ+(−44 kJ)=−77 kJ/mol of N 2 Because Δ G< 0 and Q< 1.0, the reaction is spontaneous to the right as written, so products are favored over reactants. Exercise Calculate Δ G for the reaction of nitric oxide with oxygen to give nitrogen dioxide under these conditions: T = 50°C, P NO = 0.0100 atm, P O 2 = 0.200 atm, and P NO 2 = 1.00×10−4 atm. The value of Δ G° for this reaction is −72.5 kJ/mol of O 2. Are products or reactants favored? Answer:−92.9 kJ/mol of O 2; the reaction is spontaneous to the right as written, so products are favored. Example 13 Calculate K p for the reaction of H 2 with N 2 to give NH 3 at 25°C. As calculated in Example 10, Δ G° for this reaction is −32.7 kJ/mol of N 2. Given:balanced chemical equation from Example 10, Δ G°, and temperature Asked for:K p Strategy: Substitute values for Δ G° and T (in kelvins) into Equation 18.36 to calculate K p, the equilibrium constant for the formation of ammonia. Solution: In Example 10, we used tabulated values of Δ G°f to calculate Δ G° for this reaction (−32.7 kJ/mol of N 2). For equilibrium conditions, rearranging Equation 18.36, Δ G°=−R T ln K p−Δ G°R T=ln K p Inserting the value of Δ G° and the temperature (25°C = 298 K) into this equation, ln K p=−(−32.7 kJ)(1000 J/kJ)(8.314 J/K)(298 K)=13.2 K p=5.4×10 5 Thus the equilibrium constant for the formation of ammonia at room temperature is favorable. As we saw in Chapter 15 "Chemical Equilibrium", however, the rate at which the reaction occurs at room temperature is too slow to be useful. Exercise Calculate K p for the reaction of NO with O 2 to give NO 2 at 25°C. As calculated in the exercise in Example 10, Δ G° for this reaction is −70.5 kJ/mol of O 2. Answer:2.2×10 12 Although K p is defined in terms of the partial pressures of the reactants and the products, the equilibrium constant K is defined in terms of the concentrations of the reactants and the products. We described the relationship between the numerical magnitude of K p and K in Chapter 15 "Chemical Equilibrium" and showed that they are related: Equation 18.37 K p = K(RT)Δ n where Δ n is the number of moles of gaseous product minus the number of moles of gaseous reactant. For reactions that involve only solutions, liquids, and solids, Δ n = 0, so K p = K. For all reactions that do not involve a change in the number of moles of gas present, the relationship in Equation 18.36 can be written in a more general form: Equation 18.38 Δ G° = −RT ln K Only when a reaction results in a net production or consumption of gases is it necessary to correct Equation 18.38 for the difference between K p and K.Although we typically use concentrations or pressures in our equilibrium calculations, recall that equilibrium constants are generally expressed as unitless numbers because of the use of activities or fugacities in precise thermodynamic work. Systems that contain gases at high pressures or concentrated solutions that deviate substantially from ideal behavior require the use of fugacities or activities, respectively. Combining Equation 18.26 and Equation 18.38 provides insight into how the components of Δ G° influence the magnitude of the equilibrium constant: Equation 18.39 Δ G° = Δ H° − T Δ S° = −RT ln K Notice that K becomes larger as Δ S° becomes more positive, indicating that the magnitude of the equilibrium constant is directly influenced by the tendency of a system to move toward maximum disorder. Moreover, K increases as Δ H° decreases. Thus the magnitude of the equilibrium constant is also directly influenced by the tendency of a system to seek the lowest energy state possible. Note the Pattern The magnitude of the equilibrium constant is directly influenced by the tendency of a system to move toward maximum disorder and seek the lowest energy state possible. Temperature Dependence of the Equilibrium Constant The fact that Δ G° and K are related provides us with another explanation of why equilibrium constants are temperature dependent. This relationship is shown explicitly in Equation 18.39, which can be rearranged as follows: Equation 18.40 ln K=−Δ H°R T+Δ S°R Assuming Δ H° and Δ S° are temperature independent, for an exothermic reaction (Δ H° < 0), the magnitude of K decreases with increasing temperature, whereas for an endothermic reaction (Δ H° > 0), the magnitude of K increases with increasing temperature. The quantitative relationship expressed in Equation 18.40 agrees with the qualitative predictions made by applying Le Châtelier’s principle, which we discussed in Chapter 15 "Chemical Equilibrium". Because heat is produced in an exothermic reaction, adding heat (by increasing the temperature) will shift the equilibrium to the left, favoring the reactants and decreasing the magnitude of K. Conversely, because heat is consumed in an endothermic reaction, adding heat will shift the equilibrium to the right, favoring the products and increasing the magnitude of K. Equation 18.40 also shows that the magnitude of Δ H° dictates how rapidly K changes as a function of temperature. In contrast, the magnitude and sign of Δ S° affect the magnitude of K but not its temperature dependence. If we know the value of K at a given temperature and the value of Δ H° for a reaction, we can estimate the value of K at any other temperature, even in the absence of information on Δ S°. Suppose, for example, that K 1 and K 2 are the equilibrium constants for a reaction at temperatures T 1 and T 2, respectively. Applying Equation 18.40 gives the following relationship at each temperature: ln K 1=−Δ H°R T 1+Δ S°R ln K 2=−Δ H°R T 2+Δ S°R Subtracting ln K 1 from ln K 2, Equation 18.41 ln K 2−ln K 1=ln K 2 K 1=Δ H°R(1 T 1−1 T 2) Thus calculating Δ H° from tabulated enthalpies of formation and measuring the equilibrium constant at one temperature (K 1) allow us to calculate the value of the equilibrium constant at any other temperature (K 2), assuming that Δ H° and Δ S° are independent of temperature. Example 14 The equilibrium constant for the formation of NH 3 from H 2 and N 2 at 25°C was calculated to be K p = 5.4×10 5 in Example 13. What is K p at 500°C? (Use the data from Example 10.) Given:balanced chemical equation, Δ H°, initial and final T, and K p at 25°C Asked for:K p at 500°C Strategy: Convert the initial and final temperatures to kelvins. Then substitute appropriate values into Equation 18.41 to obtain K 2, the equilibrium constant at the final temperature. Solution: The value of Δ H° for the reaction obtained using Hess’s law is −91.8 kJ/mol of N 2. If we set T 1 = 25°C = 298.K and T 2 = 500°C = 773 K, then from Equation 18.41 we obtain the following: ln K 2 K 1=Δ H°R(1 T 1−1 T 2)=(−91.8 kJ)(1000 J/kJ)8.314 J/K(1 298 K−1 773 K)=−22.8 K 2 K 1=1.3×10−10 K 2=(5.4×10 5)(1.3×10−10)=7.0×10−5 Thus at 500°C, the equilibrium strongly favors the reactants over the products. Exercise In the exercise in Example 13, you calculated K p = 2.2×10 12 for the reaction of NO with O 2 to give NO 2 at 25°C. Use the Δ H°f values in the exercise in Example 10 to calculate K p for this reaction at 1000°C. Answer:5.6×10−4 Summary For a reversible process that does not involve external work, we can express the change in free energy in terms of volume, pressure, entropy, and temperature. If we assume ideal gas behavior, the ideal gas law allows us to express Δ G in terms of the partial pressures of the reactants and products, which gives us a relationship between Δ G and K p, the equilibrium constant of a reaction involving gases, or K, the equilibrium constant expressed in terms of concentrations. If Δ G° < 0, then K or K p> 1, and products are favored over reactants. If Δ G° > 0, then K or K p< 1, and reactants are favored over products. If Δ G° = 0, then K or K p = 1, and the system is at equilibrium. We can use the measured equilibrium constant K at one temperature and Δ H° to estimate the equilibrium constant for a reaction at any other temperature. Key Takeaway The change in free energy of a reaction can be expressed in terms of the standard free-energy change and the equilibrium constant K or K p and indicates whether a reaction will occur spontaneously under a given set of conditions. Key Equations Relationship between standard free-energy change and equilibrium constant Equation 18.38: Δ G° = −RT ln K Temperature dependence of equilibrium constant Equation 18.40: ln K=−Δ H°R T+Δ S°R Calculation ofKat second temperature Equation 18.41: ln K 2 K 1=Δ H°R(1 T 1−1 T 2) Conceptual Problems Do you expect products or reactants to dominate at equilibrium in a reaction for which Δ G° is equal to 1.4 kJ/mol? 105 kJ/mol? −34 kJ/mol? The change in free energy enables us to determine whether a reaction will proceed spontaneously. How is this related to the extent to which a reaction proceeds? What happens to the change in free energy of the reaction N 2(g) + 3F 2(g) → 2NF 3(g) if the pressure is increased while the temperature remains constant? if the temperature is increased at constant pressure? Why are these effects not so important for reactions that involve liquids and solids? Compare the expressions for the relationship between the change in free energy of a reaction and its equilibrium constant where the reactants are gases versus liquids. What are the differences between these expressions? Numerical Problems Carbon monoxide, a toxic product from the incomplete combustion of fossil fuels, reacts with water to form CO 2 and H 2, as shown in the equation CO(g)+H 2 O(g)⇌CO 2(g)+H 2(g), for which Δ H° = −41.0 kJ/mol and Δ S° = −42.3 J cal/(mol·K) at 25°C and 1 atm. What is Δ G° for this reaction? What is Δ G if the gases have the following partial pressures: P CO = 1.3 atm, P H 2 O = 0.8 atm, P CO 2 = 2.0 atm, and P H 2 = 1.3 atm? What is Δ G if the temperature is increased to 150°C assuming no change in pressure? Methane and water react to form carbon monoxide and hydrogen according to the equation CH 4(g)+H 2 O(g)⇌CO(g)+3 H 2(g). What is the standard free energy change for this reaction? What is K p for this reaction? What is the carbon monoxide pressure if 1.3 atm of methane reacts with 0.8 atm of water, producing 1.8 atm of hydrogen gas? What is the hydrogen gas pressure if 2.0 atm of methane is allowed to react with 1.1 atm of water? At what temperature does the reaction become spontaneous? Calculate the equilibrium constant at 25°C for each equilibrium reaction and comment on the extent of the reaction. CCl 4(g)+6 H 2 O(l)⇌CO 2(g)+4 HCl(aq); Δ G° = −377 kJ/mol Xe(g)+2 F 2(g)⇌XeF 4(s); Δ H° = −66.3 kJ/mol, Δ S° = −102.3 J/(mol·K) PCl 3(g)+S⇌PSCl 3(l);Δ G°f(PCl 3) = −272.4 kJ/mol, Δ G°f(PSCl 3) = −363.2 kJ/mol Calculate the equilibrium constant at 25°C for each equilibrium reaction and comment on the extent of the reaction. 2 KClO 3(s)⇌2 KCl(s)+3 O 2(g); Δ G° = −225.8 kJ/mol CoCl 2(s)+6 H 2 O(g)⇌CoCl 2⋅6 H 2 O(s);Δ H°rxn = −352 kJ/mol, Δ S°rxn = −899 J/(mol·K) 2 PCl 3(g)+O 2(g)⇌2 POCl 3(g);Δ G°f(PCl 3) = −272.4 kJ/mol, Δ G°f(POCl 3) = −558.5 kJ/mol The gas-phase decomposition of N 2 O 4 to NO 2 is an equilibrium reaction with K p = 4.66×10−3. Calculate the standard free-energy change for the equilibrium reaction between N 2 O 4 and NO 2. The standard free-energy change for the dissolution K 4 Fe(CN)6⋅H 2 O(s)⇌4 K+(aq)+Fe(CN)6 4−(aq)+H 2 O(l) is 26.1 kJ/mol. What is the equilibrium constant for this process at 25°C? Ammonia reacts with water in liquid ammonia solution (am) according to the equation NH 3(g)+H 2 O(am)⇌NH 4+(am)+OH−(am). The change in enthalpy for this reaction is 21 kJ/mol, and Δ S° = −303 J/(mol·K). What is the equilibrium constant for the reaction at the boiling point of liquid ammonia (−31°C)? At 25°C, a saturated solution of barium carbonate is found to have a concentration of [Ba 2+] = [CO 3 2−] = 5.08×10−5 M. Determine Δ G° for the dissolution of BaCO 3. Lead phosphates are believed to play a major role in controlling the overall solubility of lead in acidic soils. One of the dissolution reactions is Pb 3(PO 4)2(s)+4 H+(aq)⇌3 Pb 2+(aq)+2 H 2 PO 4−(aq), for which log K = −1.80. What is Δ G° for this reaction? The conversion of butane to 2-methylpropane is an equilibrium process with Δ H° = −2.05 kcal/mol and Δ G° = −0.89 kcal/mol. What is the change in entropy for this conversion? Based on structural arguments, are the sign and magnitude of the entropy change what you would expect? Why? What is the equilibrium constant for this reaction? The reaction of CaCO 3(s) to produce CaO(s) and CO 2(g) has an equilibrium constant at 25°C of 2×10−23. Values of Δ H°f are as follows: CaCO 3, −1207.6 kJ/mol; CaO, −634.9 kJ/mol; and CO 2, −393.5 kJ/mol. What is Δ G° for this reaction? What is the equilibrium constant at 900°C? What is the partial pressure of CO 2(g) in equilibrium with CaO and CaCO 3 at this temperature? Are reactants or products favored at the lower temperature? at the higher temperature? In acidic soils, dissolved Al 3+ undergoes a complex formation reaction with SO 4 2− to form [AlSO 4+]. The equilibrium constant at 25°C for the reaction Al 3+(aq)+SO 4 2−(aq)⇌AlSO 4+(aq) is 1585. What is Δ G° for this reaction? How does this value compare with Δ G° for the reaction Al 3+(aq)+F−(aq)⇌AlF 2+(aq), for which K = 10 7 at 25°C? Which is the better ligand to use to trap Al 3+ from the soil? Answers −28.4 kJ/mol −26.1 kJ/mol −19.9 kJ/mol 1.21×10 66; equilibrium lies far to the right. 1.89×10 6; equilibrium lies to the right. 5.28×10 16; equilibrium lies far to the right. 13.3 kJ/mol 5.1×10−21 10.3 kJ/mol 129.5 kJ/mol 6 6.0 atm Products are favored at high T; reactants are favored at low T. Previous Section Table of Contents Next Section
15973	https://www.sciencedirect.com/topics/engineering/neutron-velocity	Skip to Main content Sign in Chapters and Articles You might find these chapters and articles relevant to this topic. Neutron transport equation 1.2 Definition of the ordinates and basic elements In the transport theory, neutrons are considered as point particles, which means the neutron motion state can be represented by the determined position and velocity. As shown in Fig. 1.1, the position of neutrons in space can be expressed by r. The velocity vector ν of the particles is written in terms of its solid angle as (1.1) where υ = \|ν \| is the magnitude of velocity, and its relation to the kinetic energy of neutrons E is E = mυ2/2, where m is the mass of neutrons and Ω is the unit vector of the direction of motion. Its modulus is equal to 1. Its direction is expressed by polar coordinate system through polar angle θ and azimuth angle φ. Therefore, at any moment, the state of neutron motion can be described by six independent variables such as position vector r(x,y,z), energy E, and direction of motion Ω(θ, φ). For different coordinate systems, the expressions of r and Ω are different. 1.2.1 Coordinate system (1) : Cartisian coordinate (x, y, z) Cartisian coordinate is the most commonly used coordinate system. Fig. 1.2 shows the three-dimensional cartisian coordinate and the representation of Ω in it. Coordinates of space points are determined by (x, y, z), volume element dV = dxdydz, and (1.2) Among which ex, ey, and ez are the unit vectors in the directions of three coordinate axes. And (1.3) (1.4) (1.5) (1.6) It should be noted that the polar angle θ is the angle between Ω and ex (Fig. 1.2) which is different from that shown in Fig. 1.1. (2) : Cylindrical coordinate (r, ψ, z) The cylindrical coordinate and the representation of Ω in it are shown in Fig. 1.3. Coordinates of space points are determined by (r, ψ, z), volume elementdV = rdrdψdz, and (1.7) Among which ez and er are axial and radial unit vectors, respectively. And eθ is unit vectors perpendicular to the plane (ez, er) (1.8) (1.9) (1.10) (1.11) It is apparent that the direction of er and eθ changes when the space coordinate r or ψ changes. (3) : Spherical coordinate (r, θ, φ) Fig. 1.4 shows the spherical coordinate and the representation of Ω in it, coordinates of space points are determined by (r, ψ, z). The direction of neutron motion is determined by θ′ between Ω and r (or μ = cos θ′) and angle ω, which is between the plane (r, Ω) and the plane (r, z). In fact, three-dimensional spherical coordinate system is seldom used to solve neutron transport equation. One-dimensional spherically symmetric coordinate system is often used. Because of its symmetry, the space points are determined only by coordinate r, dV = 4πr2dr. The direction of motion Ω is determined by , and is independent of ω. 1.2.2 Neutron density, angular flux, and current (1) : Neutron density In order to fully describe the distribution of neutron in the medium, the distribution of space coordinate r, energy E, and the direction of motion Ω with time t of neutron must be given. Thus, we introduce the neutron density distribution function—the neutron angular density n(r, E, Ω, t). It is defined as the number of neutrons in unit volume at point r with energy E in dE and direction Ω in dΩ at time t. Therefore, (1.12) After integrating the neutron angular density with all the three-dimensional angular directions, the neutron density n(r, E, t) independent of the angle is obtained, i.e., the total neutron density (1.13) where n(r, E, t)drdE is the total number of neutrons in volume r and energy between E and E + dE at time t and coordinate r (including all directions of motion). (2) : Neutron angular flux The neutron angular flux ϕ(r, E, Ω, t) is defined as the product of the neutron angular density and the neutron velocity (1.14) It indicates total travel length per unit time, in unit volume at point r with energy E in dE and direction Ω in dΩ at time t. Similarly, from formula (1.13), the flux of neutrons ϕ(r, E, t) independent of direction can be defined as (1.15) Therefore, the neutron flux ϕ(r, E, t) can be regarded as the sum of the intensities of infinite differential neutron beams ϕ(r, E, Ω, t) in all directions. Sometimes we call ϕ(r, E, t) which is independent of Ω as the total neutron flux or scalar flux. (3) : Neutron current We use n to describe the unit normal vector perpendicular to the dS surface at r (Fig. 1.5). Therefore, in the steady state, the number of neutrons moving in the direction of Ω with energy equal to E passing through dS in unit time is given by n(r, E, Ω)υ \|Ω ⋅ n \| dS. We specify that the positive side of n is “+” and the negative side is “−.” We also assume that the total number of neutrons passing from the “−” side in all directions to the “+” side is Jn+ dS per second, and vice versa Jn− dS, then (1.16) (1.17) The integral domain (Ω ⋅ n) > 0 (or < 0) in the formula means that only those Ω (i.e., hemispheres) with (Ω ⋅ n) > 0 (or < 0) are integral. Jn+ and Jn− along the positive and negative directions at r are called the partial neutron current. If the neutron angular flux is isotropic, then Jn+ = Jn−, that is to say the net neutron number passing through dS per second or net neutron flow per second is equal to zero. In general, if Jn+ ≠ Jn−, then the net neutron number passing through dS per second or net neutron flow per second is equal to (1.18) The above formula can be rewritten as (1.19) Among them (1.20) (1.21) Vector J(r, E) is defined as neutron current. Its projection (or component) in the direction n Jn(r, E) is equal to the number of net neutron (net flux) passing through the unit area perpendicular to n per unit time. If Jn > 0, then Jn+ > Jn−, the direction of the net neutron flow is same as that of the normal vector n. Conversely, if Jn < 0, the direction of the net neutron flow is opposite to that of the normal vector n. From formula (1.19) we can see that the net neutron current passing through dS per second is not only related to distribution of ϕ(r, E, Ω), but also to n the direction of the area element dS. Obviously, if the direction of n is the same as that of J, the value of net neutron current is the largest. View chapterExplore book Read full chapter URL: Chapter Basic nuclear structure and processes 4 Nuclear flux and reaction rates To find the rate at which the various neutron induced interactions proceed in a mixture of nuclides at a particular location in a reactor, it is necessary to determine a quantity called the neutron flux which is usually designated as “φ(r, E)”. To evaluate the neutron flux, one must first determine the number of neutrons of energy “E” crossing the surface of the differential volume element “dV” at location “r” as illustrated in Fig. 5.13. The neutron flux may then be given by: (5.10) where: neutron density at position “r” having an energy “E” traveling in direction “”. : u(E) = neutron velocity corresponding to energy “E” Physically, the neutron flux in Eq. (5.10) can be interpreted as the total track length of neutrons per cm3 within dE about “E”. This interpretation can be made more clear if the units on neutron flux are rewritten as . Reaction rates are proportional to the neutron flux level and the macroscopic reaction cross section, thus: (5.11) As an illustration, the power density at a particular location in a reactor is determined by using the fission cross section: (5.12) Also, the number of neutrons produced per unit volume can be determined by introducing the quantity “ν” which represents the number of neutrons on average emitted per fission. Typically, ν is about 2.5. (5.13) Example A neutron generator producing 1 eV monoenergetic neutrons is being used to irradiate a 239Pu sample. This irradiation produces a fission rate in the sample equivalent to a power density typical of what it would experience in an operating nuclear rocket engine. Determine the neutron flux level required in the sample to achieve the desired power density. Solution The first step in the solution is to determine the microscopic fission cross section at 1 eV for 239Pu from the National Nuclear Data Center (ENDF/B). From an examination of the plot, it may be determined that 34 b/atom. The next step in the calculation involves putting 239Pu in terms of atom density rather than mass density. From Eq. (5.9) the atom density may be determined to be: (1) The 239Pu macroscopic fission cross section may now be determined from its microscopic fission cross section and its atom density determined from Eq. (1) such that: (2) From Eq. (5.12), the power density may be evaluated using the macroscopic fission cross section determined from Eq. (2) and the neutron flux such that: (3) The neutron flux calculated from Eq. (3) may be converted to more traditional units by using appropriate conversion factors such that: (4) View chapterExplore book Read full chapter URL: Book2023, Principles of Nuclear Rocket Propulsion (Second Edition)William Emrich Chapter Reservoir Engineering and Secondary Recovery 2007, The Petroleum Engineering Handbook: Sustainable OperationsM. Ibrahim Khan, M.R. Islam 6.3.1.6 Neutron logs Neutrons exist in the nuclei of all elements except hydrogen. They have approximately the same mass as hydrogen atoms, but with no charge. When emitted from fissionable material, they possess very high velocities but are rapidly slowed down by collisions with other atoms. Atoms of nearly the same mass as the neutron are most effective in reducing neutron velocity. The neutron is greatly slowed by collisions with hydrogen atoms. Since fluids (water, oil, and gas) contain a much higher hydrogen content than rocks, it is apparent that the behavior of emitted neutrons affords a means of evaluating the hydrogen (and hence fluid) content of a formation. Inelastic scattering, elastic scattering, and absorption are the basic phenomena that occur after a fast neutron is introduced into a formation. The fast neutrons are slowed first by inelastic scattering (which takes place at a high neutron energy level) and then by elastic scattering. The neutrons eventually slow to a level of energy at which they coexist with the formation nuclei in thermal equilibrium. Neutrons in this state are called thermal neutrons. Thermal neutrons continue to scatter off the formation nuclei elastically and diffuse throughout the formation. Each thermal neutron eventually is captured by one of the nuclei. The nucleus instantaneously emits gamma rays, called capture gamma rays. Neutrons that have slowed almost to the thermal energy level yet are still energetic enough to avoid capture are known as epithermal neutrons. Several logging tools are based on these phenomena. The neutron porosity log is based on the elastic scattering of neutrons as they collide with nuclei in the formation. Each neutron scatters off a nucleus with less kinetic energy. Energy and momentum conservation in elastic collisions dictates that the presence of hydrogen in the formation dominates the slowing process. The reason for this is that the mass of the hydrogen nucleus is approximately equal to that of the incident neutron. Consequently, at a point sufficiently removed from the neutron source, formations with high hydrogen content display low concentrations of epithermal and thermal neutrons and capture gamma rays. Inversely, formations with low hydrogen content display high concentrations of epithermal and thermal neutrons and capture gamma rays. To provide a standard unit for neutron log measurements, the American Petroleum Institute (API) adopted the “API neutron unit”. One API neutron unit is defined as 1/1000 of the difference between instrument zero (i.e., tool response to zero radiation) and log deflection opposite a 6-ft zone of Indiana limestone in a neutron calibration pit at the University of Houston (Figure 6-11). The pit is 24 ft deep, with a 15-ft rathole. The pit contains three 6-ft-thick limestone zones. Each zone is made up of six 1-ft-thick octagonal blocks. The zones consist of Carthage, Indiana, and Austin limestones displaying average porosities of 1.9% and 25%, respectively. The rock is saturated with fresh water. A 6-ft layer of fresh water atop the rock blocks provides a 100% porosity reference point. A 7⅞- inch borehole extends through the pit center. View chapterExplore book Read full chapter URL: Book2007, The Petroleum Engineering Handbook: Sustainable OperationsM. Ibrahim Khan, M.R. Islam Chapter Energy Production 2018, Comprehensive Energy SystemsSümer Şahin, Yican Wu 3.14.4.2 Neutron Transport Theory The energy dependence of neutron-atom interactions is addressed in the previous chapter in details. Hence, it is indispensable to have precise information about the space and energy distribution of neutrons, i.e., the neutron spectrum, in a nuclear reactor. As neutrons are in continuous movement, their position in the matter can be determined with the help of a balance equation, called Boltzmann transport equation (BTE). BTE allows to calculate the temporal variation of the neutron density n=Φ/v as a function of space, energy, and direction with a high precision. The neutron spectrum is calculated as a function of space vector , neutron velocity vector and time variable t. It is more convenient to describe the neutron velocity vector in energy E and the solid space angle [73,74]. In the BTE, the vector space of the neutron flux Φ has seven dimensions. The BTE evaluates the variation of the neutron density in form of a balance equation containing the scattering source, neutron source, convection leakage loss and neutron–matter reaction loss terms, as follows [75–78]: (12) Following simplifications are assumed in the formulation of the BTE: • : Relativistic effects will not be considered. This is an acceptable assumption. Neutron rest mass is equivalent to 939.6 MeV (see Appendix). Fission neutrons can reach up to 10 MeV. (D,T) fusion neutron energy is 14 MeV. Although the tail of neutron energy in ADS can reach up to 1000 MeV, their average energy is around ~1 MeV. Neutron velocity in all actual nuclear energy systems are far below the relativistic range. • : Material densities do not change. Under these assumptions, the terms in the Eq. (12) are defined as follows: By defining the surface of the volume element as the neutron leakage out of the volume element will become . With the help of the divergence theorem, the surface integral of the neutron leakage out of the volume element can be expressed as the volume integral of the divergence of the neutron leakage: (13) The neutron leakage through convective neutron current can be written also in terms of the angular neutron flux The totality of the neutron loss through neutron–atom reactions, which includes all absorption and scattering reactions is For steady-state case, the time element in Eq. (12) vanishes and the neutron transport equation can be written as (14) The phase space in Eq. (10) is six-dimensional: As one can see in Eqs. (12) and (14), the neutron scattering term is strongly energy and angular dependent, i.e., it has anisotropic character, which must be measured experimentally for each nuclide individually. For the analytical-numerical treatment, the angular dependence of the neutron scattering term can be approximated by Legendre polynomials: (15) The orthogonal character of Legendre polynomials allows the calculation of scattering Kernels elegantly based on experimental values of the scattering cross-sections as (16) The cosine of the θ component of the solid angle is defined as µ=cosθ. With the help of addition theorem spherical Legendre polynomials, the term can be expressed as (17) The terms will vanish by the integration over the angle , as . Eq. (13) simplifies to (18) Finally, the Boltzmann transport for 1D plane and spherical geometry becomes; (19) Analytical solution of the BTE is given only for specific and rigorously simplified cases. Numerically, BTE allows calculating the neutron spectrum with very high precision and for complex geometries. There are deterministic and stochastically ways for the numerical solution of the BTE. The most important deterministic methods can be cited as [77,78] • : SN method, • : PL method, and • : Collision probability methods. SN is the most developed tested and worldwide used deterministic method. The formulation of the algorithms is also a good tool for a better comprehension of the subject. Hence, the numerical approach in SN will be outlined in Section 3.14.4.3. In recent years, Monte Carlo neutron transport methods have been broadly adopted due to the advantages of high simulation accuracy and flexible geometry representation. Typical Monte Carlo codes include SuperMC, MCNP, Geant4, Serpent, Open MC, etc. SuperMC (Super Multi-functional Calculation Program for Nuclear Design and safety Evaluation) developed by the FDS Team, is a general, intelligent, accurate and precise simulation software system for the nuclear design and safety evaluation of nuclear systems. It supports the comprehensive neutronics calculation, taking the radiation transport as the core and including the depletion, radiation source term/dose/biohazard, material activation and transmutation, etc. It supports the multi-physics coupling calculation including thermo-hydraulics, structural mechanics, chemistry, biology, etc. Multi-processes directly coupling accurate modeling theory and calculation methods for neutron transport of complex system were firstly developed and applied in SuperMC by FDS Team. Furthermore, novel integrated irregular modeling idea based on hybrid body-surface presentation and feature-based decomposition was proposed to support complex structures. And physical model of adaptive transition region for probability and deterministic coupling method was established to achieve accurate simulation. SuperMC supports more efficient calculation based on pre-judgment of geometry and physical features. It realizes the spanning from isolated neutron transport calculation to multi-scale whole processes neutronics simulation and makes accurate design and safety analysis possible for complex nuclear systems. SuperMC has been applied in over 30 major nuclear engineering projects such as ITER and over 60 nations. It has passed international benchmarking and QA activities of ITER IO and selected as reference code for ITER neutronics modeling. Series of ITER reference 3D neutronics basic models have been established with SuperMC for supporting ITER design and analysis. View chapterExplore book Read full chapter URL: Reference work2018, Comprehensive Energy SystemsSümer Şahin, Yican Wu Chapter BASIC ASPECTS OF TRANSPORT AND DIFFUSION THEORY 1976, Physics of High-Temperature ReactorsLUIGI MASSIMO 4.1 The neutron transport equation The behaviour of neutrons in any medium can be described by the transport equation, first used by Boltzmann for the description of the behaviour of gas molecules. Neutrons can indeed be thought as forming a kind of a gas whose treatment is simplified by the fact that neutrons do not interact with each other. The general time, position and angle-dependent form of the transport equation is (4.1) where the symbols mean: υ : neutron velocity corresponding to energy E, N : neutron angular density, σt : total neutron cross-sections (generally function of r and E), S : neutron source, r : space coordinate, Ω : unit vector in the direction of the neutron motion, E : energy, t : time, : Σs (E ′ → E, Ω′ → Ω) scattering cross-section from E′, Ω′ into E, Ω. This equation represents simply a neutron balance in a volume element dV for the neutrons having energy between E and E + dE and flight direction in the solid angle dΩ around Ω. The first term on the right side is the leakage out of dV, the second term represents the loss due to absorption and scattering, the third gives the sources due to scattering from other directions and energies and the fourth is the source term (including fission and external sources). We do not repeat here the well-known difinitions of neutron angular density, cross-sections, etc. (see refs. 1 and 2). Defining the neutron angular flux eqn. (4.1) becomes (4.2) The boundary conditions for this equation are continuity at the interface between two media; at the outer free surface ψ(r, E, Ω, t) = 0 for directions entering the system; ψ(r, E, Ω, t) = 0 for energies greater than the maximum source or fission neutron energy. For reactors this limit corresponds to 10 –15 MeV. The transport equation (4.2) without external sources takes the form: (4.3) here the last term represents the fission source which is supposed to be isotropic; χ(E) is the fission spectrum, probability that a fission neutron is generated with energy E v(E) is the average number of neutrons generated by a fission induced by a neutron having energy E. If the leakage and absorption terms are equal to the production terms (fission and slowing down) the time derivative of the left-hand side of eqn. (4.3) vanishes. In this case the reactor is said to be critical and the neutron population does not change with time. For a given material composition all terms on the right-hand side of eqn. (4.3) are fixed except the first one. This means that criticality can only be reached for a particular value of the term which represents the neutron leakage out of the systems and depends on the reactor geometry. In this way the critical geometry is defined. For a given reactor criticality can be obtained changing Σt by means of insertion or extraction of neutron absorbers. Static calculations of non-critical systems are usually performed artificially dividing the term v(E) of eqn. (4.3) by a factor keff so that the equation is always satisfied without considering the time dependence of the neutron population. The time-independent form of the Boltzmann equation is then: (4.4) View chapterExplore book Read full chapter URL: Book1976, Physics of High-Temperature ReactorsLUIGI MASSIMO Chapter NUCLEAR RADIATION, ITS INTERACTION WITH MATTER AND RADIOISOTOPE DECAY 2003, Handbook of Radioactivity Analysis (Second Edition)MICHAEL F. L'ANNUNZIATA G. Neutron Radiation The neutron is a neutral particle, which is stable only in the confines of the nucleus of the atom. Its mass, like that of the proton, is equivalent to 1 u (atomic mass unit). Unlike the particulate alpha and beta nuclear radiation previously discussed, neutron radiation is not emitted in any significant quantities from radionuclides that undergo the traditional nuclear decay processes with the exception of a few radionuclides such as 252Cf and 248Cm, which decay to a significant extent by spontaneous fission (see Section II.G.2.b). Significant quantities of neutron radiation occur when neutrons are ejected from the nuclei of atoms following reactions between the nuclei and particulate radiation. The lack of charge of the neutron also makes it unable to cause directly any ionization in matter, again unlike alpha and beta radiation. The various sources, properties, and mechanisms of interaction of neutrons with matter are described subsequently. 1. Neutron Classification Neutrons are generally classified according to their kinetic energies. There is no sharp division or energy line of demarcation between the various classes of neutrons; however, the following is an approximate categorization according to neutron energy: • : Cold neutrons < 0.003 eV • : Slow (thermal) neutrons 0.003–0.4 eV • : Slow (epithermal) neutrons 0.4–100 eV • : Intermediate neutrons 100 eV–200 keV • : Fast neutrons 200 keV-10 MeV • : High energy (relativistic) neutrons > 10 MeV The energies of neutrons are also expressed in terms of velocity (meters per second) as depicted in the terminology used to classify neutrons. A neutron of specific energy and velocity is also described in terms of wavelength, because particles in motion also have wave properties. It is the wavelength of the neutron that becomes important in studies of neutron diffraction. The values of energy, velocity, and wavelength of the neutron, as with all particles in motion, are interrelated. The velocity of neutrons increases according to the square root of the energy, and the wavelength of the neutron is inversely proportional to its velocity. Knowing only one of the properties, either the energy, velocity, or wavelength of a neutron, we can calculate the other two. We can relate the neutron energy and velocity using the kinetic energy equation (1.46) where E is the particle energy in joules (1 eV = 1.6 × 10−19 J), m is the mass of the neutron (1.67 × 10−27 kg), and v is the particle velocity in meters per second. The wavelength is obtained from the particle mass and velocity according to (1.47) where λ is the particle wavelength in meters, h is Planck's constant (6.63 × 10−34 J s), p is the particle momentum, and m and v are the particle mass and velocity as previously defined. The correlation between neutron energy, velocity, and wavelength is provided in Fig. 1.9, which is constructed from the classical Eqs. 1.46 and 1.47 relating particle mass, energy, velocity and wavelength. However, calculations involving high-energy particles that approach the speed of light will contain a certain degree of error unless relativistic calculations are used, as the mass of the particle will increase according to the particle speed. In Section IV.C of this chapter we used the Einstein equation E = mc2 to convert the rest mass of the positron or negatron to its rest energy (0.51 MeV). When gauging particles in motion the total energy of the particle is the sum of its kinetic (K) and rest energies (mc2) or (1.48) where (1.49) u is the particle speed, and u < c. If we call the particle rest mass m0, then the relativistic mass, mr, which is the speed-dependent mass of the particle is calculated as (1.50) The relativistic speed is defined as (1.51) where K is the kinetic energy, and the particle speed u is always less than c (Serway et al., 1997). The nonrelativistic speed is that described by Eq. 1.46 or . To confirm the validity of the use of nonrelativistic calculations of particle speed for the construction of Fig 1.9 let us use Eqs. 1.46 and 1.51 to compare the differences between the nonrelativistic and relativistic speeds of a neutron of 10 MeV kinetic energy. This energy was selected, because it is the highest neutron energy included in Fig. 1.9, and differences between nonrelativistic and relativistic calculations increase with particle energy. The difference between the two calculated speeds is defined by the ratio of the two or (1.52) The rest energy of the neutron, mc2, is first calculated as and since by definition, 1 eV = 1.602 × 10−19 J. From Eq. 1.52 the ratio of the nonrelativistic and relativistic speeds are calculated as The error between the nonrelativistic and relativistic calculations is small at this high neutron energy. However, if we consider higher neutron energies in excess of 10 MeV the error of making nonrelativistic calculations increases. As we observed above in the case of particle speed, we will also see that particle wavelength will also differ for nonrelativistic and relativistic calculations. In 1923 Louis Victor de Broglie first postulated that all particles or matter in motion should have wave characteristics just as photons display both a wave and particle character. We therefore attribute the wavelength of particles in motion as de Broglie wavelengths. Let us then compare calculated nonrelativistic and relativistic wavelengths. From Eq. 1.47, we can describe the nonrelativistic wavelength, λnr, as (1.53) where . For relativistic calculations the value of pc is calculated according to the following equation derived by Halpern (1988): (1.54) and the calculation for the relativistic de Broglie wavelength, λr, then becomes (1.55) We can then compare the difference between the nonrelativistic and relativistic wavelengths for the 10 MeV neutron as follows: (1.56) From the above comparison of nonrelativistic and relativistic calculations of neutron wavelength and velocity, we see that the data provided in Fig. 1.9 based on nonrelativistic calculations are valid with less than 1% error for the highest energy neutron included in that figure. However, if we consider higher energies beyond 10 MeV, where we classify the neutron as relativistic, the errors in making nonrelativistic calculations will increase with neutron energy. It will be clearly obvious to the reader that factors in Eq. 1.56 can be cancelled out readily and the equation simpliied to the following, which provides a quick evaluation of the effect of particle energy on the error in nonrelativistic calculation of the de Broglie wavelength: (1.57) where K is the particle kinetic energy in MeV and m0c2 is the particle rest energy (e.g., 939.5 MeV for the neutron and 0.511 MeV for the beta particle). For example, a nonrelativistc calculation of the wavelength of a 50-MeV neutron would have the following error: Note that the above-computed errors in nonrelativistic calculations of the de Broglie wavelength increased from 0.26% for a 10-MeV neutron to 1.31% for a 50-MeV neutron, and the error will increase with particle energy. Errors in nonrelativistic calculations are yet greater for particles of smaller mass (e.g., beta particles) of a given energy compared to neutrons of the same energy. This is due obviously to the fact that particles of lower mass and a given energy will travel at higher speeds than particles of the same energy but higher mass. This is illustrated in Fig. 1.10 where the particle speed, u, is a function of the particle kinetic energy, K, and its mass or rest energy, mc2. The particle energy in Fig. 1.10 is expressed as K/mc2 to permit the reader to apply the curves for nonrelativistic and relativistic calculations to particles of different mass. For example, from the abscissa of Fig. 1.10, the values of K/mc2 for a 2-MeV beta particle is 2 MeV/0.51 MeV = 3.9 and that for a 2-Mev neutron is 2 MeV/939.5 MeV = 0.0021. From Fig. 1.10 we see that the nonrelativistic calculation of the speed of a 2-MeV beta particle would be erroneously extreme (well beyond the speed of light), while there would be only a small error in the relativistic calculation of the speed of the massive neutron of the same energy. 2. Sources of Neutrons The discovery of the neutron had eluded humanity until as late as 1932, because of the particle's neutral charge and high penetrating power when traveling through matter. In 1932 J. Chadwick provided evidence for the existence of the neutron. He placed a source of alpha particle-radiation in close proximity to beryllium. It was known that bombarding beryllium with alpha radiation would produce another source of radiation, which had a penetration power through matter even greater than that of gamma radiation. Chadwick observed that, when a sheet of paraffin (wax) was placed in the path of travel of this unknown radiation, he could detect a high degree of ionization in a gas ionization chamber caused by protons emitted from the paraffin. This phenomenon would not occur when other materials such as metals and even lead were placed in the path of this unknown radiation. On the basis of further measurements of the proton velocities and scattering intensities, it was concluded that the unknown radiation had a mass similar to that of the proton, but with a neutral charge. Only a particle with neutral charge would have a high penetration power through matter. As noted in the previous discussion of beta particle decay, the neutron is of mass similar to that of the proton and, within the nucleus of an atom, the particle is a close union between a proton and an electron. a. Alpha Particle-Induced Nuclear Reactions It is interesting to note that the method used by Chadwick to produce neutrons by alpha particle-induced reactions, described in the previous paragraph, remains an important method of producing a neutron source, particularly when a relatively small or easily transportable neutron source is required. The source may be prepared by compressing an alpha particle-emitting radioisotope substance with beryllium metal. The nuclear reaction, which occurs between the alpha particle and the beryllium nucleus, terminates with the emission of a neutron and the production of stable carbon as follows (1.58) Several alpha particle sources are used to produce neutrons via the preceding (α, n) reaction. Among these are the alpha emitters 241Am, 242Cm, 210Po, 239Pu, and 226Ra. The alpha radiation source selected may depend on its half-life as well as its gamma-ray emissions. As noted previously in this chapter, gamma radiation often accompanies alpha decay. The use of an alpha source, which also emits abundant gamma radiation, requires additional protection for the user against penetrating gamma rays. For example, Am–Be sources are preferred over the Ra–Be sources of neutrons used in soil moisture probes (Nielsen and Cassel, 1984; O'Leary and Incerti, 1993), because the latter have a higher output of gamma radiation and require more shielding for operator protection. The energies of the neutrons emitted from these sources will vary over the broad spectrum of 0 to 10 MeV. The average neutron energy of 5.5 MeV is shown in Eq. 1.58. The neutrons produced by these sources vary in energy as a consequence of several factors, including the sharing of the liberated energy between the neutron and 12C nucleus, the varying directions of emission of neutrons from the nucleus with consequent varying energies and velocities, and the variations in kinetic energies of the bombarding alpha particles. The neutron activities available from these sources increase up to a maximum as a function of the amounts of alpha emitter and beryllium target material used. For example, as explained by Bacon (1969), the Ra-Be source, prepared by mixing and compressing radium bromide with beryllium powder, increases steadily in neutron activity (neutrons per second) for each gram of radium used as the amount of beryllium is increased to about 10 g; but no significant increase in neutron output is achieved if more beryllium is used. The maximum neutron output achieved is approximately 2 × 107 neutrons per second per gram of radium. Because alpha decay from any alpha particle-emitting source occurs by means of random events, the production of neutrons by (α, n) reactions is also a random event. Therefore, these reactions can be referred to as “not time correlated.” This is contrary to the case of neutron sources provided through fission, discussed subsequently. b. Spontaneous Fission About 100 radionuclides are known to decay by spontaneous fission (SF) with the emission of neutrons (Karelin et al., 1997) as an alternative to another decay mode, such as alpha decay. Spontaneous fission involves the spontaneous noninduced splitting of the nucleus into two nuclides or fission fragments and the simultaneous emission of more than one neutron on the average. This phenomenon occurs with radionuclides of high mass number, A ≥ 230. The radionuclide 252Cf is a good example of a commercially available spontaneous fission neutron source. It decays with a half-life of 2.65 years primarily by alpha emission (96.91% probability); the remaining of the 252Cf decay processes occur by spontaneous fission with a probability of 3.09% (Martin et al., 2000, see also Appendix A). Decay of 252Cf by spontaneous fission produces an average number of 3.7 neutrons per fission. Because the sizes of the two fragments resulting from fission are not predictable, average sizes of the two fragments are determined. Consequently, the numbers of neutrons emitted from individual fissions are not the same; and an average number of neutrons produced per fission is determined. The fission rate of 252Cf is 6.2 × 105 SFs−1 μg−1 (Isotope Products Laboratories, 1995). The neutron emission from 252Cf in units of neutrons per second per unit mass is reported to be 2.314 × 106 s−1 μg−1 with a specific activity of 0.536 mCi μg−1 (Martin et al., 2000). If we know the radionuclide specific activity and the % probability of decay by spontaneous fission, we can calculate the fission rate. For example, taking the specific activity and % probability of spontaneous fission reported above for 252Cf, we can calculate the fission rate as the product of decay rate and probability of SF per decay or which is in close agreement with the value cited above. See Section VII.A for a discussion of radioactivity units and calculations. The variations in fission fragment sizes and number of neutrons emitted per fission provide variable neutron energies over the range 0–5.5 MeV with an average neutron energy from 252Cf of approximately 2.3 MeV. Small sources of 252Cf are commercially available for a wide range of applications such as prompt-gamma neutron activation analysis of coal, cement, minerals, detection of explosives and land mines, neutron radiography and cancer therapy. These sources are described by Martin et al. (1997, 2000) among which include 50-mg sources of 252Cf providing a neutron intensity > 1011 s−1 and measuring only 5 cm in length × 1 cm diameter. They report also larger sources of mass > 100 mg of 252Cf that approach reactor capabilities for neutrons. Another standard nuclide source of neutrons is 248Cm, which provides spontaneous fission intensity of only 4.12 × 104 s−1 mg−1 and decays with a half-life of 3.6 × 105 years (Radchenko et al., 2000). The lower neutron flux intensity of this source limits its application, although it has the advantage of a very long half-life providing invariability of sample intensity with time. Some radionuclides of interest in nuclear energy and safeguards also decay by spontaneous fission. The isotopes of plutonium of even mass number, namely 238Pu, 240Pu, and 242Pu, decay principally by alpha particle-emission but can also undergo spontaneous fission to a lesser extent at rates of 1100, 471, and 800 SF s−1 g−1, respectively. The average number of neutrons emitted per fission is between 2.16 and 2.26 of broad energy spectrum (Canberra Nuclear, 1996). Because the neutrons produced with each fission occurrence are emitted simultaneously, we can refer to these emissions as “time correlated.” Other isotopes of uranium and plutonium also undergo spontaneous fission but at a much lower rate. c. Neutron-Induced Fission When the naturally occurring isotope of uranium, 235U, is exposed to slow neutrons, it can absorb the neutron to form the unstable nuclide 236U (Eq. 1.71 in Section II.G.3.c). The newly formed nucleus may decay by alpha particle and gamma ray emission with the long half-life of 2.4 × 107 years. This occurs in approximately 14% of the cases when 235U absorbs a slow neutron. However, in the remaining 86% of the cases, the absorption of a slow neutron by 235U results in the production of the unstable 236U nuclide, which takes on the characteristics of an unstable oscillating droplet. This oscillating nuclear droplet with the opposing forces of two positively charged nuclides splits into two fragments, not necessarily of equal size, with the liberation of an average energy of 193.6 MeV. The general 235U fission reaction may be illustrated by (1.59) which represents the fission of one atom of 235U by one thermal neutron n to yield the release of fission products fp of varying masses plus an average yield of ν = 2.42 neutrons and an overall average release of energy E = 193.6 MeV (Koch, 1995). Most of this energy (over 160 MeV) appears in the form of kinetic energy of the two fission fragments. The remaining energy is shared among the neutrons emitted, with prompt gamma radiation accompanying fission and beta particles and gamma radiation from decaying fission fragments and neutrinos accompanying beta decay. When a sample of 235U is bombarded with slow neutrons, the fission fragments produced are rarely of equal mass. The 236U intermediate nuclide breaks into fragments in as many as 30 different possible ways, producing, therefore, 60 different nuclide fission fragments. In a review Koch (1995) provides a list of the fission fragments and their relative abundances as produced in a typical pressurized water reactor (PWR). The most common fission fragments have a mass difference in the ratio 3 : 2 (Bacon, 1969). On the average, 2.42 neutrons are emitted per 236U fission (Koch, 1995). Neutrons emitted from this fission process vary in energy over the range 0–10 MeV with an average neutron energy of 2 MeV and are classified as fast neutrons. Because more than one neutron is released per fission, a self-sustaining chain reaction is possible with the liberation of considerable energy, forming the basis for the nuclear reactor as a principal source of neutrons and energy. In the case of 235U, slow neutrons are required for neutron absorption and fission to occur. The nuclear reactor, therefore, will be equipped with a moderator such as heavy water (D2O) or graphite, which can reduce the energies of the fast neutrons via elastic scattering of the neutrons with atoms of low atomic weight. The protons of water also serve as a good moderator of fast neutrons, provided the neutrons lost via the capture process 1H(n, γ)2H can be compensated by the use of a suitable enrichment of the 235U in the nuclear reactor fuel (Byrne, 1994). The notation 1H(n, γ)2H is a form of abbreviating a nuclear reaction according to the format Target Nucleus(Projectile, Detected Particle)Product Nucleus. It can be read as follows: The target nucleus of the isotope 1H absorbs a neutron to form the product isotope 2H with the release of gamma radiation. The previously described fission of 235U represents the one and only fission of a naturally occurring radionuclide that can be induced by slow neutrons. The radionuclides 239Pu and 233U also undergo slow neutron-induced fission; however, these nuclides are man-made via the neutron irradiation and neutron absorption of 238U and 232Th as illustrated in the following (Murray, 1993). The preparation of 239Pu occurs by means of neutron absorption by 238U followed by beta decay as follows: (1.60) (1.61) (1.62) The preparation of 233U is carried out via neutron absorption of 232Th followed by beta decay according to the following: (1.63) (1.64) (1.65) Nuclides that undergo slow neutron-induced fission are referred to as fissile materials. Although 235U is the only naturally occurring fissile radionuclide, it stands to reason that if an excess of neutrons is produced in a thermal reactor, it would be possible to produce fissile 239Pu or 233U fuel in a reactor in excess of the fuel actually consumed in the reactor. This is referred to as “breeding” fissile material, and it forms the basis for the new generation of breeder reactors (Murray, 1993). Other heavy isotopes, such as 232Th, 238U, and 237Np, undergo fission but require bombardment by fast neutrons of at least l MeV energy to provide sufficient energy to the nucleus for fission to occur. These radionuclides are referred to as fissionable isotopes. d. Photoneutron (γ, n) Sources Many nuclides emit neutrons upon irradiation with gamma or x-radiation; however, most elements require high-energy electromagnetic radiation in the range 10–19 MeV. The gamma or x-ray energy threshold for the production of neutrons varies with target element. Deuterium and beryllium metal are two exceptions, as they can yield appreciable levels of neutron radiation when bombarded by gamma radiation in the energy range of only 1.7–2.7 MeV. The target material of D2O or beryllium metal is used to enclose a β−-emitting radionuclide, which also emits gamma rays. The gamma radiation bombards the targets deuterium and beryllium to produce neutrons according to the photonuclear reactions 2H(γ, n)1H and 9Be(γ, n)8Be, respectively. The photoneutron source 124Sb + Be serves as a good example of a relatively high-yielding combination of gamma emitter with beryllium target. The 124Sb gamma radiation of relevance in photoneutron production is emitted with an energy of 1.69 MeV at 50% abundance (i.e., one-half of the 124Sb radionuclides emit the 1.69-MeV gamma radiation with beta decay). A yield of 5.1 neutrons per 106 beta disintegrations per gram of target material has been reported (Byrne, 1994). The half-life (t1/2) of 124Sb is only 60.2 days, which limits the lifetime of the photoneutron generator; nevertheless, this isotope of antimony is easily prepared in the nuclear reactor by neutron irradiation of natural stable 123Sb. e. Accelerator Sources The accelerator utilizes electric and magnetic fields to accelerate beams of charged particles such as protons, electrons, and deuterons into target materials. Nuclear reactions are made possible when the charged particles have sufficient kinetic energy to react with target nuclei. Some of the reactions between the accelerated charged particles and target material can be used to generate neutrons. When electrons are accelerated, they gain kinetic energy as a function of the particle velocity. This kinetic energy is lost as bremsstrahlung electromagnetic radiation when the accelerated electrons strike the target material. Bremsstrahlung radiation is described in Section III.F of this chapter. It is the bremsstrahlung photons that interact with nuclei to produce neutrons according to the mechanisms described in the previous section under photoneutron (γ, n) sources. The accelerated electron-generated neutrons have been reported to yield in a uranium target as many as 10−2 neutrons per accelerated electron at an electron energy of 30 MeV with a total yield of 2 × 1013 neutrons per second (Byrne, 1994). The accelerator is a good neutron source for the potential generation of nuclear fuels. Accelerated deuterons can be used to produce high neutron yields when deuterium and tritium are used as target materials according to the reactions 2H(d, n)3He and 3H(d, n)4He, respectively. In the deuterium energy range 100–300 eV it is possible to obtain neutron yields of the order of 1010 neutrons per second from these (d, n) reactions (Byrne 1994) with relatively small electrostatic laboratory accelerators. Large accelerators can provide charged particle energies > 300 MeV capable of inducing neutron sources, such as accelerated proton-induced charge exchange reactions in 3H and 7Li target nuclei according to the reactions 3H(p, n)3He and 7Li(p, n)7Be as described by Byrne (1994). Practical implications of these neutron sources for the generation of nuclear fuels were noted in the previous paragraph. Murray (1993) pointed out that a yield of as many as 50 neutrons per single 500-MeV deuteron has been predicted and that this source of neutron could be used to produce new nuclear fuels via neutron capture by 238U and 232Th according to reactions 1.60–1.65 described previously. f. Nuclear Fusion The fusion of two atomic nuclei into one nucleus is not possible under standard temperature and pressure. This is because the repulsing coulombic forces between the positive charges of atomic nuclei prevent them from coming into the required close proximity of 10−15 m before they can coalesce into one. However, as described by Kudo (1995) in a review on nuclear fusion, if temperatures are raised to 100 million degrees, nuclei can become plasmas in which nuclei and electrons move independently at a speed of 1000 km s−1, thereby overcoming the repulsing forces between nuclei. Nuclear fusion reactors or controlled thermonuclear reactors (CTRs) are under development to achieve nuclear fusion as a practical energy source. The reactors are based on maintaining plasmas through magnetic or inertia confinement as described by Dolan (1982) and Kudo (1995). Some fusion reactions also produce neutrons. The energy liberated during nuclear fusion is derived from the fact that the mass of any nucleus is less than the sum of its component protons and neutrons. This is because protons and neutrons in a nucleus are bound together by strong attractive nuclear forces discussed previously in Section II.C.1. As described by Serway et al. (1997) this energy is referred to as the binding energy (BE), that is, the energy of work required to pull a bound system apart leaving its component parts free of attractive forces described by the equation (1.66) where M is mass of the bound nucleus, the mi's are the free component particle masses (e.g., protons and neutrons), and n is the number of component particles of the nucleus. From Eq. 1.66 we can see that if it is possible to overcome the repulsive forces of protons in nuclei and fuse these into a new nucleus or element of lower mass, energy will be liberated. Nuclear fusion reactions of two types emit neutrons, and these are of prime interest in man-made controlled thermonuclear reactors. The first type is fusion between deuterium and tritium nuclei according to (1.67) and the other type involves fusion between two deuterium nuclei according to either of the following equations, which have approximately equal probabilities of occurring (Kudo, 1995): (1.68) and (1.69) The fusion reaction between deuterium and tritium or D–T reaction (Eq. 1.67) gives rise to a 14.06-MeV neutron and a 3.52-MeV alpha particle. A D–T plasma burning experiment was performed with 0.2 g of tritium fuel with the Joint European Torus (JET) reactor in November 1991; and in December 1993 a higher power D–T experimental program with 20–30 g of tritium was continued on the Tokamak Fusion Test Reactor (TFTR). These are described by JET Team (1994), Strachan et al. (1994), Hawryluk et al. (1994), and Kudo (1995). The International Thermonuclear Experimental Reactor (ITER) project was set up under the auspices of the International Atomic Energy Agency (IAEA) to develop a prototype fusion reactor by the year 2030. Fusion energy production via a commercial reactor is assumed to start around the year 2050 (Sheffield, 2001). Under development are compact neutron sources, which utilize either D–D or D–T fusion reactions. One instrument described by Miley and Sved (1997) is the inertial electrostatic confinement (IEC) device, which accelerates deuteron ions producing fusion reactions as the ions react with a pure deuterium or deuterium–tritium plasma target. The device is compact measuring 12 cm in diameter and 1 m in length and provides a neutron flux of 106–107 2.5-MeV D–D n s−1 or 108–109 14-MeV D–T n s−1. Another similar device described by Tsybin (1997) utilizes laser irradiation to create a plasma in an ion source. Compact neutron sources of these types can become competitive with other neutron sources previously described such as 252Cf and accelerator solid-target sources, because of advantages including (i) on–off capability, (ii) longer lifetime without diminished neutron flux strength, and (iii) minimum handling of radioactivity. 3. Interactions of Neutrons with Matter If a neutron possesses kinetic energy it will travel through matter much more easily than other nuclear particles of similar energy, such as alpha particles, negatrons, positrons, protons, or electrons. In great contrast to other nuclear particles, which carry charge, the neutron, because it lacks charge, can pass through the otherwise impenetrable barrier of the atomic electrons and actually collide with nuclei of atoms and be scattered in the process or be captured by the nucleus of an atom. Collision of neutrons with nuclei can result in scattering of the neutrons and recoil nuclei with conservation of momentum (elastic scattering) or loss of kinetic energy of the neutron as gamma radiation (inelastic scattering). The capture of a neutron by a nucleus of an atom may result in the emission of other nuclear particles from the nucleus (nonelastic reactions) or the fragmentation of the nucleus into two (nuclear fission). A brief treatment of the various types of neutron interactions, which are based on their scattering or capture of neutrons by atomic nuclei, is provided next. a. Elastic Scattering The elastic scattering of a neutron by collision with an atomic nucleus is similar to that of a billiard ball colliding with another billiard ball. A portion of the kinetic energy of one particle is transferred to the other without loss of kinetic energy in the process. In other words, part of the kinetic energy of the neutron can be transferred to a nucleus via collision with the nucleus, and the sum of the kinetic energies of the scattered neutron and recoil nucleus will be equal to the original energy of the colliding neutron. This process of interaction of neutrons with matter results only in scattering of the neutron and recoil nucleus. It does not leave the recoil nucleus in an excited energy state. Elastic scattering is a common mechanism by which fast neutrons lose their energy when they interact with atomic nuclei of low atomic number, such as hydrogen (1H) in light water or paraffin, deuterium (2H) in heavy water, and 12C in graphite, which may be encountered in nuclear reactor moderators. It is easy to conceptualize what would occur when particles of equal or similar mass collide; the event would result in energy transfer and scattering without any other secondary effects, similar to what occurs in billiard ball collisions. Neutron scattering is the principal mechanism for the slowing of fast neutrons, particularly in media with low atomic number. Let us consider what occurs when a neutron collides with a nucleus and undergoes elastic scattering. Figure 1.11 illustrates the direction of travel of an incident neutron with given kinetic energy (dashed line). The neutron collides with the nucleus. The nucleus is illustrated as undergoing recoil at an angle β while the neutron is scattered at an angle a to the direction of travel of the incident neutron. The kinetic energy (Ek) lost by the neutron in this collision is defined by the equation (1.70) where M is the mass of the nucleus, mn is the mass of the neutron, and β is the recoil angle of the nucleus. A derivation of Eq. 1.70 is provided by Bacon (1969). Let us look at two extreme examples of elastic collisions between a neutron and a nucleus. In the first example, it is intuitively obvious from Eq. 1.70 that for a recoil angle β = 90°, cos2 β = 0 and consequently Ek = 0. Under such a circumstance, the neutron is undeflected by the nucleus and there is no energy transfer to the nucleus. The neutron continues along its path undeflected until it encounters another nucleus. For the second case, however, let us consider the other extreme in which the recoil angle, β = 0° where we have a head-on collision of the neutron with the nucleus of an atom. In this case the maximum possible energy of the neutron is imparted to the nucleus, where cos2 β = 1. For example, Table 1.5 provides the maximum fraction of the kinetic energy calculated according to Eq. 1.70 that a neutron can lose upon collision with various atomic nuclei. As illustrated in Table 1.5, the neutron can transfer more energy to the nuclei of atoms, which have a low mass; and the highest fraction of its energy can be transferred to the nucleus of the proton, which is almost equal in mass to the neutron. Nuclides of low mass number are, therefore, good moderators for the slowing down of fast neutrons. The substances often used are light water (H2O), heavy water (D2O), paraffin (CnH2n + 2), and graphite (C). TABLE 1.5. The Maximum Fraction of the Kinetic Energy (Ek) that a Neutron Can Lose Upon Collision with the Nucleus of Various Atoms Listed in Increasing Mass in Atomic Mass Units (u) \| Nuclide \| Nuclide Mass, M \| Neutro Mass, mn \| Ek=4M mn/(M+mn)2cos2β \| --- --- \| \| 1H \| 1.007825 \| l.008665 \| 4.065566/4.066232 = 0.999 or 100% \| \| 2H \| 2.014102 \| l.008665 \| 8.126217/9.137120 = 0.89 or 89% \| \| 9Be \| 9.012182 \| l.008665 \| 36.36109/100.41737 = 0.362 or 36.2% \| \| 12C \| 12.000000 \| l.008665 \| 48.41592/169.22536 = 0.286 or 28.6% \| \| 16O \| 15.994915 \| l.008665 \| 64.53404/289.12173 = 0.223 or 22.3% \| \| 28Si \| 27.976927 \| l.008665 \| 112.87570/840.16454 = 0.134 or 13.4% \| \| 55Mn \| 54.938047 \| l.008665 \| 22l.65633/3130.0329 = 0.071 or 7.1% \| \| 197Au \| 196.96654 \| l.008665 \| 787.86616/39194.175 = 0.020 or 2.0% \| b. Inelastic Scattering We may picture a fast neutron colliding with a nucleus. The neutron is scattered in another direction as described in the previous paragraph; however, part of the neutron's kinetic energy is lost to the recoil nucleus, leaving it in an excited metastable state. Inelastic scattering can occur when fast neutrons collide with nuclei of large atomic number. The recoil nucleus may lose this energy immediately as gamma radiation or remain for a period of time in the excited metastable state. In inelastic scattering, therefore, there is no conservation of momentum between the scattered neutron and recoil nucleus. Inelastic scattering occurs mainly with fast neutron collisions with nuclei of large atomic number. Neutron scattering is a common mechanism by which fast and intermediate neutrons are slowed down to the thermal neutron energy levels. Thermal neutrons have an energy level at which they are in thermal equilibrium with the surrounding atoms at room temperature. There is an energy range for thermal neutrons as described earlier in this chapter; however, the properties of thermal neutrons are often cited at an energy calculated to be the most probable thermal neutron energy of 0.0253 eV at 20°C corresponding to a velocity of 2200 ms−1 (Gibson and Piesch, 1985). Figure 1.9 may be used to find the velocity of the neutron at energy levels over the range 0.001–10 MeV. For example, if we select the position 0.025 eV on the X axis and follow up the graph with a straight line to the upper curve, we find the value 2200 ms−1. At the thermal energy state, the mechanisms of interaction of neutrons with matter change drastically as discussed in the following. c. Neutron Capture Because of the neutral charge on the neutron, it is relatively easy for slow neutrons in spite of their low kinetic energy to “find themselves” in the vicinity of the nucleus without having to hurdle the coulombic forces of atomic electrons. Once in close proximity to nuclei, it is easy for slow neutrons to enter into and be captured by nuclei to cause nuclear reactions. The capture of thermal neutrons, therefore, is possible with most radionuclides, and neutron capture is the main reaction of slow neutrons with matter. The power of a nucleus to capture a neutron depends on the type of nucleus as well as the neutron energy. The neutron absorption cross section, σ, with units of 10−24 cm2 or “barns,” is used to measure the power of nuclides to absorb neutrons. A more detailed treatment of the absorption cross section and its units and application are given in Section II.G.4 of this chapter. However, because capture of thermal neutrons is possible with most radionuclides, references will cite the neutron cross sections of the nuclides for comparative purposes at the thermal neutron energy of 0.0253 eV equivalent to a neutron velocity of 2200 m s−1. This is also the energy of the neutron, which is in thermal equilibrium with the surrounding atoms at room temperature. For comparative purposes, therefore, Table 1.6 lists the thermal neutron cross sections for neutron capture reactions in barns (10−24 cm2) for several nuclides. The nuclides selected for Table 1.6 show a broad range of power for thermal neutron capture. Some of the nuclides listed have practical applications, which are referred to in various sections of this book. TABLE 1.6. Cross Sections σ in Barns for Thermal Neutron Capture Reactions of Selected Nuclides in Order of Increasing Magnitude \| Nuclide \| σ (barns) \| --- \| \| \| <0.000006 \| \| \| 0.00052 \| \| \| 0.00019 \| \| \| 0.0035 \| \| \| 0.332 \| \| \| 1.8 \| \| \| 2.7 \| \| \| 7.4 \| \| \| 13.3 \| \| \| 530 \| \| \| 586 \| \| \| 752 \| \| \| 940 \| \| \| 3840 \| \| \| 5330 \| \| \| 39,000 \| \| \| 61,000 \| \| \| 254,000 \| Data from Holden (1997). The capture of a slow neutron by a nucleus results in a compound nucleus, which finds itself in an excited energy state corresponding to an energy slightly higher than the binding energy of the neutron in the new compound nucleus. This energy of excitation is generally emitted as gamma radiation. Neutron capture reactions of this type are denoted as (n, γ) reactions. Two practical examples of (n, γ) neutron capture reactions were provided earlier in this chapter in the neutron irradiation of 238U and 232Th for the preparation of fissile 239Pu and 233U (Eqs. 1.60 and 1.63), respectively. Another interesting example of a (n, γ) reaction is neutron capture by 235U according to (1.71) This neutron capture reaction is interesting, because the 236U product nuclide decays by alpha emission in approximately 14% of the cases and decays by nuclear fission with emission of neutrons in the remaining 86% of the cases as discussed previously in Section II.G.2.c. The subject of neutron capture is treated in more detail in Section II.G.4, which concerns the neutron cross section and neutron attenuation in matter. d. Nonelastic Reactions Neutron capture can occur in nuclei resulting in nuclear reactions that entail the emission of nuclear particles such as protons (n, p), deuterons (n, d), alpha particles (n, α and even neutrons (n, 2n). These reactions may not occur in any specific energy range but may be prevalent at specific resonances, which are energy states of the excited compound nuclei that are specific to relatively narrow energies of the incident neutron. The effect of resonance in neutron capture by nuclei is discussed in more detail subsequently in Section II.G.4. The (n, 2n) reactions occur at very high incident neutron energies, > 10 MeV (Gibson and Piesch, 1985). The (n, p) and (n, α) reactions can occur in the slow neutron capture and reaction with nuclides of low atomic number (low Z), where the Coulomb forces of the electron shells are limited and present less a hurdle for the escape of charged particles from the confines of the atom. Some practical examples of these reactions are the (n, p) reaction used in the synthesis of 14C by slow (thermal) neutron capture by 14N (1.72) and the (n, p) and (n, α) reactions used to detect neutrons by the interaction of slow neutrons with 3He and 10B, respectively, according to Eqs. 1.73 and 1.74. (1.73) (1.74) Either of these reactions is used to detect neutrons by using gas proportional detectors containing helium or a gaseous form of boron (e.g., boron trifluoride). Slow neutrons that penetrate these detectors produce either radioactive tritium (Eq. 1.73) or alpha particles (Eq. 1.74), which produce ionization in the gas. The ionization events or ion pairs formed can be collected and counted as described in Chapter 2 to determine a neutron count rate. e. Nuclear Fission The reaction of neutron-induced fission occurs when a neutron interacts with a fissile or fissionable nucleus and the nucleus becomes unstable, taking on the characteristics of an oscillating droplet, which then fragments into two nuclides (fission fragments). At the same time there is the release of more than one neutron (2.4 neutrons on the average for 235U fission) and a relatively high amount of energy (∼194 MeV). Fission in natural 235U and man-made 233U and 239Pu is optimal at thermal incident neutron energies; whereas fission in 238U and 232Th requires neutron energies of at least 1 MeV. A more detailed treatment of nuclear fission was provided previously in Section II.G.2.c. 4. Neutron Attenuation and Cross Sections As we have seen in our previous treatment of the neutron, there are several possible interactions of neutrons with nuclei. Among these are elastic scattering, inelastic scattering, neutron capture, nonelastic reactions, and nuclear fission. As we have seen in several examples, probabilities exist for any of these interactions to occur depending on the energy of the incident neutron and the type of nuclide with which the neutron interacts. We can define this probability of interaction by the term cross section, which is a measure of the capturing power of a particular material for neutrons of a particular energy. The range of neutrons in matter is a function of the neutron energy and the cross section or capturing power of the matter or medium through which the neutrons travel. To define cross section, let us consider an incident beam of neutrons of given intensity or number (I0), which impinges on a material of unit area (e.g., cm2) and thickness dx as illustrated in Fig. 1.12. The intensity (I) of the neutron beam traveling beyond the thickness dx will be reduced according to the number of nuclei (n) per unit volume in the material and the “area of obstruction” (e.g., cm2) that the nuclei present to the oncoming beam. This area of obstruction is referred to as the cross section of the material. On the basis of the description previously given, we can write the equation (1.75) which defines the change in beam intensity (dl) with respect to absorber thickness (dx) as proportional to the beam intensity (I) times a proportionality factor, which we may call the absorption coefficient or “obstruction coefficient” that the nuclei pose to the oncoming beam. The coefficient is a function of the number of nuclei (n) in the path of the neutron beam and the stopping power of the nuclei to interact with the neutron beam or, in other words, the neutron cross section (σ) of the material through which the neutron beam travels. Equation 1.75 may be written as (1.76) Equation 1.76 is very similar to Eq. 1.117 defining the attenuation of gamma radiation in matter with the exception that the absorption coefficients and attenuation coefficients involved for neutron and gamma radiation, respectively, are very different. The negative sign of Eqs. 1.75 and 1.76 denotes the diminishing intensity of the neutron beam as a function of absorption coefficient and absorber thickness. The absorption coefficient nσ is the combined effect of the number of nuclei (n) in the neutron beam path that might impede the continued travel of neutrons and the power of the nuclei to react with the neutrons. Equation 1.76 can be integrated over the limits of beam intensity from I0 to I and absorber thickness from 0 to x as follows: (1.77) to give the equation (1.78) or (1.79) which is the most simplified expression for the calculated beam intensity (I) after passing through an absorber of thickness (x) when the absorber material consists of only one pure nuclide and only one type of reaction between the neutron beam and nuclei is possible. If, however, several types of nuclei and reactions between the neutron beam and nuclei of the absorber material are possible, we must utilize the sum of the neutron cross sections for all reactions that could take place. We can use Eq. 1.78 to calculate the half-value thickness (x1/2) or the thickness of absorber material needed to reduce the incident neutron beam intensity by one-half. If we give the initial beam intensity (I0) a value of 1 and the transmitted intensity (I) a value of 1/2, we can write (1.80) and (1.81) or (1.82) The half-value thickness for neutron beam attenuation may be written (1.83) where nσ is the number of nuclei per unit volume (cm−3) and σ the neutron cross section in cm2. The neutron cross section σ can be defined as the area in cm2 for which the number of nuclei–neutron reactions taking place is equal to the product of the number of incident neutrons that would pass through the area and the number of target nuclei. The cross section is defined in units of 10−24 cm2 on the basis of the radius of atomic nuclei being about 10−12 cm. It provides a measure of the chances for the nuclei of a material being hit by a neutron of a certain energy. The unit of 10−24 cm2 for nuclear cross sections is called the barn. Tables in reference sources of nuclear data provide the neutron cross sections in units of barns for various nuclides and nuclide energies. An example is the reference directory produced by McLane et al. (1988), which provides neutron cross section values in barns and neutron cross section curves for most nuclides over the neutron energy range 0.01 eV to 200 MeV. Let us take an example of 10-eV neutrons incident on a water barrier (i.e., neutrons traveling in water). We may use Eq. 1.83 to estimate the half-value thickness, if we ignore the less significant interactions with oxygen atoms. This is because the neutron cross section for hydrogen at 10 eV is about 20 barns (Fig. 1.13) and that of oxygen is only 3.7 barns (McLane et al., 1988), and there are twice as many hydrogen atoms as oxygen atoms per given volume of water. The half-value thickness may be calculated as follows: The value of n for the number of hydrogen nuclei per cm3 of water may be calculated on the basis of Avogadro's number of molecules per mole. If 1 mole of water is equivalent to 18.0 g and the density of water is 1.0 g cm−3, we can calculate the number of hydrogen nuclei per cm3 as By definition, 20 barns is equal to 20 × 10−24 cm2 and the half-value thickness may then be calculated as If we make the calculation for 1-MeV neutrons traversing water and use the value 4.1 barns for the neutron cross section of hydrogen nuclei at this neutron energy (McLane et al., 1988), we calculate a half-value thickness of As the examples illustrate in the case of the proton, the neutron cross section (or barns) decreases as the energy or velocity of the neutron increases. That is, the neutron reactions with nuclei obey the general rule of having some proportionality to 1/ν, where ν is the velocity of the neutron. This inverse proportionality of cross section and neutron velocity is particularly pronounced in certain regions of energy as illustrated in the total neutron cross section curves for protons and elemental boron in Figs. 1.13 and 1.14, respectively. However, this is not always the case with many nuclides at certain neutron energies where there exists a resonance between the neutron energy and the nucleus. At specific or very narrow neutron energy ranges, certain nuclei have a high capacity for interaction with neutrons. The elevated neutron cross sections at specific neutron energies appear as sharp peaks in plots of neutron cross section versus energy, such as the cross section curve illustrated in Fig. 1.15 for . These peaks are called resonances and often occur with (n, γ) reactions. The high cross sections occur when the energy of the incident neutron corresponds exactly to the quantum state of the excited compound nucleus, which is the newly formed nucleus consisting of a compound between the incident neutron and the nucleus. Most nuclides display both the 1/ν dependence on neutron cross section and the resonance effects over the entire possible neutron energy spectrum. We should keep in mind that neutron cross sections can be specific and differ in value for certain reactions, such as proton (σρ)- and alpha particle (σα)-producing reactions, fission reactions (σf), or neutron capture cross sections (σc). The total neutron cross section (σtot or σa) would be the cross section representing the sum of all possible neutron reactions at that specific neutron energy. For example, the thermal neutron cross section for 235U, which is the neutron cross section at 0.0253 eV neutron energy corresponding to a neutron velocity of 2200 ms−1 at room temperature, can be given σc = 95 barns for the neutron capture cross section, σf = 586 barns for the fission cross section, and σα = 0.0001 barns for the neutron cross section for the alpha particle-producing reaction. These neutron cross section values indicate that neutron fission would predominate at the thermal neutron energy of 0.0253 eV, although some neutron absorption would also occur. The total neutron cross section, σtot, would be the total of the three possible reactions or = 95 barns + 586 barns + 0.0001 barns = 681 barns. In our treatment of slow neutron capture by 235U in Section II.G.3, illustrated by Eq. 1.71, we noted that about 14% of the slow neutron captures by 235U nuclei result in the formation of 236U and gamma radiation and the remaining 86% of the slow neutron captures result in nuclear fission. This is exactly what is predicted by the thermal neutron cross section values just provided; that is, for 235U and 5. Neutron Decay We have seen that fast neutrons may lose their energy through elastic and inelastic collisions with other nuclei, and if these neutrons do not undergo other reactions with nuclei (e.g., fission), they may lose sufficient energy to reach thermal equilibrium with surrounding atoms and possibly be captured by atomic nuclei. The question remains of what would happen to a free neutron that is not absorbed by any atomic nucleus. Earlier in this chapter (Section II.B) we discussed the transformation of the neutron within nuclei of radioactive atoms, which have a neutron/proton ratio too high for stability. In these unstable nuclides the neutron breaks up into a proton, negatron (negative electron), and antineutrino. However, within the confines of a stable nucleus, that is, one that does not have an n/p imbalance, there is no transformation of the neutron. If the neutron can transform itself in unstable nuclei, it stands to reason that the neutron might be unstable outside the protective boundaries of the stable nucleus. This is just the case, as A. H. Snell and L. C. Miller demonstrated in 1948 followed by further studies by Robson (1950a,b) and Snell et al. (1950) that when neutrons were in free flight in a vacuum, they would indeed decay with a lifetime in the range of 9–25 minutes with a release of 0.782 MeV of energy. More recent and accurate measurements of neutron decay demonstrate the lifetime to be 885.4 ± 0.9 s (Abele, 2000; Arzumanov et al., 2000; Pichlmaier et al., 2000; Snow et al., 2000). The decay of elementary particles is characterized in terms of lifetime. The lifetime, usually symbolized as τ, is related to the term half-life, t1/2, the mean time it takes for one-half of the particles to decay (Sundaresan, 2001) according to the relationship (1.84) The free neutron decays according to the scheme (1.85) The 0.782 MeV of energy released in the neutron decay corresponds to the difference in mass of the neutron (1.0086649 u) and the sum of the masses of the products of the neutron decay, the proton (1.0072765 u) plus the electron (0.0005485 u), or 1.0078250 u. Using Einstein's equation of equivalence of mass and energy (Section IV.C of this chapter), this mass difference of 0.0008399 u can be converted to the equivalent of 0.782 MeV of energy. This calculation provides additional evidence for the decay of the neutron into a proton and an electron. The neutron, therefore, outside the protective confines of a stable nucleus, has a very short lifetime. View chapterExplore book Read full chapter URL: Book2003, Handbook of Radioactivity Analysis (Second Edition)MICHAEL F. L'ANNUNZIATA Review article Method of characteristics – A review with applications to science and nuclear engineering computation 2015, Progress in Nuclear EnergyMatthew Eklund, ... Tatjana Jevremovic 3.1.2 Study case #5: neutron transport equation relevant to nuclear reactor physics and neutron transport Solving for neutron transport in an accurate and timely efficient way is paramount in nuclear reactor design and performance analysis. Neutron flux within a reactor core, combined with the geometry and material composition determine the amount of heat generated and the resultant thermal power output of the reactor. As described in Section 3.1.1, the AGENT code is based on a synergism of the MOC and the theory of R-functions to calculate neutron flux and reaction rates at different energies within a unit cell, an assembly or full core, in 2D or 3D. There are several methods to solve the neutron transport equation; the MOC provides an efficient, rapid solution by reducing the PDEs to ODEs. The PDE that predicts the behavior of neutrons in a system is the neutron transport equation which has roots in the Boltzmann transport equation (originally devised to study the behavior of gases). It takes into account the gain and loss of neutrons to a system. Usually the solution to neutron transport with MOC in complex reactor cores is divided into three levels of complexity. The unit cell (first level of complexity) is the smallest representative region in a nuclear reactor core. The second level of complexity is reactor assembly (composed of a number of unit cells arranged in a square or hexagonal array). And finally the third level of complexity is the full core. Solutions can be found in 2D or 3D. The most common approach to 3D full core model with MOC is to separate the solutions into 2D along assembly radial configurations, and couple this solution through neutron leakage in between them thus adding the third dimension to the solution. The following is a brief description of the solution as developed in the ANEMONA and AGENT codes, including the method of ray tracing (Jevremovic et al., 2001; Hursin et al., 2006). The ‘solution-rays,’ representing neutron trajectories, are positioned along many angular and polar directions to pass through the unit cell or reactor assembly. The location at which the ray enters the cell or assembly takes into account a neutron flux into the cell, while the location at which the ray exits the cell accounts for neutron leakage and outgoing neutron flux. The distance between rays is designated the ray separation. A smaller ray separation results in more rays therefore expected higher accuracy of the MOC solution. The locations at which a ray intersects with a zone boundary are calculated to determine weighted flux values at the boundary edges. The flux in a particular zone is calculated by taking the weighted average of flux values on each characteristic ray intersecting through the zone. The weighted values result in a scalar flux value. Cross sections for the unit cell or assembly are determined upfront for each material zone. Fig. 8 shows a square unit cell consisting of fuel, cladding, and moderator with rays separated at angle m, cell boundary divided into a set of edges where the average of the incoming and outgoing neutron flux are calculated amongst the rays that collect within the cell. The higher accuracy will require smaller zone size (meaning more zone subdivisions), more rays (meaning smaller ray separation), more angles (along azimuthal angles in the x−y plane and polar angles from the z axis) and finer energy group structure (meaning more energy groups). Following is a description of the application of MOC steady-state solution to 3D full core model. AGENT methodology is used as an example to illustrate the coupling of 2D radial to 1D axial solution in obtaining a 3D model of complex reactor cores based on MOC. Fig. 9 presents a visual scheme utilized by the AGENT code. First, the unit cell geometry and materials are specified and the cross sections are calculated accordingly from microscopic cross section libraries. An arrangement of multiple 2D unit cells (where each unit cell may be composed of different materials and/or geometries) is combined and extruded to create a 3D assembly. The AGENT code then simultaneously and synergistically performs calculations in the 2D radial and 1D axial direction using separate 2D and 1D MOC solvers, respectively. The former is prepared by separating the 3D assembly into separate 2D planes. The latter decomposes the 3D assembly into separate 1D pins. Each solver then employs a separate MOC scheme which is designed to determine outgoing neutron flux and transverse neutron leakage values. Each solver then receives and transmits the interfacing neutron leakage values one with another. A separate set of characteristic rays are used in the 2D and 3D MOC solvers. The 2D solver equations are developed followed by the equations for the 1D solver, beginning with the time-dependent neutron transport equation. The equation as taken from Duderstadt and Hamilton, 1976 is reproduced: (42) where: : t = time (s) : v = neutron velocity (cm/s) : Ψ = angular neutron flux (neutrons/cm2-s) : = angular direction vector of neutrons : Σ = total macroscopic cross section (cm−1) : = position vector (cm) : E = neutron energy (eV) : Q = neutron source term (neutrons/cm3-s). Examining the left-hand side of Eq. (42), the first term represents the change in neutrons within the system over a period of time. The second term represents the loss or gain of neutrons in the system as they exit or enter the system boundaries, respectively. The third term takes into account all interactions between neutrons and other particles within the system. The term on the right-hand side represents the neutron source, or the amount of neutrons produced within the system itself. While neutron flux will change during the startup and shutdown of a nuclear reactor, more often the steady-state conditions of the reactor are needed. This determines all partial differential terms in Eq. (42) to become zero and forces all terms to become time-independent as follows: (43) The interaction between neutrons and other nuclei are highly dependent on the neutron velocity or energy. Each term in Eq. (43) is dependent on neutron energy. Since a neutron's energy can be a single value within a continuum from 0 eV to over 1 MeV, the computational resources necessary to solve the neutron transport equation for every neutron's energy is very extensive. To further simplify this equation, thus reducing computation time and power requirements, neutron energies are separated into energy “bins”, or discrete energy distributions (as opposed to a continuous energy distribution). This results in what is known as the multi-group neutron transport equation: (44) where the g subscripts represent the specific energy group. The first term on the right-hand side replaces the continuous energy integral with a summation from the first energy group to the last energy group G. The neutron sources in the neutron transport equation are, namely, the fission and the scattering source. Fission is considered as an isotropic source while scattering may be approximated as an isotropic source. Therefore the overall source takes the form: (45) with the fission Fg and isotropic scattering Sg terms equal to: (46) (47) where, with respect to energy group g': : = scalar flux of group g' (cm2-s) : = isotropic scattering cross section from group g' to group g (cm−1) : ν = average number of neutrons per fission : = fission cross section of group g' (cm−1) : = fission spectrum for group g : keff = effective neutron multiplication factor. The radial solution is taken on 2D dimensional planes. The thicknesses of the planes are assumed to be infinite (termed the Infinite Thick Plane Approximation) and thus the multi-group 2D transport equation with leakage on the axial direction takes the form: (48) where : θ = polar angle : α = azimuthal angle : = neutron transverse leakage term (neutrons/cm3) and the 2D superscript indicates the value was calculated using the 2D solver. The MOC is applied in Eq. (48) with the characteristic rays to be straight lines in the direction of , reducing Eq. (48) to an ODE. Thus, the equation is discretized along the characteristic ray s: (49) whose solution is obtained by spatial and angular discretization. In order to achieve an accurate solution to neutron transport using the MOC, the region of interest must be subdivided into small zones. It is assumed that each subdivided zone has a constant scalar flux, cross sections, and material properties and is independent of the discretized polar and azimuthal angles of the characteristic rays. Thus, the source term will be constant and the cross sections only depend on the energy group and zone number i. Regarding spatial variables, Eq. (49) is discretized using the azimuthal and polar angles (denoted by subscripts az and pol, respectively), while there are numerous parallel rays (indexed as k) separated by a distance δA as shown in Fig. 8. The angular flux is calculated at characteristic ray line k, zone i, azimuthal angle α (denoted by subscript az), and polar angle θ (denoted by subscript pol). The intersection of the ray k having azimuthal angle α and polar angle θ with the zone i generates a segment, denoted as saz,pol,i,k. The projection of a segment to 2D is precalculated in the AGENT methodology by: (50) The radial neutron transport equation after using MOC for the ray k can be written as: (51) with (52) and : Ψg,az,pol,i,k = angular neutron flux of the gth energy group along the ray k that has azimuthal and polar angles α and θ, respectively : Qg,i = isotropic source term of energy group g and zone i (neutrons/cm3-s) : = transverse leakage term from axial solver of energy group g, azimuthal angle α, polar angle θ, and zone i (neutrons/cm3-s). Since there are two terms on the left-hand side of Eq. (51) dependent on angular neutron flux, Ψ, a simple separation of variables and integration is insufficient. Instead, an exponential factor is added to both sides, since the following mathematical equation applies: (53) where λ is a constant, y is dependent variable and x is an independent variable. Integrating this equation, it follows: (54) where x2 and x1 are the upper and lower bounds, respectively. This allows for a simple division to solve for dependent variable, y. Multiplying Eq. (51) with an integration factor on both sides: (55) (56) Integrating both sides, it follows: (57) (58) Solving for the flux at ray segment s′ gives: (59) The average scalar flux is obtained using a weighting function, ω: (60) where is the average angular neutron flux (units: neutrons/cm2-s). The equations for the 1D axial solver are now introduced. Xiao (2009) presented in detail the equations for the 1D axial MOC solver used in AGENT. Within this solver, it is assumed that the pin cells expand at an infinite radius and are homogeneous. This assumption is termed the Infinite Homogenized Pin-Cell Approximation. The equations for the cross sections used in each axial cell are as follows: 1D Transport cross section: (61) 1D Scattering cross section: (62) 1D Nu-fission cross section: (63) where: : Σg,i = transport cross section for group g and zone i (cm−1) : ϕg,i = scalar neutron flux of group g and zone i solved by the 2D radial solver : = scattering cross section from group g' to group g within zone i (cm−1) : Σf,g,i = fission cross section of group g within zone i (cm−1) : Ai = 2D area of zone i (cm2) : = summation for all zones within pin cell p : ν = number of neutrons released per fission. The direction vector for the characteristic rays, , is then discretized along the azimuthal and polar angles, α and θ, respectively. With this assumption, the 1D transport equation becomes (64) where: : = angular neutron flux of energy group g, azimuthal angle α, polar angle θ, and pin cell p (neutrons/cm2-s) : = isotropic source term of energy group g and pin cell p (neutrons/cm3-s) : = transverse leakage term from radial solver of energy group g, azimuthal angle α, polar angle θ, and pin cell p (neutrons/cm3-s). The isotropic source term is then (65) where: : χg,p = fission spectrum for group g and pin cell p and where the 1D superscript and p and g subscripts apply as previously. The solution of the 1D axial equations is now discussed. Due to the fact that the axial neutron leakage term, , is independent of azimuthal angle, α, the az subscript may be removed from this term only. Note that this only applies under the Infinite Homogenized Pin-cell Approximation, explained previously. The solution for the angular flux at location s on the characteristic ray k using the discretized 2D radial transport equation is (66) where: : Ψg,az,pol,i,k = angular flux of energy group g, azimuthal angle α, polar angle θ, zone i, and characteristic ray k (neutrons/cm2-s) : Σg,i = transport cross section of energy group g and zone i (cm−1). Solving for the average angular flux along segment length saz,pol,i,k with azimuthal angle α, polar angle θ, zone i, and characteristic ray k: (67) (68) (69) (70) This equation may be simplified by introducing the Greek symbol delta (Δ) for the change in angular neutron flux along the segment saz,pol,i,k: (71) which may be reduced to (72) In order to determine the average angular zone flux, the sum of the average angular neutron flux multiplied by the segment length saz,pol,i,k and ray separations δA (units: cm) is weighted by the sum of the latter two variables along all characteristic rays k: (73) The Infinite Thick Plane Approximation is reiterated: (74) Eq. (72) is now combined with Eq. (73): (75) (76) Now combining Eq. (74) with Eq. (76): (77) To simplify, the term (units: cm2), which represents the area of zone i, is replaced with the true area Ai: (78) The average scalar zone flux of group g and zone i, represented as ϕg,i, may be rewritten by employing the Flat Zone Approximation (where it is assumed that each zone has a constant scalar flux, cross sections and material properties): (79) (80) where the weights of azimuthal and polar angles are represented by ωaz and ωpol, respectively. These values are independent and are set to 1 in AGENT by default. The radial leakage term is now introduced. Using the Homogenized Radial Leakage Approximation, where it is assumed that radial leakage is homogeneous in all directions, the radial leakage term for group g and pin cell p in AGENT is calculated as (81) (82) where is the side-averaged angular flux at side i on the rectangular boundary edges (starting on the point nearest to the origin in the first quadrant of the x−y plane and rotating counter-clockwise to all 4 sides) which is used to calculate the axial neutron leakage term. The terms hx and hy are the dimension on the x and y axes, respectively, at which the neutron transport equation is integrated on the x−y plane. The solution of 1D MOC equations in 2D/1D coupling is now discussed. The methodology for solving the 1D MOC transport equation, Eq. (64), is similar to that done with the 2D MOC equation. Employing the previous assumptions (Infinite Homogenized Pin-cell Approximation and Homogenized Radial Leakage Approximation), Eq. (64) is independent of all azimuthal angles. Eq. (64) may, then, be solved along a polar direction: (83) The calculation for the average angular flux of segment spol,p for polar angle θ and pin cell p is: (84) where: (85) and the calculation for the scalar flux for pin cell p and group g: (86) (87) (88) where: : hz,p = length of pin cell p in z direction (cm). Note that Eq. (84) and Eq. (86) were developed and reduced in procedures similar to those used for Eq. (78) and Eq. (79), respectively. The axial neutron leakage, , calculated for group g, polar angle θ, and zone i is calculated as: (89) where: : , = angular flux at top and bottom of pin cell, respectively, solved by the 1D axial solver (neutrons/cm2-s). The zones i which share the same pin cell will also have the same leakage term. With every MOC solution, there is a need to optimize the solution in between these parameters (usually called the MOC resolution parameters) combined with the computational expense. When surveying at the unit cell and assembly level, the solution for multiplication factor and flux and reaction rates, starts to converge around so called ‘best estimate’; in other words, more refine resolution parameters beyond the set of those providing converged values would increase computational time and the solution will not change importantly when compared to the ‘best estimate’. Of note is that there is assumed significant heterogeneity in the radial direction as opposed to the axial direction (Xiao, 2009). If this assumption is valid for a particular system, AGENT's 2D/1D coupling method is extremely effective. Thus, there is a higher number of characteristic rays used in the radial directions, x and y, than in the axial direction z, as shown in Fig. 9. For a given ray separation, the number of characteristic rays in the x direction, represented as nx, and in the y direction, represented as ny, are in the order of n. These measurements of proportionality are represented as nx ∼ O(n) and ny ∼ O(n), respectively. Conversely, the number of characteristic rays in the axial or z direction is represented as nz and is on the order of n (nz ∼ O(n)) only if the level of heterogeneity in the reactor design requires it. Otherwise, if the radial direction is assumed to be of higher heterogeneity than the axial direction, the number of characteristic rays in the z direction may be one order lower than that for the x and y directions, i.e. nz ∼ 1. If such is the case, the total number of rays in 3D, nray3D, would be on the order of n2, i.e. nray3D ∼ O(n2). Using the coupled 2D/1D MOC solvers, the solution for the neutron flux in the 3D assembly is calculated. Since the extent of the computational resources required is proportional to the number of characteristic rays used in MOC, it is more effective to use a fewer number of characteristic rays for the 3D MOC than for the 2D MOC solver. In the future, when allocated computational resources become less of an issue, more characteristic rays may be used without hesitancy. View article Read full article URL: Journal2015, Progress in Nuclear EnergyMatthew Eklund, ... Tatjana Jevremovic Chapter Neutron Diffraction and (, )-Based Techniques for Cultural Heritage 2019, Nanotechnologies and Nanomaterials for Diagnostic, Conservation and Restoration of Cultural HeritageGiulia Festa, ... Roberto Senesi 4.1 Introduction 4.1.1 Neutrons Neutrons are electrically neutral subatomic particles, constituents of the nuclei whose physical properties are listed in Table 4.1. The weak decay of a free neutron (n) is described as , where p is the proton, e the electron, and is the electronic antineutrino, with a lifetime of about 15 min. Such a lifetime is long enough to enable the use of a neutron beam as a probe for the investigation of materials. Because of their characteristics, neutrons are an excellent nondestructive and noninvasive probe, a fundamental feature for the application of neutron techniques in cultural heritage. With the advent of nuclear reactors and spallation neutron sources, thermal and epithermal neutrons became an important probe of the structure and dynamic phenomena in condensed matter systems (Eisenbarth and Hille, 1977; Emeleus and Simpson, 1960). Table 4.1. Fundamental Properties of Neutrons \| Fundamental Properties \| \| --- \| \| Mass \| 1.675 × 10−27 kg \| \| Charge \| 0 \| \| Spin \| 1/2 \| \| Magnetic dipole moment \| μn = −1.913 μN = 6.0311 × 10−5 meV/Oersted \| Neutrons for matter investigation are mostly produced in large-scale facilities where they are generated through different processes: (1) nuclear fission, (2) photo-production, or (3) spallation. In the case of nuclear fission, neutrons are produced in the core of reactors employing U235, where neutron capture by fissile nucleus is the basis of the production process. After neutron capture, it induces the nucleus to break into two fragments emitting neutrons, 2–3 neutrons per fission event. In the photo-production process, high-energy photons hit a target with higher energy than the threshold energy of the (γ, n) reaction for the target nuclei, emitting neutrons. Such threshold ranges from 7 to 8 MeV for high Z materials (W, Pb, Fe) to 16–18 MeV for low Z elements (C, O). A photo-neutron energy spectrum is characterized by an evaporation peak in the low energy range and a relatively weak (10% of the integrated intensity) direct-reaction component in the several-MeV energy range with a resultant mean energy of (700 KeV–1 MeV) (Ongaro et al., 2000). In the spallation process, a high energy proton produced by an accelerator with a constant repetition rate (i.e., 50 Hz ISIS website) hits the nucleus of heavy atoms of the target such as tungsten (W), lead (Pb), or tantalum (Ta), producing an intranuclear cascade and leaving the target nuclei in an excited state (see Fig. 4.1). As an example, protons with energy E ∼ 1 GeV, hitting a tantalum target, will produce an average of 25 spallation neutrons for each incident proton. Neutrons produced by these processes have energies that span between a few millielectronvolts (meV) to several hundred Megaelectronvolts (MeV). The study of the matter properties through neutron beams is based on the interaction of the incident particle with the sample. The interaction probability depends on the investigation process such as scattering, neutron radiative absorption, and so on. In the case of scattering and (n, γ) processes, the probability of interaction is higher in the range of thermal-epithermal energy, meaning between the meV to the thousands of kiloelectronvolt (keV). For this reason, the energy of the neutrons produced using the aforementioned processes needs to be reduced before they can be used for matter investigation experiments: such reduction is obtained through a slowdown process. A slowdown process is achieved through multiple inelastic scattering in moderators made of hydrogenated materials, which make neutrons thermalize. In this process, neutrons enter the moderator and, after a number of collisions, their average energy will equal the thermal energy of the moderator nuclei. The moderator temperature will then determine the energy spectrum of the emitted neutrons at the moderator surface. Beams produced at reactors and spallation sources show different characteristics. Beams produced at spallation sources have a higher percentage of high-energy neutrons compared to the ones produced in reactors. Furthermore, while reactor beams are continuous with the only exception of Dubna, spallation neutrons can be produced in pulses. For neutron sources with neutron-flux time structure (as in the case of pulsed sources), the time each neutron takes to cover a certain distance is defined as time of flight (TOF) and is measured in the experiments. From the TOF, the neutron velocity, and therefore its wavelength, is obtained. Because thermalized neutrons obtained at spallation sources need a time interval typically in the order of microseconds (μs) to move along a path of an experimental beamline, their TOF can be determined with high precision. Neutrons interact with atoms through nuclear forces or magnetic-dipole of unpaired electrons. Nuclear forces have a very short range, in the order of picometers (10−12 m). Therefore, neutrons can travel long distances (from several mm to some cm) through heavy element materials, such as metals, without substantial attenuation. Moreover, elastic cross section is not directly dependent on the atomic number, unlike X-rays and electrons where it is proportional to the atomic number (Z). 4.1.2 Neutron Diffraction Neutron diffraction is a powerful technique based on the elastic scattering of neutrons. It is remarkably nondestructive and particularly suited for the investigation of the crystal structures and their relative amount in the samples under study. Thanks to the particle-wave dualism, neutrons can be represented using both formalisms. When neutron waves are scattered from the periodic structure of a crystal lattice, the scattered waves interfere with each other in a unique way, depending on the different crystal structure and size of the irradiated material (metal, ceramic, rock, etc.). In this way, the crystal structure of investigated objects is determined. In cultural heritage applications, the identification of crystalline structures present in a sample and their relative concentration, for example, help to discover the processes employed in ancient metals manufacturing or to determine material compositions and working conditions of key components in ceramics. Neutron diffraction is based on the elastic scattering, where the energy of incident neutrons is the same as the energy of the scattered neutrons. Diffraction is described through Bragg's law: nλ = 2dhklsin θ, where λ is the wavelength of the incident neutron beam, dhkl is the distance between lattice planes in a crystalline structure, n is a positive integer, and θ is the angle between incident beam direction and the crystalline plane. In the case of cultural heritage, samples are not single crystals but can be represented as poly-crystalline powders such as in metals, ceramic, rocks, and so forth. Neutron diffraction can be carried out at both reactors and spallation neutron sources. Diffractometers at spallation sources are primarily based on the TOF technique, by analyzing neutron energy of a white beam. The energy is: (4.1) where mnis the neutron's mass, v is neutron velocity, and L is the length of the flight path of the neutron beam from the moderator to the detector, passing through the sample and scattered at an angle 2θ. A diffraction process is described as the reflection of an incident beam by crystal planes (hkl). In the case of TOF measurements, Bragg's law can be rewritten as a relation between the TOF of neutrons scattered from a set of planes in the sample and the spacing between these planes, dhkl: (4.2) where θ0 is a fixed scattering angle. An example of diffraction pattern is shown in Fig. 4.2. Diffraction patterns show Bragg peaks: peak positions are directly related to the crystal lattice dimensions and geometry, while peak intensities are determined by the atomic arrangement in the unit cell and are related to the structure factor by: (4.3) where b is the neutron scattering length and depends on the atoms in the sample; (h, k, l) are Miller indexes of the lattice planes, and (x, y, z) define the atomic position in the unit cell. Diffraction patterns are related to structural information at different scales: phase structure describes compounds content; crystal structure is related to the atomic arrangement of each phase; grain structure represents size, shape, and orientation of grains (the latter is also referred as texture); and microstructure describes structural deviation from an ideal crystal within a grain. Quantitative phase analysis is based on the following assumptions: (1) each phase exhibits a unique set of Bragg diffraction peaks, (2) the intensity of the whole set of peaks in a phase is proportional to the phase weight-fraction, and (3) the measured diffraction pattern is the sum of all single-phase patterns. In quantitative phase analysis, the identification of the phases that are present in a sample is obtained by comparing experimental diffraction patterns with the peak sets of known compounds, reported in databases such as PDF, ICSD (ICSD, 2004), and CRYSTMET (CRYSTMET, 2002). Structure phase information (such as lattice parameters, space group, atomic position) is generally stored in crystallographic information file (CIF) format, the standard data exchange file format used in crystallography (Hall et al., 1991). After phase identification, experimental diffraction patterns are fitted through the Rietveld method obtaining quantitative multiphase analysis (Siano et al., 2002; Kockelmann and Kirfel, 2001). The core of such a procedure is a least-square fitting of the calculated pattern, which provides the phase composition and the adjustment of the structure parameters. For complex and multiphase objects, such as ceramics, for example, more than six or seven phases can be identified. Microstructural characterization and preferred orientation parameters through refinement of peak broadening and intensity deviations with respect to an isotropic powder are also obtained. Fitting procedures are performed through dedicated software such as GSAS (Larson and Von Dreele, 1994), MAUD (Ferrari and Lutterotti, 1994), and FULLPROF (Rodríguez-Carvajal, 2001). 4.1.3 Neutron Activation Analysis and Prompt Gamma Activation Analysis Prompt gamma activation analysis (PGAA) and neutron activation analysis (NAA) are neutron techniques based on the radiative capture of neutrons, a nuclear reaction, which takes place for each isotope of every element (except 4He). The so-called (n, γ)-reaction occurs when a nucleus absorbs a neutron and a new compound nucleus is formed with subsequent γ emission. The exciting energy is equal to the sum of the binding energy and the kinetic energy of the absorbed neutron. The decay of a compound nucleus takes place in about 10−16 s while the nucleus reaches its ground state in 10−9–10−12 s, emitting 2–4 prompt gammas in a cascade. A compound nucleus is generally unstable and decays emitting delayed gammas. Both prompt gammas and delayed gammas are characteristic of the emitting nuclide. The intensity of the gamma signal, moreover, is proportional to the number of atoms of a given nuclide present in the irradiated area and is used for quantitative analysis. In radiative capture of neutrons, both neutrons and gammas particles have a high penetration power in metals, rocks, ceramics, and so on, and information about the bulk is gained. The neutron beam typically illuminates the sample and the resulting measurements are given by an average composition of the illuminated area. NAA is based on the conversion of stable nuclei into radiative nuclei by neutron irradiation and the detection of emitted radiation during the unstable radioactive isotopes of new formation. NAA is used for the isotopic analysis of materials. In this nuclear method the sample is irradiated for a period ranging from 10 s to 3 min to several hours so as to activate nuclei, and gamma emissions are recorded after irradiation following radioactive decay (NA, 1974; Bode, 2017). Each radioactive nuclide is characterized by different decay constants and the type and energy of emitted radiation. Although the emitted radiation could be of a different nature, gamma rays are the best characteristics (penetration power, uniqueness for each isotope, and selectivity) to be detected for elemental and isotopic analysis. In fact, NAA is recognized as a reference method with high sensitivity, accuracy, and reliability for quantitative and qualitative analysis of major, minor, and trace elements present in the investigated samples. NAA is employed with success in several disciplines, including cultural heritage, archaeometry, surface geochemistry, and environmental studies. It has been used to investigate archaeological ceramics, stones from monumental architecture, historical monuments, and metal objects (Dias et al., 2007; Hamidatou et al., 2013). In PGAA, gamma detection is recorded during irradiation: a neutron beam illuminates the sample while gamma radiation is detected in general orthogonally to the neutron beam direction, through a germanium detector. The complexity of a prompt gamma spectrum depends on the nuclear level structure of the irradiated sample, in particular the number of levels between capture state and ground state. In light elements, such as 7Li or H, the number of levels below capture state is restricted to one or a few, and the spectrum shows just one or a few lines with high intensity that correspond to the direct transition. More complex spectra are expected for heavier elements because the number of levels involved during the decay of the compound nucleus increases (Molnar, 2004) (Fig. 4.3). PGAA is mostly applicable to elements with high neutron capture cross sections, elements that decay too rapidly to be measured by decay analysis and elements that produce only stable isotopes, while NAA is useful for the majority of elements that produce radioactive nuclides. 4.1.4 Neutron Resonance Capture Analysis The presence of resonance structures in neutron induced reaction cross sections is the basis of neutron resonance capture analysis (NRCA). An (n, γ) cross section presents peaks, called resonances, if the isotope is irradiated with neutrons in the epithermal energy range (Postma and Schillebeeckx, 2005, 2009). These resonances are due to the neutron absorption that can induce the transition of the nucleus to an excited state and the subsequent deexcitation to the ground state, which produces a prompt gamma cascade. In NRCA the energy of the absorbed neutron is obtained indirectly through the measurement from the TOF using a time-resolved gamma-sensitive detector. The energy of resonances indeed provides qualitative information about the irradiated isotopes, while the intensity of the observed resonances provides quantitative information about these isotopes and thus can be used to investigate the bulk composition of the irradiated object. An example of NRCA resonance spectrum as a function of TOF is shown in Fig. 4.4. Data analysis, in the case of cultural heritage samples with irregular and large shapes, is performed through an empirical approach. The net area of the peaks is recorded and, after a correction of the self-shielding due to neutron attenuation inside the sample, is considered as directly proportional with the areal density of the detected nuclide (Postma and Schillebeeckx, 2009). Following the approach described in Postma, the ratio of the nuclide's masses (indicated as WX and WY) is obtained from the ratio of the resonances net areas (Na and Nb) as: (4.4) where Fa and Fb are the corrections for the self-shielding and Ka,b is a calibration factor obtained through the measurements with standard samples of known composition. View chapterExplore book Read full chapter URL: Chapter Neutron Diffraction and (, )-Based Techniques for Cultural Heritage 2019, Nanotechnologies and Nanomaterials for Diagnostic, Conservation and Restoration of Cultural HeritageGiulia Festa, ... Roberto Senesi 4.1.1 Neutrons Neutrons are electrically neutral subatomic particles, constituents of the nuclei whose physical properties are listed in Table 4.1. The weak decay of a free neutron (n) is described as , where p is the proton, e the electron, and is the electronic antineutrino, with a lifetime of about 15 min. Such a lifetime is long enough to enable the use of a neutron beam as a probe for the investigation of materials. Because of their characteristics, neutrons are an excellent nondestructive and noninvasive probe, a fundamental feature for the application of neutron techniques in cultural heritage. With the advent of nuclear reactors and spallation neutron sources, thermal and epithermal neutrons became an important probe of the structure and dynamic phenomena in condensed matter systems (Eisenbarth and Hille, 1977; Emeleus and Simpson, 1960). Table 4.1. Fundamental Properties of Neutrons \| Fundamental Properties \| \| --- \| \| Mass \| 1.675 × 10−27 kg \| \| Charge \| 0 \| \| Spin \| 1/2 \| \| Magnetic dipole moment \| μn = −1.913 μN = 6.0311 × 10−5 meV/Oersted \| Neutrons for matter investigation are mostly produced in large-scale facilities where they are generated through different processes: (1) nuclear fission, (2) photo-production, or (3) spallation. In the case of nuclear fission, neutrons are produced in the core of reactors employing U235, where neutron capture by fissile nucleus is the basis of the production process. After neutron capture, it induces the nucleus to break into two fragments emitting neutrons, 2–3 neutrons per fission event. In the photo-production process, high-energy photons hit a target with higher energy than the threshold energy of the (γ, n) reaction for the target nuclei, emitting neutrons. Such threshold ranges from 7 to 8 MeV for high Z materials (W, Pb, Fe) to 16–18 MeV for low Z elements (C, O). A photo-neutron energy spectrum is characterized by an evaporation peak in the low energy range and a relatively weak (10% of the integrated intensity) direct-reaction component in the several-MeV energy range with a resultant mean energy of (700 KeV–1 MeV) (Ongaro et al., 2000). In the spallation process, a high energy proton produced by an accelerator with a constant repetition rate (i.e., 50 Hz ISIS website) hits the nucleus of heavy atoms of the target such as tungsten (W), lead (Pb), or tantalum (Ta), producing an intranuclear cascade and leaving the target nuclei in an excited state (see Fig. 4.1). As an example, protons with energy E ∼ 1 GeV, hitting a tantalum target, will produce an average of 25 spallation neutrons for each incident proton. Neutrons produced by these processes have energies that span between a few millielectronvolts (meV) to several hundred Megaelectronvolts (MeV). The study of the matter properties through neutron beams is based on the interaction of the incident particle with the sample. The interaction probability depends on the investigation process such as scattering, neutron radiative absorption, and so on. In the case of scattering and (n, γ) processes, the probability of interaction is higher in the range of thermal-epithermal energy, meaning between the meV to the thousands of kiloelectronvolt (keV). For this reason, the energy of the neutrons produced using the aforementioned processes needs to be reduced before they can be used for matter investigation experiments: such reduction is obtained through a slowdown process. A slowdown process is achieved through multiple inelastic scattering in moderators made of hydrogenated materials, which make neutrons thermalize. In this process, neutrons enter the moderator and, after a number of collisions, their average energy will equal the thermal energy of the moderator nuclei. The moderator temperature will then determine the energy spectrum of the emitted neutrons at the moderator surface. Beams produced at reactors and spallation sources show different characteristics. Beams produced at spallation sources have a higher percentage of high-energy neutrons compared to the ones produced in reactors. Furthermore, while reactor beams are continuous with the only exception of Dubna, spallation neutrons can be produced in pulses. For neutron sources with neutron-flux time structure (as in the case of pulsed sources), the time each neutron takes to cover a certain distance is defined as time of flight (TOF) and is measured in the experiments. From the TOF, the neutron velocity, and therefore its wavelength, is obtained. Because thermalized neutrons obtained at spallation sources need a time interval typically in the order of microseconds (μs) to move along a path of an experimental beamline, their TOF can be determined with high precision. Neutrons interact with atoms through nuclear forces or magnetic-dipole of unpaired electrons. Nuclear forces have a very short range, in the order of picometers (10−12 m). Therefore, neutrons can travel long distances (from several mm to some cm) through heavy element materials, such as metals, without substantial attenuation. Moreover, elastic cross section is not directly dependent on the atomic number, unlike X-rays and electrons where it is proportional to the atomic number (Z). View chapterExplore book Read full chapter URL: Book2019, Nanotechnologies and Nanomaterials for Diagnostic, Conservation and Restoration of Cultural HeritageGiulia Festa, ... Roberto Senesi Chapter Hardmetals 1.13.2.2 Use of Neutrons Neutrons enable study of the volumetric residual thermal microstresses in cemented carbides due to their much greater penetration power in most engineering materials because they are uncharged particles. Two types of neutron sources are now utilized. Reactor sources produce beams via nuclear fission from which fixed wavelengths are extracted with monochromators. The use of position-sensitive linear or area detectors enables whole peaks to be collected without scanning, increasing throughput. Reactor spectrometers are similar in concept to X-ray instruments. Pulsed neutron sources, also called spallation sources, produce neutrons by the impact of accelerated particles on a heavy-metal target, rather than by fission. The resultant neutron beam has a range of velocity (energy) and, hence, wavelength. In this case, the diffraction angle is fixed and there is a range of wavelengths in the incident beam, typically from 0.05 to 0.50 nm. The energies are resolved using time-of-flight methods. This enables whole diffraction patterns to be recorded simultaneously, without the requirement of sample motion. The neutron energy, velocity and wavelength are related by (8) where E is the kinetic energy of the neutron, v is the neutron velocity, m is the neutron mass, h is the Planck's constant and λ is the wavelength (Bacon, 1975). The relation between time-of-flight t, distance traveled L, and wavelength λ is given by (9) More and more neutron measurements are being done at pulsed sources. Advantages include (1) the ability to record whole patterns from all phases simultaneously with no sample or detector movement; (2) the ability to measure in two orthogonal directions simultaneously, as shown in Figure 3; (3) superior data for low-symmetry structures; and (4) greater ease of use for mechanical/temperature stages. A listing of both types of neutron facilities is given by Krawitz (2011). A few words about the flux (number of neutrons per unit area per unit time at a specified distance) of sources are in order, as the low flux in neutron beams relative even to laboratory X-rays is often cited as a limitation of the method. While it is true that incident beam flux is low relative to laboratory and, especially, synchrotron X-rays, this argument does not account for the "effective flux" represented by the measurement of whole diffraction patterns of all crystalline phases, and in two orthogonal directions simultaneously. These features represent a significant gain in effective flux over the traditional peak-scanning mode of recording used for many years on neutron powder instruments. Also, detector efficiency has improved. The performance of hundreds of measurements on engineering materials is testament to the practical utility of the method. Thus, the answer to the question "Can useful measurements be made in reasonable times?" is yes, within limits that allow many aspects to be studied-aspects that cannot be addressed as effectively in other ways. Neutrons measure in three dimensions, and therefore fundamentally differ from traditional X-ray stress measurements. A major implication of the three-dimensional character of neutron strain measurements is that stress-free interplanar spacings are required to convert measured d-spacings to strains. Useful discussions of stress-free reference values in neutron strain measurements can be found in the literature (Tanaka, Akinawa, & Hayashi, 2002; Withers, Preuss, Steuwer, & Pang, 2007). For WC-based cemented carbides, the procedure is to use loose WC powder as the stress-free reference material and to directly determine strain in WC. The powder should be contained in a can having a shape as close as possible to that of the samples. WC is an ideal standard material because it is stoichiometric and does not take binder elements into solution during sintering. To obtain thermal residual stress values, strain is converted to stress using (10) where pWC is the average diffraction hydrostatic thermal stress of WC, KWC is the bulk modulus, and is the strain measured in one direction by use of Eqn (10). The stress in the binder phase is obtained using Eqn (3). If the strain is not the same in all directions, then a stress state must be assigned that matches the observed data. Examples can be found in Krawitz (2001). A general diffraction text that discusses diffraction stress measurements (Krawitz, 2001) and more specialized neutron (Hutchings et al., 2005; Kisl & Howard, 2008), and neutron and synchrotron stress measurements (Reimers, Pyzalla, Schreyer, & Clemens, 2008) is also available. A standard source book on residual stresses and their measurement by diffraction is by Noyan and Cohen (1987). View chapterExplore book Read full chapter URL:
15974	https://blog.csdn.net/qq_33060405/article/details/144655150	直线与圆的关系_圆跟直线中x+y为什么一定要与圆相切-CSDN博客博客下载学习社区 GitCode InsCodeAI 会议搜索 AI 搜索登录登录后您可以：复制代码和一键运行与博主大V深度互动解锁海量精选资源获取前沿技术资讯立即登录会员·新人礼包消息历史创作中心创作直线与圆的关系原创已于 2024-12-23 01:07:22 修改·559 阅读 · 3 · 0· CC 4.0 BY-SA版权版权声明：本文为博主原创文章，遵循CC 4.0 BY-SA版权协议，转载请附上原文出处链接和本声明。文章标签： #解析几何于 2024-12-22 23:56:42 首次发布解析几何专栏收录该内容 22 篇文章订阅专栏直线与圆的关系是几何学中的一个重要主题，涉及到直线与圆的相交、切线和不相交等情况。以下是对直线与圆关系的详细分析： 1. 圆的方程 3. 直线与圆的关系直线与圆的关系可以分为以下几种情况： 1. 相交直线与圆相交的情况是指直线与圆有两个交点。要判断直线与圆是否相交，可以将直线方程代入圆的方程，得到一个关于 (x) 或 (y) 的二次方程。 2. 切线直线与圆相切的情况是指直线与圆有一个交点。此时，直线是圆的切线。条件：如上所述，判别式 (D = 0) 表示直线与圆相切。 3. 不相交直线与圆不相交的情况是指直线与圆没有交点。条件：如上所述，判别式 (D < 0) 表示直线与圆不相交。 4. 切线的方程如果已知圆心 ((h, k)) 和圆的半径 (r)，以及直线的斜率 (m)，可以求出切线的方程。切线的斜率与圆心到切点的连线的斜率是负倒数关系。切线方程：如果直线与圆相切，切点 ((x_0, y_0)) 满足圆的方程，并且可以通过切线的斜率来求得。 5. 应用直线与圆的关系在许多领域都有应用，包括：计算机图形学：用于碰撞检测和图形渲染。物理学：用于分析运动轨迹和力的作用。工程学：用于设计和分析结构。总结直线与圆的关系是几何学中的基本概念，通过分析直线与圆的交点，可以判断它们的相对位置关系。这些关系在数学、物理和工程等领域都有广泛的应用。直线与圆的关系在游戏中的应用案例说明直线与圆的关系在游戏开发中有许多实际应用，尤其是在碰撞检测、路径规划、物理模拟和图形渲染等方面。以下是一些具体的应用案例说明： 1. 碰撞检测在游戏中，碰撞检测是确保物体之间相互作用的关键。直线与圆的关系可以用于检测角色与圆形物体（如球体、敌人、道具等）之间的碰撞。案例： 2. 射击游戏中的弹道计算在射击游戏中，子弹的轨迹通常可以用直线表示，而目标（如敌人）可能是圆形的。通过计算直线与圆的交点，可以判断子弹是否击中目标。案例：子弹与敌人：当玩家发射子弹时，可以将子弹的路径表示为一条直线。通过将直线方程代入敌人的圆形方程，可以判断子弹是否击中敌人。如果计算得到的判别式 (D) 大于或等于零，则表示子弹击中敌人。 3. 路径规划在游戏中，角色的移动路径可能会受到圆形障碍物的影响。通过分析直线与圆的关系，可以实现角色的避障行为。案例：角色避障：当角色在游戏中移动时，如果检测到前方有圆形障碍物，可以通过计算角色的移动路径（直线）与障碍物（圆）的关系，决定角色的移动方向。如果直线与圆相交，角色可以选择绕过障碍物，改变移动路径。 4. 物理模拟在物理引擎中，直线与圆的关系可以用于模拟物体的运动和相互作用。案例：弹跳效果：在一个包含圆形物体（如球）的游戏中，当球与地面（视为一条直线）碰撞时，可以通过计算碰撞点和法线方向来实现弹跳效果。通过计算碰撞点的法线方向，可以确定球的反弹方向，从而实现真实的物理效果。 5. 角色控制与移动在角色控制中，直线与圆的关系可以用于实现平滑的运动和转向。案例：角色转向：在一个角色扮演游戏中，角色可能需要朝向某个目标（如敌人或道具）。可以通过计算角色当前位置与目标位置之间的直线，判断角色的转向。如果目标在角色的圆形视野范围内（视为一个圆），则角色可以转向目标并移动。总结直线与圆的关系在游戏开发中具有广泛的应用，涉及到碰撞检测、射击计算、路径规划、物理模拟和角色控制等多个方面。通过利用这些几何关系，开发者能够创建更加真实和互动的游戏体验。直线与圆的关系在游戏开发中应用碰撞检测的c#代码实现在游戏开发中，使用 C# 实现直线与圆的碰撞检测是一个常见的任务。下面是一个简单的示例，展示如何在 Unity 引擎中实现这一功能。我们将创建一个方法来检测一条直线（由两个点定义）是否与一个圆（由圆心和半径定义）相交。 C# 代码示例 ```csharp using UnityEngine; public class CollisionDetection : MonoBehaviour { // 圆的中心和半径 public Vector2 circleCenter; public float circleRadius; // 直线的两个端点 public Vector2 lineStart; public Vector2 lineEnd; void Update() { // 检测碰撞 if (IsLineIntersectingCircle(lineStart, lineEnd, circleCenter, circleRadius)) { Debug.Log("直线与圆相交！"); } else { Debug.Log("直线与圆不相交。"); } } // 检测直线与圆的碰撞 bool IsLineIntersectingCircle(Vector2 lineStart, Vector2 lineEnd, Vector2 circleCenter, float circleRadius) { // 计算直线的方向向量 Vector2 lineDir = lineEnd - lineStart; Vector2 lineToCircle = circleCenter - lineStart; // 计算直线的长度的平方 float lineLengthSquared = lineDir.sqrMagnitude; // 计算投影系数 float t = Vector2.Dot(lineToCircle, lineDir) / lineLengthSquared; // 限制 t 的范围在 [0, 1] 之间 t = Mathf.Clamp01(t); // 计算最近点 Vector2 closestPoint = lineStart + t lineDir; // 计算最近点到圆心的距离 float distanceToCircle = Vector2.Distance(closestPoint, circleCenter); // 检查距离是否小于等于圆的半径 return distanceToCircle <= circleRadius; } } ``` AI写代码 csharp 运行 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 代码说明变量定义： circleCenter：圆心的坐标。 circleRadius：圆的半径。 lineStart 和 lineEnd：直线的两个端点。 Update 方法：在每一帧中调用 IsLineIntersectingCircle 方法来检测直线与圆的碰撞，并输出结果。 IsLineIntersectingCircle 方法：计算直线的方向向量 lineDir。计算从直线起点到圆心的向量 lineToCircle。计算直线的长度的平方 lineLengthSquared。计算投影系数 t，并将其限制在 [0, 1] 之间，以确保最近点在直线段上。计算最近点 closestPoint。计算最近点到圆心的距离 distanceToCircle。如果距离小于等于圆的半径，则返回 true，表示直线与圆相交。使用方法将此脚本附加到 Unity 场景中的一个 GameObject。在 Inspector 面板中设置 circleCenter、circleRadius、lineStart 和 lineEnd 的值。运行游戏，查看控制台输出，判断直线与圆的关系。总结这个示例展示了如何在 Unity 中使用 C# 实现直线与圆的碰撞检测。通过这种方法，开发者可以在游戏中实现各种碰撞检测功能，增强游戏的互动性和真实感。如果你有其他问题或需要更复杂的实现，欢迎进一步讨论！关注博主即可阅读全文确定要放弃本次机会？福利倒计时 : : 立减 ¥ 普通VIP年卡可用立即使用你一身傲骨怎能输关注关注 3点赞踩 0 收藏觉得还不错? 一键收藏 0评论分享复制链接分享到 QQ 分享到新浪微博扫一扫打赏打赏打赏举报举报专栏目录计算几何——直线、线段和圆的交点 Meloor的博客 05-13 5600 汇总篇：计算几何汇总首先，求出圆心O在直线 AB上的投影点E 然后求出AB的方向向量e=AB/\|AB\| 令EC=ED=base base = sqrt(r^2-OE^2) EC = -base e ED = basee C=E + EC D=E + ED 相切时求得的C=D,CDE重合如果A为切点，那么ACDE重合 #include<cm... [计算几何] (二维)圆与直线的交点 looeyWei的博客 09-26 4708 给出圆心O的坐标, 和半径r, 再给出点A,B的坐标构成直线 AB, 求出圆与直线 AB交点的坐标如下图 Step1: 首先求出圆心c在直线 l 上的投影点pr的坐标可通过求解向量p1pr(p1pr的长度 p1p2的单位向量) Step2: 计算向量p1p2的单位向量e, 再勾股定理求出base的长度, 进而求出向量base Step3: 最后,以pr作为起点, 向正or负... 参与评论您还未登录，请先登录后发表或查看评论 js过圆外一点的直线与圆相切的切点坐标计算 9-17 并且因为点P0是圆上的点,=> (x0- a)2 + (y0- b)2= R2 解方程就可以求出P0(x0,y0)坐标,当然这样求法太复杂,我没有耐心解下去了,我们换个思维求我们把圆换成这样,相当于把整个系统做了一次相对位移,移动到0点 x2 + y2= R2 那么经过点P0(x0,y0)这个圆的的切线方程就是 x0x1 + y0y1= R2 并且x20 + y20= R 直线和曲线相切,曲线和曲线相切 9-24 函数y=x + my=x + m的图像是动态的直线, 在同一个坐标系中做出两个函数的图像,由图像可知, 由图可知,直线和椭圆相交的一个位置是过点(2,0)(2,0),代入求得m=−2m=−2; 另一个相交的临界位置是直线和函数y=f(x)=3−34x2−−−−−−−√y=f(x)=3−34x2在第二象限的部分相切,... 求圆和直线之间的交点 u013013797的博客 12-03 1万+ 求圆和直线之间的交点 / 求圆和直线之间的交点直线方程：y = kx + b 圆的方程：(x - m)² + (x - n)² = r² x1, y1 = 线坐标1, x2, y2 = 线坐标2, m, n = 圆坐标, r = 半径 / public getInsertPointBetweenCircleAnd... 物理引擎探究(7)---第一次碰撞 ARTELE的博客 01-31 598 0.简介碰撞是比较难写的部分，但是也不难写，掌握了规律就好办不少，这次实现一个最简单的碰撞判断。 1.圆圈与直线的碰撞上一次实现了直线，严谨来说是线段，圆圈与一条线段的碰撞，方法有很多，不要拘泥于我接下来要介绍的一种方法。首先我利用圆圈的圆心，将圆心O投影到线段所在直线上作为点C，线段的端点是A和B，如果C在AB之间，则向量CA 与 CB的夹角应该是180度，此时计算圆心到C点距离，如果... ACIS内核和parasolid内核的来龙去脉与比较 9-23 以后在使用中发现函数样条曲线存在一些问题,例如同样一组型值点,如果对型值点的坐标系施加不同的旋转变换,拟合出来的曲线形状会有所不同;又如用它拟合螺旋线,它不能处理同一个(x, y)值的点对应有多个不同z值解的情况。因此后来改用参数形式的曲线表达式,例如从P0到P1点作一条直线,写成参数形式的矢量方程是 ... 圆的相切相交相离公式及圆与圆的位置关系【同步练习】 9-23 ⊙C2:(x﹣2)2 +(y + 1)2=9, 两圆圆心距为d= =5, r1=6, r2=3, r1﹣r2<d<r1 + r2,所以两圆相交. 故选:C. 6、若圆 C1:(x﹣1)2 + y2=1 与圆 C2:x2 + y2﹣8x + 8y + m=0 相切,则m等于( ) A、16B、7C、﹣4或16D、7或16 正确答案 C 直线与圆、二次曲线交点计算 pipi_xia的博客 10-27 708 此时，通过对二次根判别式进行求解，当然如果只是要做根判别计算的话，可以免去平方根计算，提高计算效率。若等于0，则有一个交点；若大于0，则有2个交点。同理1.2 我们可以在考虑定以域后自行去推理。在我们求出t值后，需要做t>0判断即可。几何求交（二）：直线和圆的交点热门推荐主要研究图形学相关领域 06-08 1万+ 几何求交：直线和圆的交点一次函数的斜率公式_初中数学知识点公式和规律速记口诀!为了孩子成绩... 9-21 对角互补记心间,外角等于内对角,四边形定内接圆;直角相对或共弦,试试加个辅助圆;若是证题打转转,四点共圆可解难;要想证明圆切线,垂直半径过外端,直线与圆有共点,证垂直来半径连,直线与圆未给点,需证半径作垂线;四边形有内切圆,对边和等是条件;如果遇到圆与圆,弄清位置很关键,两圆相切作公切,两圆... 探索性思维——How to Solve It 9-6 现在,这三个中面和△ABC 所在平面交于该三角形的三个中线。这三条中线交于一点(这里利用了前面较简单的类比问题的结果),而这点和D 点一样,也是三中面的公共点。连结这两个公共点的直线就是这三个中面的公共线。我们证明了六个中面中通过顶点D 的三个中面有一条公共直线。对于通过顶点A/B/C 的三个中... 高一数学直线与圆关系习题课PPT学习教案.pptx 10-03 高一数学直线与圆关系习题课PPT学习教案.pptx 2 直线与圆的位置关系——小学生学习课件 11-23 ### 2 直线与圆的位置关系——小学生学习课件 #### 知识点解析 ##### 一、点与圆的位置关系点与圆的位置关系可以通过点到圆心的距离与圆的半径之间的比较来确定： 1. 点在圆外：当点到圆心的距离大于圆的半径... 简化为“圆弧-直线”的侧方停车路径_泊车位姿用直线圆弧连接 9-28 如下图所示,采用“圆弧-直线”的单步平行泊车路径为直线 C0C1、圆弧C1C2、直线 C2C3、圆弧C3C4和直线 C4C5,当泊车空间较小时,直线 C0C1、C2C3、C4C5的长度可以缩短为0,此时路径由两段圆弧相切组成。 (图片来源于参考文献) 2)两步平行泊车当泊车空间较小时,无法实现单步平行泊车,需采用两步泊车方式,如下图所示,车... 数学的意义与数学教育的价值_数学的教育价值具体实例 9-23 至于函数y=f(x),则是更进一步的抽象.在几何中的点、直线、圆、平面同样是对现实世界中事物的抽象,同样是人们为描述现实生活中某些事物而创造的一种语言.比如,在世界地图上,北京可以看成一个点,而在中国地图中,天安门可以看成一点.因此,数学中的“点”实际上就是我们所考察的事物位置的抽象,它没有大小,没有... 部编4 第4讲直线与圆、圆与圆的位置关系新题培优练.doc 09-09 这些题目涉及的是平面几何中的直线与圆以及圆与圆的位置关系，主要考察了直线与圆的相交情况、圆的方程、点到直线的距离、圆的几何性质以及圆与圆之间的位置关系（如外切、相交）等问题。下面对这些知识点进行详细... 公开课初中数学直线与圆的位置关系 PPT课件.pptx 10-07 《初中数学：直线与圆的位置关系》在初中数学的学习中，直线与圆的位置关系是一个重要的知识点，它涉及几何的基本概念和性质，是理解和解决相关问题的基础。本篇内容将详细解析这一主题。首先，我们要了解点与圆... WPF入门到跪下第六章图形-绘制_wpf strokedasharray 9-26 MX,Y:M指令用于坐标移动,也就是指定绘制的起始点。(将所在容器的左上角作为零点来计算结束点) mX,Y:m指令用于坐标移动,也就是指定绘制的起始点。(将上一个结束点作为零点来计算结束点) 直线指令 LX,Y:L表示绘制直线,指定绘制的结束点(将所在容器的左上角作为零点来计算结束点)。 lX,Y:l表示绘制直线,指定... 新人教A数学必修二直线与圆的位置关系 PPT学习教案.pptx 10-05 《新人教A数学必修二直线与圆的位置关系》的学习教案深入探讨了高中数学中的重要概念，即直线与圆之间的关系。这部分内容是高中数学几何学的重要组成部分，对于理解和解决相关问题至关重要。首先，直线与圆的位置... [算法]直线与圆的交点程序设计图解AI 01-15 1万+ / Created by apple on 2017/1/15. / //求直线与圆的交点 /函数参数说明: cx:圆 X轴坐标 cy:圆 y轴坐标 r:圆半径 stx:起点直线的X轴坐标 sty:起点直线的轴坐标 edx:终点直线的X轴坐标 edy:终点直线的Y轴坐标返回值:交点坐标(x,y) / function getPoint(cx,cy,r 计算几何之求圆与直线的交点龙进的博客 02-17 4085 求圆与直线的交点的方法是：求圆心c在直线 l上的投影点pr 求出直线 l上的单位向量e 根据r和pr的长度来计算出圆内线段部分的一半base 用pr±basee即得到答案题目：CGL_7_D AC代码： #include #include #include using namespace std; #define COUNTER_CLOCKWISE -1 //逆时针 #define CLOCK.. 圆与直线的交点肘子的博客 06-03 1万+ 只过了一组数据，未更新。项目极简说明这是一个目前处于开发初期阶段尚未完全成熟的软件项目其可执行文件位于dist文件夹中用户需自行探索许多功能若遇到进程未响应情况请耐心等待因为程序未采用多进程技术.zip 09-28 项目极简说明这是一个目前处于开发初期阶段尚未完全成熟的软件项目其可执行文件位于dist文件夹中用户需自行探索许多功能若遇到进程未响应情况请耐心等待因为程序未采用多进程技术.zip 【无人机路径规划】基于NMOPSO算法的城市场景三维航迹优化：多目标{完整资源下载，分享}导航变量求解 09-28 内容概要：本文研究了基于城市场景下无人机三维路径规划的导航变量多目标粒子群优化算法（NMOPSO），结合Matlab代码实现，重点解决高维多目标优化问题。通过引入导航变量并考虑无人机运动学约束，构建适用于复杂城市环境的三维路径规划模型，利用NMOPSO算法实现对飞行路径的安全性、效率与能耗等多重目标的协同优化。文中详细阐述了算法设计流程、适应度函数构造、Pareto最优解集处理方法，并通过仿真实验验证了该算法在规避障碍物、降低路径成本和提升收敛性能方面的有效性。; 适合人群：具备一定Matlab编程基础，从事无人机路径规划、智能优化算法研究或相关领域科研工作的研究生、科研人员及工程技术人员；熟悉多目标优化理论并希望进行算法仿真实践的高年级本科生亦可参考；使用场景及目标：①应用于复杂城市环境中无人机三维路径规划问题的研究与仿真；②为多目标粒子群优化算法（MOPSO）的改进与实际落地提供技术参考；③服务于智能交通、物流配送、应急救援等需要自主导航无人机系统的场景；阅读建议：建议结合提供的Matlab代码进行逐模块调试与实验分析，重点关注导航变量建模与算法收敛性表现，同时可拓展至其他智能优化算法（如NSGA-II）的对比研究，以深化对多目标优化机制的理解。自动表面波两工位色散分析（Matlab GUI）.zip 最新发布 09-28 1.版本：matlab2014a/2019b/2024b 2.附赠案例数据可直接运行。 3.代码特点：参数化编程、参数可方便更改、代码编程思路清晰、注释明细。 4.适用对象：计算机，电子信息工程、数学等专业的大学生课程设计、期末大作业和毕业设计。编队控制考虑一组多个水面飞行器（SV）中的路径跟踪问题.zip 09-28 1.版本：matlab2014a/2019b/2024b 2.附赠案例数据可直接运行。 3.代码特点：参数化编程、参数可方便更改、代码编程思路清晰、注释明细。 4.适用对象：计算机，电子信息工程、数学等专业的大学生课程设计、期末大作业和毕业设计。兼职美工任务管理-JAVA-基于springBoot兼职美工任务管理系统设计与实现（论文 + 开题 + PPT） 09-28 基于Web的兼职美工任务管理系统采用BS架构模式开发，vue前端，后台springboot，mysql数据库开发 idea平台 1、前台功能（兼职美工使用）（1）待处理任务查看与下载。查看所分配的处理任务，下载待处理素材资料；（2）美工作品提交。当完成处理任务后提交出美工作品及相关信息；（3）任务查询统计。按时间和任务类别查询历史任务和统计个人工作量；（4）成员管理，注册、登录和个人信息等修改功能。（5）美工评价 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15975	https://ocw.mit.edu/courses/18-02-multivariable-calculus-fall-2007/c081c812df4a020e26de273c7cea957c_lec_week11.pdf	MIT OpenCourseWare 18.02 Multivariable Calculus Fall 2007 For information about citing these materials or our Terms of Use, visit: p 18.02 Lecture 26. – Tue, Nov 13, 2007 Spherical coordinates (ρ, φ, θ). ρ = rho = distance to origin. φ = ϕ = phi = angle down from z-axis. θ = same as in cylindrical coordinates. Diagram drawn in space, and picture of 2D slice by vertical plane with z, r coordinates. Formulas to remember: z = ρ cos φ, r = ρ sin φ (so x = ρ sin φ cos θ, y = ρ sin φ sin θ). 2 2 ρ = p x2 + y2 + z = √ r2 + z . The equation ρ = a deﬁnes the sphere of radius a centered at 0. On the surface of the sphere, φ is similar to latitude, except it’s 0 at the north pole, π/2 on the equator, π at the south pole. θ is similar to longitude. φ = π/4 is a cone (asked using ﬂash cards) (z = r = x2 + y2). φ = π/2 is the xy-plane. Volume element: dV = ρ2 sin φ dρ dφ dθ. To understand this formula, ﬁrst study surface area on sphere of radius a: picture shown of a “rectangle” corresponding to Δφ, Δθ, with sides = portion of circle of radius a, of length aΔφ, and portion of circle of radius r = a sin φ, of length rΔθ = a sin φΔθ. So ΔS ≈ a2 sin φ ΔφΔθ, which gives the surface element dS = a2 sin φ dφdθ. The volume element follows: for a small “box”, ΔV = ΔS Δρ, so dV = dρ dS = ρ2 sin φ dρdφdθ. Example: recall the complicated example at end of Friday’s lecture (region sliced by a plane inside unit sphere). After rotating coordinate system, the question becomes: volume of the portion of unit sphere above the plane z = 1/ √ 2? (picture drawn). This can be set up in cylindrical (left as exercise) or spherical coordinates. For ﬁxed φ, θ we are slicing our region by rays straight out of the origin; ρ ranges from its value on the plane z = 1/ √ 2 to its value on the sphere ρ = 1. Spherical coordinate equation of the plane: z = ρ cos φ = 1/ √ 2, so ρ = sec φ/ √ 2. The volume is: Z 2π Z π/4 Z 1 ρ2 sin φ dρ dφ dθ. 0 0 √ 1 2 sec φ (Bound for φ explained by looking at a slice by vertical plane θ = constant: the edge of the region is at z = r = √ 1 2 ). 2π 5π Evaluation: not done. Final answer: . 3 − 6 √ 2 Application to gravitation. Gravitational force exerted on mass m at origin by a mass ΔM at (x, y, z) (picture shown) is given by F ~ = G ΔM m , dir(F ~) = hx, y, zi, i.e. F ~ = G ΔM m hx, y, zi. (G = gravitational \| \| ρ2 ρ ρ3 constant). If instead of a point mass we have a solid with density δ, then we must integrate contributions to gravitational attraction from small pieces ΔM = δ ΔV . So ZZZ ZZZ F ~ = Gm hx, y, zi δ dV, i.e. z-component is Fz = Gm z δ dV, . . . ρ3 ρ3 R R If we can set up to use symmetry, then Fz can be computed nicely using spherical coordinates. General setup: place the mass m at the origin (so integrand is as above), and place the solid so that the z-axis is an axis of symmetry. Then F ~ = h0, 0, Fzi by symmetry, and we have only one 1 R RR RR RR 2 component to compute. Then ZZZ ZZZ ZZZ z ρ cos φ Fz = Gm ρ3 δ dV = Gm ρ3 δ ρ2 sin φ dρ dφ dθ = Gm δ cos φ sin φ dρ dφ dθ. R R R Example: Newton’s theorem: the gravitational attraction of a spherical planet with uniform density δ is the same as that of the equivalent point mass at its center. 18.02 Lecture 27. – Thu, Nov 15, 2007 Handouts: PS10 solutions, PS11 Vector ﬁelds in space. At every point in space, F ~ = P ˆ ı + Qˆ j + Rk ˆ, where P, Q, R are functions of x, y, z. Examples: force ﬁelds (gravitational force F ~ = −chx, y, zi/ρ3; electric ﬁeld E, magnetic ﬁeld B); velocity ﬁelds (ﬂuid ﬂow, v = v(x, y, z)); gradient ﬁelds (e.g. temperature and pressure gradients). Flux. Recall: in 2D, ﬂux of a vector ﬁeld F ~ across a curve C = C F ~ n ˆ ds. · In 3D, ﬂux of a vector ﬁeld is a double integral: ﬂux through a surface, not a curve! F ~ vector ﬁeld, S surface, n ˆ unit normal vector: Flux = F ~ n ˆ dS. · Notation: dS ~ = n ˆ dS. (We’ll see that dS ~ is often easier to compute than n ˆ and dS). Remark: there are 2 choices for n ˆ (choose which way is counted positively!) Geometric interpretation of ﬂux: As in 2D, if F ~ = velocity of a ﬂuid ﬂow, then ﬂux = ﬂow per unit time across S. Cut S into small pieces, then over each small piece: what passes through ΔS in unit time is the contents of a parallelepiped with base ΔS and third side given by F ~. Volume of box = base × height = (F ~ n ˆ) ΔS. · • Examples: 1) F ~ = xˆ ı + yˆ j + zk ˆ through sphere of radius a centered at 0. n ˆ = a 1 hx, y, zi (other choice: −a 1 hx, y, zi; traditionally choose n ˆ pointing out). F ~ n ˆ = hx, y, zi · n ˆ = 1 (x2 + y2 + z2) = a, so F ~ n ˆdS = a dS = a (4πa2). · a S S · ZZ ZZ RR q q 3 z 2) Same sphere, H ~ = zk ˆ: H ~ n ˆ = a 2 . · z2 Z 2π Z π a2 cos2 φ Z π 4 H ~ dS ~ = dS = a 2 sin φ dφdθ = 2πa3 cos2 φ sin φ dφ = πa3 . · a a 3 S S 0 0 0 Setup. Sometimes we have an easy geometric argument, but in general we must compute the surface integral. The setup requires the use of two parameters to describe the surface, and F ~ n ˆ dS · must be expressed in terms of them. How to do this depends on the type of surface. For now, formulas to remember: 0) plane z = a parallel to xy-plane: n ˆ = ±k ˆ, dS = dx dy. (similarly for planes // xz or yz-plane). 1) sphere of radius a centered at origin: use φ, θ (substitute ρ = a for evaluation); n ˆ = a 1 hx, y, zi, dS = a2 sin φ dφ dθ. 2) cylinder of radius a centered on z-axis: use z, θ (substitute r = a for evaluation): n ˆ is radially out in horizontal directions away from z-axis, i.e. n ˆ = a 1 hx, y, 0i; and dS = a dz dθ (explained by drawing a picture of a “rectangular” piece of cylinder, ΔS = (Δz) (aΔθ)). 3) graph z = f(x, y): use x, y (substitute z = f(x, y)). We’ll see on Friday that n ˆ and dS separately are complicated, but n ˆ dS = h−fx, −fy, 1i dx dy. 18.02 Lecture 28. – Fri, Nov 16, 2007 Last time, we deﬁned the ﬂux of F ~ through surface S as F ~ n ˆ dS, and saw how to set up in · various cases. Continue with more: Flux through a graph. If S is the graph of some function z = f(x, y) over a region R of xy-plane: use x and y as variables. Contribution of a small piece of S to ﬂux integral? Consider portion of S lying above a small rectangle Δx Δy in xy-plane. In linear approximation it is a parallelogram. (picture shown) The vertices are (x, y, f(x, y)); (x + Δx, y, f(x + Δx, y)); (x, y + Δy, f(x, y + Δy)); etc. Linear approximation: f(x + Δx, y) ' f(x, y) + Δx fx(x, y), and f(x, y + Δy) ' f(x, y) + Δy fy(x, y). So the sides of the parallelogram are hΔx, 0, Δx fxi and h0, Δy, Δy fyi, and ˆ ı ˆ j k ˆ ΔS ~ = (Δx h1, 0, fxi) × (Δy h0, 1, fyi) = ΔxΔy 1 0 fx 0 1 fy = h−fx, −fy, 1iΔxΔy. So dS ~ = ±h−fx, −fy, 1idx dy. (From this we can get n ˆ = dir(dS ~) = h−fx, −fy, 1i and dS = \| dS ~ \| = fx 2 + fy 2 + 1 dx dy. The f2 + f2 + 1 x y conversion factor √ between dS and dA relates area on S to area of projection in xy-plane.) · · · Example: ﬂux of F ~ = zk ˆ through S = portion of paraboloid z = x2 + y2 above unit disk, • oriented with normal pointing up (and into the paraboloid): geometrically ﬂux should be > 0 (asked using ﬂashcards). We have n ˆ dS = h−2x, −2y, 1i dx dy, and ZZ ZZ ZZ Z 2π Z 1 F ~ dS ~ = z dx dy = (x 2 + y 2) dx dy = r 2 r dr dθ = π/2. · S S S 0 0 Parametric surfaces. If we can describe S by parametric equations x = x(u, v), y = y(u, v), z = z(u, v) (i.e. ~ r = ~ r(u, v)), then we can set up ﬂux integrals using variables u, v. To ﬁnd dS ~, RR 4 consider a small portion of surface corresponding to changes Δu and Δv in parameters, it’s a parallelogram with sides ~ r(u + Δu, v) − ~ r(u, v) ≈ (∂~ r/∂u) Δu and (∂~ r/∂v) Δv, so ∂~ r ∂~ r ∂~ r ∂~ r ΔS ~ = ± ∂u Δu × ∂v Δv , dS ~ = ± ∂u × ∂v du dv. (This generalizes all formulas previously seen; but won’t be needed on exam). Implicit surfaces: If we have an implicitly deﬁned surface g(x, y, z) = 0, then we have a (non unit) normal vector N = rg. (similarly for a slanted plane, from equation ax + by + cz = d we get N = ha, b, ci). Unit normal n ˆ = ±N/\|N\|; surface element ΔS = ? Look at projection to xy-plane: ΔA = ΔS cos α = (N · k ˆ/\|N\|) ΔS (where α = angle between slanted surface element and horizontal: projection shrinks one direction by factor cos α = (N · k ˆ)/\|N\|, preserves the other). Hence dS = \|N\| dA, and n ˆ dS = \|N\|n ˆ dx dy = ± N dx dy. N k ˆ N k ˆ N k ˆ · · · (In fact the ﬁrst formula should be dS = \|N\| dA, I forgot the absolute value). \|N · k ˆ\| Note: if S is vertical then the denominator is zero, can’t project to xy-plane any more (but one could project e.g. to the xz-plane). Example: if S is a graph, g(x, y, z) = z − f(x, y) = 0, then N = hgx, gy, gzi = h−fx, −fy, 1i, N k ˆ = 1, so we recover the formula dS ~ = h−fx, −fy, 1idx dy seen before. · Divergence theorem. (“Gauss-Green theorem”) – 3D analogue of Green theorem for ﬂux. If S is a closed surface bounding a region D, with normal pointing outwards, and F ~ vector ﬁeld deﬁned and diﬀerentiable over all of D, then ZZ ZZZ F ~ dS ~ = F dV, where div (Pˆ j + Rk ˆ) = Px + Qy + Rz. div ~ ı + Qˆ · S D Example: ﬂux of F ~ = zk ˆ out of sphere of radius a (seen Thursday): div F ~ = 0 + 0 + 1 = 1, so F ~ dS ~ = 3 vol(D) = 4πa3/3. S · Physical interpretation (mentioned very quickly and verbally only): div F ~ = source rate = ﬂux generated per unit volume. So the divergence theorem says: the ﬂux outwards through S (net amount leaving D per unit time) is equal to the total amount of sources in D.
15976	https://phywe-itemservice.s3.eu-central-1.amazonaws.com/sites/DMS-Phywe/PROD/de-DE/item/phy_itemtestinstruction/P9/P9170300/P9170300_en.pdf	P9170300 Nature & technology Substances in everyday use Thermal expansion of water and spirit  Difficulty level  Group size  Preparation time  Execution time easy 1 10 minutes 10 minutes This content can also be found online at: Robert-Bosch-Breite 10 37079 Göttingen Tel.: 0551 604 - 0 Fax: 0551 604 - 107 info@phywe.de www.phywe.de P9170300 Teacher information Application Test setup In this experiment, the students observe the behavior of the liquid columns in two capillary tubes when the respective screw-top glass is heated. One of the glasses is filled mostly with water, the other mostly with alcohol. They discovered that not only water but also alcohol expands when heated, but that this expansion varies in magnitude. In addition, they should recognize which of the two liquids would be more suitable for the most sensitive thermometer and why. They thus learn how a thermometer works in a simple and intuitive way. 2/10 Robert-Bosch-Breite 10 37079 Göttingen Tel.: 0551 604 - 0 Fax: 0551 604 - 107 info@phywe.de www.phywe.de P9170300 Other teacher information (1/2) Prior Principle Students know that heating the vessel causes the liquid inside to expand. They should be familiar with the rough operating principle of a thermometer. In this experiment, the students are to identify the different degrees of expansion of water and alcohol when the screw-top glass is heated by taking their own measurements. From this, they are to consider which liquid would be more suitable for use in a thermometer. Notice:To help students distinguish between the two transparent liquids, provide colored water Other teacher information (2/2) When heated, spirit expands more than water. To build a thermometer that is as sensitive as possible, spirit would therefore be a better choice than water. Students fill one screw-top jar with water and the other screw-top jar with approximately the same amount of alcohol Students warm the glasses with their hands and observe the rise of the water columns You consider what is the relationship between the height of the water column and the expansion of the medium in the glass Learning Tasks 3/10 Robert-Bosch-Breite 10 37079 Göttingen Tel.: 0551 604 - 0 Fax: 0551 604 - 107 info@phywe.de www.phywe.de P9170300 Safety instructions When handling methylated spirits, students must be informed about the high flammability Before carrying out the test, the students should be made aware of the need for caution when handling the screw tubes and especially the capillary tubes The general instructions for safe experimentation in science education apply to this experiment. Student Information 4/10 Robert-Bosch-Breite 10 37079 Göttingen Tel.: 0551 604 - 0 Fax: 0551 604 - 107 info@phywe.de www.phywe.de P9170300 Motivation Whether you're sick and need to know if you have a fever, or just want to know how warm it is outside, in some situations in everyday life we rely on thermometers. But do you know how they work? A thermometer consists of a thin glass tube that is often filled with a colored liquid to facilitate reading. The level of the liquid column allows us to determine the temperature by matching it with the scale. The temperature therefore causes the liquid to expand or contract, which we can then read as a change in temperature on the scale. In this experiment, we want to investigate what property a liquid should have so that it can be used in a thermometer. Room temperature Fever thermometer Tasks How do water and methylated spirit behave when heated to the same temperature? Fill one screw-top glass with 40 ml of water and the other with the same amount of alcohol. Then adjust the capillary tubes Warm both glasses with your hands and observe the rise of the liquid columns in the capillary tubes Consider what your observations tell you about the expansion of water and alcohol and which of the two liquids would be more suitable for a sensitive thermometer What do you think will expand more when you heat it? Spirit Both expand equally Water 5/10 Robert-Bosch-Breite 10 37079 Göttingen Tel.: 0551 604 - 0 Fax: 0551 604 - 107 info@phywe.de www.phywe.de P9170300 Equipment Position Equipment Item no. Quantity 1 Set student experiments heat for 13 experiments, TESS beginner nature and technology NT-WAE 15235-88 1 6/10 Robert-Bosch-Breite 10 37079 Göttingen Tel.: 0551 604 - 0 Fax: 0551 604 - 107 info@phywe.de www.phywe.de P9170300 Structure (1/2) Fig. 1 Using the beaker and the funnel, fill one screw-top glass with about 40 ml of dyed water, as in Fig. 1. Fill the second glass equally with 40 ml of alcohol. Slide a gasket onto both capillary tubes so that the tube protrudes about 4 cm on one side, and a cap (Fig.2). Fig. 2 Structure (2/2) Fig. 3 Carefully place the tubes on the jars and screw them tight with the cap. The liquid rises in the tube, be careful that it does not overflow or splash! To start the measurement, carefully move both tubes until the liquid in both is at the same level above the cap (Fig. 3). 7/10 Robert-Bosch-Breite 10 37079 Göttingen Tel.: 0551 604 - 0 Fax: 0551 604 - 107 info@phywe.de www.phywe.de P9170300 Procedure Fig. 4 Warm the two glasses with your hands for some time (Fig. 4) and observe the water levels in the two capillary tubes. Make sure you don't wiggle the caps or tubes! Measure the height of the rise in the glass with alcohol and in the glass with water and note the time after you have measured the values. Then go to the protocol and answer the questions about the experiment Report 8/10 Robert-Bosch-Breite 10 37079 Göttingen Tel.: 0551 604 - 0 Fax: 0551 604 - 107 info@phywe.de www.phywe.de P9170300 Task 1 In which glass did the water level in the capillary tube rise more after the same time? In a jar with spirit. It rises a both equally strong. In a glass with plenty of water. Summarize what you learned in this experiment. When the screw-top glasses are heated, expands more than . Thus, spirit can make temperature changes visible because its rise in the capillary is in the process. On a thermometer, you want to be able to read the temperature accurately, so you should choose a whose changes greatly with temperature. Check smaller liquid expansion water spirit greater Task 2 9/10 Robert-Bosch-Breite 10 37079 Göttingen Tel.: 0551 604 - 0 Fax: 0551 604 - 107 info@phywe.de www.phywe.de P9170300 Slide Score/Total Slide 8: Presumption expansion Slide 14: Water level in capillary tube Slide 15: Heating water and spirit 0/1 0/3 0/6 Total 0/10  Solutions  Repeat 10/10 Robert-Bosch-Breite 10 37079 Göttingen Tel.: 0551 604 - 0 Fax: 0551 604 - 107 info@phywe.de www.phywe.de
15977	https://sites.math.washington.edu/~king/coursedir/m444a03/notes/12-05-isom-from-segments.html	Isometry Constructions from Triangles and Segments Isometry from a triangle \| Orientation-Preserving from Segment \| Orientation-reversing from segment 1. How to construct the data of an isometry from the image of a triangle Given two triangles ABC and A'B'C', we say that a transformation T takes ABC to A'B'C' if T(A) = A' and T(B) = B', and T(C) = C'. If ABC and A'B'C' are congruent triangles, we know from a fundamental theorem on isometries that there is exactly one isometry T that takes ABC to A'B'C'. Note: This statement takes into account the order of the vertices. If T takes ABC to A'B'C', then it does not take ABC to C'A'B'. However, in specific cases, one needs to determine from the two triangles exactly what isometry is T. What type of isometry is it and what is its defining data. Construction Method 1: (Use the proof of the fundamental theorem) A fundamental theorem is proved that a triangle ABC can be transformed to a congruent triangle A'B'C' by the composition of one, two or three line reflections. Thus the isometry T that takes ABC to A'B'C' is this composition. The proof is a step by step construction of up to three lines of reflection. One can make this construction and then use the theorems about composition of isometries to determine exactly what isometry is T. However, this method is long and indirect, especially for glide reflections. Construction Method 2: (Use midpoints) Given two congruent triangles ABC and A'B'C', construct the midpoints A'' of AA', B'' of BB', C'' of CC'. Then the arrangement of the midpoints will tell what kind of isometry T takes ABC to A'B'C'. If T is the identity or a translation, A''B''C'' is a triangle congruent to ABC. If T is a rotation by angle x but not a halfturn, then A''B''C'' is a triangle similar to ABC (the scaling factor is cos(x/2). If T is a half turn, the 3 midpoints are all the same point, which is the center of the halfturn. If T is a line reflection, the 3 midpoints are collinear and at most 2 coincide, since A, B, C are not collinear. Thus there is one line through the midpoints, and this is the mirror line of T. If T is a glide reflection, the 3 midpoints are collinear and at most 2 coincide, since A, B, C are not collinear. Thus there is one line through the midpoints, and this is the invariant line of T. \| \| \| --- \| \| \| \| \| Line Reflection \| Glide Reflection \| The proof that the points A'', B'' and C'' cannot all be the same for a line reflection or a glide reflection follows from the definitions of these transformations. For T = a line reflection in a line m, A'', B'', C'' are the feet of the perpendiculars to m through A, B, C. Thus they can only coincide if A, B, C are all on the same line perpendicular to m. For T = glide reflection with glide vector v, then if A, B, C are the feet of the perpendiculars to m through A, B, C, then A'', B'', C'' are the translations of these points by vector (1/2)v. Thus the midpoints A'', B'', C'' coincide if and only if the feet A, B, C coincide, which means again that A, B, C are all on the same line perpendicular to m. These cases allow us to deduce the nature of the isometry from the midpoints A''B''C''. If the points coincide with a point P, T is a halfturn with center P. If the points are collinear, but do not all coincide, then T is orientation reversing. Complete the construction of the geometric data is spelled out in this section below: Construction Method: Line reflection or glide reflection from 2 congruent segments If the points form a triangle, then T is orientation preserving, and the data can be constructed by the description in this section below: Construction Method: Rotation or translation from 2 congruent segments 2. How to construct the data of a rotation or translation from the image of a segment or a point Fundamental Theorem \| From Image of a Segment \| From Image of a Point knowing Angle Given two segments AB and CD, we say that a transformation T takes AB to CD if T(A) = C and T(B) = D. Note: This statement takes into account the order of the endpoints. If T takes AB to CD, then it does not take AB to DC. Theorem (Fundamental Theorem for Orientation-Preserving Isometries) Given two congruent segments AB and CD, there is exactly one orientation-preserving isometry T that takes AB to CD. Proof (also a construction): This consists of the first two steps of the Second Fundamental Theorem of Isometries. When A and C are distinct, let m1 be the perpendicular bisector of AC. In the special case where A = C, let m1 = line AB (actually, any line through A will do). Let R1 denote line reflection in m1. \| \| \| --- \| \| For this reflection, R1(A) = C. Let B' = R1(B). Segment CD is congruent to CB' since R1 is an isometry. If B' is not D, let m2 be the perpendicular bisector of B'D. Since \|CB'\| = \|CD\|, the point C is on m2. In the special case B' = D, let m2 = line CD. Let R2 be reflection in m2. In either case, R2(B') = D and R2(C) = C. So setting T = R2R1, T(A) = C and T(B) = D. \| \| T is the composition of two line reflections. T is a rotation with center O if the two lines m1 and m2 intersect in one point O; it is a translation if m1 and m2 are parallel. If the lines are the same, then T is the identity. Finally, it is claimed that T is the only orientation-preserving isometry that takes AB to CD. To see this, assume that S = R4R3 also takes AB to CD. Then S-1 = R3R4 takes CD to AD. Thus S-1T = R3R4R2R1 takes AB to AB. This means that S-1T (A) = A and S-1T (B) = B are two fixed points of the isometry. But the isometry S-1T is an orientation-preserving isometry, since it is the product of 4 line reflections. This means that it is a rotation, a translation, or the identity. But the only one of these isometries with two fixed points is the identity, so S-1T = I, and so T = S. Thus there is only one orientation-preserving isometry that takes AB to CD. Construction Method: Rotation or translation from 2 congruent segments Given two congruent segments AB and CD, how can one construct the unique orientation-preserving isometry T that takes AB to CD? In practice, the cases of interest are when T is a rotation or a translation, since if A = C and B = D, no construction is necessary for the identity. General Case: Assume A is not C and B is not D. Construct the lines m = perpendicular bisector of AC and n = perpendicular bisector of BD. \| \| \| --- \| \| If T is a rotation, the lines m and n intersect at O, the center of the rotation. For in this case A and C = T(A) are both on the same circle with center O, so the bisector m of the chord AC passes through O. For the same reason, n passes through O. The angle of the rotation is angle AOC. In the special case of a halfturn, O is also the midpoint of AC and BD as well as the intersection of m and n. If T is a translation, then AC and BD are parallel, so m and n are also parallel. The translation vector is AC. \| \| Special cases: Assume A = C but B and D are distinct. In this case, A is already the center of a rotation that will take B to D. The rotation angle is angle BAD. The case when B = D is handled in the same way. If A = C and B = D, then T = identity. Construction Method: Rotation from image of one point plus the rotation angle Suppose that T is known to be a rotation by angle x. If we are given two distinct points A and C, such that T(A) = C, is this enough to determine T? Since the rotation angle of T is given, if the center O can be constructed, then T will be known completely. From the given information T(A) = C, it follows that O is on the line m, the perpendicular bisector of AC. Also, angle AOC = x. The angle x is an oriented, or signed angle. There are several cases: Angle x = 0 degrees. This means T = I and so A = T(A) = C. This violates the assumption that A and C are distinct. Angle x = 180 degrees. In this case T is a halfturn and O = midpoint AC. This completes the construction. For angle x, 0 < x < 180. This will be discussed below For angle x, -180 < x < 0. This is much like case 3, and will be discussed below. Cases 3 and 4: In each case, 0 < \|x\| < 180 is the unoriented angle AOC. \| \| \| --- \| \| Imagine that O has been constructed. Then OA = OC, since T is a rotation with center O. Also, angle AOC = x. This means that triangle OAC is an isosceles triangle with (unoriented) angle O = \|x\|, and angle A = angle C = y = (180-\|x\|)/2 = 90 – (\|x\|/2). So to construct such a triangle, it is only necessary to construct a ray AD so that angle CAD = y. Intersect this ray with m to get P. \| \| Reflect P in line AC to get Q. Then both triangles PAC and QAC are isosceles triangles with (unoriented) angles \|x\|, y, y. One of the angles APC or AQC is +\|x\| and the other is -\|x\|. So one is the solution to case 3 and one is the solution to case 4. It is clear from this construction that there is only one T for each (oriented) x. \| \| \| --- \| \| To construct the angle y, it is only necessary to construct an isosceles triangle with angles \|x\|, y, y and copy the angle y. For example, suppose that angle FGH = \|x\| is given. Then choose some point K on ray GF and let the point L on GH be that point with \|GL\| = \|GK\|. This means that GKL is an isosceles triangle with angles \|x\|, y, y. This completes the construction, and also proves the theorem that for any distinct A and C and any oriented, non-zero angle, there is a unique rotation that takes A to C with this angle of rotation \| \| Other constructions: Translation or Halfturn from image of one point Given two distinct points A and C, there is a unique Halfturn H that takes A to C. The construction of the center O of the halfturn is simple. O is the midpoint of AC. For any other point B, H(B) is the point D for which O is the midpoint of BD. Given two distinct points A and C, there is a unique translation T that takes A to C. The translation vector is AC. Thus for any B not on line AC, T(B) is the point D such that ACDB is an parallelogram. If B is on line AC, then this alternative criterion for parallelogram still works. D is the point so that the midpoint of AD = the midpoint of BC. 3. How to construct the data of a line reflection or glide reflection from the image of a segment or a point Fundamental Theorem \| From Image of a Segment \| From Image of a Point Knowing Vector Recall our definition: Given two segments AB and CD, we say that a transformation T takes AB to CD if T(A) = C and T(B) = D. Note: This statement takes into account the order of the endpoints. If T takes AB to CD, then it does not take AB to DC. Theorem (Fundamental Theorem for Orientation-Reversing Isometries) Given two congruent segments AB and CD, there is exactly one orientation-reversing isometry U that takes AB to CD. Proof: This follows from the theorem for orientation-preserving isometries Given AB and CD, there is exactly one orientation-preserving isometry T that takes AB to CD. Let R = line reflection in AB. Then U = RT is an orientation-reversing isometry that takes AB to CD. This shows that U exists. To see that there is only one such U, suppose that V is another orientation-reversing isometry that takes AB to CD. Then RV is an orientation preserving isometry that takes AB to CD, and so RV must = T. Then V = RRV = RT. But by definition, RT = U. QED Construction Method: Line reflection or glide reflection from 2 congruent segments Given two congruent segments AB and CD, how can one construct the unique orientation-reversing isometry U that takes AB to CD? From the definitions of line reflection and a theorem about glide reflection, for any orientation-reversing U and for any point A and C = U(A), the midpoint of AC is on the mirror line (if U is a line reflection) or the invariant line (if U is a glide reflection). Thus if U takes AB to CD, let M = midpoint AC and N = midpoint BD. \| \| \| --- \| \| Let A' and B' be the reflections of A and B in line MN. If U is a line reflection, then the segments AC and BD are both perpendicular to MN and in fact MN is the perpendicular bisector of these two segments. In this case A' = C and B' = D. \| \| \| If U is a glide reflection, MN is the invariant line. Neither of the segments AC and BD will be perpendicular to MN. The glide vector = A'C = B'D. To construct the image of some other point P, reflect P in line MN to get P' and then translate by the glide vector. \| \| Construction Method: Line reflection or glide reflection from the image of one point plus glide vector Suppose that U is known to be either (a) a line reflection or (b) a glide reflection with given glide vector EF. Given two distinct points A and C, how can one construct the unique orientation-reversing isometry U that takes A to C? First, if U is a line reflection, then let m be the perpendicular bisector of AC. Then U is line reflection in m. Second, if EF is given, let A' be the translation of C by vector FE = -EF. By the definition of the glide vector, then the perpendicular bisector n of AA' must be the invariant line of U and U = TEFRn.
15978	https://www.cs.utexas.edu/~bajaj/graphics23/cs354/lect06.pdf	Department of Computer Sciences Graphics – Fall 2003 (Lecture 6) Vector and Aﬃne Algebra • Diﬀerence of points (x1, 1) −(x0, 1) = (x1 −x0, 0) (x 0,1) (x 1 (x 1 0,0) x __ w ,1) The University of Texas at Austin 1 Department of Computer Sciences Graphics – Fall 2003 (Lecture 6) • Aﬃne combination of points (1 −t)(x1, 1) + t(x0, 1) = ((1 −t)x1 + tx0, 1) (x 0,1) (x 1,1) (x 0,1) (x 1,1) (1 __ t) t + w The University of Texas at Austin 2 Department of Computer Sciences Graphics – Fall 2003 (Lecture 6) • Linear combinations of vectors a(v0, 0) + b(v1, 0) = (av0 + bv1, 0) (v 0,1) (v w ,1) 1 + (v 0,1) ,1) b a 1 (v The University of Texas at Austin 3 Department of Computer Sciences Graphics – Fall 2003 (Lecture 6) Linear Transformations Vector space V • Linear combinations of vectors in V are in V • For ⃗ u, ⃗ v ∈V – ⃗ u + ⃗ v ∈V – α⃗ u ∈V for any scalar α – In general, P i αi⃗ ui ∈V for any scalars αi • Linear transformations – Let T : V0 7→V1, where V0 and V1 are vector spaces – Then T is linear iﬀ ∗T(⃗ u + ⃗ v) = T(⃗ u) + T(⃗ v) ∗T(α⃗ u) = αT(⃗ u) ∗In general, T (P i αi⃗ ui) = P i αiT (⃗ ui) The University of Texas at Austin 4 Department of Computer Sciences Graphics – Fall 2003 (Lecture 6) Example of linear tranformation for vectors u = α1u1 + α2u2 2 1 u u u T (u) = w = T (α1u1 + α2u2) = T (α1u1) + T (α2u2) = α1T (u1) + α2T (u2) 1 2 w w w The University of Texas at Austin 5 Department of Computer Sciences Graphics – Fall 2003 (Lecture 6) Aﬃne Transformations Aﬃne space A = (V, P) • For ⃗ u ∈V and P ∈P P + ⃗ u ∈P • Deﬁne point subtraction: – For P, Q ∈P and ⃗ u ∈V, if P + ⃗ u = Q, then Q −P ≡⃗ u – So in general we have P i αiPi is a vector iﬀP i αi = 0 • Deﬁne point blending: – For P, P1, P2 ∈P and scalar α, if P = P1+α (P2 −P1) then P ≡(1 −α) P1+ αP2 – This can also be written P ≡α1P1 + α2P2 where α1 + α2 = 1 – So in general we have P i αiPi is a point iﬀP i αi = 1 • Geometrically, we have \|P −P0\| \|P −P1\| = d1 d2 or P = d1P1+d2P2 d1+d2 • Vectors can always be combined linearly P i αi⃗ ui • Points can be combined linearly P i αiPi iﬀ – The coeﬃcients sum to 1, giving a point (“aﬃne combination”) – The coeﬃcients sum to 0, giving a vector (“vector combination”) The University of Texas at Austin 6 Department of Computer Sciences Graphics – Fall 2003 (Lecture 6) – Example aﬃne combination: P (t) = P0 + t(P1 −P0) = (1 −t)P0 + tP1 1 2 1 3 4 1 2 2 3 P P P P P P P P P – This says any point on the line is an aﬃne combination of the line segment’s endpoints. • Aﬃne transformations – Let T : A0 7→A1 where A0 and A1 are aﬃne spaces – T is said to be an aﬃne transformation iﬀ ∗T maps vectors to vectors and points to points ∗T is a linear transformation on the vectors ∗T(P + ⃗ u) = T(P ) + T(⃗ u) – Properties of aﬃne transformations ∗T preserves aﬃne combinations: T(α0P0 + · · · + αnPn) = α0T(P0) + · · · + αnT(Pn) The University of Texas at Austin 7 Department of Computer Sciences Graphics – Fall 2003 (Lecture 6) where P i αi = 0 or P i αi = 1 ∗T maps lines to lines: T((1 −t)P0 + tP1) = (1 −t)T(P0) + tT(P1) ∗T is aﬃne iﬀit preserves ratios of distance along a line: P = d0P0 + d1P1 d0 + d1 ⇒T(P ) = d0T(P0) + d1T(P1) d0 + d1 ∗T maps parallel lines to parallel lines (can you prove this?) – Example aﬃne transformations ∗Rigid body motions (translations, rotations) ∗Scales, reﬂections ∗Shears The University of Texas at Austin 8 Department of Computer Sciences Graphics – Fall 2003 (Lecture 6) Matrix Representation of Transformations • Let A0 and A1 be aﬃne spaces. Let T : A0 7→A1 be an aﬃne transformation. Let F0 = ( ⃗ i0,⃗ j0, O0) be a frame for A0. Let F1 = ( ⃗ i1,⃗ j1, O1) be a frame for A1. • Let P = x ⃗ i0 + y⃗ j0 + O0 be a point in A0. The coordinates of P relative to A0 are (x, y, 1). This can also be represented in vector form as P = h ⃗ i0 ⃗ j0 O0 i   x y 1   The University of Texas at Austin 9 Department of Computer Sciences Graphics – Fall 2003 (Lecture 6) • What are the coordinates (x′, y′, 1) of T(P ) relative to F1? – An aﬃne transformation is characterized by the image of a frame in the domain. T(P ) = T(x ⃗ i0 + y⃗ j0 + O0) = xT( ⃗ i0) + yT(⃗ j0) + T(O0) – T( ⃗ i0) must be a linear combination of ⃗ i1 and ⃗ j1, say T( ⃗ i0) = t1,1⃗ i1 + t2,1⃗ j1. – Likewise T(⃗ j0) must be a linear combination of ⃗ i1 and ⃗ j1, say T(⃗ j0) = t1,2⃗ i1 + t2,2⃗ j1. – Finally T(O0) must be an aﬃne combination of ⃗ i1, ⃗ j1, and O1, say T(O0) = t1,3⃗ i1 + t2,3⃗ j1 + O1. The University of Texas at Austin 10 Department of Computer Sciences Graphics – Fall 2003 (Lecture 6) – Then by substitution we get T(P ) = x(t1,1⃗ i1 + t2,1⃗ j1) + y(t1,2⃗ i1 + t2,2⃗ j1) + t1,3⃗ i1 + t2,3⃗ j1 + O1 = h t1,1⃗ i1 + t2,1⃗ j1 t1,2⃗ i1 + t2,2⃗ j1 i t1,3⃗ i1 + t2,3⃗ j1 + O1   x y 1   = h ⃗ i1 ⃗ j1 O1 i   t1,1 t1,2 t1,3 t2,1 t2,2 t2,3 0 0 1     x y 1   Using MT to denote the matrix, we see that F0 = F1MT • Let T(P ) = P ′ = x′⃗ i1 + y′⃗ j1 + O1 The University of Texas at Austin 11 Department of Computer Sciences Graphics – Fall 2003 (Lecture 6) In vector form this is P ′ = h ⃗ i1 ⃗ j1 O1 i   x′ y′ 1   = h ⃗ i1 ⃗ j1 O1 i   t1,1 t1,2 t1,3 t2,1 t2,2 t2,3 0 0 1     x y 1   So we see that   x′ y′ 1  =   t1,1 t1,2 t1,3 t2,1 t2,2 t2,3 0 0 1     x y 1   We can write this in shorthand – p′ = MTp • MT is the matrix representation of T – The ﬁrst column of MT represents T( ⃗ i0) – The second column of MT represents T(⃗ j0) – The third column of MT represents T(O0) The University of Texas at Austin 12 Department of Computer Sciences Graphics – Fall 2003 (Lecture 6) • Translation – Points are transformed as x′ y′ 1 T = [x y 1]T + [∆x ∆y 0]T. – Vectors don’t change. – Thus translation is aﬃne but not linear. If it were linear, we would have T(P + Q) = T(P ) + T(Q), but point addition is undeﬁned. – Translation can be applied to sums of vectors and vector-point sums. – Matrix formulation:   x′ y′ 1  =   1 0 ∆x 0 1 ∆y 0 0 1     x y 1  =   x + ∆x y + ∆y 1     x′ y′ 0  =   1 0 ∆x 0 1 ∆y 0 0 1     x y 0  =   x y 0   – Shorthand for the above matrix: T (∆x, ∆y) The University of Texas at Austin 13 Department of Computer Sciences Graphics – Fall 2003 (Lecture 6) • Scale – Linear transform — applies equally to points and vectors – Points transform as x′ y′ 1 T = [xSx ySy 1]T. – Vectors transform as x′ y′ 0 T = [xSx ySy 0]T. – Matrix formulation:   x′ y′ 1  =   Sx 0 0 0 Sy 0 0 0 1     x y 1  =   xSx ySy 1     x′ y′ 0  =   Sx 0 0 0 Sy 0 0 0 1     x y 0  =   xSx ySy 0   – Shorthand for the above matrix: S(Sx, Sy) – Note that this is origin sensitive. – How do you do reﬂections? The University of Texas at Austin 14 Department of Computer Sciences Graphics – Fall 2003 (Lecture 6) • Rotate – Linear transform — applies equally to points and vectors – Points transform as x′ y′ 1 T = [x cos(θ) −y sin(θ) x sin(θ) + y cos(θ) 1]T. – Vectors transform as x′ y′ 0 T = [x cos(θ) −y sin(θ) x sin(θ) + y cos(θ) 0]T. – Matrix formulation:   x′ y′ 1  =   cos(θ) −sin(θ) 0 sin(θ) cos(θ) 0 0 0 1     x y 1  =   x cos(θ) −y sin(θ) x sin(θ) + y cos(θ) 1     x′ y′ 0  =   cos(θ) −sin(θ) 0 sin(θ) cos(θ) 0 0 0 1     x y 0  =   x cos(θ) −y sin(θ) x sin(θ) + y cos(θ) 0   – Shorthand for the above matrix: R(θ) – Note that this is origin sensitive. The University of Texas at Austin 15 Department of Computer Sciences Graphics – Fall 2003 (Lecture 6) • Shear – Linear transform — applies equally to points and vectors – Points transform as x′ y′ 1 T = [x + αy, y + βx, 1]T. – Vectors transform as x′ y′ 0 T = [x + αy, y + βx, 0]T. – Matrix formulation:   x′ y′ 1  =   1 α 0 β 1 0 0 0 1     x y 1  =   x + αy y + βx 1     x′ y′ 0  =   1 α 0 β 1 0 0 0 1     x y 0  =   x + αy y + βx 0   – Shorthand for the above matrix: Sh(α, β) The University of Texas at Austin 16 Department of Computer Sciences Graphics – Fall 2003 (Lecture 6) • Composition of Transformations – Now we have some basic transformations, how do we create and represent arbitrary aﬃne transformations? – We can derive an arbitrary aﬃne transform as a sequence of basic transformations, then compose the transformations – Example — scaling about an arbitrary point [xc yc 1]T 1. Translate [xc yc 1]T to [0 0 1] (T (−xc, −yc)) 2. Scale x′ y′ 1 T = S(Sx, Sy) [x y 1]T 3. Translate [0 0 1]T back to [xc yc 1] (T (xc, yc)) – The sequence of transformation steps is T (−xc, −yc) ◦S(Sx, Sy) ◦T (xc, yc) The University of Texas at Austin 17 Department of Computer Sciences Graphics – Fall 2003 (Lecture 6) – In matrix form this is   x′ y′ 1   =   1 0 xc 0 1 yc 0 0 1     Sx 0 0 0 Sy 0 0 0 1     1 0 −xc 0 1 −yc 0 0 1     x y 1   =   Sx 0 xc(1 −Sx) 0 Sy yc(1 −Sy) 0 0 1     x y 1   – Note that the matrices are arranged from right to left in the order of the steps. – The order is important (why)? The University of Texas at Austin 18 Department of Computer Sciences Graphics – Fall 2003 (Lecture 6) • Three Dimensional Transformations – A point is p = [x y z 1], a vector ⃗ v = [x y z 0] – Translation: T (∆x, ∆y, ∆z) =     1 0 0 ∆x 0 1 0 ∆y 0 0 1 ∆z 0 0 0 1     – Scale: S(Sx, Sy, Sz) =     Sx 0 0 0 0 Sy 0 0 0 0 Sz 0 0 0 0 1     – Rotation: Rz(Θ) =     cos(θ) −sin(θ) 0 0 sin(θ) cos(θ) 0 0 0 0 1 0 0 0 0 1     The University of Texas at Austin 19 Department of Computer Sciences Graphics – Fall 2003 (Lecture 6) Extra: Example of Invariance of Projective Transformation, The Cross Ratio Deﬁnition: O C B A A B C B A l l l The University of Texas at Austin 20 Department of Computer Sciences Graphics – Fall 2003 (Lecture 6) A B C D x = CA CB DA DB CA CB DA DB = C′A′ C′B′ D′A′ D′B′ The University of Texas at Austin 21 Department of Computer Sciences Graphics – Fall 2003 (Lecture 6) area OCA = 1 2h · CA = 1 2OA · OC sin ̸ COA area OCB = 1 2h · CB = 1 2OB · OC sin ̸ COB area ODA = 1 2h · DA = 1 2OA · OD sin ̸ DOA area ODB = 1 2h · DB = 1 2OB · OD sin ̸ DOB Hence CA CB DA DB = CA CB · DB DA = OA · OC sin ̸ COA OB · OC sin ̸ COB · OB · OD sin ̸ DOB OA · OD sin ̸ DOA = sin ̸ COA sin ̸ COB · sin ̸ DOB sin ̸ DOA The University of Texas at Austin 22 Department of Computer Sciences Graphics – Fall 2003 (Lecture 6) O A B C D A B C D h Invariance of cross-ratio under central projection The University of Texas at Austin 23 Department of Computer Sciences Graphics – Fall 2003 (Lecture 6) A B C D A B C D ’ ’ ’ ’ 8 Invariance of cross-ratio under parallel projection The University of Texas at Austin 24 Department of Computer Sciences Graphics – Fall 2003 (Lecture 6) (ABCD) > 0 -r r r r A B C D (ABCD) < 0 -r r r r A B C D Sign of cross-ratio (ABCD) = CA CB DA DB = x3 −x1 x3 −x2 x4 −x1 x4 −x2 = x3 −x1 x3 −x2 · x4 −x2 x4 −x1 The University of Texas at Austin 25 Department of Computer Sciences Graphics – Fall 2003 (Lecture 6) -r r r r r x 0 A B C D x1 -x2 -x3 -x4 -Cross-ratio in terms of coordinates. The University of Texas at Austin 26 Department of Computer Sciences Graphics – Fall 2003 (Lecture 6) Reading Assignment and News Chapter 4 pages 181 - 201, of Recommended Text. Please also track the News section of the Course Web Pages for the most recent Announcements related to this course. ( The University of Texas at Austin 27
15979	https://math.stackexchange.com/questions/3649791/maximum-area-of-the-triangle-formed-by-the-chord	conic sections - Maximum area of the triangle formed by the chord - 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 Maximum area of the triangle formed by the chord Ask Question Asked 5 years, 5 months ago Modified5 years, 3 months ago Viewed 2k times This question shows research effort; it is useful and clear 2 Save this question. Show activity on this post. A chord of negative slope from the point P(264−−−√,0)264,0) is drawn to the ellipse x 2+4 y 2=16 x 2+4 y 2=16 . This chord intersects the ellipse at A and B.(O is the origin). Find the maximum area of the △△ AOB and the slope of the line AB when the area is maximum. My trial ------ Let the chord intersect the ellipse at two points (4cos(α α),2sin(α α)) and (4cos(β β),2sin(β β)) This would lead to the chord equation ---->x 4 x 4 c o s(α+β 2)c o s(α+β 2) + y 2 y 2 sin(α+β 2)α+β 2)= cos(α−β 2)α−β 2). This gives 264√4 264 4 c o s(α+β 2)c o s(α+β 2)= cos(α−β 2)α−β 2). And I've found length of OA and OB.... and tried differentiating to get maximum area. But it is ending up in vain and more over too lengthy. The answer given is maximum area =4, and slope of the line AB is (−1 8 2√)−1 8 2). Any help/hint will be appreciated. Thanks in advance triangles conic-sections polar-coordinates Share Share a link to this question Copy linkCC BY-SA 4.0 Cite Follow Follow this question to receive notifications edited Apr 29, 2020 at 13:13 karthikeya kurellakarthikeya kurella asked Apr 29, 2020 at 11:08 karthikeya kurellakarthikeya kurella 146 9 9 bronze badges Add a comment\| 1 Answer 1 Sorted by: Reset to default This answer is useful 3 Save this answer. Show activity on this post. Let O A O A be a semi-diameter of the ellipse and the tangent at B B be parallel to it. It is evident (see figure below) that the area of any other triangle with the same base O A O A and its third vertex on the ellipse cannot be greater tan the area of triangle O A B O A B. Two semi-diameters like O A O A and O B O B are called conjugate semi-diameters and by the second theorem of Apollonius the area of a triangle formed by any two conjugate semi-diameters is always 1 2 a b 1 2 a b, where a a and b b are the semi-axes of the ellipse. Hence no calculus is needed to answer the given question: maximum area is 1 2 a b=4 1 2 a b=4 and the chord must be drawn so that points A A and B B are the endpoints of two conjugate semi-diameters. To this end it is useful to know that if A=(4 cos α,2 sin α)A=(4 cos⁡α,2 sin⁡α) then O B O B is conjugate to O A O A if B=(4 cos(π/2+α),2 sin(π/2+α))=(−4 sin α,2 cos α).B=(4 cos⁡(π/2+α),2 sin⁡(π/2+α))=(−4 sin⁡α,2 cos⁡α). The value of α α can then be found by imposing line A B A B to pass through point P P. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications edited Jun 12, 2020 at 21:41 answered Jun 12, 2020 at 9:03 Intelligenti paucaIntelligenti pauca 56.2k 4 4 gold badges 50 50 silver badges 91 91 bronze badges 1 i guess this can be done without knowing anything about semi diameters i mean (4cos(x1),4sin(x1)),(4cos(x2),4sin(x2)) and then finding the area using the shoelace theorem we get that the area is 4sin(x2-x1) so max value is 4 and x2=x1+pi/2 ,and similarly the remaining can be done as suggested above inoxb911 –inoxb911 2023-03-04 12:02:24 +00:00 Commented Mar 4, 2023 at 12:02 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 triangles conic-sections polar-coordinates 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 Report this ad Related 5Property of ellipses involving normals at the endpoints of a focal chord and the midpoint of that chord 1Ellipse and chord length 1How do I derive the equation of the chord of a conic in polar coordinates? 2Locus of the midpoint of the chord of contact of orthogonal tangents. 0From a point perpendicular tangents are drawn on the ellipse x 2+2 y 2=2 x 2+2 y 2=2. The chord of contact touches a circle concentric with ellipse... 2Variable pairs of chords at right angles are drawn through a point P P (with eccentric angle π/4 π/4) on the ellipse. 1P Q:2 x+y+6=0 P Q:2 x+y+6=0 is a chord hyperbola x 2−4 y 2=4 x 2−4 y 2=4, and R=(α,β)R=(α,β) with α 2+β 2−1≤0 α 2+β 2−1≤0, such that area of triangle P Q R P Q R is minimum. 3Maximum distance between mid-point of chord of ellipse Hot Network Questions Gluteus medius inactivity while riding alignment in a table with custom separator Why is a DC bias voltage (V_BB) needed in a BJT amplifier, and how does the coupling capacitor make this possible? Why are LDS temple garments secret? Suggestions for plotting function of two variables and a parameter with a constraint in the form of an equation Checking model assumptions at cluster level vs global level? 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15980	https://www.sciencedirect.com/science/article/abs/pii/S0936655516303004	Management of Early-stage Hodgkin Lymphoma: A Practice Guideline - ScienceDirect Skip to main contentSkip to article Journals & Books Access throughyour organization Purchase PDF Patient Access Other access options Search ScienceDirect Article preview Abstract Introduction Section snippets References (25) Cited by (2) Clinical Oncology Volume 29, Issue 1, January 2017, Pages e5-e12 Guidelines Management of Early-stage Hodgkin Lymphoma: A Practice Guideline Author links open overlay panel J.Herst∗, M.Crump†, F.G.Baldassarre‡, J.MacEachern§, J.Sussman¶, D.Hodgson†, M.C.Cheung\|\| Show more Add to Mendeley Share Cite rights and content Highlights •A clinical practice guideline for early stage Hodgkin lymphoma is proposed. •The recommendations, based on a systematic review, have been reviewed by an external panel. •Evidence quality was evaluated with the Cochrane Risk of Bias tool and we used GRADE. •Combined modality therapy or chemotherapy alone are options for early-stage Hodgkin lymphoma. •PET scanning was not considered a good tool to identify patients for whom IFRT could be omitted. Abstract In the past, treatment for patients with early-stage Hodgkin lymphoma consisted mainly of radiotherapy. Now, chemotherapy alone and chemoradiotherapy are treatment options. These guidelines aim to provide recommendations on the optimal management of early-stage Hodgkin lymphoma. We conducted a systematic review searching MEDLINE, EMBASE, the Cochrane Library and other literature sources from 2003 to 2015, and applied the Grading of Recommendations Assessment, Development and Evaluation (GRADE). Two authors independently reviewed and selected studies, and appraised the evidence quality. The document underwent internal and external review by content, methodology experts, a patient representative and clinicians in Ontario. We have issued recommendations for patients with classical Hodgkin lymphoma and with nodular lymphocyte predominant Hodgkin lymphoma; with favourable and unfavourable prognosis; and for the use of positron emission tomography to direct treatment. We have provided our interpretation of the evidence and considerations for implementation. Examples of recommendations are: ‘Patients with early-stage classical Hodgkin lymphoma should not be treated with radiotherapy alone’; ‘chemotherapy plus radiotherapy or chemotherapy alone are recommended treatment options for patients with early-stage non-bulky Hodgkin lymphoma’; ‘The Working Group does not recommend the use of a negative interim positron emission tomography scan alone to identify patients with early-stage Hodgkin lymphoma for whom radiotherapy can be omitted without a reduction in progression-free survival’. Through the use of GRADE, recommendations were geared towards patient important outcomes and their strength reflected the available evidence and its interpretation from the patients’ point of view. Introduction Hodgkin lymphoma is notable among malignancies in being associated with high cure rates with contemporary therapies. Current 5 year survival rates for patients with early-stage Hodgkin lymphoma are in the range of 94–96% . With improvements in therapy and disease control, awareness has gradually increased of the potential long-term effects of therapy. Chemotherapy may be associated with delayed toxicity such as infertility and second malignancies. Radiation therapy may be associated with long-term risks such as cardiovascular effects and in-field second malignancies. With high contemporary cure rates and a gradual appreciation of the long-term toxicity of therapy, attention has shifted towards a de-escalation of therapy in an attempt to minimise late adverse effects. The most recent manifestation of this trend has been the incorporation of positron emission tomography (PET) into treatment protocols in an attempt to identify patients who may be candidates for less intensive therapy, while still being able to preserve excellent results of treatment. The attempt to strike a balance between the competing risks of effective therapy and the minimisation of late effects has resulted in substantial variation in practice in the management of Hodgkin lymphoma, particularly early-stage Hodgkin lymphoma. This guideline on the optimal management of early-stage Hodgkin lymphoma was undertaken by the Early-Stage Hodgkin Lymphoma guideline development group, which was convened at the request of the Cancer Care Ontario (CCO) Hematology Disease Site Group (HDSG) in collaboration with the CCO Program in Evidence-based Care (PEBC). The PEBC is a provincial initiative of CCO supported by the Ontario Ministry of Health and Long-Term Care (OMHLTC). All work produced by the PEBC is editorially independent from the OMHLTC. Access through your organization Check access to the full text by signing in through your organization. Access through your organization Section snippets Materials and Methods The goal of this guideline was to provide recommendations for the management of patients with early-stage Hodgkin lymphoma. A Working Group was formed for this project as part of the CCO HDSG. The HDSG includes specialists in haematology, radiation oncology, pharmacy, health research methodology and a patient representative. The HDGS reviewed, provided feedback on drafts and approved this guideline at multiple meetings during its development. The draft guideline was reviewed by experts external Evidence-based Recommendations Recommendation 1 (A)Patients with early-stage classical Hodgkin lymphoma should not be treated with radiotherapy alone. (B)In patients with early-stage nodular lymphocyte predominant Hodgkin lymphoma (NLPHL) it is reasonable to use involved field radiotherapy (IFRT) alone. However, no phase III clinical trials have focused exclusively on NLPHL. Therefore, no strong evidence base for such treatment, or for relative dosage, is currently available, and this recommendation is based on expert opinion of the Discussion and Conclusions Early-stage Hodgkin lymphoma was first treated with EFRT. Radiotherapy as a single modality treatment is no longer used because it was shown to be inferior to treatment with combination chemo- and radiotherapy in trials conducted before our literature search cut-off. In light of the equivalent efficacy and reduction in toxicity offered by more limited radiotherapy fields, IFRT in combination with chemotherapy has now supplanted EFRT (either alone or with chemotherapy). INRT or ISRT are now Conflict of Interest M. Crump was consultant for Eli Lilly in 2013, and has been the principal investigator in a trial involving one of the objects of study; J. Sussman received grants and research support as a principal investigator in a trial involving one of the objects of study. All other authors declared no conflict of interest. Acknowledgements The Hematology Disease Site Group would like to thank Cancer Care Ontario for funding this guideline and the following individuals: Chika Agbassi, Melissa Brouwers, Elizabeth Chen, Joseph Connors, Roxanne Cosby, David MacDonald, Sheila McNair, Hans Messersmith, Ur Metser, Janet Rowe, Marko Simunovic, Crystal Su, Shail Verma and Kristy Yiu for providing feedback on draft versions. Recommended articles References (25) R.H. Hamed et al. A randomized trial of brief treatment of early-stage Hodgkin lymphoma: is it effective? Hematol Oncol Stem Cell Ther (2012) V. Pavone et al. ABVD plus radiotherapy versus EVE plus radiotherapy in unfavorable stage IA and IIA Hodgkin's lymphoma: results from an Intergruppo Italiano Linfomi randomized study Ann Oncol (2008) K. Behringer et al. Omission of dacarbazine or bleomycin, or both, from the ABVD regimen in treatment of early-stage favourable Hodgkin's lymphoma (GHSG HD13): an open-label, randomised, non-inferiority trial Lancet (2015) R.M. Meyer et al. ABVD alone versus radiation-based therapy in limited-stage Hodgkin's lymphoma N Engl J Med (2012) A. Engert et al. Two cycles of doxorubicin, bleomycin, vinblastine, and dacarbazine plus extended-field radiotherapy is superior to radiotherapy alone in early favorable Hodgkin's lymphoma: final results of the GHSG HD7 trial J Clin Oncol (2007) P.A. Ganz et al. Health status and quality of life in patients with early-stage Hodgkin's disease treated on Southwest Oncology Group Study 9133 J Clin Oncol (2003) M. Rancea et al. Hodgkin's lymphoma in adults: diagnosis, treatment and follow-up Dtsch Arztebl Int (2013) R.M. Meyer et al. Randomized comparison of ABVD chemotherapy with a strategy that includes radiation therapy in patients with limited-stage Hodgkin's lymphoma: National Cancer Institute of Canada Clinical Trials Group and the Eastern Cooperative Oncology Group J Clin Oncol (2005) J. Radford et al. Results of a trial of PET-directed therapy for early-stage Hodgkin's lymphoma N Engl J Med (2015) J. Thomas et al. Results of the EORTC-GELA H9 randomized trials: the H9-F trial (comparing 3 radiation dose levels) and H9-U trial (comparing 3 chemotherapy schemes) in patients with favorable or unfavorable early stage hodgkin’s lymphoma (HL) J.M. Raemaekers et al. Omitting radiotherapy in early positron emission tomography-negative stage I/II Hodgkin lymphoma is associated with an increased risk of early relapse: clinical results of the preplanned interim analysis of the randomized EORTC/LYSA/FIL H10 trial J Clin Oncol (2014) View more references Cited by (2) Hodgkin’s lymphoma in adults – diagnosis and treatment 2024, Family Medicine and Primary Care Review ### A comparison of selected dosimetric parameters of two Hodgkin Lymphoma radiotherapy techniques with reference to potential risk of radiation induced heart disease 2018, Polish Journal of Medical Physics and Engineering View full text © 2016 The Royal College of Radiologists. Published by Elsevier Ltd. All rights reserved. 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15981	https://www.insidemathematics.org/sites/default/files/materials/bikes%20and%20trikes.pdf	Bikes and Trikes This problem gives you the chance to: • solve number problems in a real context The cycle shop on Main Street sells bikes (two wheels) and trikes (three wheels). 1. Yesterday, Sarah counted all of the cycles in the shop. There were seven bikes and four trikes in the shop. How many wheels were there on these eleven cycles? __ Show your calculation. 2. Today, Sarah counted all of the wheels of all of the cycles in the shop. She found that there were 30 wheels in all. There were the same number of bikes as there were trikes. How many bikes were there? __ How many trikes were there? __ Show how you figured it out. Copyright © 2006 by Mathematics Assessment Page 8 Bikes and Trikes Test 4 Resource Service. All rights reserved. 8 Bikes and Trikes Rubric The core elements of performance required by this task are: • solve number problems in a real context Based on these, credit for specific aspects of performance should be assigned as follows points section points 1. Gives correct answer: 26 wheels Shows work such as: 7 x 2 and 4 x 3 14 + 12 = Accept repeated addition or diagrams 1 2 3 2. Gives correct answers: 6 bikes and 6 trikes Gives correct explanation such as: 6 bikes = 12 wheels 6 trikes = 18 wheels in all 30 wheels May list or draw diagrams 1 bike and 1 trike = 2 + 3 = 5 wheels 2 bikes and 2 trikes = 4 + 6 or 2 x 5 = 10 wheels 3 bikes and 3 trikes = 6 + 9 or 3 x 5 = 15 wheels 4 bikes and 4 trikes = 8 + 12 or 4 x 5 = 20 wheels 5 bikes and 5 trikes = 10 + 15 or 5 x 5 = 25 wheels 6 bikes and 6 trikes = 12 + 18 or 6 x 5 = 30 wheels 2 3 5 Total Points 8 Grade Four – 2006 pg. (c) Noyce Foundation 2006. To reproduce this document, permission must be granted by the Noyce Foundation: info@noycefdn.org. 83 Bikes and Trikes Work the task and examine the rubric. How does this task get students to focus on multiple groups and using multiplication in context? What strategies do you think students would use to solve this task? How often do students in your class get problems with multiple constraints? What strategies do you have them use to keep track of what they know and what they are trying to find out? Look at student work in part 1. What might be the thinking of students who gave an answer of 22? Of 28? Of 14? What strategies did your students use to think about and solve this part of the task? Did your students use: Calculations or number sentences only Repeated addition Diagrams or pictures Combination calculations & diagrams Labeling of answers Other In part two, the students needed to keep track of several constraints: the number of wheels on a bike and on a trike, the number of cycles had to be equal, and altogether there had to be 30 wheels. How many of your students put: 6,6, 4,7 Combination that yields correct # wheels, such as 9, 4 or 12, 2 15,10 Other What might be the errors in logic for some of these errors? Now look at strategies. Did your students use: Calculations like: 6x2=12, 6x3=18, 12+18=30 Counting by 5’s or dividing by 5’s Making a table Drawing or Diagrams Other What might have confused students about the mathematics of the task? What types of experiences do students need? Grade Four – 2006 pg. (c) Noyce Foundation 2006. To reproduce this document, permission must be granted by the Noyce Foundation: info@noycefdn.org. 84 Looking at Student Work on Bikes and Trikes Many students seem to understand the multiple groups and combining groups. For them, the task is just a series of multiplication and addition steps. See the work of Student A. Student A Grade Four – 2006 pg. (c) Noyce Foundation 2006. To reproduce this document, permission must be granted by the Noyce Foundation: info@noycefdn.org. 85 Student B uses pictures to think about labeling the computations in part one and then transitions into using a diagram in part two. The student knows that for the bikes and trikes to be equal there will be sets consisting of a bike and a trike. The diagram shows that for each set there are 5 wheels. This allows the student to find how many sets of 5 fit into 30 wheels. Student B Grade Four – 2006 pg. (c) Noyce Foundation 2006. To reproduce this document, permission must be granted by the Noyce Foundation: info@noycefdn.org. 86 Student C shows similar thinking about the 5 wheels for each set, but solves the task by making a table. The numbers on the left also make equivalent fractions, 2/3=4/6=6/9. At later grades this would be a useful connection to help students make. Student C Grade Four – 2006 pg. (c) Noyce Foundation 2006. To reproduce this document, permission must be granted by the Noyce Foundation: info@noycefdn.org. 87 Student D also uses a table to think about all the combinations of equal bikes and trikes. Then the student can check for combinations that yield 30 wheels. Student D Student E and F both use diagrams to find the number of bikes and trikes that yield 30 wheels. For part two the diagrams look similar. The difference is in how Student E makes sense of the constraint, equal number of each, and matches the 2’s and 3’s. Grade Four – 2006 pg. (c) Noyce Foundation 2006. To reproduce this document, permission must be granted by the Noyce Foundation: info@noycefdn.org. 88 Student E Student F Grade Four – 2006 pg. (c) Noyce Foundation 2006. To reproduce this document, permission must be granted by the Noyce Foundation: info@noycefdn.org. 89 Scores for this task, clustered around 8, 3, and 0. Students with a score of 3,like Student F, struggled with part three of the task. Student F was able to find a solution that yielded the right number of wheels, but missed the constraint “same number of each”. There were other types of conceptual errors around part two of the task. Student G did not make sense of the 30 wheels and just used listed the information given about bikes and trikes in previous part of the task. Student G Student H did not understand that the two types of bikes combined needed to equal 30 wheels and solved the simpler problem of how many bikes equal 30 wheels and how many trikes equal 30 wheels. Grade Four – 2006 pg. (c) Noyce Foundation 2006. To reproduce this document, permission must be granted by the Noyce Foundation: info@noycefdn.org. 90 Student H Many students, like Student I, found equal amounts of bikes and trikes to yield 30 “cycles” instead of 30 “wheels”. Student I Grade Four – 2006 pg. (c) Noyce Foundation 2006. To reproduce this document, permission must be granted by the Noyce Foundation: info@noycefdn.org. 91 Why are some students scoring no points on the task? Are there any understandings to build on? What misconceptions do they need to overcome? Look at Student J. The student adds the bikes and trikes together and multiplies by 2 wheels per cycle. The student doesn’t pick up on the difference in wheels between the two types of cycles. In the part two, the student seems to have learned an underline strategy to help identify what is being asked, but then mistakes 30 for bikes instead of wheels. Student J Student K shows similar thinking. The student draws out the 11 bicycles (The top dot may represent the seat or handle bars on the bike) and counts the wheels. In part 2, Student K splits 30 cycles into two equal parts. Grade Four – 2006 pg. (c) Noyce Foundation 2006. To reproduce this document, permission must be granted by the Noyce Foundation: info@noycefdn.org. 92 Student K Student L seems to have the strategy of underlining important information. The student sees two wheels and three wheels and adds them together to get 5 wheels, ignoring the other information in the story. In part two the student says, “Read the directions, it says there are 30.” The student doesn’t perceive that a question is being asked that requires calculations. Grade Four – 2006 pg. (c) Noyce Foundation 2006. To reproduce this document, permission must be granted by the Noyce Foundation: info@noycefdn.org. 93 Student L Student M uses numbers and number sentences that appear unrelated to the context of the problem. Where do you start to help this student understand basic ideas about operation and quantity? Student M Grade Four – 2006 pg. (c) Noyce Foundation 2006. To reproduce this document, permission must be granted by the Noyce Foundation: info@noycefdn.org. 94 Student N makes no sense of the first part of the task or the meaning of multiplication. The student multiplies 4 trikes by 7 bikes to get 28_(?). However, the student does have some understanding of the context, because in part two the solution would yield 30 wheels. It is important not to generalize about all the students with the same low score, because what they are making sense of varies widely and requires different intervention strategies. Student N Grade Four – 2006 pg. (c) Noyce Foundation 2006. To reproduce this document, permission must be granted by the Noyce Foundation: info@noycefdn.org. 95 Fourth Grade 4th Grade Task 5 Bikes and Trikes Student Task Use multiplication and division to solve problems about wheels per bike and total wheels in a bike shop. Core Idea 3 Patterns, Functions, and Algebra Understand patterns and use mathematical models to represent and to understand qualitative relationships. • Find results of a rule for a specific value. • Use inverse operations to solve multi-step problems • Use concrete, pictorial, and verbal representations to solve problems involving unknowns. • Understand and use the concept of equality. Mathematics of the task: • Ability to add and multiply • Ability to work with equal-sized groups of objects • Ability to use multiple constraints • Ability to begin reasoning proportionally Based on teacher observations, this is what fourth graders knew and were able to do: • Knew multiplication facts • Could draw diagrams or make number sentences to help them solve the task • Knew the difference between bikes and trikes Areas of difficulty for fourth graders: • Tracking all the constraints in part two • Confusing wheels and cycles • Making sense of the entire set of information before beginning computation Strategies used by successful students: • Making diagrams • Labeling answers to keep track of what each calculation represented • Counting or dividing by 5’s (seeing the incremental number of wheels) • Making tables Grade Four – 2006 pg. (c) Noyce Foundation 2006. To reproduce this document, permission must be granted by the Noyce Foundation: info@noycefdn.org. 96 MARS Test Task 5 Frequency Distribution and Bar Graph, Grade 4 Task 5 – Bikes and Trikes Mean: 3.81 StdDev: 3.18 MARS Task 5 Raw Scores The maximum score available for this task is 8 points. The minimum score for a level 3 response, meeting standards, is 3 points. Most students, 70%, understood the process for finding the total number of wheels in part one, but some may have made computational errors. More than half the students, 65%, could solve part one with no computational errors. 30% of the students could make sense of the entire task including finding equal numbers of bikes and trikes to make 30 wheels. 27% of the students scored no points on this task. All of the students in the sample with this score attempted the task. Grade Four – 2006 pg. (c) Noyce Foundation 2006. To reproduce this document, permission must be granted by the Noyce Foundation: info@noycefdn.org. 97 Bikes and Trikes Points Understandings Misunderstandings 0 All the students in the sample with this score attempted the task. Students were confused by the constraints. They may have added bikes and trikes before multiplying, added just the wheels, or multiplied bikes times trikes. 2 Students understood the process for finding the total number of wheels in part one, but made computational errors. Basic addition and multiplication errors. 3 Students could find the number of wheels for 7 bikes and 3 trikes using drawings/diagrams (7%), or multiplication and addition (32%). Students did not understand some of the constraints in part 3. They confused 30 wheels for 30 cycles. They worried about getting 30 wheels and forgot about getting the “same number” of each kind. Some solved the simpler problem. How many bikes make 30 wheels? How many trikes make 30 wheels? 5 Students could find the equal number of bikes and trikes to make 30 wheels, but could not solve anything in part 1 of the task. 8 Students could deal with multiple constraints and think about equal, repeated groups: groups of bikes with two-wheels or groups of trikes with three-wheels. They could find the total number of wheels for a given number of bikes and trikes or work backwards from the number of wheels to the number of cycles. Grade Four – 2006 pg. (c) Noyce Foundation 2006. To reproduce this document, permission must be granted by the Noyce Foundation: info@noycefdn.org. 98 Implications for Instruction Students need to look to recognize contexts, which involve multiple groups, as multiplication situations. Students need practice picking out key pieces of information and organizing their work to solve problems. The grappling with setting up the problem and then discussing ways other students set up the problem helps students build an understanding of the meaning of the operation of multiplication. Learning to make diagrams (students should be transitioning from pictures to diagrams), using a bar model, or working with a number line help students to “see” the action of the story problem. Another important tool for making sense of calculations is the use of labels. It is not that labels are a rule to please the teacher. When working several steps at once, labels can help the student know what has been found and think about what still needs to be done. In looking at student work, many of them quit before the final step. Could putting a post-it with a quick note about labels help them see that they aren’t done yet? How would using labels help students think more carefully in part 2? Ideas for Action Research Learning from Good Mistakes: Sometimes looking at a mistake can help uncover some deeper mathematics and confront students’ misconceptions and can promote good, productive discussions. The process of trying to reconcile what is in the mistake and what would work to solve the problems, helps students to firm up their ideas and cement their learning. It also allows students to see their own logic errors and revise their thinking. Consider posing the following problem to your class after everyone has had a chance to try and make sense of this problem by themselves: Cynthia thinks that it is important to add the number of wheels together. Three wheels plus 2 wheels equal 5. Could this help her solve part one? Why or why not? Could it help her solve part two? Why or why not? How are these two situations different? Why can we use it in one part but not both? Then have students think about: Conner thinks there is an add and multiply part to the problem. He does 7+4 = 11 and then 11 x2 = 22. What would be the labels for the 7, 4, and 11? What does it mean when we multiply the 11 by 2? What is being found? Is this what we want to know? Why or why not? Make an overhead transparency of the work of Student E and F. Both students have solutions that yield an answer of 30 wheels. Can they both be right? After students have discussed the question, state that both drawings look almost alike. Why did one give the correct answer for this task and one give only a partially correct solution? What is different about the two? This important idea of what is the same and what is different helps students think about diagram literacy and ways to use diagrams productively in the future. © 2012 Noyce Foundation Performance Assessment Task Bikes and Trikes Grade 4 The task challenges a student to demonstrate understanding of concepts involved in multiplication. A student must make sense of equal sized groups of objects. A student must recognize and abide by the multiple constraints of the problems. A student must determine how best to find the unknown number in a word problem using inverse operations and/or pictorial or verbal representations. A student must understand and use the concept of equality. Common Core State Standards Math - Content Standards Operations and Algebraic Thinking Use the four operations with whole numbers to solve problems. 4.OA.2 Multiply or divide to solve word problems involving multiplicative comparison, e.g., by using drawings and equations with a symbol for the unknown number to represent the problem, distinguishing multiplicative comparison from additive comparison. Common Core State Standards Math – Standards of Mathematical Practice MP.2 Reason abstractly and quantitatively. Mathematically proficient students make sense of quantities and their relationships in problem situations. They bring two complementary abilities to bear on problems involving quantitative relationships: the ability to decontextualize—to abstract a given situation and represent it symbolically and manipulate the representing symbols as if they have a life of their own, without necessarily attending to their referents—and the ability to contextualize, to pause as needed during the manipulation process in order to probe into the referents for the symbols involved. Quantitative reasoning entails habits of creating a coherent representation of the problem at hand; considering the units involved; attending to the meaning of quantities, not just how to compute them; and knowing and flexibly using different properties of operations and objects. MP.4 Model with mathematics. Mathematically proficient students can apply the mathematics they know to solve problems arising in everyday life, society, and the workplace. In early grades, this might be as simple as writing an addition equation to describe a situation. In middle grades, a student might apply proportional reasoning to plan a school event or analyze a problem in the community. By high school, a student might use geometry to solve a design problem or use a function to describe how one quantity of interest depends on another. Mathematically proficient students who can apply what they know are comfortable making assumptions and approximations to simplify a complicated situation, realizing that these may need revision later. They are able to identify important quantities in a practical situation and map their relationships using such tools as diagrams, two-way tables, graphs, flowcharts and formulas. They can analyze those relationships mathematically to draw conclusions. They routinely interpret their mathematical results in the context of the situation and reflect on whether the results make sense, possibly improving the model if it has not served its purpose. Assessment Results This task was developed by the Mathematics Assessment Resource Service and administered as part of a national, normed math assessment. For comparison purposes, teachers may be interested in the results of the national assessment, including the total points possible for the task, the number of core points, and the percent of students that scored at standard on the task. Related materials, including the scoring rubric, student work, and discussions of student understandings and misconceptions on the task, are included in the task packet. Grade Level Year Total Points Core Points % At Standard 4 2006 8 3 66 %
15982	https://www-users.cse.umn.edu/~stant001/PAPERS/sieve.pdf	SIEVED PARTITION FUNCTIONS AND Q-BINOMIAL COEFFICIENTS Frank Garvan and Dennis Stanton Abstract. The q-binomial coeﬃcient is a polynomial in q. Given an integer t and a residue class r modulo t, a sieved q-binomial coeﬃcient is the sum of those terms whose exponents are congruent to r modulo t. In this paper explicit polynomial identities in qt are given for sieved q-binomial coeﬃcients. As a limiting case, gener-ating functions for the sieved partition function are found as multidimensional theta functions. A striking corollary of this representation is the proof of Ramanujan’s congruences mod 5, 7, and 11 by exhibiting symmetry groups of orders 5, 7, and 11 of explicit quadratic forms. We also verify the Subbarao conjecture for t = 3, t = 5, and t = 10. 1. Introduction. The q-binomial coeﬃcient (1.1) N k q = (1 −qN) · · · (1 −qN−k+1) (1 −q) · · · (1 −qk) = X i≥0 aiqi is a polynomial in q with integer coeﬃcients. In this paper we shall consider the following families of polynomials formed from (1.1). Let t be a positive integer and consider the terms in (1.1) with residue class r modulo t: (1.2) X i≥0 ati+rqti+r. We refer to (1.2) as a sieved q-binomial coeﬃcient. We give explicit formulas (Theorems 1 and 2 of §2) for the sieved q-binomial coeﬃcient as polynomials in qt. Some limiting cases (Theorems 3A, 3B and 3C of §3) are expressions for sieved partition functions as multidimensional theta functions. In §4 the symmetry groups of the quadratic form of the theta function are computed. Applications of these groups to congruences for the partition function are given in §5. 1991 Mathematics Subject Classiﬁcation. Primary 05A19; Secondary 11P76. Key words and phrases. q-binomial coeﬃcient, partitions. Institute for Mathematics and its Applications, University of Minnesota, Minneapolis, MN 55455. School of Mathematics, University of Minnesota, Minneapolis, MN 55455. This work was partially supported by NSF grant DMS:8700995, and a grant from the Minnesota Supercomputer Institute. Typeset by A MS-T EX 1 2 FRANK GARVAN AND DENNIS STANTON To set the notation and explain the method, we shall now do the t = 2 case. For an integer N and complex numbers a and q, \|q\| < 1 let (a; q)N = N−1 Y i=0 (1 −aqi). We shall also allow N = ∞and note that the generating function for all partitions of i, p(i), is ∞ X i=0 p(i)qi = (q; q)−1 ∞= (q)−1 ∞. We next let t = 2 and r = 0, so that we want the even terms in the q-binomial coeﬃcient. Recall the terminating q-binomial theorem (1.3) (−x; q)N = N X k=0 N k q q( k 2)xk. Suppose N is even. Since (1.4) (−x; q)N + (−x; −q)N = (−x; q2)N/2{(−xq; q2)N/2 + (xq; q2)N/2}, (1.3) clearly implies (1.5) N k q q( k 2) + N k −q (−q)( k 2) = 2 k/2 X m=0 N/2 2m q2 N/2 k −2m q2 q4m2+2( k−2m 2 ). We obtain the following proposition. Proposition 1. If N is even, then k/2 X m=0 N/2 2m q2 N/2 k −2m q2 q4m2+2( k−2m 2 )−( k 2) =        X i a2iq2i for k ≡0 or 1 mod 4 X i a2i+1q2i+1 for k ≡2 or 3 mod 4. The corresponding cases for the diﬀerence in (1.4) are the following. Proposition 2. If N is even, then k/2 X m=0 N/2 2m + 1 q2 N/2 k −2m −1 q2 q(2m+1)2+2( k−2m−1 2 )−( k 2) =        X i a2iq2i for k ≡2 or 3 mod 4 X i a2i+1q2i+1 for k ≡0 or 1 mod 4. Propositions 1 and 2 are evaluations of special well-poised 4φ3’s in the theory of basic hypergeometric series. It is easy to give the relevant versions of Propositions 1 and 2 for N odd, but we do not do so here. SIEVED PARTITION FUNCTIONS AND Q-BINOMIAL COEFFICIENTS 3 2. Sieved q-binomial coeﬃcients. In this section we follow the outline of §1 for a general t, and ﬁnd an expression for the sieved q-binomial coeﬃcients (1.2). We must replace the −q in (1.4) by a sum over all t th roots of unity, so let ω = exp(2πi/t). It is clear from (1.3) that (2.1) t−1 X i=0 (−x; ωiq)N = N X k=0 t−1 X i=0 N k ωiq (ωiq)( k 2)xk. If N is a multiple of t, it is easy to see that (2.2) t−1 X i=0 (−x; ωiq)N = (−x; qt)N/t t−1 X i=0 t−1 Y k=1 (−xωkiqk; qt)N/t The coeﬃcient of xk in (2.2) yields t−1 X i=0 N k ωiq (ωiq)( k 2) = X n1,...,nt t Y i=1 N/t ni qt t−1 X p=0 θp where θ = ω Pt i=2(i−1)ni. The result is the following theorem. Note that the residue class chosen from the q-binomial coeﬃcient depends upon the value of k 2 mod t. Theorem 1. Let N be a multiple of t, and k 2 ≡t −r mod t, 1 ≤r ≤t. If ai is the coeﬃcient of qi in the q-binomial coeﬃcient N k q , then ∞ X i=0 ati+rqti+r = X n1,...,nt t Y i=1 N/t ni qt qt(Pt i=1 ( ni 2 )+m)−( k 2), where the summation parameters satisfy n1 + · · · + nt = k and n2 + 2n3 + · · · + (t − 1)nt = tm, for some integer m. The residue class r in Theorem 1 is determined by k and t. For a given N,k, and t, one would like a version for any residue class r. This can be done by inserting the appropriate roots of unity in the left side of (2.1). We give such a version for k ≡0 mod 2t. Theorem 2. Let N be a multiple of t, k be a multiple of 2t, and 1 ≤r ≤t −1. If ai is the coeﬃcient of qi in the q-binomial coeﬃcient N k q , then ∞ X i=0 atiqti+t−r = X n1,...,nt t Y i=1 N/t ni qt qt(Pt i=1 ( ni 2 )+m)−( k 2), where the summation parameters satisfy n1 + · · · + nt = k and n2 + 2n3 + · · · + (t − 1)nt = tm + r, for some integer m. 4 FRANK GARVAN AND DENNIS STANTON 3. Sieved partition functions. Theorems 1 and 2 are polynomial identities which imply generating functions for sieved partition functions. In this section we take the appropriate limits to obtain these identities: Theorems 3A, 3B, and 3C. To see this, recall that the q-binomial coeﬃcient (1.1) is the generating function for partitions which lie inside a k × (N −k) rectangle. If N →∞, we obtain all partitions with at most k parts. Next if k →∞, we ﬁnd all partitions. We will apply this sequence of limits to Theorems 1 and 2 for our results on the sieved partition function. First we take the t = 2 case to demonstrate the technique. Let k ≡0 mod 4 in Proposition 1. If N →∞, we ﬁnd that the right side becomes k/2 X m=0 q8m2−4km+2m+k2/2−k/2 (q2; q2)2m(q2; q2)k−2m . The exponent of q is a quadratic polynomial in m whose minimum occurs at m = k/4 −1/8, 2(m −k/4) + 8(m −k/4)2. Thus replacing m by m + k/4 and letting k →∞, we ﬁnd (3.1) ∞ X i=0 p(2i)qi = 1 (q)2 ∞ ∞ X m=−∞ q4m2+m. For k ≡2 mod 4, Proposition 1 implies (3.2) ∞ X i=0 p(2i + 1)qi = 1 (q)2 ∞ ∞ X m=−∞ q4m2−3m. We do the same sequence of steps for Theorems 1 and 2, which are t−1 fold sums-the summation parameters being m, n3, . . . , nt. The N →∞limit of both theorems is easily computed by replacing the q-binomial coeﬃcients by a single product in the denominator. For the k →∞limit, one must again ﬁnd the minimum value of the quadratic function of m, n3, . . . , nt in the exponent of q, and then shift the domain of these summation parameters so that the minimum occurs at the origin. For Theorem 2, it can be shown that a minimum value occurs at m = (t −1)k/2t, ni = k/t, i ̸= r + 2, . . . , nr+2 = k/t −1, or at m = (t −1)k/2t, ni = k/t, i ̸= r + 1, . . . , nr+2 = k/t −1. Explicitly computing the Taylor series of the exponent of q at these values, shifting the parameters, and taking the limit on k gives the following two theorems. Theorem 3A. Let 1 ≤r ≤t −2. Then ∞ X n=0 p(tn + t −r)qn = 1 (q)t ∞ ∞ X m,n3,...,nt=−∞ qQ(m,n3,...,nt)+LA(m,n3,...,nt), where Q(m, n3, . . . , nt) = t2m2+ t X i=3 (i2−3i+3)n2 i + t X i=3 (3−2i)tmni+ X 3≤i<j≤t (2ij−3i−3j+5)ninj, LA(m, n3, . . . , nt) = t X i=3,i̸=r+2 (1 −i)ni + (−r −2)nr+2 + (1 + t)m. SIEVED PARTITION FUNCTIONS AND Q-BINOMIAL COEFFICIENTS 5 Theorem 3B. Let 2 ≤r ≤t −1. Then ∞ X n=0 p(tn + t −r)qn = 1 (q)t ∞ ∞ X m,n3,...,nt=−∞ qQ(m,n3,...,nt)+LB(m,n3,...,nt), where Q(m, n3, . . . , nt) is given in Theorem 3A, and LB(m, n3, . . . , nt) = t X i=3,i̸=r+1 (i −2)ni + (r −2)nr+1 + (1 −t)m. Note that for diﬀerent r, the right sides of Theorems 3A (3B) diﬀer only in the special linear term related to r. A linear change of variables shows that Theorems 3A and 3B are equivalent where they overlap, 2 ≤r ≤t −2. The result which follows from Theorem 1 for r = 0 is Theorem 3C. We have ∞ X n=0 p(tn)qn = 1 (q)t ∞ ∞ X m,n3,...,nt=−∞ qQ(m,n3,...,nt)+LC(m,n3,...,nt), where Q(m, n3, . . . , nt) is given in Theorem 3A, and LC(m, n3, . . . , nt) = m. We note that Kolberg has also obtained representations for the generating functions for p(tn + r). His representations are in terms of t × t determinants of theta-functions and are diﬀerent from our results. For the cases t = 2, 3, 5, 7 his determinants simplify to nice linear combinations of certain inﬁnite products. For example, Kolberg’s [5, (3.1)] is (3.3) ∞ X n=0 p(3n)qn = (q3; q3)∞(q9; q9)2 ∞(q4; q9)2 ∞(q5; q9)2 ∞ (q)4 ∞ −q (q9; q9)2 ∞ (q)3 ∞(q4; q9)∞(q5; q9)∞ . We can ﬁnd similar identities by diagonalizing the quadratic form Q and applying the Jacobi triple product identity [1, p.21]. For example, let t = 3 in Theorem 3C ∞ X n=0 p(3n)qn = 1 (q)3 ∞ ∞ X m,s=−∞ q3s2+m−9ms+9m2. There are several ways to diagonalize this form. One is to note that Q(m + s, 2s) and Q(m + s, 2s + 1) are diagonal, and we ﬁnd the following evaluation. ∞ X n=0 p(3n)qn = (q18; q18)∞(q6; q6)∞ (q)3 ∞ (−q7; q18)∞(−q11; q18)∞(−q3; q6)2 ∞ + 2q2(−q2; q18)∞(−q16; q18)∞(−q6; q6)2 ∞ (3.4) There are also versions of (3.4) for 3n + 1 and 3n + 2. Equating the right sides of (3.3) and (3.4) gives rise to a surprising theta-function identity. 6 FRANK GARVAN AND DENNIS STANTON 4. Symmetry groups. It is natural to ask if Theorems 3A, 3B, or 3C imply congruences for the partition function p(n); for example the Ramanujan congruences for p(5n + 4), p(7n + 5), and p(11n + 6). In this section we ﬁnd the symmetry groups for Theorems 3A-3C which will imply congruence theorems in §5. How would Ramanujan’s congruences follow from Theorems 3A and 3B? We take as an example p(5n+4) ≡0 mod 5. Consider Q(m, n3, n4, n5)+LA(m, n3, n4, n5) from Theorem 3A with t = 5 and r = 1. Suppose there exists an aﬃne transforma-tion T : Z4 →Z4, Tx = Mx + τ, such that (1) T 5 = identity, (2) T has no ﬁxed points in Z4, (3) T preserves the form Q + LA. Then each orbit of T on the vectors (m, n3, n4, n5) must consist of ﬁve vectors. This clearly implies Ramanujan’s congruence. Thus we should compute the symmetry group G of the form Q + LA, and see if G contains a cycle of order ﬁve with property (2). For any of the forms in Theorems 3A, 3B, and 3C, we deﬁne G to be the set of aﬃne transformations which preserve the form and the lattice Zt−1. Table 1 lists these groups. We let Zn denote the cyclic group of order n, and Dn denote the dihedral group of order 2n. 3n Z2 4n Z2 5n D6 6n Z2 7n D8 3n+1 Z2 4n+1 Z2 5n+1 D4 6n+1 Z2 7n+1 D6 3n+2 Z2 4n+2 Z2 5n+2 D4 6n+2 Z2 7n+2 D8 4n+3 Z2 5n+3 D6 6n+3 Z2 7n+3 D6 5n+4 D5 6n+4 Z2 7n+4 D6 6n+5 Z2 7n+5 D7 7n+6 D8 Table 1 The groups in Table 1 were found in the following way. Let L be the linear form LA, LB, or LC for Theorems 3A, 3B, and 3C respectively. Any element of G must permute the set of vectors {(m, n3, . . . , nt)} on which Q + L is constant. Let the Q + L = 0 term correspond to a set V with p(r) such vectors, including the zero vector. (The ﬁve vectors for p(5n + 4) are given in §5.) If V contains at least t −1 independent vectors, then any permutation of V induces at most one aﬃne transformation of Zt−1, which may or may not preserve Q + L. For small values of t the permutations can be checked by hand; large values of t require a computer. The following proposition is in agreement with Table 1. Proposition 3. For any r and t, the symmetry group of the forms Q+LA, Q+LB, or Q + LC contains Z2. Proof. Let x denote the column vector (m, n3, . . . , nt), and let L be a linear form on x. Let m(Q) be the symmetric matrix such that Qx = transpose(x)m(Q)x. It is easy to verify that Tx = Mx + τ has order two, and preserves Q + L and the lattice Zt−1 if, and only if, (1) M 2 = I, (2) M and τ have integral entries, (3) Mτ = −τ, (4) M preserves Q, QMx = Qx, SIEVED PARTITION FUNCTIONS AND Q-BINOMIAL COEFFICIENTS 7 (5) transpose(τ) = (LM −L)(2m(Q))−1. We now give the (t −1) × (t −1) matrix M and the vector τ. The matrix (2m(Q))−1 can be found explicitly, so that the veriﬁcations of (1)-(5) are tedious. M is independent of the residue class r. The only non-zero entries in rows 2 through t −3 are the -1’s at (i, t −1 −i). M =           1 0 0 · · · 0 0 0 0 0 0 · · · −1 0 0 . . . . . . . . . ... . . . . . . . . . 0 0 −1 · · · 0 0 0 0 −1 0 · · · 0 0 0 −t 2 3 · · · t −3 t −2 t −1 t −1 −2 · · · −(t −4) −(t −3) −(t −2)           The vector τ does depend upon the residue class r and the form L. For r = 0 and L = LC, it is easy to see that τ = 0. For Theorem 3B, τ has at most three non-zero entries: (1) if 2 ≤r ≤t −3 and 2r + 1 ̸= t, the non-zero entries are τr = τt−1−r = 1 and τt−1 = −1, (2) if 2r + 1 = t, the non-zero entries are τr = 2 and τt−1 = −1, (3) if r = t −2, the non-zero entries are τt−2 = 1 and τt−1 = −1, (4) if r = t −1, then τ = 0. The ﬁnal residue class is r = 1 in Theorem 3A. In this case the non-zero entries are τ2 = 1, τt−3 = 1, and τt−2 = −1. □ 5. Congruences. The groups listed in §4 can be used to derive congruence theorems. We do so in this section, proving the Ramanujan congruences and verifying the Subbarao conjecture for t = 3 and t = 5. The basic idea is that elements of G of order k should give us information about p(tn + r) mod k. This information is given in the next proposition. Proposition 4. Let g ∈G have order pa, for some prime p. If FP is the ﬁxed point set of g in Zt−1, then ∞ X n=0 p(tn + r)qn ≡ 1 (q)t ∞ X (m,n3,...,nt)∈F P qQ(m,n3,...,nt)+LA(m,n3,...,nt) mod p. For the ﬁrst application of Proposition 4 and Table 1 we take p(5n + 4) ≡0 mod 5. Table 1 shows that the symmetry group for 5n + 4 contains an element g of order ﬁve. Thus we need FP(g) = ∅. Here is the explicit construction of g = T. The ﬁve vectors which correspond to p(4) in Theorem 3A are (−1, 1, −1, −1), (0, 1, 1, −1), (0, 2, −1, 0), (0, 0, 0, 0), and (0, 1, 0, 0). We then take the unique aﬃne transformation which cycles these ﬁve vectors in this order; it is Tx = Mx + τ, where the matrix M and the vector τ are M =    4 −1 −2 −3 −1 0 1 0 2 −1 −2 −2 5 −1 −2 −3   , τ =    0 1 0 0   . 8 FRANK GARVAN AND DENNIS STANTON It is easy to verify that T has no ﬁxed points in Z4. The transformation T satisﬁes another nice property, (5.1) x + Tx + T 2x + T 3x + T 4x = (−1, 5, −1, 2). Clearly (5.1) implies that we could insert −5m, n3,−5n4, or −5n5/2 on the right side of Theorem 3A, which directly proves Ramanujan’s congruence. Corollary 1. We have ∞ X n=0 p(5n + 4)qn = −5 (q)5 ∞ ∞ X m,n3,n4,n5=−∞ mqQ(m,n3,n4,n5)+LA(m,n3,n4,n5), where Q and LA are given in Theorem 3A. The proofs for p(7n + 5) and p(11n + 6) are similar. Note that corresponding dihedral groups contain a seven cycle and an eleven cycle. The corresponding aﬃne transformations again have no ﬁxed points. The analogues of the right side of (5.1) are (−2, 1, 7, −1, −2, −3) and (−5, 3, 2, 1, 0, 10, −2, −3, −4, −5), so we obtain the next two corollaries. Corollary 2. We have ∞ X n=0 p(7n + 5)qn = 7 (q)7 ∞ ∞ X m,n3,...,n7=−∞ n3qQ(m,n3,...,n7)+LA(m,n3,...,n7), where Q and LA are given in Theorem 3A. Corollary 3. We have ∞ X n=0 p(11n + 6)qn = 11 (q)11 ∞ ∞ X m,n3,...,n11=−∞ n5qQ(m,n3,...,n11)+LA(m,n3,...,n11), where Q and LA are given in Theorem 3A. The next application is to consider the mod 2 behavior of p(tn + r). We take the element g of order two in the proof of Proposition 3, for which it can be shown that dimension(FP(g)) = ⌊t/2⌋. Thus, from Proposition 4, the generating function for p(tn + r) mod 2 is a ⌊t/2⌋-fold sum. For t = 3 the sets FP(g) are FP(g) =      {(2m, 3m) : m ∈Z} for r = 0, {(2m, 3m) : m ∈Z} for r = 1, {(m, 3m −1) : m ∈Z} for r = 2. An explicit computation of the form and the Jacobi triple product identity give the next proposition. SIEVED PARTITION FUNCTIONS AND Q-BINOMIAL COEFFICIENTS 9 Proposition 5. We have ∞ X n=0 p(3n + 2)qn ≡ 1 (q)3 ∞ ∞ X m=−∞ q9m2−8m+1 ≡q(q18; q18)∞(−q; q18)∞(−q17; q18)∞ (q)3 ∞ mod 2, ∞ X n=0 p(3n + 1)qn ≡ 1 (q)3 ∞ ∞ X m=−∞ q9m2−4m ≡(q18; q18)∞(−q5; q18)∞(−q13; q18)∞ (q)3 ∞ mod 2, ∞ X n=0 p(3n)qn ≡ 1 (q)3 ∞ ∞ X m=−∞ q9m2+2m ≡(q18; q18)∞(−q7; q18)∞(−q11; q18)∞ (q)3 ∞ mod 2. For t = 4 or t = 5 the set FP(g) is two-dimensional, and cannot be evaluated by the Jacobi triple product identity. However, there are other available involutions in the appropriate dihedral group for t = 5. One may hope that for each residue class r, there is at least one involution g such that dimension(FP(g)) = 1. This is true for t = 5, r ̸= 4, and we obtain product identities analogous to Proposition 5. For p(5n + 4), each of the ﬁve involutions has a two-dimensional ﬁxed point set. However, the generating function for FP of one these involutions is obviously diagonalizable to give the mod 2 version of Ramanujan’s generating function for p(5n + 4). We next use these identities and Proposition 5 to verify the following conjecture for t = 3 and t = 5. A variant of this conjecture is given in [7, p. 854, §5]. Its history is given in §6. Subbarao’s Conjecture. For 0 ≤r ≤t −1, p(tn + r) is inﬁnitely often even, and inﬁnitely often odd. Theorem 4. Subbarao’s conjecture holds for t = 3 and t = 5. Before proving Theorem 4, we give a technical lemma for establishing Subbarao’s Conjecture. Lemma 1. Let 0 ≤r ≤t −1, and let Q1(n) (Q2(y)) be non-negative quadratic functions in one (several) variable(s), with Q1(n) strictly increasing. Suppose ∞ X n=0 qQ1(n) ∞ X n=0 p(tn + r)qn ≡ X y qQ2(y) mod 2. If there exists an integer s, an odd integer j, and an integer 0 ≤i ≤j −1 such that (1) the equation Q2(y) = Q1(jm + i) + s has no integral solutions y, (2) p(st + r) ≡1 mod 2, then Subbarao’s Conjecture holds for r and t. Proof. If we ﬁnd the coeﬃcient of qM we see that ∞ X n=0 p(t(M −Q1(n)) + r) ≡\|{y : Q2(y) = M}\| mod 2. Next let M = Q1(k)+s, and assume that p(tn+r) mod 2 is eventually the constant c. Then for large values of k, p(ts + r) + kc ≡\|{y : Q2(y) = Q1(k) + s}\| mod 2. 10 FRANK GARVAN AND DENNIS STANTON By (1), there is no solution y for the right side if k ≡i mod j, so p(ts+r)+kc ≡0 mod 2. If c = 0, this contradicts (2); if c = 1, then (2) implies k must be odd. However we can take k to be even, by taking k = 2mj + i for i even, or taking k = 2mj + j + i for i odd. □ Proof of Theorem 4. Here is an outline of our proof. We will use Propositions 4 and 5 for the generating function identity assumed in Lemma 1. Then we must ﬁnd an odd residue class which is missed by the quadratic function function on the right side, and hit by the quadratic function on the left side. Hopefully this class will yield a viable choice for s in (2). For the generating functions, Proposition 4 implies (5.2) (q)t ∞ ∞ X n=0 p(tn + r)qn ≡ X x∈F P (g) q(Q+L)x mod 2, for an appropriate form L. Thus we need to eliminate (q)t ∞to apply Lemma 1. Jacobi’s identity will do this for t = 3 and t = 5, (q)3 ∞= ∞ X n=0 (−1)n(2n + 1)qn(n+1)/2 ≡ ∞ X n=0 qn(n+1)/2 mod 2. For t = 5 we use (q)5 ∞= (q)6 ∞/(q)∞≡(q2; q2)3 ∞/(q)∞ mod 2 and the Euler Pentagonal Number Theorem [1, 1.3.1]. For t = 3 (5.2) becomes (5.3) ∞ X n=0 qn(n+1)/2 ∞ X n=0 p(3n + r)qn ≡ X x∈F P (g) q(Q+L)x mod 2 while the t = 5 version of (5.2) is (5.4) ∞ X n=0 qn(n+1) ∞ X n=0 p(5n + r)qn ≡ ∞ X n=−∞ qn(3n+1)/2 X x∈F P (g) q(Q+L)x mod 2. We now apply Lemma 1 with Q1(n) = n(n + 1)/2 and Q2(x) = (Q + L)(x) for t = 3; and Q1(n) = n(n + 1) and Q2(x, n) = n(3n + 1)/2 + (Q + L)(x) for t = 5. Here is the list of the appropriate values for s, j, and i, and the form Q + L in Lemma 1. (1) 3n: Q + L = 9m2 + 2m, s = 0, i = 3, j = 5, (2) 3n + 1: Q + L = 9m2 −4m, s = 0, i = 1, j = 5, (3) 3n + 2: Q + L = 9m2 −8m + 1, s = 1, i = 4, j = 5, (4) 5n: Q + L = 5m2 + 2m, s = 0, i = 6, j = 49, (5) 5n + 1: Q + L = 10m2 + 4m, s = 1, i = 3, j = 49, (6) 5n + 2: Q + L = 10m2 + 8m + 1, s = 1, i = 0, j = 49, (7) 5n + 3: Q + L = 5m2 + 4m, s = 2, i = 4, j = 49. The remaining case is p(5n + 4), which has been done by Kolberg . □ SIEVED PARTITION FUNCTIONS AND Q-BINOMIAL COEFFICIENTS 11 6. Remarks. Ramanujan [3, p.288-289] gave elementary proofs of the congruences for p(5n+4) and p(7n + 5), while Winquist gave such a proof for p(11n + 6). Our proof is also elementary, and extends to mod 2 results. Unfortunately we are unable to give a geometric interpretation for the elements in the symmetry group G. It would be very interesting to have some geometric intuition for these elements, as opposed to our computational approach. For ex-ample, one could hope for reasonable version of a fundamental domain for the ﬁve cycle in 5n + 4 which would give Ramanujan’s generating function [5, (4.21)]. Such a fundamental domain, together with a bijection proving Theorems 3A, 3B, and 3C, may be related to the crank of Andrews and Garvan . For 11n + 6 the group G is again the dihedral group of order 22. One may also hope for an interpretation of Atkin’s congruence theorem [1, p.160]. The ﬁrst proof of a result analogous to Subbarao’s conjecture is due to Kolberg , who proved the t = 1 case. Kolberg and Subbarao did the t = 2 case (see [7, p. 854], where Subbarao also mentions the t = 4 case). One can try to apply Lemma 1 to prove the Subbarao Conjecture for values of t besides 3 or 5. For t = 4, where the ﬁxed point set is two dimensional, a slight variation of Lemma 1 gives the Subbarao Conjecture for t = 16. Recently Hirschhorn and Subbarao did this case. We could not ﬁnd appropriate residue classes for the three variable quadratic functions which occur for t = 6. However, for t = 7, the eight cycle for r = 0, 2, or 6 (see Table 1) has a one dimensional ﬁxed point set, and we ﬁnd a one variable quadratic function. Lemma 1, with modulus 169, gives the Subbarao conjecture in these cases. In fact since Q1(n) is always divisible by 8, we also get the t = 56 case, when r ≡0, 2 or 6 mod 7. Similar reasoning shows that we can establish the t = 10 case from our t = 5 theorem. Kolberg has done the 7n+5 case. No other cases are known. Proposition 6. Subbarao’s conjecture holds for t = 10, and t = 56 and r ≡0, 2, or 6 mod 7. For t = 5 Kolberg gave the generating function for p(5n + r) as a sum of two inﬁnite products [5, (4.17)-(4.21)]. The mod 2 versions for 5n+1, 5n+2 and 5n+4 agree with our results. Note that 3 divides the order of the groups for 5n + 2 and 5n + 3. We could ﬁnd a mod 3 result which agrees with Kolberg’s mod 3 versions for 5n + 2 and 5n + 3. References 1. G. E. Andrews, The Theory of Partitions, Encyclopedia of Mathematics and Its Applications, Vol. 2 (G.-C. Rota ed.), Addison-Wesley, Reading, Mass., 1976, (Reissued: Cambridge Univ. Press, London and New York, 1985).. 2. G. E. Andrews and F. G. Garvan, Dyson’s crank of a partition, Bull. Amer. Math. Soc. 18 (1988), 167-171. 3. G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers, Oxford, London, 1979. 4. M. D. Hirschhorn and M. V. Subbarao, On the parity of p(n), preprint. 5. O. Kolberg, Some identities involving the partition function, Math. Scand. 5 (1957), 77-92. 6. , Note on the parity of the partition function, Math. Scand. 7 (1959), 377-378. 7. M. V. Subbarao, Some remarks on the partition function, Amer. Math. Monthly 73 (1966), 851-854. 8. L. Winquist, Elementary proof of p(11m + 6) ≡0 (mod 11), J. Comb. Th. 6 (1969), 56-59.
15983	https://www.youtube.com/watch?v=l3wwvVAlsu0	Graphing Quadratic Inequalities Algebra \| Vertex Form Kevinmathscience 862000 subscribers 8 likes Description 361 views Posted: 24 Mar 2024 Graphing Quadratic Inequalities Algebra. This algebra video shows how to graph quadratic inequalities in vertex form. It is ideal for learners doing algebra in the American High School Curriculum but is also suited to all other learners. Transcript: in this lesson we're going to be learning how to graph quadratic inequalities but we're going to do it in vertex form okay so maybe you've already watched my videos on how to graph quadratic inequalities but in that lesson maybe we used standard form but now we're going to use vertex form just a reminder we're not going to look at any intercepts there are going to be lessons that we'll do at a later stage where we learn about the X intercepts y intercepts now we're just using symmetry to draw the graphs and just a quick reminder in case you haven't watched my previous lesson on standard form remember with inequalities the inequalities are when you have symbols like this for example or sometimes they look like this and so there are two things that you need to remember when they give it to you um first of all the first thing we always need to look at is whether we are going to draw using a solid line or sometimes we're going to use a dash so sometimes we're going to draw quadratic inequality with the solid line but then sometimes it's going to be a dash so the way that we we the way that we determine whether we should use a solid line or a dash is by looking at whether we have it like this or whether we have it like this so when there is a little line at the bottom like that you see the line at the bottom then if it looks like that you're going to use a solid line when it's like this so like this or like this then you're just going to use a dash okay whoopsie Kevin Dash now the next thing we need to look at is whether we are going to shade above or below you know like when you have an inequality uh maybe you've seen um a straight line inequality okay now sometimes you're going to shade above and then at other times you're going to shade below now the same thing is true with quadratics which we're going to learn how to do now um a quadratic typically looks something like that so we can also shade them above or we can shade them below and that's what we need to look at Now quickly so the way that you determine that it all depends on the following so if you have something like this Y is bigger than 3x^2 - 6X + 1 now this is the crocodile method that I sometimes speak about maybe you've heard me speak about it before the way it works is you look at the inequality can you see that it's a crocodile's mouth you can think of it like a crocodile's mouth look at that it's got all these little teeth over here um yeah so that's like a crocodile's mouth now this crocodile if you heard this before I'm so sorry but this crocodile is a very hungry crocodile and it's looking for a lot of food because it hasn't eaten for a very very long time and it's it it needs to find as much food as it can so this crocodile always likes to eat big pieces of food okay it's a such a lame example but it works so if you look at the crocodile's mouth what is that crocodile trying to eat right now is it trying to eat the this or is it trying to eat this side over here well I hope that you can see that the mouth is trying to eat this side over here so what that means is that the Y is the big side and this here is the small side because remember this crocodile is extremely hungry so it's looking for the big food because maybe this crocodile is not going to have food for another like 3 or 4 days so when you have a situation where um the Y is the big one then you're going to shade above when you have a situation where the Y is the small one then you're going to shade below it's as simple as that so here is our first example so I hope that you've already watched the lesson where I taught you how to draw graphs quadratics using vertex form because remember there's standard form and there's vertex form so when you draw a quadratic remember quadratic looks something like this so it has a turning point you see that it has a turning turning point or we call it some some of you might call it the vertex so step one when you draw a quadratic is to go find that step one is to go find the vertex now maybe you've seen me where I use this formula x = minus b/ 2 a we're not going to use that here why Kevin I don't understand bro you told me we should use that that only gets used when the equation is written in standard form that's when it looks like this a X2 + B X plus c you know the normal quadratic that's when you use that formula but when it's got this format over here where you've got the bracket to the power of two this is vertex form so it's actually easier the the reason it's easier is the following you need to know what this part does to the graph this part here that has the square that part moves the graph horizontally or left and right but you need to know exactly what this one does so whenever it's a minus that means right remember that minus means right Kevin that doesn't make sense my bro minus usually means left and then plus means right in like if we were thinking normally about this like logically you know if you go to the right um that's Plus in values and if you go to the left that's minusing I know guys so this has been confusing students for 7,432 years so I don't know I random numbers but yeah it's okay so that's means right plus means left you just got to remember that okay um then so that's this part okay um that's that part over there uh so let's say this is the horizontal I'm not going to fit that in okay so that's that part now this part here this is vertical this is vertical so up and down but now this one is normal this one is pretty easy so minus means down that seems logical plus means up you just got to remember that okay so now what you got to remember is that a ver a quadratic you can imagine that it starts over here this is like the basic quadratic this is called a parents quadratic it's it's the most basic quadratic you can ever get and it has a Vertex at 0 0 so imagine that then all of a sudden we're going to move four places to the right okay so we're going to go four places to the right and then we're going to go um oh we didn't even say here so we're going to go one place down one down and why did I say four to the right because there's a minus and when there's a minus you go right okay so we're going to go one down so um let's go one down there we go so where would the vertex be well think about it if you go four to the right You' be at four and if you go one down you'd be at NE 1 as easy as that okay so that's step one complete vertex so Kevin you're saying we don't have to use any formulas to find the vertex not for this one guys um why because this is called vertex form it already gives you the vertex that's why we call it vertex form now the next step is to use our table of values if you've watched any of my videos on drawing quadratics recently um you would remember this this is now step two and this is now um yeah it's our final step so table of values this is an easy step so you just put X and Y and then you make a table and then you want to make five little parts of oh that's going to look great there boom boom okay so um so you want five of them over here uh in the middle one okay so the middle of five we all know we have a middle of five we have a middle finger ha um so um we're going to go put um the vertex over there so X is four whoa given X is four and Y is1 okay now what you do now we use symmetry remember we've spoken about this before if you have a quadrant IC they are perfectly symmetrical Kevin what does symmetrical mean it means it's the same on the left and the right so that means that if you go one place to the left of the of the Symmetry and one place to the right then these y values are going to be identical and if you go two places to the left and two places to the right the same thing happens okay so what I want you to do is go one place to the right five one place to the left for the X values there we go then I want you to go another one place to the right and another one place to the left wonderful now what you're going to do is you are going to choose either these two or you can choose these two it doesn't matter but just don't choose all of them you can if you want but it's just going to take you a lot longer than what you need to take I'm going to choose these two so um I'm going to start with this one that's a x value so I can now find the Y value by just plugging it into this equation notice we're not even worrying about the inequality sign right now now that only comes in at the end so if I know the x value is two then I can now find the Y value cuz I can literally just take this x value of two plug it in there and that's how we do it so we can now say that Y is equal to 3 and then X is 2 there we go go ahead type this all in on your calculator and it gives us 11 okay then let's do the three so all I'm going to do now is just replace this two with a three and we end up with a two okay so now you don't have to go do these ones you just use symmetry so check this out when you go one to the left and one to the right the Y values would be the same so that's also a two then if you go whoa Kevin what is going on let's put that stuff back then if you go um two places to the left and two places to the right then their y values are also the same so we can just put 11 now we just go draw now this is awkward why am I drawing the graph over here when clearly the elephant in the room is like standing right over here so that's weird so 4 - one I did have a good weekend I will I won't lie so maybe I'm still recovering I don't mean like drinking and stuff I just mean it was an amazing weekend uh I'll quickly tell you what I did um so I'm currently in South Africa and this weekend there was a a festival like a music festival um sort of a trance kind of festival and it started on Friday and promise you did not sleep 1 minute on Friday night like we literally bunch of my friends and I we stayed up and the sun was coming up those are amazing moments just like chatting and talking absolute nonsense and suddenly the sun's coming up so that was amazing then at around 12:00 the next day uh lunchtime I decided I have to go sleep because like I'm not going to make it the next night if I don't go sleep now but the problem is is we're in a tent right um and it is so hot like it is um okay so in South Africa we use degrees but I I'm pretty good at converting you know I'm sort of into numbers if you ever noticed so it was about 100 uh Fahrenheit so um really hot and so what happens is you're in the tent and the and there's no shade so the tent's just beat I mean the sun's just beating down on this tent and you're like oh like you can't sleep you wake up and you're like sweating and yeah not great so couldn't sleep luckily there was like this lake so I got up went and had a swim woke up a little bit um then at 3:00 I like I couldn't keep my eyes open so I went back to the tent and I did another 1 hour of sleep um and then you wake up and you get some food and everything so it's like 4:00 in the afternoon but you know at these festivals that music in the background just isn't stopping it's just B the whole time it's like pumping through your veins um and then then um yeah it was cool like people were just dancing during the daytime uh on like this beach sand it was amazing and then and then it was uh night time and suddenly you just have more energy you're awake again and yeah so then we were partying like absolute crazy dance floor was packed it was completely outdoor there was no indoors it was just outdoor it was beautiful and um then what happened yeah so then we're partying and then partied all all the way through until about half 3 in the morning um just meeting lots of people chatting to them it's just amazing amazing amazing and then I decided okay I'm going to quickly go sleep now just just go close my eyes for a bit then that that was amazing cuz it was nice and cold um cuz it's weird when you're at these like outdoor um festivals where you're like on like open land um it's super hot in the daytime but nice and cold at night well like not nice and cold but if you have like a nice blanket and stuff it's amazing so I slept but only slept for an hour then woke up half past 4 um headed back to the dance floor and there were still so many people dancing and that was it we then me and my one friend we were just completely like ripped it out on the dance floor dancing um until until sunrise again um and then and then yeah stayed there until about 12: so yeah just a crazy awesome awesome awesome weekend and so yeah I can't remember what the point of the story was oh it's because I was drawing over here so yeah forgive me I did not sleep as much as a normal human sleeps uh this weekend so in total is about 3 or 4 hours of sleep so and here it is Monday morning and here I am making some lessons and some stuff for you because this is what I love to do so um onto the topic at hand so 2 and 11 2 and 11 um is over here there we go there we go there's 2 and 11 boom three and two three and two um what's funny though is when you go to these festivals um there's that constant beat in the background and I just keep in my head it just keeps going the whole time and then four and minus one and then five and two 5 and two is over here and then 6 and 11 6 and 11 is over here see how we've got the Symmetry that's happening um those two are the same and those two are the same okay so now we just going to okay now wait we can't just draw this thing cuz now we have to remember that this is a inequality so we need to look at two things right first one is we need to know whether we're going to go solid solid line or dashed so that's all about looking at whether there is a little line underneath like that or whether it's completely open like that so um this one's completely open so that means we're going to use Dash so uh let's quickly try to do this there we go the next thing we need to know is whether we are going to shade above we're going to shade above or shade below and this is where you look at the whole crocodile thing right so like you got to see what the crocodile is eating the big mouth or the small mouth whoa the big mouth or the small mouth no um you got to look at the crocodile's mouth and see if it's eating the big or or what the crocodile's eating okay so this is the crocodile that's its mouth look at that what is it trying to eat is it trying to eat this side or is it trying to eat this side what it's trying to eat the Y so what that means is that the Y is the big side and then this side is the small side and if you can remember correctly when the Y is the big side then you shade above okay so we're going to shade above the the the the line so give me a few seconds there we go so something like that okay and so we're just going to do um maybe one more example maybe two I I'll keep you posted up of like during the next example we can see if if we feel like we need to do another one here's our next example so remember step one is all about the vertex okay I think we only need to do one more example because if you've watched my other lessons on quadratics they're very repetitive um it becomes very repetitive the only difference with this one and the previous lesson on standard form is literally how to get the vertex the rest is the same um so so yeah uh Step One is vertex so remember um you don't use minus b/ 2 a that's only if we were busy with standard form but this isn't standard form this is vertex form so all you do is you look at this part and you look at this part so remember that this part here that has the X remember X is a x is left and right okay um whereas this part is the vertical so we're going to look at this part here and remember that with that uh negative means we're going right and plus means we're going left so what we doing here is we're going two to the right so we're going to go two right and then we're going to go four up four up so remember you always start at the z0 position that's where a Vertex originates now we're going to go two to the right boom and then we're going to go four up and there we have it so we're going to put a little point there and that's going to be um our vertex wonderful step two is the table of values okay so we're going to go X and Y and then we're going to put um the vertex in the middle so what was the vertex 2 and four and then you're going to go one to the right and go one to the left just for the X values then go another one to the right and go another one to the left okay wonderful now what we're going to go do is realize here that we have X values okay so we can go get the Y values but you don't have to do it for all four of these you only have to do it for either these two or these two don't go do it for for all of them you can if you want but it's just going to be a waste of your time there's a better way because we use symmetry so what I'm going to do is I'm going to choose these two over here but obviously you can't do both of them at the same time um so I'm going to start with this so I'm going to take that x value okay and I'm going to plug it into the x value over there or the X place and so that's going to be um I'm just going to say equals and that's going to be Z take away two uh let's just do this -2 0 to^ 2 + 4 and if you had to go work this out you get -4 -4 okay -4 now we're going to go use a one in the X's place so I'm just going to take this zero out and replace it with a one okay and if you had to go calculate that one now you get two so I'm going to put a two over there now you don't have to go any further you don't have to go do these ones you can if you want it's not wrong um but it's unnecessary because of symmetry remember um we've said this many times quadratics have symmetry so if you have your vertex right in the middle then if you go one place to the right and one place to the left the Y values are going to be exactly the same and if you go two places to the right and two places to the left the Y values are the same you can go three places four places it doesn't matter as long as you go the same distance on the right and the same distance on the left the Y values would be exactly the same so what that then means is if this is the vertex then if you go one to the right and one to the left their y values would be identical and then if you go two to the right and two to the left their y values are identical and that's it now we just go plot the points quickly so Z I'm plotting this one 0 and minus 4 there you are 1 and two uh 1 and two there you are 3 and two there we go and then four - 4 there we go now we need to make sure whether we need to use um we need to know a few things now whether we're going to use shaded solid lines or whether we're going to shade above or below so those are the two things we need to quickly check so let's quickly see whether we're going to use a solid or a dash how do we do that again well the way that we do that is we look at the symbol over here if it has a little line underneath then it's solid if it doesn't then it's a dash so this one's definitely a dash okay so there we go and then um there we go okay so we have our we have that the next thing we need to know is whether we are going to shade above or below above or below above or below and this is where the whole crocodile mouth comes in crocodile okay so remember that the this is the crocodile you see its mouth now what is that mouth trying to eat is that mouth is this CR trying to eat um this side over here or is it trying to eat this side over here well if you look at the mouth the mouth is trying to eat this side now the remember that the crocodile always likes to eat the big side so if that's the big side then that makes this the small side the Y is the small side now remember our little rules when the Y is the small side then you shade below if the Y was the big one then you would shade above so we're going to shade below the line so that's all of this over here so let's quickly shade that for you and there we have it so that's it guys that's how we do um vertex form uh graphing quadratic inequalities
15984	https://conceptually.org/concepts/occams-razor	Conceptually Occam's Razor Cut through the crap with a tool from your mate, Occam. What is Occam’s razor? Definition and explanation If Occam’s razor brings to mind images of stubbled gentlemen and shaving cream, you’re not actually that far off! Occam’s razor (also known as the ‘law of parsimony’) is a philosophical tool for ‘shaving off’ unlikely explanations. Essentially, when faced with competing explanations for the same phenomenon, the simplest is likely the correct one. Namesake William of Occam said the best explanation of any phenomenon is the one that makes the fewest assumptions. A statement that includes many ‘ifs’ should trigger mental alarm bells: you should consider Occam’s razor and investigate it further. Some argue that the scientific method was built upon the principles of Occam’s razor. Underdetermination says that for any theory in science there will always be at least one other rival theory that could conceivably be correct, so the scientific method uses Occam’s razor in order to circumvent this issue and choose a working hypothesis. Remember, however, that Occam’s razor is a heuristic, a rule of thumb, to suggest which hypothesis is most likely to be true. It doesn’t prove or disprove, it simply leads you down the path that’s most likely to be correct. Also, ‘simplicity’ is often subject to heavy debate, so you and I might come to different conclusions when faced with a decision between the same 2 hypotheses. Examples of Occam’s razor This principle is popular among skeptics, a group of people inclined to keep an open mind and believe only what we can sense or what can be proven scientifically. But there are plenty of examples in our everyday lives too. Occam’s razor tells us that we shouldn’t get sucked into a whirlpool of paranoia after scrolling through WebMD. “You have a headache?”, “Oh no… you might have the Black Death!” Sure, it’s true that one of the symptoms of the Black Death is a headache but, using Occam’s razor, it’s obviously much more likely that you’re dehydrated or suffering from a common cold. Many Creationists use Occam’s razor to argue the existence of God. “Isn’t the simplest explanation of how the Earth was created that God created it?” they say. But atheists might counter that the existence of a divine being who created the world in just seven days is much less simple (and relies on more assumptions) than the big bang theory - a great example of how simplicity is in the eye of the beholder. Occam’s razor is also often used to debunk conspiracy theories. Faced with the disappointing mess that is modern politics, how likely is it REALLY that reptilian aliens have infiltrated our government? The simpler explanation is a combination of corruption, incompetence and structural inefficiency. Share on Facebook Share on Twitter Also check out How Occam's Razor Works, Josh Clark Conjunction fallacy, Wikipedia An Intuitive Explanation of Solomonoff Induction, LessWrong Get one concept every week in your inbox Written by Frankie Andersen-Wood
15985	https://www.aafp.org/pubs/afp/issues/2021/1100/p476.html	KEVIN R. HERRICK, MD, PhD, JENNIFER M. TERRIO, DDS, AND CRISPIN HERRICK, DDS Am Fam Physician. 2021;104(5):476-483 Author disclosure: No relevant financial affiliations. Medical consultations before dental procedures present opportunities to integrate cross-disciplinary preventive care and improve patient health. This article presents recommendations related to patients with certain medical conditions who are planning to undergo common dental procedures, such as cleanings, extractions, restorations, endodontic procedures, abscess drainage, and mucosal biopsies. Specifically, prophylactic antibiotics are not recommended for preventing prosthetic joint infections or infectious endocarditis except in certain circumstances. Anticoagulation and antiplatelet therapies typically should not be suspended for common dental treatments. Elective dental care should be avoided for six weeks after myocardial infarction or bare-metal stent placement or for six months after drug-eluting stent placement. It is important that any history of antiresorptive or antiangiogenic therapies be communicated to the dentist. Ascites is not an indication for initiating prophylactic antibiotics before dental treatment, and acetaminophen is the analgesic of choice for patients with liver dysfunction or cirrhosis who abstain from alcohol. Nephrotoxic medications should be avoided in patients with chronic kidney disease, and the consultation should include the patient's glomerular filtration rate. Although patients undergoing chemotherapy may receive routine dental care, it should be postponed when possible in those currently undergoing head and neck radiation therapy. A detailed history of head and neck radiation therapy should be provided to the dentist. Multimodal, nonnarcotic analgesia is recommended for managing acute dental pain. Integrating patients' medical and dental health care is important because there are correlations between periodontal disease and some medical conditions, such as diabetes mellitus, coronary artery disease, hypertension, kidney disease, and rheumatoid arthritis.1–7 Medical consultations before dental procedures present opportunities to integrate cross-disciplinary preventive care and provide recommendations for treatment considerations before, during, and after a dental visit. Although dentists are ultimately responsible for the treatments they provide, they need the patient's complete medical information and often consult physicians when planning common dental procedures, such as cleanings, extractions, restorations (e.g., fillings, crowns, bridges, implants), endodontic procedures, abscess drainage, or mucosal biopsies.8 \| Clinical recommendation \| Evidence rating \| Comments \| --- \| A medical consultation in preparation for a dental procedure should include the patient's medical conditions, treatment plans, and current levels of management; any resuscitation directives; and any history of therapy with bisphosphonates or other antiresorptive drugs, antiangiogenic drugs, or head and neck radiation.7,9,10,34 \| C \| Consensus of expert opinion \| \| A history of orthopedic joint replacement is not an automatic indication for prophylactic antibiotics, and physicians should consider discontinuing routine procedural antibiotic prophylaxis after discussing risks and benefits with patients.13,41 \| C \| Consensus guidelines \| \| For simple cleanings or single tooth extractions, it is reasonable to continue oral anticoagulation and antiplatelet therapies at therapeutic doses.14 \| C \| Consensus of expert opinion \| \| Consider postponing elective dental treatments for six weeks after myocardial infarction or bare-metal stent placement and for six months after drug-eluting stent placement.14,21 \| C \| Consensus of expert opinion \| \| Consider optimizing a patient's oral health before initiation of chemotherapy or head and neck radiation therapy to avoid adverse sequelae.34,44 \| C \| Consensus of expert opinion, in the absence of clinical trials \| \| Recommend multimodal analgesia for management of acute dental pain, if not contraindicated.35,36 \| B \| Expert guidelines supported by clinical trials \| A medical consultation in preparation for a dental procedure should detail the patient's medical conditions, treatment plans, and current levels of management.7 A medical history, including allergies and use of herbal remedies and prescribed and over-the-counter medications, should be provided. It is essential to include any history of bisphosphonate use or cancer treatments.9 A relevant psychiatric history, including special needs, and the patient's resuscitation wishes or advance directive may be helpful.10 eFigure A is a sample consultation report form to assist physicians when evaluating patients before dental procedures (see template). Table 1 summarizes key concepts discussed in this article.1,5,9,11–39 \| Condition \| Concepts \| --- \| \| Antibiotic prophylaxis \| \| \| Infectious endocarditis \| The American Heart Association recommends considering antibiotic prophylaxis only when dentogingival manipulations are planned for selected patients at highest risk of complications (Table 2).11 \| \| Prosthetic joints \| Periprocedural antibiotic prophylaxis does not affect the incidence of prosthetic knee or hip infections.12 Because dentists usually do not provide an antibiotic prescription, patients electing to use prophylaxis may need to obtain a prescription from their physician.13 \| \| Cardiovascular \| \| \| Anticoagulation and antiplatelet therapies \| For patients undergoing dental procedures, evidence supports continuing antiplatelet and anticoagulation medications in therapeutic doses.14–39 Dentists routinely manage bleeding in these patients using simple topical treatments, pressure packs, or sutures.19,20 \| \| Coronary artery disease \| Patients can consider delaying elective dental procedures for six weeks after myocardial infarction or bare-metal stent placement or six months after drug-eluting stent placement.21,22 Patients are considered at low cardiac risk when undergoing dental procedures if they have no active cardiac conditions and can perform at least 4 metabolic equivalants.23 \| \| Hypertension \| Many dentists routinely measure blood pressure before dental procedures, but it is unclear whether a high preprocedural office-based measurement should postpone treatment.23 With a lack of evidence-based guidance, many dentists postpone elective dental procedures when the patient's blood pressure exceeds 160/100 mm Hg.23 \| \| Metabolic \| \| \| Diabetes mellitus \| Patients with diabetes do not have an increased risk of infection from tooth extraction.24 Patients with diabetes have an increased risk of periodontal disease.1 \| \| Hepatic disease and cirrhosis \| The need for perioperative platelet transfusion when platelet counts are below 50 × 103 per μL (50 × 109 per L) is being challenged.25,26 The preferred analgesic for patients with dental pain and compensated hepatic dysfunction is acetaminophen, and nonsteroidal anti-inflammatory drugs are usually avoided.27,28 Ascites is not an indication for initiating prophylactic antibiotics before dental treatment.28 It is reasonable to provide the dentist with a recent complete blood count, prothrombin time, and international normalized ratio.25,29 \| \| Osteoporosis \| A history of bisphosphonate use increases a patient's lifetime risk of osteonecrosis of the jaw and should be reported to the dentist.9,30 Patients should optimize their oral health before initiating bisphosphonates if possible.30 Perioperative bisphosphonate holidays have not been shown to be beneficial.31 \| \| Renal insufficiency and dialysis \| Daily oral care and semiannual dental checkups reduce mortality in patients receiving dialysis.5 Patients who have stage I to IV chronic renal failure or are undergoing peritoneal dialysis should avoid nephrotoxic medications, and renal dosage adjustments should be considered.32 For patients receiving extracorporeal dialysis, scheduling dental procedures between dialysis days can prevent patient fatigue and complications of heparin.33 Dentists should be provided with a recent glomerular filtration rate to determine the severity of renal disease and medication dosing adjustments.5,33 \| \| Other situations \| \| \| Cancer \| Patients should obtain dental treatments before chemotherapy or radiation when possible.34 Patients undergoing chemotherapy without radiation may receive routine dental care, but a complete blood count should be performed on the day of the planned treatment.34 Routine dental care should be postponed in patients receiving head and neck radiation therapy.34 A detailed history of head and neck radiation therapy, antiresorptive agents, or antiangiogenic agents should be communicated to the dentist.9,30,34 \| \| Pain and narcotics \| Multimodal analgesia is superior to monomodal analgesia in the management of acute dental pain; combining acetaminophen with nonsteroidal anti-inflammatory drugs is highly effective.35,36 Initiating opioids may lead to addiction and should be considered only when other modalities fail.37,38 If used, opioid therapy should be limited to less than three to seven days, and precautions should be taken to help prevent long-term use or overdose.37,39 \| Antibiotic Prophylaxis INFECTIOUS ENDOCARDITIS For decades, the American Heart Association recommended prophylactic antibiotics for patients with cardiac conditions that might increase the risk of contracting infectious endocarditis during dental procedures. However, because studies have found that bacteremia occurs routinely with common activities such as chewing, brushing, and flossing and there is a lack of evidence that procedural prophylaxis is effective,40 the American Heart Association now recommends considering it only when dentogingival manipulations are planned for selected patients at highest risk of complications (Table 2).11 \| \| \| Prosthetic heart valves or heart valve repairs using prosthetic material \| \| History of infectious endocarditis \| \| Unrepaired cyanotic congenital heart disease or repaired congenital heart disease, with residual shunts or valvular regurgitation at the site of or adjacent to the prosthetic patch or device \| \| Cardiac transplant with valve regurgitation due to a structurally abnormal valve \| PROSTHETIC JOINTS A high-quality prospective, case-control study found that antibiotic prophylaxis does not affect the incidence of prosthetic knee or hip infections.12 A 2013 joint guideline from the American Dental Association (ADA) and the American Academy of Orthopaedic Surgeons suggests that physicians consider discontinuing routine procedural antibiotic prophylaxis after discussing risks and benefits of antibiotic prophylaxis with patients.41 Since 2015, the ADA has recommended against routine prophylaxis for patients with prosthetic joints.13 Because dentists usually follow ADA guidelines, they often do not provide an antibiotic prescription; patients electing to use prophylaxis may need to obtain the prescription from their physician.13 Cardiovascular Conditions ANTICOAGULATION AND ANTIPLATELET THERAPIES For simple cleanings or single tooth extractions, evidence supports continuing antiplatelet and anticoagulation medications at a therapeutic international normalized ratio (INR) because the indications for these medications usually outweigh the risks of dental complications.14 Studies have demonstrated that patients taking a vitamin K antagonist at a therapeutic INR, direct oral anticoagulants, or daily aspirin are not at increased risk of uncontrollable bleeding after outpatient oral surgeries.15–19 Another study of patients on nonaspirin antiplatelet therapy, alone or in combination with aspirin, demonstrated a negligible increased risk of bleeding after invasive dental treatments.20 Dentists routinely manage such perioperative bleeding using simple topical treatments, pressure packs, or sutures.19,20 CORONARY ARTERY DISEASE Elective dental treatments have traditionally been deferred for patients with unstable angina and postponed for six weeks after myocardial infarction or bare-metal stent placement or for six months after drug-eluting stent placement.14,21,22 However, evidence suggests that, when necessary, invasive procedures (e.g., extractions) may be safely performed shortly after myocardial infarction or in patients with unstable angina, particularly if dentists provide effective anesthesia, control postoperative pain, and take measures to reduce perioperative anxiety.22 Patients are considered at low cardiac risk when undergoing dental procedures if they have no active cardiac conditions and can perform at least 4 metabolic equivalents.23 HYPERTENSION Dental treatments are rarely contraindicated in patients with hypertension who have no other significant symptoms or comorbidities.23 Many dentists routinely measure blood pressure before dental procedures, but it is unclear whether a high preprocedural office-based blood pressure measurement should postpone treatment.23 With a lack of evidence-based guidance, many dentists postpone elective dental procedures when the patient's blood pressure exceeds 160/100 mm Hg to avoid the possibility of a hypertensive crisis.23 By documenting a history of adequate blood pressure control, physicians can help patients avoid the setback of canceled dental appointments.23 Metabolic Conditions DIABETES People with type 1 or 2 diabetes have an increased risk of periodontal disease, increasing their need for preventive dental care.1 Although people with diabetes have a well-documented increased risk of infection in general, research suggests that this does not extend to tooth extractions, regardless of the level of glycemic control.24 Thus, diabetes alone is not an indication for using antibiotics in patients undergoing routine extractions.24 Studies are not as clear about dental implant procedures, and research is ongoing to investigate whether the success of implants is influenced by glycemic control.42 It may be helpful for physicians to include a recent A1C level in the medical consultation. HEPATIC DISEASE AND CIRRHOSIS Cirrhosis is associated with coagulopathy, thrombocytopenia, renal failure, anemia, ascites, and spontaneous bacterial peritonitis. A dental assessment and treatment are critical in preparing patients with cirrhosis for liver transplantation by optimizing oral health and minimizing the risk of oral infections.25,29 The traditional practice of performing a perioperative platelet transfusion for patients with platelet counts below 50 × 103 per μL (50 × 109 per L) has been challenged by studies showing that dental extractions are safe when platelet counts are as low as 10 × 103 per μL (10 × 109 per L).25,26 Studies also show that the risk of hemorrhage during dental extraction in patients with cirrhosis is not predicted by laboratory values.25 Given the lack of consensus, it may be reasonable for physicians to include a recent complete blood count, prothrombin time, and INR in the medical consultation for a dental extraction.25,29 The preferred analgesic for patients with dental pain and compensated hepatic dysfunction or cirrhosis who abstain from alcohol is acetaminophen, sometimes limited to 2 g per day.27,28 Nonsteroidal anti-inflammatory drugs (NSAIDs), except aspirin when prescribed for cardiovascular disease, are usually avoided because of platelet, kidney, and gastric risks.28,29 Certain patients with ascites and a history of spontaneous bacterial peritonitis are given long-term daily prophylactic antibiotics.28,29 However, dental procedures are not an indication to initiate procedural prophylaxis in patients with ascites.28 OSTEOPOROSIS Medication-related osteonecrosis of the jaw is a serious complication associated with the use of bisphosphonates and other antiresorptive agents or antiangiogenic cancer treatments (Table 3).9,30 A history of exposure to these drugs can increase the patient's lifetime risk of osteonecrosis of the jaw by up to 100-fold, depending on the treatment regimen and indication for which it was prescribed.9 Dental extraction and implant procedures are known to trigger osteonecrosis; therefore, the dentist may need to modify the care plan in these patients.30 \| Class \| Representative medications \| Typical indications \| --- \| Antiangiogenic agents \| \| \| \| Monoclonal antibodies or nucleic acids targeting vascular endothelial growth factor \| Bevacizumab (Avastin) Pegaptanib (not available in the United States) Ramucirumab (Cyramza) Ranibizumab (Lucentis) \| Age-related macular degeneration Breast, cervical, colorectal, and gastric cancers Glioblastoma Non–small cell lung cancer Renal cell cancer \| \| Tyrosine kinase inhibitors \| Axitinib (Inlyta) Bosutinib (Bosulif) Cabozantinib (Cabometyx) Dasatinib (Sprycel) Erlotinib (Tarceva) Imatinib (Gleevec) Nilotinib (Tasigna) Sorafenib (Nexavar) Sunitinib (Sutent) \| Certain leukemias Colorectal cancer Gastrointestinal stromal tumor Giant cell tumor of bone Hepatocellular cancer Medullary thyroid cancer Myelodysplastic syndrome or myeloproliferative diseases Pancreatic neuroendocrine tumor Renal cell cancer \| \| Antiresorptive agents \| \| \| \| Bisphosphonates \| Alendronate (Fosamax) Ibandronate (Boniva) Risedronate (Actonel) Zoledronic acid (Reclast) \| Bone metastases from solid tumors, such as breast, lung, and prostate cancers Multiple myeloma Osteoporosis Paget disease of bone \| \| RANK ligand inhibitors \| Denosumab (Prolia) \| Bone metastases from solid tumors, such as breast, lung, and prostate cancers Giant cell tumor of bone Osteoporosis \| \| Sclerostin inhibitor \| Romosozumab (Evenity) \| Osteoporosis \| For patients with a history of treatment with the drugs listed in Table 3, details such as drug name, dosage, route of administration, indication, and treatment duration should be communicated to the dentist.9,30 Patients should optimize their oral health before initiating bisphosphonates if possible.30 For those already taking bisphosphonates, current literature has not shown a benefit to discontinuing the treatment perioperatively.31 RENAL INSUFFICIENCY AND DIALYSIS Chronic renal disease is associated with poor oral health.33 Daily oral care and semiannual dental checkups have been shown to reduce mortality in patients receiving dialysis.5 It is good practice to evaluate for and address oral infections before renal transplantation.43 Patients who have stage I to IV chronic renal failure with or without peritoneal dialysis usually do not need special accommodations for dental procedures other than the avoidance of nephrotoxic medications, such as NSAIDs, and consideration of renal dosage adjustments.32 It is helpful to provide the dentist with a recent glomerular filtration rate to determine the severity of renal disease.5,33 For patients receiving extracorporeal dialysis, scheduling dental procedures between dialysis days can prevent patient fatigue and complications of heparin. Providing a current INR (if taking warfarin), hemoglobin level, and platelet count may help the dentist assess the risk of bleeding.33 Other Situations CANCER When considering cancer treatment, it is recommended to involve the patient's dentist early to provide timely preventive or proactive dental care and avoid sequelae.9,30,34,44 Patients currently undergoing chemotherapy without radiation may receive routine dental care. The dentist should be provided with a current medication list, and a complete blood count should be performed on the day of the planned treatment to ensure the absolute neutrophil count is greater than 1,000 per μL (1 × 109 per L) and platelet count is greater than 50 × 103 per μL.34 Any history of antiangiogenic treatments (Table 39,30) should be disclosed. Routine dental care should be postponed in patients currently undergoing head and neck radiation therapy.34 A history of head and neck radiation is associated with numerous oral complications, such as increased dental caries, xerostomia, and osteoradionecrosis of the jaw, and dentists should be provided with a detailed oncology history (i.e., type of radiation treatment, dose, and specific anatomic site). This information is important because most cases of osteoradionecrosis of the jaw occur at sites exposed to greater than 60 Gy of radiation, and perioperative hyperbaric oxygen or other precautions may be indicated.34,45 PAIN AND NARCOTICS Studies show that multimodal analgesia is more effective than monomodal analgesia in the management of acute dental pain; combining acetaminophen with NSAIDs is highly effective.35,36 Despite a declining trend, opioids are often prescribed for dental pain.46 Long-term opioid use often begins with treatment of acute pain; therefore, the risks of opioid addiction, including death, should be discussed with patients considering opioids.37–39 Absent compelling study data, expert opinion and the Centers for Disease Control and Prevention recommend that dentists consider adding the minimal effective doses of immediate-release opioids, for less than three to seven days, when multimodal and nonpharmacologic therapies fail.37,38 Accessing the state prescription drug monitoring program, required in some jurisdictions, can help verify medication histories and help ensure that benzodiazepines are not being used concurrently.37 Physicians should consider prescribing a naloxone antidote for patients at risk of an opioid overdose.37 Communicating any history of addiction or contraindications to NSAIDs or acetaminophen can help the dentist select an appropriate analgesic regimen. Data Sources: PubMed literature searches were completed using combinations of the following terms: analgesia, anemia, antibiotic prophylaxis, anticoagulation, antiplatelet, antithrombotic, ascites, atrial fibrillation, cancer, chemotherapy, cirrhosis, clearance, consult, coronary artery disease, dental care, dental clearance, dental procedure, diabetes, dialysis, end stage renal disease, epilepsy, guidelines, heart valve, hemoglobin A1C, hepatic disease, hypertension, infective endocarditis, joint prosthesis, joint replacement, myocardial infarction, narcotic, nitrous oxide, osteonecrosis of the jaw, pain, periodontitis, periprocedural management, prostheses, radiation therapy, recommendations, renal insufficiency, seizure, spontaneous bacterial peritonitis, thrombocytopenia, and venous thromboembolism. References therein were also reviewed. Literature searches included clinical trials, UpToDate, Cochrane reviews, randomized controlled trials, clinical trials, and systematic reviews. Search dates: March 1, 2020, to July 19, 2020, and May 15, 2021. The authors thank Frank Perez for his editorial contribution. Continue Reading Copyright © 2021 by the American Academy of Family Physicians. This content is owned by the AAFP. A person viewing it online may make one printout of the material and may use that printout only for his or her personal, non-commercial reference. This material may not otherwise be downloaded, copied, printed, stored, transmitted or reproduced in any medium, whether now known or later invented, except as authorized in writing by the AAFP. See permissions for copyright questions and/or permission requests. Copyright © 2025 American Academy of Family Physicians. All Rights Reserved.
15986	https://napacenter.org/gross-motor-development/	Gross Motor Skills by Age: Developmental Milestones & Examples Skip to content Open toolbar Accessibility Tools Increase Text Decrease Text Grayscale High Contrast Negative Contrast Light Background Links Underline Readable Font Reset 888-711-NAPA SEARCH Generic selectors [x] Exact matches only [x] Search in title [x] Search in content [x] Post Type Selectors [x] Search in posts [x] Search in pages Accessibility MenuMENU About Us About NAPA Meet the Team Cody’s Story Resources Supported By Research Programs Overview Intensive Therapy Pediatric Physical Therapy Pediatric Occupational Therapy Pediatric Speech Therapy Developmental Feeding Therapy Early Intervention Program Weekly Therapy Telehealth Therapy Tools Overview NeuroSuit SpiderCage DMI Vitalstim Redcord Functional Estim Galileo Theratogs Locations Los Angeles Boston Austin Denver Chicago Charlotte Sydney Melbourne Brisbane London Pop-Ups Get Started Intensive Registration Process Intake Forms Billing and Insurance Grant Options Frequently Asked Questions COVID-19 & Illness Clinic Policies MyNAPA Portal Log In Join our Team Careers Students Volunteer Contact Us About Us About NAPA Meet the Team Cody’s Story Resources Supported By Research Programs Overview Intensive Therapy Pediatric Physical Therapy Pediatric Occupational Therapy Pediatric Speech Therapy Developmental Feeding Therapy Early Intervention Program Weekly Therapy Telehealth Therapy Tools Overview NeuroSuit SpiderCage DMI Vitalstim Redcord Functional Estim Galileo Theratogs Locations Los Angeles Boston Austin Denver Chicago Charlotte Sydney Melbourne Brisbane London Pop-Ups Get Started Intensive Registration Process Intake Forms Billing and Insurance Grant Options Frequently Asked Questions COVID-19 & Illness Clinic Policies MyNAPA Portal Log In Join our Team Careers Students Volunteer Contact Us Follows us: Understanding Gross Motor Skills: Development, Milestones, and Examples May 28th, 2025 \| by Cait Parr, PT, DPT Cait Parr, PT, DPT May 28th, 2025 You’ve probably heard your therapist, pediatrician, or friend who’s a parent talk about gross motor skills. But what exactly are they, and why are they important? In this blog, NAPA pediatric physical therapist Cait is here to discuss the progression of gross motor skill development, provide gross motor skills examples by age, and share tips for supporting your child in reaching gross motor milestones. Let’s dive in! What Are Gross Motor Skills? Gross motor skills involve the large muscles of the body, allowing for big movements like crawling, walking, running, jumping, and climbing. They also encompass higher-level coordination skills such as skipping, balancing, and throwing or catching a ball. Movement is important to all of us, but especially to our developing little ones. Babies, toddlers, and children learn so much through movement, and gross motor skills are a significant aspect of their overall development. From crawling and walking to jumping and climbing, these skills help children explore their environment, build strength, and gain confidence. Jump to a Section: Gross Motor Skills for Infants (Newborn – 18 months) Gross Motor Skills for Toddlers/Children (2 years old – 5 years old) Addressing Concerns About Delayed Gross Motor Development Gross Motor Development: Babies Learn From Head to Toe As babies grow, they first develop control in their neck (head control) and trunk (sitting balance) and then they learn to control their shoulders, elbows, wrists, and finally, their fingers. The same goes for the lower body, starting at the hips first, then learning to control their legs, feet, and eventually toes. Babies develop gross motor skills from head to toe, meaning upper-body control comes before lower-body control. Developmental Progression of Gross Motor Milestones Each child develops at their own pace, so these gross motor milestones by age listed below are approximate. As gross motor development happens at these approximate ages and stages, they build upon each other. For example, a baby needs to be able to pull up to stand before they can walk. We start with foundational skills and work through a developmental progression. For instance, we know that oftentimes, rolling leads to crawling andcrawling leads to walking. Gross Motor Skills Examples by Age We typically see a range of child development for each gross motor milestone, where kids may develop that skill in the few months before or after their peers. If you notice your child continuing to struggle with the development of an age-appropriate gross motor milestone, please see your pediatrician to request a PT evaluation. Find an overview of examples of gross motor skills by age below! Gross Motor Skills for Infants Newborn to 2 months: Head lag with pull to sit (head lags behind body when being pulled up into a sitting position) Lifts head and can turn to both sides while on the belly (View our guide for mastering tummy time!) Kicks both legs and moves both arms equally while on back Turns head to both sides while on back 3-4 months : Raises head in line with trunk when pulled to sit Pushes up on forearms and turns head side to side while on belly Rolls from belly to back (Use these 3 tips to help teach your baby to roll over!) 5 months : Brings feet to mouth laying on back Rolls from back to belly Pushes up on hands with arms extended while on belly Pivots in a circle on the belly 6-8 months: Catches self with loss of balance in sitting Crawls on belly Reaches for toys to play in sitting Sits independently 9-11 months: Crawls on hands and knees Cruises around furniture Moves between lying down and sitting upright without help Pulls to a standing position with one foot leading Walks with two hands held 11-12 months: Walks with one hand held Stands independently for a few seconds 13-14 months: Crawls up stairs Stands up from the floor without support Walks independently: Yes, walking is a gross motor skill! (Peek at our tricks used to help children who are on the verge of independent walking!) Squats and stands back up without support 15-18 months: Walks upstairs with hands or rails to help Crawls downstairs on the belly, feet first Can kick a ball forward Gross Motor Skills for 2-Year-Olds: In addition to the skills listed above, gross motor skills for 2-year-olds include: Walks and runs fairly well Kicks a ball with either foot Walks up and down stairs alone Jumps in place (both feet off the ground) Gross Motor Skills for 3-Year-Olds: Examples of gross motor skills for 3-year-olds include: Can balance on one foot for a few seconds Catches a large ball (Find 5 activities to improve hand-eye coordination here) Jumps forward 10-24 inches Rides a tricycle Gross Motor Skills by 4 Years Old: Runs, jumps, and climbs well Hops on one foot Catches a ball Somersaults Gross Motor Skills by 5 Years Old: Skips and jumps rope Starts to skate and swim Rides bicycle with or without training wheels Gross Motor vs. Fine Motor Skills While gross motor skills involve large muscle groups (arms, legs, trunk), fine motor skills involve the small muscles (hands and fingers) used for tasks like writing, buttoning, or grasping small objects. Gross motor skills develop first and lay the foundation for fine motor control. Gross motor skills develop first and lay the foundation for fine motor control. Encouraging Gross Motor Development in Children Gross motor skill development helps children to build strength and confidence in their bodies. Kids also enjoy the same benefits of exercise and physical activity as adults do, which is important for a healthy lifestyle, no matter your age. Developing gross motor skills helps a child grow in the ability to do more complex skills, such as navigating a new playground environment or playing a team sport. As kids gain control of their bodies, they start to build up strength. Little ones need lots of opportunities to practice movement because that’s how they learn and grow! Little ones need lots of opportunities to practice movement because that’s how they learn and grow! Looking for activities to encourage movement? Find our therapists’ favorite gross motor skill activities here: 15 Activities to Build Gross Motor Skills What Are the 3 Different Types of Gross Motor Movements? 1. Locomotion, which means movement! Anything a child does to get from one spot to another is locomotion. Examples of gross motor skills in the locomotion category can include rolling, belly crawling, crawling on hands and knees, scooting, walking, running, climbing, leaping, jumping, and hopping. 2. Stationary skills,which refer to movement in a stationary place. Gross motor skills that are stationary include head control, sitting balance, standing on one or both legs, rising, falling, bending, stretching, pushing, pulling, swinging, swaying, twisting, and turning. 3. Manipulation,which means moving objects in a variety of ways. Think about all the things a child can do with a ball – they can roll, throw, catch, kick, stop, or bat a ball. All of these actions are manipulative gross motor skills. What If My Child’s Gross Motor Development Is Delayed? When a child’s gross motor development is delayed,pediatric physical therapyis often prescribed to help a child work towards gaining gross motor skills. How Can Physical Therapy Support Gross Motor Development? Physical therapy for kids focuses on developing a wide range of foundational skills to help children maximize their gross motor potential, including: Balance Coordination Muscular Strength and Endurance Motor Learning and Planning Body Awareness Sensory Processing Coordination Postural Control Muscle Tone (Addressing low muscle tone or high muscle tone) Crossing the Mid-Line (moving arms or legs across the middle of the body to perform a task) View this post on Instagram A post shared by NAPA Center (@napacenter) Explore, Learn, and Grow — Find Resources & Activities in the NAPA Blog: 15 Activities to Build Gross Motor Skills for Summer 35 Fine Motor Activities: Our Ultimate List Gross Motor Toys Chosen by Our Therapists 25 Obstacle Course Ideas to Improve Gross Motor Skills What to Expect From a Pediatric PT Evaluation Your Guide to Infant Physical Therapy Occupational Therapy vs. Physical Therapy: What’s the Difference? About the Author Cait received her Doctorate in Physical Therapy from Washington University in St. Louis and a Bachelor’s degree in Pediatric Health and Motor Development from Pepperdine. During her 10-week clinical rotation at NAPA, we fell in love with Cait (and the feeling was mutual.) Cait is passionate about the power of play and strives to educate and empower all of her families. When she’s not working with the superstar NAPA kids, Cait enjoys swimming, surfing, traveling,and organizing/leading physical therapy service trips. About NAPA Center NAPA Center is a global leader in pediatric therapy. We offer a multidisciplinary approach integrating physical, occupational, speech and feeding therapy. We specialize in individualized intensive therapy programs designed to accelerate progress and help children achieve meaningful developmental milestones. With clinics throughout the United States, United Kingdom, and Australia, we aim to bring hope to families worldwide by providing access to the best and most innovative therapies all under one roof. Book a Free Phone Screening with a Physical Therapist If you’re concerned that your child may be experiencing delays in gross motor skills, we’re here to help. Schedule a complimentary 30-minute phone screening with a NAPA physical therapist to discuss your concerns, learn about our therapy programs, and receive personalized recommendations for your child. Welcome to the NAPA Family! NAPA US Clinic Locations: Austin \| Boston \| Charlotte \| Chicago \| Denver \| Los Angeles NAPA International Clinic Locations: Australia – Brisbane, Melbourne, Sydney \| UK – London Book Phone Screening → locations / contact us TAGS:Blogs, Popular, PT SHARE: Related Posts ##### Blogs SLP #### Speech Delay vs. Autism: Identifying the Differences ##### Activities Blogs #### 10 Outdoor Gross Motor Activities ##### Blogs PT #### What Is Serial Casting and How Can It Help My Child? PHONE 888-711-NAPA EMAIL info@napacenter.org Stay in the NAPA know. Sign up for our newsletter to receive important updates! 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15987	https://www.youtube.com/watch?v=UtNe_WulKAI	Comparing Ratios and Proportions With Inequality MK Learningcenter 2840 subscribers 2 likes Description 1186 views Posted: 15 May 2020 This video will compare ratios and proportions with inequality. Transcript: welcome back guys last video we learned how to identify ratio rate and propulsion and also that we comparing one to another and also it doesn't have to be always number we know that we can come up with some object and we can actually play with it and know that we can express in three different ways as a ratio so now this video I'm going to learn actually how we can form or is it forming proportion given two fractions okay so and also I'll try to show you if we can figure out which fraction is bigger which one fraction is smaller on much friction is equal so we're gonna figure that out also it using inequality sign less than greater than and equal okay so guys here you go so let's stick with the first one here so the best and easiest toy okay to check out based and easiest way to check out in order to fraction is equal we just cross product remember there is a technique you have to do you have to start with these three okay we have to start with these three this three down here okay you multiply this first okay so if you do multiply this first okay or you can reduce the next one to the right it's up to you or you can say well my first number is here I'll actually you know what mister generalize that first so some we say you know what this is the agent this is 64 can you get 64 so if we multiply by 8 and multiply by 8 here we go okay so multiply by 8 and multiply by 8 that gives me 24 also on the top so yeah that means this is a proportion or I can clearly just cross multiply which is always it works okay how do you do that you have to do this first the three first numerator so we can say 3 times 64 equal to 8 times 24 okay so if you're looking to this year I think jump into any calculator you have available right any calculator you have an you long multiplication is up to you and I'm going to show you there three times 64 he's gonna give you 192 well you see something weird because I didn't clear the memory okay now you will see actually nothing is going so we have three times 64 and also we have 8 times 24 okay exactly same see that they're exactly same so on clearly I see that would give me 192 192 is equal to 192 so guess what they are equal so that's form proportion yes it is a proportion proportion okay so that's how we come up with that this is a proportion so check not is a proportion and also yeah you cross multiply now we do the same thing on the next one again you can do anyway one but the best thing is just cross multiply and see if they are equal so you do the first one first again you want to do this one first so clearly again 3 times 9 is 27 is equal to 6 times 20 is 120 easy to calculate this and I can clearly see they're not equal so we say this time not proportion trophy or k1 not a proportion okay now let's take a look on the next one falling moon so the same thing here we're gonna do with a cross multiply okay well you can use this one cross multiply just like that remember I have to do this first okay that is the first thing I have to do three times 64 that is what we must follow the first neon order so 3 times 64 is equal to 8 times 24 okay so again you want to check if you have calculator hand you can check it out 3 times 64 and this gives you 192 Oh seems like we using the same cushion so I pay adjusted sorry about that seems like we have a same push in here so I'll have to come up with a different example here just give me a little bit time I'm going to change this to more than maybe nine okay I'm gonna put down this was it nine all right let's make it nice okay so now we're gonna have a different number okay all right so basically we do nine times 64 24 times 8 so it's gonna give us nine times 64 equal to eight times 24 and you do the same oh yeah you check out with your little calculator you have every level right nine times 64 which gives you five seventy six and eight times 24 that would give you 192 so clearly first one gives me 576 second one is giving me 192 therefore they're not equal as you clearly see so it's not forming the proportion not a proportion not a proportion why because left hand side the left hand side is not equal to the right hand side this always works this way okay now also we want to check out which one is bigger which one is smaller okay any thumbs up so I'm gonna go back to the beginning actually and I'm gonna actually add just a little bit here okay I'm gonna keep everything same just add just a little bit here you see this one I'm gonna actually put down you know take off those okay and I'm using this also so I don't have to recreate it again okay now if I ask a question use less than and not equal to or maybe equal to okay or maybe equal to use this symbol in between them use them okay so well 192 is clearly equal to 192 so we know they are equal okay now next 127 well what symbol we do here we take what symbol let's take a look here 27 is smaller one is big number so 27 is less than so clearly we come up with the first one here less than okay less than clearly it's less than right and the next one we do is basically checking out here 576 of course is greater than as you clearly see it's going to be greater than so this symbol you're going to work out here so why because that number is bigger here look this bigger so that is the greater than symbol themself to here okay so that is how we compare also I hope you liked this video and you know how to come up with which fraction is bigger which version is smaller and also all that is forming a proportion or not thank you very much
15988	https://flexbooks.ck12.org/cbook/ck-12-math-analysis-concepts/section/4.10/primary/lesson/powers-and-roots-of-complex-numbers-mat-aly/	Skip to content Elementary Math Grade 1 Grade 2 Grade 3 Grade 4 Grade 5 Math 6 Math 7 Math 8 Algebra I Geometry Algebra II Math 6 Math 7 Math 8 Algebra I Geometry Algebra II Probability & Statistics Trigonometry Math Analysis Precalculus Calculus What's the difference? 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Learn. Interact. eXplore. CCSS Math Concepts and FlexBooks aligned to Common Core NGSS Concepts aligned to Next Generation Science Standards Certified Educator Stand out as an educator. Become CK-12 Certified. Webinars Live and archived sessions to learn about CK-12 Other Resources CK-12 Resources Concept Map Testimonials CK-12 Mission Meet the Team CK-12 Helpdesk FlexLets Know the essentials. Pick a Subject Donate Sign Up 4.10 Powers and Roots of Complex Numbers Written by:Raja Almukkahal \| Larame Spence \| Fact-checked by:The CK-12 Editorial Team Last Modified: Aug 01, 2025 Manually calculating (simplifying) a statement such as: (14−17i)5 or 4√(3−2i) in present (rectangular) form would be a very intensive process at best. Fortunately you will learn in this lesson that there is an alternative: De Moivre's theorem. De Moivre's theorem is really the only practical method for finding the powers or roots of a complex number, but there is a catch... What must be done to a complex number before De Moivre's theorem can be utilized? Powers and Roots of Complex Numbers Powers of Complex Numbers How do we raise a complex number to a power? Let’s start with an example: (−4−4i)3=(−4−4i)⋅(−4−4i)⋅(−4−4i) In rectangular form, this can get very complex. What about in r cis θ form? (−4−4i)=4√2 cis(5π4) So the problem becomes 4√2 cis(5π4)⋅ 4√2 cis(5π4)⋅ 4√2 cis(5π4) and using our multiplication rule from the previous section, (−4−4i)3=(4√2)3 cis(15π4) Notice, (a + bi)3 = r3cis 3 θ In words: Raise the r-value to the same degree as the complex number is raised and then multiply that by cis of the angle multiplied by the number of the degree. Reflecting on the example above, we can identify De Moivre's Theorem: \| Let z = r(cos θ + i sin θ) be a complex number in rcisθ form. If n is a positive integer, znis zn = rn (cos(nθ) + i sin(nθ)) \| It should be clear that the polar form provides a much faster result for raising a complex number to a power than doing the problem in rectangular form. Roots of Complex Numbers You probably noticed long ago that when an new operation is presented in mathematics, the inverse operation often follows. That is generally because the inverse operation is often procedurally similar, and it makes good sense to learn both at the same time. This is no exception: The inverse operation of finding a power for a number is to find a root of the same number. Recall from algebra that any root can be written as x1/n Given that the formula for De Moivre’s theorem also works for fractional powers, the same formula can be used for finding roots: z1/n=(a+bi)1/n=r1/ncis(θn) Examples Example 1 Earlier, you were asked what should be done to a complex number before you can use De Moivre's theorem on it. A complex number operation written in rectangular form, such as: (13−4i)3 must be converted to polar form before utilizing De Moivre's theorem. Example 2 Find the value of (1+√3i)4. r=√(1)2+(√3)2=2 tan θref=√31, and θ is in the 1st quadrant, so θ=π3 Using our equation from above: z4=r4 cis 4θ z4=(2)4 cis 4π3 Expanding cis form: z4=16(cos(4π3)+i sin(4π3)) =16((−0.5)−0.866i) Finally we have z4 = -8 - 13.856i Example 3 Find √1+i. First, rewriting in exponential form: (1 + i)½ And now in polar form: √1+i=(√2 cis(π4))1/2 Expanding cis form, =(√2(cos(π4)+i sin(π4)))1/2 Using the formula: =(21/2)1/2(cos(12⋅π4)+i sin(12⋅π4)) =21/4(cos(π8)+i sin(π8)) In decimal form, we get =1.189( 0.924 + 0.383i) =1.099 + 0.455i To check, we will multiply the result by itself in rectangular form: (1.099+0.455i) ⋅ (1.099+0.455i)=1.0992+1.099(0.455i)+1.099(0.455i) +(0.455i)2 =1.208+0.500i+0.500i+0.208i2 =1.208+i−0.208 or =1+i Example 4 Find the value of x:x3=(1−√3i). First we put 1−√3i in polar form. Use x=1, y=−√3 to obtain r=2, θ=5π3 let z=(1−√3i) in rectangular form z=2 cis (5π3) in polar form x=(1−√3i)1/3 x=[2cis(5π3)]1/3 Use De Moivre’s theorem to find the first solution: x1=21/3cis(5π/33) or 21/3cis(5π9) Leave answer in cis form to find the remaining solutions: n = 3 which means that the 3 solutions are 2π3 radians apart or x2=21/3cis(5π9+2π3) and x3=21/3cis(5π9+2π3+2π3) NOTE: It is not necessary to add 2π3 again. Adding 2π3 three times equals 2π. That would result in rotating around a full circle and to start where it all began- that is the first solution. The three solutions are: x1=21/3cis(5π9) x2=21/3cis(11π9) x3=21/3cis(17π9) Each of these solutions, when graphed will be 2π3 apart. Check any one of these solutions to see if the results are confirmed. Checking the second solution: x2=21/3cis(11π9) =1.260[cos(11π9)+i sin(11π9)] =1.260[−0.766−0.643i] =−0.965−0.810i Does (-0.965 – 0.810i)3 or (-0.965 – 0.810i) (-0.965 – 0.810i) (-0.965 – 0.810i) =(1−√3i)? Example 5 What are the two square roots of i? Let z=√0+i. r=1, θ=π/2 or z=[1× cisπ2]1/2 Utilizing De Moivre’s theorem: z1=[1× cisπ4] or z2=[1× cis5π4] z1=1(cosπ4+i sinπ4) or z2=1(cos5π4+i sin5π4) z1=0.707+0.707i or z2=−0.707−0.707i Check for z1 solution: (0.707 + 0.707i)2 = i? 0.500 + 0.500i + 0.500i + 0.500i2 = 0.500 + i + 0.500(-1) or i Example 6 Calculate 4√(1+0i). What are the four fourth roots of 1? Let z = 1 or z = 1 + 0i. Then the problem becomes find z1/4 = (1 + 0i)1/4. Since r=1 θ=0, z1/4=[1×cis 0]1/4 with z1=11/4(cos 04+i sin 04) or 1(1+0) or 1 That root is not a surprise. Now use De Moivre’s to find the other roots: z2=11/4[cos(0+π2)+i sin(0+π2)] Since there are 4 roots, dividing 2π by 4 yields 0.5π or 0 + i or just i z3=11/4[cos(0+2π2)+i sin(0+2π2)] which yields z3 = -1 Finally, z4=11/4[cos(0+3π2)+i sin(0+3π2)] or z4=−i The four fourth roots of 1 are 1, i, -1 and -i. Example 7 Calculate (√3+i)7. To calculate (√3+i)7 start by converting to rcis form. First, find r. Recall r=√√32+12. r=√3+1 r=2 If cosθ=√32 and sinθ=12 then θ=30o and is in quadrant I. Now that we have trigonometric form, the rest is easy: (√3+i)7=[2(cos30o+isin30o)]7 ..... Write the original problem in rcis form 27[(cos(7⋅30o)+isin(7⋅30o)] ..... De Moivre's theorem 128[−√32+−12i] ..... Simplify (√3+i)7=−64√3−64i ..... Simplify again ∴(√3+i)7=−64√3−64i Review Perform the indicated operation on these complex numbers: Divide: 2+3i1−i Multiply: (−6−i)(−6+i) Multiply: (√32−12i)2 Find the product using polar form: (2+2i)(√3−i) Multiply: 2(cos 40∘+i sin 40∘)∙4(cos 20∘+i sin 20∘) Multiply: 2(cos π8+i sin π8)∙2(cos π10+i sin π10) Divide: 2(cos 80∘+i sin 80∘)÷6(cos 200∘+i sin 200∘) Divide: 3 cis(130∘)÷4 cis(270∘) Use De Moivre’s theorem. [3(cos 80∘+i sin 80∘)]3 [√2(cos 5π16+i sin 5π16)]4 (√3−i)6 Identify the 3 complex cube roots of 1+i Identify the 4 complex fourth roots of −16i Identify the five complex fifth roots of i Review (Answers) Click HERE to see the answer key or go to the Table of Contents and click on the Answer Key under the 'Other Versions' option. \| Image \| Reference \| Attributions \| --- Student Sign Up Are you a teacher? Having issues? Click here By signing up, I confirm that I have read and agree to the Terms of use and Privacy Policy Already have an account? Save this section to your Library in order to add a Practice or Quiz to it. (Edit Title)35/ 100 This lesson has been added to your library. \|Searching in: \| \| \| Looks like this FlexBook 2.0 has changed since you visited it last time. We found the following sections in the book that match the one you are looking for: Go to the Table of Contents No Results Found Your search did not match anything in . Student Sign Up Are you a teacher? 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15989	https://www.wyzant.com/resources/answers/893362/maclaurin-series	maclaurin series \| Wyzant Ask An Expert Log inSign up Find A Tutor Search For Tutors Request A Tutor Online Tutoring How It Works For Students FAQ What Customers Say Resources Ask An Expert Search Questions Ask a Question Wyzant Blog Start Tutoring Apply Now About Tutors Jobs Find Tutoring Jobs How It Works For Tutors FAQ About Us About Us Careers Contact Us All Questions Search for a Question Find an Online Tutor Now Ask a Question for Free Login WYZANT TUTORING Log in Sign up Find A Tutor Search For Tutors Request A Tutor Online Tutoring How It Works For Students FAQ What Customers Say Resources Ask An Expert Search Questions Ask a Question Wyzant Blog Start Tutoring Apply Now About Tutors Jobs Find Tutoring Jobs How It Works For Tutors FAQ About Us About Us Careers Contact Us Subject ZIP Search SearchFind an Online Tutor NowAsk Ask a Question For Free Login CalculusTrigonometryLinear AlgebraAlgebra Ana A. asked • 04/23/22 maclaurin series Find the first three nonzero terms in the Maclaurin series of f(x)= tan x. Follow •1 Comment •1 More Report Mark M. You post 9 problems. Do you have a specific question or is this an attempt to get your work done for you? Report 04/23/22 1 Expert Answer Best Newest Oldest By: Raymond B.answered • 04/24/22 Tutor 5(2) Math, microeconomics or criminal justice See tutors like this See tutors like this Maclaurin or Taylor series for f(x) = tanx = the series for sinx divided by the series for cosx tanx = x + x^3/3 + 2x^5/15 +17x^7/315 + ... 1st 3 terms = x + (1/3)x^3 + (2/15)x^5 + ... or 0 + x + 0 = x for 1st 3 terms but 1st 3 non-zero terms are: tanx = f(0) + f'(0)x/1! + f"(0)x^2/2! + fiv(x^3)/3! + fv(x^4)/4! + fvi(x^5)5! + ... = 0 + x + 0 + x^3/3 + 0 + 2x^5/15 + ... = x + x^3/3 + 2x^5/15 + ... = x +x^3/3 + 2x^5/15 + ... tanx)' = 1+tan^2(x) (tanx)" = 2tanx +2tan^3(x) (tanx)"' = 2+ 8tan^2(x) + 6tan^4(x) (tanx)iv =16tanx + 40tan^3(x) +246tan^4(x) (tanx)v = 16+ 136tan^2(x) +240tan^4(x) = 120tan^6(x) all even derivatives = 0 when x=0 only the odd derivatives are non zero terms it helps to review derivatives of trig functions Upvote • 0Downvote Add comment More Report Still looking for help? 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15990	https://www.sciencedirect.com/science/article/pii/S0007091217332142	Differentiating the cellular and humoral components of neuromuscular blocking agent-induced anaphylactic reactions in patients undergoing anaesthesia - ScienceDirect Skip to main contentSkip to article Journals & Books ViewPDF Download full issue Search ScienceDirect Outline Background Methods Results Conclusions Key words Methods Results Discussion Conflict of interest Funding Acknowledgement References Show full outline Cited by (17) Figures (5) Tables (2) Table 1 Table 2 British Journal of Anaesthesia Volume 106, Issue 5, May 2011, Pages 665-674 Clinical Practice Differentiating the cellular and humoral components of neuromuscular blocking agent-induced anaphylactic reactions in patients undergoing anaesthesia Author links open overlay panelR.C.Aalberse 1, I.Kleine Budde 1, M.Mulder 1, S.O.Stapel 1, W.Paulij 2, F.Leynadier 3, M.W.Hollmann 4 Show more Outline Add to Mendeley Share Cite rights and content Under an Elsevier user license Open archive Background The significance of IgE antibodies to neuromuscular blocking agent (NMBA)-induced anaphylactic reactions during anaesthesia is unclear. We investigated the relevance of IgE to rocuronium using an in vitro technique. Methods Serum samples from 61 patients with anaphylactic reactions during anaesthesia were investigated. On the basis of clinical history, allergy to NMBA was considered likely in 48 patients, further assessed using intradermal skin tests for several commonly used NMBAs, including rocuronium, vecuronium, and succinylcholine. To determine the presence of rocuronium IgE in human serum, a rocuronium-human serum albumin (rocHSA) conjugate was coupled to a solid phase and a radioallergosorbent test performed. The biological effects of patient serum NMBA-IgE on histamine release were investigated using in vitro sensitized basophils from healthy blood donors. Results IgE to rocuronium was found in 23 of 48 serum samples (48%) with NMBA allergy, although only two of these were able to sensitize basophils to release histamine in response to rocHSA. IgE-responsiveness in the basophil test was only observed with conjugated rocHSA and not with unconjugated rocuronium or the other NMBAs evaluated. However, unconjugated rocuronium inhibited the histamine release induced by rocHSA. Correlation between skin-test reactivity to rocuronium and IgE to rocHSA was low (P> 0.1). In contrast, striking correlation between IgE to rocuronium and skin-test reactivity to succinylcholine was found (P<0.001). Conclusions Our results indicate that NMBA-related anaphylaxis requires not only IgE NMBA reactivity, but also altered cellular reactivity in the patient. The latter may be demonstrable by testing basophils from the patient, a skin test with (steroidal) NMBA, or both. Previous article in issue Next article in issue Key words neuromuscular block, allergy neuromuscular block, rocuronium neuromuscular block, succinylcholine Editor’s key points •Neuromuscular blocking agents are the main cause of anaphylaxis during anaesthesia. •Both IgE and cellular components appear to be involved but skin-prick and serum tests give inconsistent results. •Only two of 23 rocuronium IgE-positive serum samples were able to sensitize stripped basophils. •There was a correlation between IgE to rocuronium and skin-test reactivity to succinylcholine. •The authors conclude that both IgE reactivity and altered cellular reactivity are required for anaphylaxis. The incidence of anaphylactic reactions (immune- or non-immune-mediated) in anaesthesia has been reported to be between one in 6000 and one in 20 000 procedures,12 with neuromuscular blocking agents (NMBAs) implicated as the main cause.1, 2, 3, 4 However, the real incidence remains inadequately characterized. Skin-prick or intradermal tests are commonly used to identify the probable drug involved, while drug-induced activation of patient basophils has successfully been used to confirm the diagnosis.5 Indeed, NMBA-induced basophil reactivity was reported in 17 of 47 patients (36%) with proven NMBA anaphylaxis, with positive predictive value increasing to 86% (six of seven patients) for those tested within 3 yr.6 NMBA-induced anaphylaxis is associated with an IgE-dependent mechanism in ∼60% of cases.3 A French study of the prevalence of NMBA sensitivity found that 24 of 258 patients (9.3%) were reactive to NMBA; 12 had positive skin-prick tests and the serum from 14 contained IgE antibodies but only two subjects were positive in both tests.7 The prevalence of reactivity to quaternary ammonium groups, the key antigenic determinants of NMBAs, is substantially higher in the general population than the incidence of anaphylactic reactions during anaesthesia, suggesting that an NMBA-induced anaphylactic reaction requires a combination of humoral (NMBA-reactive IgE antibody) and cellular factors (altered mast cell reactivity). Allergic reactivity to conventional allergens, such as grass pollen allergens, can be transferred via serum.8 An in vitro alternative is the stripped basophil histamine release assay.9 In this assay, basophils are stripped of their own IgE, and then sensitized by the patient serum, thereby avoiding the effects of changed reactivity in the patient’s cells. The aim of the current study was to evaluate the role of cellular factors and IgE to NMBA in anaphylactic events during surgery using the stripped basophil test in conjunction with the results of intradermal skin tests with NMBAs. Methods The study was approved by the local medical ethics committee at the Hôpital Tenon, Paris, France, and conducted in accordance with the Declaration of Helsinki. Serum samples were collected at the Hôpital Tenon, Paris, France, and other hospitals from patients with anaphylactic reaction during surgery [defined according to clinical criteria, ranging from cutaneous signs (flushing, rash, urticaria) with hypotension associated with unexplained tachycardia to life-threatening symptoms, including cardiovascular collapse, bronchospasm, hypoxia, angiooedema, and cardiac arrest with or without cutaneous symptoms]. In patients agreeing to return for a follow-up skin test, diagnosis of NMBA involvement was evaluated based on clinical data (i.e. timing of event relative to NMBA administration and known allergy to other products used during surgery) and skin tests by an investigator who was blinded to serology results. All patients agreed to serum being collected for analysis, which was obtained during the return visit for skin tests. Samples were immediately frozen at −80°C and stored for future analysis. For the intradermal skin tests, a volume of 0.03–0.05 ml of test solution was injected into the forearm and the wheal and erythema diameter measured 20 min later. The test was judged to be positive if the diameter of the wheal was at least 80% of the positive control (histamine 10 mg ml−1). NMBA intradermal skin tests were performed at the following dilutions: succinylcholine 1/100, 1/1000 (Celocurine®, Sanofi-Synthelabo, Paris, France; 1/100=200 µg ml−1, 554 µM), rocuronium 1/100, 1/1000 (Esmeron®, MSD, Oss, The Netherlands; 1/100=100 µg ml−1, 164 µM), pancuronium 1/10, 1/100, 1/1000 (Pavulon®, MSD; 1/10=200 µg ml−1, 273 µM), vecuronium 1/10, 1/100, 1/1000 (Norcuron®, MSD; 1/10=200 μg ml−1, 314 μM), and atracurium 1/10, 1/100, and 1/1000 (Tracrium®, GlaxoSmithKline, Greenford, UK; 1/100=100 μg ml−1, 80.4 μM). The higher concentrations were only tested if there was a negative skin test with a more dilute solution. The results were expressed as –log 10 of the threshold concentration, i.e. 2+ indicated a skin test positive at 1/100 but negative at 1/1000 dilution and 3+ indicated a skin test positive at 1/1000 dilution. Steroidal drugs were grouped into one class and any patients scoring a positive reaction to any member of this class were considered as fulfilling the inclusion criterion for the study. Conjugation of rocuronium bromide To couple rocuronium to a solid phase suitable for IgE measurements, rocuronium bromide (supplied as a powder by MSD) was coupled to human serum albumin (HSA) (18 mol per mol HSA) via the 3α-hydroxyl group of the androstane structure using carbodiimide-mediated esterification to produce the rocuronium bromide conjugate, rocHSA. As a negative control, an androstane derivative of rocuronium, without the morpholinyl and propenyl pyrrolidinium group (17b-acetyloxy-3a-hydroxy-5a-androstane, supplied by MSD) was also conjugated to HSA (15.6 mol per mol HSA), referred to as androstaneHSA. Radioallergosorbent test The CAP system (Pharmacia, Uppsala, Sweden) was used to determine the serum concentration of IgE to succinylcholine. IgE to rocHSA was determined using the radioallergosorbent test (RAST) as, at the time of the study, no suitable reagent was available for the CAP system, although one has subsequently been developed.10 For the RAST, rocHSA and androstaneHSA (equivalent to 5 mg HSA) were added to 300 mg cyanogen bromide-activated Sepharose (Pharmacia, Uppsala, Sweden), and suspended at 2 mg ml−1 in phosphate-buffered saline (PBS) with 0.3% HSA, 0.01 M EDTA, 0.05% sodium azide (NaN 3), and Tween 20 (PBS-AT). This results in the binding of >80% of the added HSA conjugate. The RAST was performed as described previously.11 Briefly, serum (50 µl) from patients and control subjects without history of anaphylaxis was incubated overnight with rocHSA- or androstaneHSA-Sepharose (250 µl). IgE serum levels >2500 IU ml−1 can produce false-positive results; therefore, if the total IgE level was >2500 IU ml−1, the serum was mixed with human serum with low total IgE (<2 IU ml−1) to decrease the IgE concentration to 2000 IU ml−1. After washing five times, 125 I-labelled sheep anti-human IgE (see below) was added together with 500 μl PBS, with 100 µg ml−1 HSA, 20 µg ml−1 sheep immunoglobulin (purified using caprylic acid as described by McKinney and Parkinson),12 0.1% Tween 20, and 0.06% NaN 3. The suspension was then incubated overnight. After washing four times, the Sepharose-bound radioactivity was determined, expressed as a percentage of the total counts added. Negative control monoclonal mouse/human chimeric IgE antibody to the house dust mite allergen Der p 2 (IgE 2000 IU ml−1)13 did not result in non-specific IgE binding for any NMBA. Affinity-purified sheep anti-human IgE (M1294, Sanquin, Amsterdam, The Netherlands) was labelled using the 125 I chloramine-T method,14 using HSA as a carrier protein instead of bovine serum albumin. Labelled anti-IgE was diluted in PBS with 0.3% HSA, 0.01 M EDTA, and 0.05% NaN 3 and stored at –20°C. After correction for radioactivity bound to the buffer control, the results were converted to international units of IgE per millilitre using chimeric IgE antibody as a reference.13 If applicable, these results were corrected for non-specific binding to a negative-control Sepharose. If the corrected IgE value was ≤0.3 IU ml−1 or <1.5 times the baseline value, the results were considered to be negative. If the blank-corrected IgE value was between >0.3 and <1.0 IU ml−1, the result was classified as borderline, but for the χ 2 statistic, IgE antibody results <1 IU ml−1 were scored as negative. A blank-corrected IgE value of ≥1.0 IU ml−1 was considered positive. Positive RAST to rocHSA was confirmed at least once. RAST inhibition The specificity of NMBA-IgE positive sera (rocHSA and succinylcholine) was tested in an inhibition RAST. Before addition of rocHSA-Sepharose (0.5 or 2 mg test), serum was incubated with 100 µl of an NMBA-containing agent [rocuronium, succinylcholine (Suxamethonium, Pharmachemie), rocHSA, and androstaneHSA] diluted with an equal volume of 0.1 M phosphate buffer (pH 7.2) for 2 h. The RAST was repeated as described above. To exclude RAST inhibition by a non-specific effect, grass pollen-positive serum was pre-incubated with the NMBA and subsequently tested in the grass pollen RAST. No non-specific inhibition was found. On the basis of the standard error, a substance was considered to inhibit a serum RAST if pre-incubation of the serum and substance resulted in a decrease in the raw RAST score by 2% of added radioactivity. Histamine release bioassay To circumvent effects due to variable activity of nicotinic acetylcholine receptors on basophilic leucocytes, the biological effects of NMBA-IgE were investigated using in vitro sensitized basophils from selected, consenting healthy blood donors at the Department of Plasmapheresis, Sanquin (Amsterdam, The Netherlands). The buffy coats were derived (Haemonetics Plasma Collection System, Haemonetics Corporation, Braintree, MA, USA) using 1000 ml of blood from each donor. The basophils were enriched (2–5% purity) using Percoll centrifugation (1.078 g cm−3) and the histamine release assay was performed as described previously.1516 Three assay runs were performed, in which each serum sample was tested twice. Lactic acid buffer (pH 3.9) was used to remove IgE from the surface of the basophils and the cells were then sensitized17 by incubation (37°C, 90 min) with a ‘sensitization mixture’ containing 150 µl human serum, 4 mM EDTA, and 10 µg ml−1 heparin in a total volume of 1 ml. Because of the limited amount of serum available, sensitization with serum from the anaphylactic patient group was performed using a total volume of 333 µl. The sensitized cells (2.5–0.8×10 6 in a final volume of 350 µl) were challenged in the presence of 1 mM CaCl 2 by adding rocHSA, free rocuronium, or a control and incubated for 60 min at 37°C. For each serum sample, an allergen, anti-IgE, or both were used as a positive control for sensitization of IgE. The negative control was HEPES buffer with interleukin-3 [IL-3; 20 mM HEPES, 132 mM NaCl, 6 mM KCl, 1 mM CaCl 2, 1 mM MgSO 4, 1.2 mM K 2 HPO 4, 5.5 mM glucose, 5 mg ml−1 HSA, pH 7.4 and 0.6 nM IL-3 (Pepro-Tech Inc., Rocky Hill, NJ, USA)]. The controls gave good responses (histamine release >15%). The histamine release reaction was stopped by the addition of 750 µl of ice-cold normal saline. Positive serum samples were confirmed in four further experiments. To assess inhibition of histamine release, rocHSA or allergen extract was diluted in different concentrations of NMBA [rocuronium, vecuronium, pancuronium, mivacurium (Mivacron®, GlaxoSmithKline), or succinylcholine] adjusted to pH 7.4. Histamine release was determined by fluorimetric analysis18 and calculated as the percentage of total cellular histamine content, determined by perchloric acidcell lysis. Histamine release values were corrected for non-specific histamine release; the lowest value either from basophils sensitized by IgE-deficient serum or from the serum of an NMBA-negative atopic subject. If percentage histamine release after correction was below 5%, the reactivity was considered negative; between 5% and 10%, the result was classified as borderline and additional testing was needed. A score of >10% was interpreted as definite IgE reactivity. The background fluorescence signal of the investigational product in buffer was subtracted from the histamine release of the sample before this percentage was determined. If this signal was >10% of the total histamine fluorescence signal in the cell sample, a specific response <10% was considered negative. Both atracurium (di)besilate and cisatracurium (di)besilate (Nimbex®, GlaxoSmithKline) showed an unacceptably high background in our histamine assay and were not evaluated. Statistical analysis Associations were analysed using χ 2 with Yates’ correction for continuity, and by Spearman’s correlation. For the χ 2 statistic, IgE antibody results were scored as negative if the antibody level was <1 IU ml−1. Results Patients allergic to NMBA Of 63 patients assessed, 61 provided serum samples and sufficient information to ascertain that an anaphylactic reaction had occurred during anaesthesia. In total, 58 patients had skin tests for succinylcholine, vecuronium, and/or pancuronium; of these, 26 (45%), 27 (47%), and 25 (43%), respectively, were graded as ≥2+. Skin tests for rocuronium were performed in 42 patients, 13 (31%) of whom were graded as ≥2+. On the basis of intradermal skin testing and clinical evaluation, allergy to NMBA was considered likely in 48 of the 61 patients (79%) (Table1). Most (81%) were female, mean age at skin testing was 49.8 yr (range 28–75 yr), and the time between anaphylactic reaction and skin testing ranged from 0 to 19 yr. During surgery, the NMBAs given were: vecuronium (n=14), succinylcholine (n=10), rocuronium (n=10), pancuronium (n=3), atracurium (n=4), gallamine (n=3), mivacurium (n=1), and unknown NMBA (n=3). NMBA involvement was insufficiently substantiated in the remaining 13 patients (21%) for the following reasons: no NMBA was used, lack of response to the intradermal tests, or known allergy to latex or an antibiotic. Table 1. Clinical data for the 48 patients considered likely to have an allergy to an NMBA, based on skin tests and clinical evaluation. NA, not evaluated or data not available; IDT, intradermal test; Atra, atracurium; Gall, gallamine; Miv, mivacurium; Panc, pancuronium; Roc, rocuronium; Sux, succinylcholine; Vec, vecuronium. Results expressed as –log 10 of the threshold concentration, i.e. 2 indicates a skin test that was positive at the 1/100 dilution but negative at the 1/1000 dilution and 3 indicates a skin test that was positive at the 1/1000 dilution. Lack of response in the intradermal test is depicted as 0; †International units of IgE per ml Case number S1 S3 S4 S5 S6 S7 S9 S11 S12 S13 S14 S16 S17 S18 S19 S20 NMBA used Roc Atra Roc Vec Vec Sux Atra Sux Roc Gall Miva NA Sux NA Sux Vec Year of surgery 1996 1998 1999 1994 1995 1997 1999 NA 1997 1980 1999 NA 1980 1980 1999 1994 Time between anaphylaxis and skin testing (yr)0 0 0 0 1 0 0 NA 0 18 0 NA 19 18 0 0 IDT rocuronium NA 0 0 NA 0 0 0 0 2 0 0 0 0 0 0 NA IDT vecuronium 2 1 1 3 2 0 2 1 2 1 1 0 0 0 0 1 IDT pancuronium 1 0 1 3 2 0 2 0 2 0 1 0 0 2 0 1 IDT succinylcholine 0 3 0 0 2 2 0 2 2 0 2 2 2 0 2 0 IDT gallamine 0 3 0 3 2 3 0 3 2 3 2 2 0 0 0 0 IDT atracurium 0 0 0 0 0 0 2 0 0 0 0 0 NA 0 0 0 IgE† rocHSA 3.5 1.0 0.1 0.1 5.8 8.0 0.1 0.1 1.1 0.1 0.2 1.7 0.8 0.1 4.1 0.1 IgE† succinylcholine 32.0 32.6 1.1 0.1 5.4 31.8 0.1 0.1 11.2 0.5 17.8 1.9 9.9 0.1 3.1 0.5 IgE† phosphorylcholine 5.6 16.0 0.6 0.3 6.5 16.4 0.1 0.1 4.8 0.1 20.1 6.9 13.3 0.1 2.4 0.1 IgE† airborne allergen 1.2 0.1 0.2 0.1 0.1 24.1 4.5 0.1 0.1 0.1 0.1 90.0 0.1 1.6 0.1 0.1 Total IgE†2027 211 7 18 671 1383 80 36 347 48 101 743 179 69 567 165 Case number S21 S22 S24 S27 S28 S29 S30 S31 S32 S35 S36 S37 S38 S39 S43 S44 NMBA used Vec Sux Sux Vec Vec Roc Roc Gall Atra NA Vec Vec Panc Roc Panc Vec Year of surgery NA 1996 1998 1991 1996 1997 1998 NA 1999 NA 1996 1997 1998 1999 1984 1997 Time between anaphylaxis and skin testing (yr)NA 0 1 4 0 1 0 NA 0 NA 0 1 0 0 13 1 IDT rocuronium NA 0 0 0 2 3 3 0 0 0 3 0 3 3 2 0 IDT vecuronium 3 0 1 3 2 3 3 0 1 0 3 2 3 2 3 3 IDT pancuronium 3 0 2 2 2 3 3 0 1 0 2 2 2 1 3 3 IDT succinylcholine 0 3 2 0 2 0 0 0 0 2 2 0 2 0 2 0 IDT gallamine 0 2 NA 0 2 0 2 2 0 2 3 0 2 2 0 0 IDT atracurium 0 0 0 0 0 0 0 0 0 0 0 0 3 0 0 0 IgE† rocHSA 0.1 34.9 2.1 0.1 0.1 2.3 76.9 0.1 0.1 4.1 4.1 0.1 1.0 2.7 1.5 0.1 IgE† succinylcholine 4.2 34.5 34.6 0.1 0.1 2.7 1.0 0.1 6.7 2.5 6.0 2.6 21.3 0.3 1.2 1.6 IgE† phosphorylcholine 4.9 20.1 11.3 0.1 0.1 5.8 5.0 0.1 0.1 4.9 7.3 3.6 4.0 0.5 1.6 0.6 IgE† airborne allergen 1.9 0.5 0.1 1.9 0.1 1.4 0.5 0.1 0.1 0.1 1.7 0.1 0.8 1.0 0.1 0.1 Total IgE†540 5423 43 56 32 194 1091 6 177 205 133 28 330 256 74 18 Case number S45 S46 S47 S48 S49 S50 S51 S52 S53 S54 S55 S58 S59 S61 S62 S63 NMBA used Roc Vec Vec Sux Panc Sux Vec Vec Gall Roc Vec Sux Roc Atra Roc Sux Year of surgery 1998 1994 1996 1981 1997 NA NA 1995 1997 1995 NA 1987 1999 1998 1999 1988 Time between anaphylaxis and skin testing (yr)0 1 0 18 1 NA NA 1 0 0 NA 9 0 1 0 9 IDT rocuronium 2 NA NA 0 0 0 NA NA 0 3 NA NA 2 NA 3 0 IDT vecuronium 1 3 3 0 0 0 2 0 0 2 2 2 3 NA 3 0 IDT pancuronium 1 3 1 0 3 0 0 2 1 1 2 1 3 NA 1 0 IDT succinylcholine 2 3 0 2 0 3 0 0 0 3 0 0 2 NA 0 3 IDT gallamine 2 3 2 0 0 0 0 0 2 3 0 2 0 NA 0 2 IDT atracurium 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 IgE† rocHSA 2.0 8.2 0.1 24.6 11.6 7.3 0.1 0.1 0.1 0.1 0.2 0.3 5.0 0.1 0.7 17.6 IgE† succinylcholine 2.6 21.5 8.5 17.9 8.8 13.0 0.1 0.3 0.3 2.9 0.5 0.6 0.8 0.1 1.2 67.9 IgE† phosphorylcholine 1.6 12.9 5.1 11.8 3.3 35.2 0.1 0.1 0.1 1.3 0.4 0.3 2.5 0.1 0.6 29.2 IgE† airborne allergen 0.1 0.1 26.6 0.1 16.9 0.1 0.2 0.1 0.2 0.1 0.2 47.1 0.1 1.1 0.1 0.1 Total IgE†160 156 157 2849 3465 1511 51 3 59 99 75 275 39 83 383 725 Prevalence of NMBA-specific IgE and its relation to skin-test reactivity After correction (androstaneHSA: highest IgE level 1.1 IU ml−1), 23 of the 48 serum samples (48%) were positive for IgE to rocHSA (Table1; Fig.1). Several samples (six of 99) from the control group were also positive for IgE to rocHSA (Fig.1). Thirty-one of the 48 serum samples (65%) were positive for IgE to succinylcholine, and 28 to phosphorylcholine (58%) (Table1). Correlation between the concentration of IgE to rocHSA and IgE to succinylcholine showed that of the 48 patients, 21 were positive to rocHSA and succinylcholine, 10 only to succinylcholine, and two only to rocHSA (Fig.2). 1. Download: Download high-res image (121KB) 2. Download: Download full-size image Fig 1. Concentration of IgE antibodies to rocHSA in a reference group of atopic patients without a history of perioperative allergic reactions and in patients who experienced an anaphylactic event while receiving an NMBA during surgery. Individual concentrations (circles) and geometric mean (horizontal lines) with standard error of the geometric mean (sem; vertical bars) for the two groups are presented. A blank-corrected IgE value of ≥1 IU ml−1 was considered positive. 1. Download: Download high-res image (104KB) 2. Download: Download full-size image Fig 2. Correlation between the concentration of IgE to rocHSA and IgE to succinylcholine in the serum of patients who experienced an anaphylactic event while receiving an NMBA during surgery. The two closed circles indicate serum samples S1 and S30, obtained from two separate patients, which were able to sensitize basophils to release histamine in response to rocHSA. Basophil histamine release Of the samples positive/borderline for IgE to rocHSA in the RAST (i.e. samples with IgE >0.3 IU ml−1; n=25), only two, S1 and S30, were able to sensitize basophils to release histamine to rocHSA. Basophils sensitized with S1 required a high concentration of rocHSA (optimal concentration 232 µM; calculation based on rocuronium) to release histamine, whereas the optimal concentration for basophils sensitized with S30 was substantially lower (0.2–2.0 µM rocHSA) (Fig.3). 1. Download: Download high-res image (136KB) 2. Download: Download full-size image Fig 3. Sensitization of donor basophils to release histamine. (a) and (b) are two representatives of a total of five experiments. S1 and S30 are serum samples from two separate patients. RocHSA is rocuronium bromide conjugated to human serum albumin. The positive control was anti-IgE and the negative control was HEPES buffer with IL-3. Basophils sensitized with serum samples, S1 and S30, did not release histamine in response to unconjugated rocuronium, vecuronium, pancuronium, succinylcholine, or mivacurium. The positive controls used for sensitization of IgE (allergen and/or anti-IgE) were associated with a good response (histamine release >15%). Neither rocuronium nor the rocHSA conjugate had an effect on non-sensitized basophils or on activation of basophils sensitized by IgE (with a specificity unrelated to rocuronium) and challenged with the appropriate allergen. Inhibition of histamine release The histamine release induced by rocHSA from basophils sensitized by serum samples S30 (Fig.4) and S1 could be inhibited by unconjugated rocuronium. Notably, to inhibit the histamine release in response to rocHSA, a 10-fold higher concentration of unconjugated rocuronium was needed for basophils sensitized by serum sample S1 compared with cells sensitized by S30. Vecuronium and pancuronium also abolished histamine release induced by rocHSA from basophils sensitized by S30, whereas neither mivacurium nor succinylcholine had inhibitory effects. AndrostaneHSA did not influence histamine release for either serum sample. There was no evidence of non-specific inhibition, since unconjugated rocuronium did not inhibit the reaction induced by mite allergen in control bioassays. 1. Download: Download high-res image (144KB) 2. Download: Download full-size image Fig 4. Inhibition of histamine release from basophils sensitized by serum sample S30 (one of only two serum samples able to sensitize basophils to release histamine in response to rocHSA) upon challenge of the basophils with unconjugated rocHSA at concentrations of 0.2, 2, 20, or 200 μmol litre−1. Specificity of IgE to rocuronium determined by RAST inhibition IgE cross-reactivity between rocHSA and succinylcholine was investigated in four serum samples (S1, S30, S48, S63) based on their rocHSA IgE content and whether they were positive (S1, S30) or negative (S48, S63) in the histamine release assay in response to rocHSA (Table2). Table 2. Effect of pre-incubation of serum samples with unconjugated rocuronium and succinylcholine on IgE binding to rocHSA Sepharose. International units of IgE per ml; †Stripped basophils were preincubated with the serum samples, and then tested for their ability to release histamine in response to rocHSA. Only two samples, S1 and S30, were able to sensitize basophils to release histamine in response to rocHSA; ‡Inhibition of rocHSA-Sepharose binding in the RAST by pre-incubation for 2 h with succinylcholine at a final concentration of 18.5 mM or unconjugated rocuronium at a final concentration of 5.5 mM \| Empty Cell \| Serum sample tested \| \| Empty Cell \| S1 \| S30 \| S48 \| S63 \| \| Serum concentration of rocHSA IgE \| 3.5 \| 76.9 \| 24.6 \| 17.6 \| \| Serum concentration of succinylcholine IgE \| 32.0 \| 1.0 \| 17.9 \| 67.9 \| \| Histamine release assay† \| + \| + \| − \| − \| \| Per cent inhibition of rocHSA IgE binding in the RAST following preincubation with succinylcholine‡ \| 48 \| 59 \| 57 \| 27 \| \| Per cent inhibition of rocHSA IgE binding in the RAST following preincubation with unconjugated rocuronium‡ \| 68 \| 83 \| 99 \| 83 \| More efficient RAST inhibition by succinylcholine was expected in serum sample S1 (low rocHSA IgE, high succinylcholine IgE levels) compared with serum sample S30 (high rocHSA IgE, low succinylcholine IgE levels). However, in all four serum samples, inhibition with unconjugated rocuronium (tested at 5.5 mM) was greater than with succinylcholine (tested at 18.5 mM) (Table2). Pre-incubation with androstaneHSA did not affect IgE binding in the tested samples. There was no evidence of non-specific RAST inhibition, with pre-incubation with grass pollen-positive serum. Level of association between NMBA-specific IgE and skin-test reactivity The association between the presence of IgE to rocHSA and skin-test reactivity to rocuronium was low: 12 of 30 patients (40%) in the skin-test-negative group were positive or borderline positive for IgE to rocHSA vs nine of 13 patients (69%) in the skin-test-positive group (P>0.1). The association between IgE to rocHSA and skin-test reactivity to succinylcholine (Spearman’s ρ=0.53; 95% confidence interval: 0.29–0.71, P<0.001) was higher than with skin test reactivity to rocuronium (Spearman’s ρ=0.13; 95% confidence interval: −0.21 to 0.43, P>0.1; Fig5). The two patients (serum samples S1 and S30) with a positive basophil test to rocHSA were exceptional in this respect, because both had a negative skin test with succinylcholine (despite the presence of IgE reactive with succinylcholine). Whereas unconjugated rocuronium did not activate basophils sensitized by S1 and S30 to release histamine upon challenge with rocHSA, the skin test with unconjugated rocuronium (1/1000 dilution) was positive in the patient from whom serum sample S30 was derived (no rocuronium skin test was performed in the patient for serum sample S1). 1. Download: Download high-res image (161KB) 2. Download: Download full-size image Fig 5. Correlation between level of skin-test reactivity to rocuronium (a) or succinylcholine (b) and concentration of IgE to rocHSA (upper panels) and IgE to succinylcholine (lower panels). The horizontal lines and vertical bars represent the medians and inter-quartile range. The closed circles indicate the serum samples, S1 and S30, obtained from two separate patients, which were able to sensitize basophils to release histamine in response to rocHSA. For S1, no skin test results for rocuronium were available, thus only S30 is shown in (a). roc, rocuronium; sux, succinylcholine. Discussion In this study, we examined the mechanisms involved in NMBA-induced anaphylactic reactions, and also the underlying mechanisms of tests utilized in current diagnoses. The likely cause of the anaphylactic reaction was determined using intradermal skin tests for several commonly used NMBAs, using the dilutions recommended in guidelines to avoid false-positive results.19, 20, 21 None of the four patients in whom atracurium was implicated had a positive response at the recommended 1/1000 dilution. Nevertheless, these patients were included in the study because they had a positive response to at least two other NMBAs and/or because the timing of the event relative to atracurium suggested NMBA involvement and other potential causes were excluded. Rocuronium and succinylcholine were each used in only 10 of the 48 cases in our study. However, in agreement with previous reports of cross-reactivity between NMBAs,19 about half of the samples were positive for IgE to rocHSA and 65% for IgE to succinylcholine. Most of the positive samples (21/33) had IgE to both rocHSA and succinylcholine and only two to rocHSA alone. A substantial fraction of our natural antibody repertoire is reactive with quaternary ammonium groups, including acetylcholine-related substances such as NMBA and phosphorylcholine, a far more ubiquitous compound. These antibodies are inducible by exposure to environmental factors, microbial (e.g. pneumococcal), and parasitic antigens, and may be cross-reactive with some commonly used chemicals (detergents, shampoos, toothpastes),1922 and other quaternary ammonium ion-containing drugs.23 As anaphylactic responses to NMBAs have occurred in patients who have not received an NMBA before,24 it is likely that most NMBA-reactive antibodies are not initially induced by NMBA exposure.23 Indeed, we found IgE to rocuronium in samples taken before rocuronium was introduced to the market. Cross-reactive antibodies are expected to have a relatively low affinity for NMBA, offering a plausible explanation for the lack of biological relevance in some cases. Succinylcholine and rocHSA IgE cross-reactivity was assessed in our study using four samples selected for their IgE content. Even in samples with a much greater succinylcholine IgE content than rocHSA IgE content, pre-incubation with unconjugated rocuronium resulted in a greater inhibition of IgE binding in the rocHSA RAST than pre-incubation with succinylcholine. This suggests that, while there is cross-reactivity between the NMBA IgE, there are differences in sensitivity. Surprisingly, we showed that correlation between skin test reactivity to rocuronium and presence of IgE to rocHSA was low. In contrast, skin test reactivity to succinylcholine was closely associated with IgE reactivity (not only to succinylcholine, but also to rocuronium). Interestingly, only two of the IgE-positive serum samples were able to sensitize stripped basophils to release histamine in response to rocHSA, and neither sample sensitized basophils to release histamine in response to unconjugated rocuronium or the other NMBAs. This indicates that the stripped basophil test is unsuitable for diagnosis of NMBA-related anaphylactic reactions. This is not because of an intrinsic lack of diagnostic sensitivity of the test, as histamine release in the bioassay has been shown to have good correlation with IgE levels in patients with allergy to house dust mites.9 On the basis of this previous experience, we expected 15 of the 48 current samples to have a positive response to rocHSA in the stripped basophil bioassay. The reason for this discrepancy is not clear, especially given the close association between IgE serology and skin test reactivity to succinylcholine, but it suggests that IgE to NMBAs does not have the same properties as IgE to protein allergens. Differences in affinity cannot be the sole explanation for the discrepancies between IgE antibody measurements and biological reactivity, particularly for those patients with a positive skin test to rocuronium without obvious specific IgE. The two serum samples with a positive histamine release result cannot be used as an illustration of an NMBA-induced allergic reaction because we found basophil activation only with conjugated rocuronium (rocHSA), whereas non-conjugated rocuronium was ineffective and actually inhibited activation by the conjugate. This indicates that rocuronium acts as a monovalent hapten, whereas anaphylactic activity requires the NMBA to be at least divalent.5,25, 26, 27 The current study (Fig.1) and others22 showed a substantially increased prevalence of NMBA-reactive IgE in patients with anaphylaxis during surgery compared with the general population and patients with other allergies, and thus it is likely to play some pathogenic role. We hypothesize that some NMBA-reactive cellular component is hyperactive, which is sufficient in some cases to cause an IgE-independent, non-immune-mediated anaphylactic reaction. In other cases, collaboration with NMBA-reactive IgE may be needed. This cellular hyperactivity is likely to be expressed in tissue cells and blood basophils, as suggested by the high prevalence of basophil activation in patients with anaphylaxis during surgery.6 One interesting possibility for the contrasting correlation data is that the skin test to succinylcholine reflects immunological hyper-reactivity to acetylcholine-related structures, whereas skin-test reactivity to pancuronium and vecuronium (and similar NMBAs, but not succinylcholine) reflects cellular hyper-reactivity. In future studies, it will be important simultaneously to compare in vivo sensitized basophils and stripped normal basophils sensitized with IgE from serum obtained from the same patients. Measurement of tryptase levels in the current study would also have provided important information on whether the anaphylactic reaction in these patients involved mast cell activation. Unfortunately, this was not possible because all the serum samples were obtained some weeks after the anaphylactic reaction and tryptase levels are only meaningful if samples are obtained within a few hours of the event. In conclusion, this study provides further information on mechanisms underlying the current diagnostic procedures for suspected NMBA-induced anaphylactic reactions. The analytical protocol for NMBA-related anaphylactic events should (in addition to measurement of plasma tryptase) include two types of diagnostic test: (i) a test to establish the presence of IgE reactive to succinylcholine; and (ii) a test to establish cellular hyper-reactivity. The first test could be a skin test with succinylcholine, IgE serology, or both while the second test may comprise a skin test with a steroidal NMBA (e.g. pancuronium), and/or an ex vivo cellular test using basophils obtained from the patient before, or a few weeks after, the anaphylactic event. Conflict of interest W.P. is an employee of MSD, Oss, The Netherlands. R.C.A., S.S., I.K.B., and M.M. are employees of Sanquin, Amsterdam, The Netherlands. For the duration of this project, R.C.A. was a paid consultant to MSD. Funding The project was funded by a grant from MSD, Oss, The Netherlands, to Sanquin, which covered the salary of I.K.B. and the costs of the assays. Editorial support was provided by Valerie Moss of Prime Medica, Knutsford, Cheshire, UK, funded by Merck, Summit, NJ, USA. Acknowledgement The authors would like to thank Professor David Levy (Paris) for his help in preventing information getting lost in translation. Recommended articles References 1MM Fisher, BA Baldo The incidence and clinical features of anaphylactic reactions during anesthesia in Australia Ann Fr Anesth Reanim, 12 (1993), pp. 97-104 View PDFView articleView in ScopusGoogle Scholar 2MC Laxenaire Epidemiology of anesthetic anaphylactoid reactions. Fourth multicenter survey (July 1994–December 1996) Ann Fr Anesth Reanim, 18 (1999), pp. 796-809 View PDFView articleView in ScopusGoogle Scholar 3JH Laake, JA Røttingen Rocuronium and anaphylaxis—a statistical challenge Acta Anaesthesiol Scand, 45 (2001), pp. 1196-1203 doi:10.1034/j.1399-6576.2001.451004.x View in ScopusGoogle Scholar 4D Vervloet, S Durham Adverse reactions to drugs Br Med J, 316 (1998), pp. 1511-1514 CrossrefView in ScopusGoogle Scholar 5Société Françaises d’Anesthésie et de Réanimation Reducing the risk of anaphylaxis during anaesthesia. Abbreviated text Ann Fr Anesth Reanim, 21 (Suppl 1) (2002), pp. 7s-23s Google Scholar 6V Kvedariene, S Kamey, Y Ryckwaert, et al. Diagnosis of neuromuscular blocking agent hypersensitivity reactions using cytofluorimetric analysis of basophils Allergy, 61 (2006), pp. 311-315 doi:10.1111/j.1398-9995.2006.00978.x CrossrefView in ScopusGoogle Scholar 7F Porri, C Lemiere, J Birnbaum, et al. Prevalence of muscle relaxant sensitivity in a general population: implications for a preoperative screening Clin Exp Allergy, 29 (1999), pp. 72-75 doi:10.1046/j.1365-2222.1999.00453.x View in ScopusGoogle Scholar 8C Prausnitz, H Küstner Studien über die Überempfindlichkeit (Study of hypersensitivity) Zbl Bakt Mik Hyg, 86 (1921), pp. 160-169 Google Scholar 9I Kleine Budde, PG de Heer, JS van der Zee, RC Aalberse The stripped basophil histamine release bioassay as a tool for the detection of allergen-specific IgE in serum Int Arch Allergy Immunol, 126 (2001), pp. 277-285 doi:10.1159/000049524 View in ScopusGoogle Scholar 10DG Ebo, L Venemalm, CH Bridts, et al. Immunoglobulin E antibodies to rocuronium. A new diagnostic tool Anesthesiology, 107 (2007), pp. 253-259 doi:10.1097/01.anes.0000270735.40872.f2 CrossrefView in ScopusGoogle Scholar 11JS Van der Zee, H de Groot, P van Swieten, HM Jansen, RC Aalberse Discrepancies between the skin test and IgE antibody assays: study of histamine release, complement activation in vitro, and occurrence of allergen-specific IgG J Allergy Clin Immunol, 82 (1988), pp. 270-281 doi:10.1016/0091-6749(88)91011-1 View PDFView articleView in ScopusGoogle Scholar 12MM McKinney, A Parkinson A simple, non-chromatographic procedure to purify immunoglobulins from serum and ascites fluid J Immunol Methods, 96 (1987), pp. 271-278 doi:10.1016/0022-1759(87)90324-3 View PDFView articleView in ScopusGoogle Scholar 13J Schuurman, GJ Perdok, TE Lourens, PW Parren, MD Chapman, RC Aalberse Production of a mouse/human chimeric IgE monoclonal antibody to the house dust mite allergen Der p 2 and its use for the absolute quantification of allergen-specific IgE J Allergy Clin Immunol, 99 (1997), pp. 545-550 doi:10.1016/S0091-6749(97)70083-6 View PDFView articleView in ScopusGoogle Scholar 14N Kageyama-Yahara, Y Suehiro, T Yamamoto, M Kadowaki IgE-induced degranulation of mucosal mast cells is negatively regulated via nicotinic acetylcholine receptors Biochem Biophys Res Commun, 377 (2008), pp. 321-325 doi:10.1016/j.bbrc.2008.10.004 View PDFView articleView in ScopusGoogle Scholar 15K Ishizaka, H Tomioka, T Ishizaka Mechanisms of passive sensitization. I. Presence of IgE and IgG molecules on human leukocytes J Immunol, 105 (1970), pp. 1459-1467 CrossrefView in ScopusGoogle Scholar 16EF Knol, TW Kuijpers, FP Mul, D Roos Stimulation of human basophils results in homotypic aggregation. A response independent of degranulation J Immunol, 151 (1993), pp. 4926-4933 CrossrefView in ScopusGoogle Scholar 17JJ Pruzansky, LC Grammer, R Patterson, M Roberts Dissociation of IgE from receptors on human basophils. I. Enhanced passive sensitization for histamine release J Immunol, 131 (1983), pp. 1949-1953 CrossrefView in ScopusGoogle Scholar 18RP Siraganian Refinements in the automated fluorometric histamine analysis system J Immunol Methods, 7 (1975), pp. 283-290 doi:10.1016/0022-1759(75)90025-3 View in ScopusGoogle Scholar 19P Dewachter, C Mouton-Faivre, CW Emala Anaphylaxis and anesthesia: controversies and new insights Anesthesiology, 111 (2009), pp. 1141-1150 doi:10.1097/ALN.0b013e3181bbd443 View in ScopusGoogle Scholar 20PM Mertes, MC Laxenaire, A Lienhart, et al., Working Group for the SFAR; ENDA; EAACI Interest Group on Drug Hypersensitivity Reducing the risk of anaphylaxis during anaesthesia: guidelines for clinical practice J Investig Allergol Clin Immunol, 15 (2005), pp. 91-101 View in ScopusGoogle Scholar 21PM Mertes, DA Moneret-Vautrin, F Leynadier, MC Laxenaire Skin reactions to intradermal neuromuscular blocking agent injections: a randomized multicenter trial in healthy volunteers Anesthesiology, 107 (2007), pp. 245-252 doi:10.1097/01.anes.0000270721.27309.b3 CrossrefView in ScopusGoogle Scholar 22E Florvaag, SG Johansson, H Oman, et al. Prevalence of IgE antibodies to morphine. Relation to the high and low incidences of NMBA anaphylaxis in Norway and Sweden, respectively Acta Anaesthesiol Scand, 49 (2005), pp. 437-444 doi:10.1111/j.1399-6576.2004.00591.x CrossrefView in ScopusGoogle Scholar 23BA Baldo, MM Fisher, NH Pham On the origin and specificity of antibodies to neuromuscular blocking (muscle relaxant) drugs: an immunochemical perspective Clin Exp Allergy, 39 (2009), pp. 325-344 doi:10.1111/j.1365-2222.2008.03171.x CrossrefView in ScopusGoogle Scholar 24T Lobera, MT Audicana, MD Pozo, et al. Study of hypersensitivity reactions and anaphylaxis during anesthesia in Spain J Investig Allergol Clin Immunol, 18 (2008), pp. 350-356 View in ScopusGoogle Scholar 25AL de Weck Immunochemical particularities of anaphylactic reactions to compounds used in anesthesia Ann Fr Anesth Reanim, 12 (1993), pp. 126-130 View PDFView articleView in ScopusGoogle Scholar 26BA Baldo, MM Fisher Mechanisms in IgE-dependent anaphylaxis to anesthetic drugs Ann Fr Anesth Reanim, 12 (1993), pp. 131-140 View PDFView articleView in ScopusGoogle Scholar 27A Didier, D Cador, P Bongrand, et al. Role of the quaternary ammonium ion determinants in allergy to muscle relaxants J Allergy Clin Immunol, 79 (1987), pp. 578-584 doi:10.1016/S0091-6749(87)80152-5 View PDFView articleView in ScopusGoogle Scholar Cited by (17) Allergenic and Mas-Related G Protein-Coupled Receptor X2-Activating Properties of Drugs: Resolving the Two 2023, Journal of Allergy and Clinical Immunology in Practice Citation Excerpt : Anti-rocuronium sIgE was also depicted in sera obtained long before this aminosteroid NMBA entered the market.23 In other words, although the presence of an IDHR upon first exposure is sometimes perceived to be an argument against an sIgE-dependent mechanism (eg, in FQs),25 this likely does not apply to all NMBAs21-23 or any situation in which patients apparently react to an in vivo created neoepitope and in whom evidence for an sIgE-mediated reaction is demonstrable (eg, the sugammadex–rocuronium inclusion complex).26 Except that a patient might be genuinely naive to a drug (and likely sensitized by a structural homolog24), a previous exposure may have gone unnoticed or forgotten. Show abstract Since the seminal description implicating occupation of the Mas-related G protein-coupled receptor X2 (MRGPRX2) in mast cell (MC) degranulation by drugs, many investigations have been undertaken into this potential new endotype of immediate drug hypersensitivity reaction. However, current evidence for this mechanism predominantly comes from (mutant) animal models or in vitro studies, and irrefutable clinical evidence in humans is still missing. Moreover, translation of these preclinical findings into clinical relevance in humans is difficult and should be critically interpreted. Starting from our clinical priorities and experience with flow-assisted functional analyses of basophils and cultured human MCs, the objectives of this rostrum are to identify some of these difficulties, emphasize the obstacles that might hamper translation from preclinical observations into the clinics, and highlight differences between IgE- and MRPGRX2-mediated reactions. Inevitably, as with any subject still beset by many questions, alternative interpretations, hypotheses, or explanations expressed here may not find universal acceptance. Nevertheless, we believe that for the time being, many questions remain unanswered. Finally, a theoretical mechanistic algorithm is proposed that might advance discrimination between MC degranulation from MRGPRX2 activation and cross-linking of membrane-bound drug-reactive IgE antibodies. ### Consensus clinical scoring for suspected perioperative immediate hypersensitivity reactions 2019, British Journal of Anaesthesia Show abstract Grading schemes for severity of suspected allergic reactions have been applied to the perioperative setting, but there is no scoring system that estimates the likelihood that the reaction is an immediate hypersensitivity reaction. Such a score would be useful in evaluating current and proposed tests for the diagnosis of suspected perioperative immediate hypersensitivity reactions and culprit agents. We conducted a Delphi consensus process involving a panel of 25 international multidisciplinary experts in suspected perioperative allergy. Items were ranked according to appropriateness (on a scale of 1–9) and consensus, which informed development of a clinical scoring system. The scoring system was assessed by comparing scores generated for a series of clinical scenarios against ratings of panel members. Supplementary scores for mast cell tryptase were generated. Two rounds of the Delphi process achieved stopping criteria for all statements. From an initial 60 statements, 43 were rated appropriate (median score 7 or more) and met agreement criteria (disagreement index <0.5); these were used in the clinical scoring system. The rating of clinical scenarios supported the validity of the scoring system. Although there was variability in the interpretation of changes in mast cell tryptase by the panel, we were able to include supplementary scores for mast cell tryptase. We used a robust consensus development process to devise a clinical scoring system for suspected perioperative immediate hypersensitivity reactions. This will enable objectivity and uniformity in the assessment of the sensitivity of diagnostic tests. ### Three cases of suspected sugammadex-induced hypersensitivity reactions 2012, British Journal of Anaesthesia Show abstract Neuromuscular blocking agents have been implicated in 60–70% of anaphylactic events associated with anaesthesia. We report two cases of probable hypersensitivity reaction to sugammadex and an additional suspected but less supported case of possible immune-mediated reaction or other adverse reaction. The patients were given a bolus of sugammadex 100 mg immediately before extubation. In all three patients, a possible allergic reaction was suspected within 4 min of sugammadex administration, but with different degrees of severity. Skin testing was positive in two of these patients. Hypersensitivity to sugammadex unaccompanied by cardiovascular or respiratory symptoms might be missed during the course of anaesthesia. Careful monitoring for possible allergic responses is required in patients who have received sugammadex. ### Reclassifying anaphylaxis to neuromuscular blocking agents based on the presumed Patho-Mechanism: IgE-Mediated, pharmacological adverse reaction or “innate hypersensitivity”? 2017, International Journal of Molecular Sciences ### The Developmental History of IgE and IgG4 Antibodies in Relation to Atopy, Eosinophilic Esophagitis, and the Modified TH2 Response 2016, Current Allergy and Asthma Reports ### Meyler’s Side Effects of Drugs: The International Encyclopedia of Adverse Drug Reactions and Interactions, Sixteenth Edition 2015, Meyler S Side Effects of Drugs the International Encyclopedia of Adverse Drug Reactions and Interactions View all citing articles on Scopus Copyright © 2011 The Author(s). Published by Elsevier Ltd. All rights reserved. Recommended articles Centrifugal microfluidics for sorting immune cells from whole blood Sensors and Actuators B: Chemical, Volume 245, 2017, pp. 1050-1061 Zeta Tak For Yu, …, Jianping Fu ### Suitability of a preserved human cadaver model for the simulation of facemask ventilation, direct laryngoscopy and tracheal intubation: a laboratory investigation British Journal of Anaesthesia, Volume 116, Issue 3, 2016, pp. 417-422 Z.Szűcs, …, E.Tassonyi View PDF ### Dexmedetomidine: review, update, and future considerations of paediatric perioperative and periprocedural applications and limitations British Journal of Anaesthesia, Volume 115, Issue 2, 2015, pp. 171-182 M.Mahmoud, K.P.Mason View PDF ### Peripheral blood cultured mast cells: Phenotypic and functional outcomes of different culture protocols Journal of Immunological Methods, Volume 492, 2021, Article 113003 Jessy Elst, …, Didier G.Ebo ### A new MAS-related G protein-coupled receptor X2 cell membrane chromatography analysis model based on HALO-tag technology and its applications Talanta, Volume 268, Part 1, 2024, Article 125317 Qianqian Jia, …, Langchong He ### Secretory leukocyte protease inhibitor modulates FcεRI-dependent but not Mrgprb2-dependent mastocyte function in psoriasis International Immunopharmacology, Volume 122, 2023, Article 110631 Patrycja Kwiecinska, …, Beata Grygier View PDF Show 3 more articles Article Metrics Citations Citation Indexes 17 Captures Mendeley Readers 25 View details About ScienceDirect Remote access Contact and support Terms and conditions Privacy policy Cookies are used by this site.Cookie settings All content on this site: Copyright © 2025 Elsevier B.V., its licensors, and contributors. 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15991	https://artofproblemsolving.com/wiki/index.php/2022_AMC_10A_Problems/Problem_15?srsltid=AfmBOopojW5oB34-GDqMr3ElyVJrG4aeZ0avmnp2EWmvEOIBVKiyUZpy	Art of Problem Solving 2022 AMC 10A Problems/Problem 15 - 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 2022 AMC 10A Problems/Problem 15 Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search 2022 AMC 10A Problems/Problem 15 Contents 1 Problem 2 Diagram 3 Solution 1 (Inscribed Angle Theorem) 4 Solution 2 (Brahmagupta‘s Formula) 5 Solution 3 (Circumradius's Formula) 6 Solution 4 (Complete Bash Using Brahmagupta’s and Parameshvara’s Formulas) 7 Video Solution 1 8 Video Solution 2 9 Video Solution by TheBeautyofMath 10 See Also Problem 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 Diagram ~MRENTHUSIASM Solution 1 (Inscribed Angle Theorem) Opposite angles of every cyclic quadrilateral are supplementary, so We claim that We can prove it by contradiction: If then and are both acute angles. This arrives at a contradiction. If then and are both obtuse angles. This arrives at a contradiction. By the Inscribed Angle Theorem, we conclude that is the diameter of the circle. So, the radius of the circle is The area of the requested region is Therefore, the answer is ~MRENTHUSIASM Solution 2 (Brahmagupta‘s Formula) When we look at the side lengths of the quadrilateral we see and which screams out because of Pythagorean triplets. As a result, we can draw a line through points and to make a diameter of See Solution 1 for a rigorous proof. This can also be shown using the Law of Cosines: Since and it follows that Since the diameter is we can see the area of the circle is just from the formula of the area of the circle with just a diameter. Then we can use Brahmagupta Formula where are side lengths, and is semi-perimeter to find the area of the quadrilateral. You can find the answer to this quickly by noting that . So now the area of the region we are trying to find is Therefore, the answer is ~Gdking ~Oinava ~South Solution 3 (Circumradius's Formula) We can guess that this quadrilateral is actually made of two right triangles: has a ratio in the side lengths, and is a triangle. (See Solution 1 for a proof.) Next, we can choose one of these triangles and use the circumradius formula to find the radius. Let's choose the triangle. The area of the triangle is equal to the product of the side lengths divided by times the circumradius. Therefore, . Solving this simple algebraic equation gives us . Plugging in the values, we have . Rewriting this gives us . Therefore, adding these values gets us ~orenbad Solution 4 (Complete Bash Using Brahmagupta’s and Parameshvara’s Formulas) To start, we can apply a couple formulas to find the circumradius and the area of the cyclic quadrilateral, and then just subtract the area of the quadrilateral from the area of the circle, and that is our final answer. This formula is really tedious with many calculations so don't do this in a contest. To find the circumradius we have: If we let , that means that the semi-perimeter is just . Note that stands for the semi-perimeter. The calculations will take very long so Im just going to skip to the answer of as our circumradius. This means that the area of the circle is . Now applying Brahmagupta's formula to find the area of a cyclic quadrilateral which is just: . Again we will use the same representations as these letters from before, so . Plugging in the numbers we have that the area is . To express our answer in the form that was asked at the top we have: , so our final answer is -jb2015007 Video Solution 1 ~Education, the Study of Everything Video Solution 2 Video Solution by TheBeautyofMath ~IceMatrix See Also 2022 AMC 10A (Problems • Answer Key • Resources) Preceded by Problem 14Followed by Problem 16 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. 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.
15992	https://www.askiitians.com/forums/Algebra/explain-why-a-b-a-b-4ab-tell-the-answer_269229.htm	Explain why (a-b)² = (a+b)² - 4abTell the answer as fast as you could - askIITians help@askiitians.com 1800-150-456-789 Live Courses Resources Exam Info Forum Notes Upcoming Examination Login BACK Algebra>Explain why (a-b)² = (a+b)² - 4ab Tell th... Explain why (a-b)² = (a+b)² - 4ab Tell the answer as fast as you could because I'm studying right now. Harry Styles , 4 Years ago Grade 10 1 Answers Sahithi Last Activity: 4 Years ago Given, (a-b)2= (a+b)2– 4ab (1) Let’s take R.H.S first and we will compare it with L.H.S. (a+b)2= a 2+ b 2+2ab(2) By adding (-4ab) to (2) on both sides we get (a+b)2– 4ab = a 2+ b 2+2ab – 4ab =a 2+ b 2– 2ab =(a-b)2 Hence prooved LIVE ONLINE CLASSES Prepraring for the competition made easy just by live online class. Full Live Access Study Material Live Doubts Solving Daily Class Assignments Start your free trial Other Related Questions on algebra What is a binomial expression? algebra 1 Answer Available Last Activity: 1 Year ago Can we divide two vectors? If this is possible then how? algebra 1 Answer Available Last Activity: 1 Year ago How do you write “the product of 18 and q” as an algebraic expression? algebra 1 Answer Available Last Activity: 1 Year ago दहावीच्या आय सी एस इ विद्यार्थ्यांनी पुढे काय करायला पाहिजे algebra 1 Answer Available Last Activity: 2 Years ago Find the least number of square tiles by which the floor of a room of dimensions 16.58 m and 8.32 m can be covered completely. why HCF and not LCM as “least” keywor dis given? algebra 1 Answer Available Last Activity: 2 Years ago View all Questions We help you live your dreams. Get access to our extensive online coaching courses for IIT JEE, NEET and other entrance exams with personalised online classes, lectures, study talks and test series and map your academic goals. Company About US Privacy Policy Terms and condition Course Packages Contact US +91-735-322-1155 info@askiitians.com AskiiTians.com C/O Transweb B-30, Sector-6 Noida - 201301 Tel No. +91 7353221155 info@askiitians.com , 2006-2024, All Rights reserved Type a message here... Free live chat⚡ by ·
15993	https://math.stackexchange.com/questions/85775/geometry-problem-line-intersecting-a-semicircle	Skip to main content Geometry problem: Line intersecting a semicircle Ask Question Asked Modified 13 years, 9 months ago Viewed 1k times This question shows research effort; it is useful and clear 3 Save this question. Show activity on this post. Suppose we have a semicircle that rests on the negative x-axis and is tangent to the y-axis.A line intersects both axes and the semicircle. Suppose that the points of intersection create three segments of equal length. What is the slope of the line? I have tried numerous tricks, none of which work sadly. calculus geometry Share CC BY-SA 3.0 Follow this question to receive notifications edited Nov 26, 2011 at 15:11 Quixotic 22.9k3434 gold badges133133 silver badges225225 bronze badges asked Nov 26, 2011 at 14:01 AndrewAndrew 3122 bronze badges 2 Are the three segments of equal length on the line or the semi-circle? user02138 – user02138 11/26/2011 14:14:50 Commented Nov 26, 2011 at 14:14 1 It would be helpful if you can post an image... Pedja – Pedja 11/26/2011 14:18:53 Commented Nov 26, 2011 at 14:18 Add a comment \| 2 Answers 2 Reset to default This answer is useful 4 Save this answer. Show activity on this post. In this kind of problem, it is inevitable that plain old analytic geometry will work. A precise version of this assertion is an important theorem, due to Tarski. If "elementary geometry" is suitably defined, then there is an algorithm that will determine, given any sentence of elementary geometry, whether that sentence is true in Rn. So we might as well see what routine computation buys us. We can take the equation of the circle to be (x+1)2+y2=1, and the equation of the line to be (what else?) y=mx+b. Let our semicircle be the upper half of the circle. Substitute mx+b for y in the equation of the circle. We get (1+m2)x2+2(1+mb)x+b2=0.(∗) Let the root nearest the origin be r1, and the next one r2. Note that the line meets the x-axis at x=−b/m. From the geometry we can deduce that −r2=−2r1 and b/m=−3r1, and therefore r1=−b3mandr2=−2b3m. By looking at (∗) we conclude that −bm=−2(1+mb)1+m2and2b29m2=b21+m2. Thus the algebra gives us the candidates m=±27−−√. (Of course, the first equation was not needed.) Sadly, we should not always believe what algebraic manipulation seems to tell us. I have checked out the details for the positive candidate for the slope, and everything is fine. Our line has equation y=27−−√x+214√5. Pleasantly, the points r1 and r2 turn out to have rational coordinates. However, the negative candidate is not fine. That can be checked by looking at the geometry. But it is also clear from the algebra, which has been symmetrical about the x-axis. The algebra was not told that we are dealing with a semicircle, not a circle. So naturally it offered us a mirror symmetric list of configurations. We conclude that the slope is 27−−√. Share CC BY-SA 3.0 Follow this answer to receive notifications edited Nov 26, 2011 at 22:25 answered Nov 26, 2011 at 15:55 André NicolasAndré Nicolas 515k4747 gold badges583583 silver badges1k1k bronze badges 2 I never thought to use Vieta. (The solution I was working on was quite messier.) Neat! J. M. ain't a mathematician – J. M. ain't a mathematician 11/26/2011 16:13:24 Commented Nov 26, 2011 at 16:13 I prefer to break symmetry late, if at all. As a problem-solving strategy, it works more often than it should. André Nicolas – André Nicolas 11/26/2011 16:51:51 Commented Nov 26, 2011 at 16:51 Add a comment \| This answer is useful 3 Save this answer. Show activity on this post. Let the line meet the axes at A(−a,0) and B(0,b), and the semi-circle at A′ and B′ (with A′ the closer to A, and B′ the closer to B). Let O be the origin, and define d:=\|AA′\|=\|A′B′\|=\|B′B\|>0 as the common length of the segments. The Power of a Point Theorem, applied to point B, tells us that \|BB′\|⋅\|BA′\|⟹d⋅(2d)⟹2d2===\|BO\|2b2b2 Also, Pythagoras tells us that a2+b2=(3d)2=9d2 Eliminating b, we have a2=7d2 so that the slope is ba=2–√d7–√d=27−−√ NOTE. If we cared for the actual value of d, we could leverage the Power of Point A (writing P for the point (−2,0)): \|AP\|⋅\|AO\|⟹(2−a)⋅a⟹(2−d7–√)⋅d7–√⟹27–√9====\|AA′\|⋅\|AB′\|2d22d2d Share CC BY-SA 3.0 Follow this answer to receive notifications edited Nov 26, 2011 at 22:26 answered Nov 26, 2011 at 18:18 BlueBlue 84.2k1414 gold badges128128 silver badges266266 bronze badges 2 1 You made a typo in the last line before the Note... David Mitra – David Mitra 11/26/2011 18:55:53 Commented Nov 26, 2011 at 18:55 @David: Jumping Jiminy! A bit of fractional dyslexia. Fixed. Blue – Blue 11/26/2011 22:27:22 Commented Nov 26, 2011 at 22:27 Add a comment \| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions calculus geometry See similar questions with these tags. 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15994	https://gocardless.com/en-us/guides/posts/markup-vs-margin-whats-the-difference/	Markup vs. Margin: What’s the Difference? \| GoCardless Skip to content Go to GoCardless Blog Homepage Why GoCardless? Product Partners Pricing Resources Log inSign up USA English Close menu Open site navigation sidebar Why GoCardless? 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Written by GoCardless The GoCardless content team comprises a group of subject-matter experts in multiple fields from across GoCardless. The authors and reviewers work in the sales, marketing, legal, and finance departments. All have in-depth knowledge and experience in various aspects of payment scheme technology and the operating rules applicable to each. The team holds expertise in the well-established payment schemes such as UK Direct Debit, the European SEPA scheme, and the US ACH scheme, as well as in schemes operating in Scandinavia, Australia, and New Zealand. See full bio Last edited Feb 2022 — 2 min read Table of contents What’s the difference between profit margin and markup? Gross margin vs. markup: how do they work? Markup vs. margin formula When to use markup vs. margin It’s a basic sales principle: to make a profit, businesses must price products at a high enough level to cover costs. Both “margin” and “markup” are related to this principle but they’re not interchangeable, as the meanings vary slightly. What’s the difference between profit margin and markup? The main difference between profit margin and markup is that margin is equal to sales minus the cost of goods sold (COGS), while markup is a product’s selling price minus its cost price. Margin is equal to sales minus the cost of goods sold (COGS). Markup is equal to a product’s selling price minus its cost price. Confusing profit margin vs. markup can lead to accounting and sales errors. For example, you might end up either under- or overpricing your products, which can cut away into your profits. Understanding the two terms is essential to know if you’re pricing your products most effectively. Gross margin vs. markup: how do they work? Let’s take a closer look at the differences between gross margin vs. markup, starting with margin. To calculategross margin, you must subtract the cost of goods sold from an item’s sale price. For example, imagine that a product costs $50 to produce, and sells for $80. This means that it has a margin of $30. Another option is to express this as a percentage calculating margin divided by sales. The margin percentage is therefore 37.5%. By contrast, markup refers to the difference between a product’s selling price and its cost price. It’s looking at the same transaction but from a different angle. Using the same sale above, the item at a cost price of $50 is marked up by $30 to its final sale price of $80. Expressed as a percentage calculated by dividing markup by product cost, the markup percentage is 60%. From looking at these two examples of markup vs. margin, it’s easy to see why the terms are often confused. In terms of dollar amount, both the margin and markup are $30. However, you can see that the markup percentage is higher than the margin percentage. The basis for the markup percentage is cost, while the basis for margin percentage is revenue. The cost figure should always be lower than the revenue figure, so markup percentages will be higher thanprofit margins. Markup vs. margin formula We can express this basic concept in a markup vs. margin formula below: Margin ÷ Cost of Goods = Markup Percentage For example, if you want to earn a profit margin of $5 on a product with a cost price of $8, you can plug these numbers into the formula to arrive at the markup percentage: $5 Margin ÷ $8 Cost = 62.5% Markup Percentage You can then multiple the markup percentage by the cost price to arrive at a sales price of $13. You can also use these profit margin vs. markup formulas when expressing the figures in percentages. Profit margin percentage formula: ((Sale price – Cost price) ÷ Sale Price)(100) Markup percentage formula: ((Sale price – Cost price) ÷ Cost Price)(100) When to use markup vs. margin If you want to decide on the rightselling price to achieve a certain profit, you should use the markup percentage as in the example below. However, if you’re looking at performance, you’ll want to look at margins to assess past sales. Choosing a markup percentage can be complicated. You should take various factors including competitor costs, distribution, marketing, and the supply chain to choose a reasonable value. By taking these factors into consideration, you can ideally maximize profit. We can help GoCardless helps you automate payment collection, cutting down on the amount of admin your team needs to deal with when chasing invoices. Find out how GoCardless can help you with ad hoc payments or recurring payments. Over 100,000 businesses use GoCardless to get paid on time. Learn more about how you can improve payment processing at your business today. Get StartedLearn More Related topics Accountants All Categories Recommended for you Everything you need to know about advanced billing Learn the ins and outs of advanced billing with our helpful guide. 2 min readAccountants Everything you need to know about advanced billing 2 min readAccountants Top three retail accounting software 2 min readAccountants Eight steps to changing accounting software 3 min readAccountants When would you use progress invoicing? 3 min readAccountants What is progress billing and how does it benefit your small business? 3 min readAccountants What is revenue leakage and how to prevent it 2 min readAccountants Interested in automating the way you get paid? 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15995	https://dnr.maryland.gov/wildlife/documents/lath_leaf_activities.pdf	Leaf Activities By: Kerry Wixted No matter where you are, there are always leaves available! You can use leaves you find outside or those found in the refrigerator or on house plants. Early Learners (preK-2nd grade) 1. Collect leaves and sort by shape, size, color, etc. Teach your student descriptive words and use the Growing Up WILD Leaf Observation Sheet. 2. Create a leaf graph on a poster board or other surface. 3. Perform the life cycle of a tree dance. 4. Read Leaf Man by Lois Ehlert. 5. Create a Leaf Man or leaf critter using leaves and craft materials. 6. Grab a bowl of water, leaves, and pennies. Predict how many pennies each leaf will hold. Test your prediction. Retest with a new leaf. See if the way pennies are added to the leaves matters. What leaves hold the most pennies and why? 7. For additional ideas, check out Project Learning Tree’s Leaf Activities for Young Learners. 8. For Fall Leaf Songs and Fingerplay, check out this video: Upper Elementary (3rd-5th) 1. Create a leaf bingo sheet and look for different types of leaves around the neighborhood. 2. Go through the refrigerator and pantry. List what types of leaves we eat. 3. Identify the parts of a leaf. Here are worksheet examples. 4. Collect leaves and examine the different vein patterns. The veins help carry water and food from the leaves to the roots and back. Use paint or chalk markers to trace the vein patterns. How many different vein patterns can you find? 5. Conduct an experiment to learn how water moves through leaves: 6. Watch the video on Why Do Leaves Change Color. 7. Conduct a science experiment to see the different pigments in leaves (this works best with fresh leaves): 8. Press leaves between newspapers and a heavy, flat weight. Once dry, create leaf bookmarks, leaf lanterns, and other leaf crafts. 9. Make leaf rubbings. Middle School and High School (6th-12th) 1. Did you know the cherry blossoms in D.C. bloomed almost 2 weeks earlier than average? The bloom times for many species are changing with the climate. Become a community scientist. Look for trees and other plants that are beginning to leaf out. Use applications like Seek or iNaturalist to identify plants and then record the phenology (timing of leaf out, flowers, etc) in Project BudBurst. You can also use Nature’s Notebook. 2. Create a leaf collection and identify the leaves you collected. Use our Deciduous tree guide and our conifer tree guide to help with identification. Here is an example of a sixth grade leaf project with writing requirements. 3. If you have access to a microscope or another magnifying device, go on a stomata safari. Some phones like iPhones have magnifying features built in. 4. Look up Chinese Leaf Carving. Attempt to create artwork using this technique. 5. Use leaves to create Cubism Leaves.
15996	https://brainly.in/question/55728869	What is the 'triple point' temperature of water in kelvin ? A.273.16K B. C. D. - Brainly.in Skip to main content Ask Question Log in Join for free For parents For teachers Honor code Textbook Solutions Brainly App Gopalkrishna1349 10.03.2023 Chemistry Secondary School answered What is the 'triple point' temperature of water in kelvin ?A.273.16K B. C. D. 2 See answers See what the community says and unlock a badge. 0:00 / -- Read More Answer No one rated this answer yet — why not be the first? 😎 tejasbhoir263 tejasbhoir263 Answer: Correct answer is: Option A Explanation: The 'triple point' temperature of water is 273 K. The triple point of a substance is the temperature and pressure at which the three phases (gas, liquid, and solid) of that substance coexist in thermodynamic equilibrium. The triple point of water is 273.16 K at 611.73 Pa. This is the basis of the Kelvin scale and it is equal to 0.01 °C, which is the freezing point of water. Explore all similar answers Thanks 0 rating answer section Answer rating 5.0 (1 vote) Find Chemistry textbook solutions? See all Class 12 Class 11 Class 10 Class 9 Class 8 Class 7 Class 6 Preeti Gupta - All In One Chemistry 11 3080 solutions Selina - Concise Chemistry - Class 9 1071 solutions 1500 Selected Problems In Chemistry for JEE Main & Advanced 2928 solutions Lakhmir Singh, Manjit Kaur - Chemistry 10 1797 solutions Lakhmir Singh, Manjit Kaur - Chemistry 9 1137 solutions NCERT Class 11 Chemistry Part 1 431 solutions NEET Exam - Chemistry 360 solutions Chemistry 643 solutions Selina - Concise Chemistry - Class 8 487 solutions Selina - Chemistry - Class 7 394 solutions SEE ALL Advertisement Answer No one rated this answer yet — why not be the first? 😎 zera18 zera18 Answer: A.273.16K Explanation: The triple point of water is used to define the Kelvin(K), the base unit of thermodynamic temperature in the International System of Units (SI). The triple point of water is 273.16 K, 0.01∘ C, or 32.018∘ F Explore all similar answers Thanks 0 rating answer section Answer rating 0.0 (0 votes) Advertisement Still have questions? Find more answers Ask your question New questions in Chemistry Write the structural formula of 3- ethyl-2, 5-dimethyl hexan-3-ol Anti iron gel how to prepare for neede rotation 90 degree Copper Sulphate give the valency 1. Re write the correct functions of the given table Sl.No. 1 23 Forms of water Character Solid Liquid Gaseous When water is heated it convert into 1. Re write the correct functions of the given table Forms of water Sl.No. 1 Solid 2 Liquid Gaseous 23 Character When water is heated it convert into PreviousNext Advertisement Ask your question Free help with homework Why join Brainly? ask questions about your assignment get answers with explanations find similar questions I want a free account Company Careers Advertise with us Terms of Use Copyright Policy Privacy Policy Cookie Preferences Help Signup Help Center Safety Center Responsible Disclosure Agreement Get the Brainly App ⬈(opens in a new tab)⬈(opens in a new tab) Brainly.in We're in the know (opens in a new tab)(opens in a new tab)(opens in a new tab)(opens in a new tab)
15997	https://www.math.uh.edu/~irina/MATH1300/Notes1300/1300S11.pdf	1 1.1 The Real Number System Types of Numbers: The following diagram shows the types of numbers that form the set of real numbers. Definitions 1. The natural numbers are the numbers used for counting. 1, 2, 3, 4, 5, . . . A natural number is a prime number if it is greater than 1 and its only factors are 1 and itself. A natural number is a composite number if it is greater than 1 and it is not prime. Example: 5, 7, 13,29, 31 are prime numbers. 8, 24, 33 are composite numbers. 2. The whole numbers are the natural numbers and zero. 0, 1, 2, 3, 4, 5, . . . 3. The integers are all the whole numbers and their additive inverses. No fractions or decimals. . . . , -3, -2, -1, 0, 1, 2, 3, . . . An integer is even if it can be written in the form n 2 , where n is an integer (if 2 is a factor). An integer is odd if it can be written in the form 1 2 − n , where n is an integer (if 2 is not a factor). Example: 2, 0, 8, -24 are even integers and 1, 57, -13 are odd integers. 4. The rational numbers are the numbers that can be written as the ratio of two integers. All rational numbers when written in their equivalent decimal form will have terminating or repeating decimals. 1 5 , 3.25, 0.8125252525 …, 0.6 , 2 ( 1 2 = ) 2 5. The irrational numbers are any real numbers that can not be represented as the ratio of two integers. The numbers usually are imperfect roots. Pi is also an irrational number. Irrational numbers when written in their equivalent decimal form have non-terminating and non-repeating decimals. The square root of a prime number is irrational. 13 , 2.236067978 … , π ( 142 . 3 ≈ ) , 2 ( 414 . 1 ≈ ) , 3 ( 732 . 1 ≈ ) 6. A real number is either a rational or an irrational number. A real number is positive if it is greater than 0, negative if it is less than 0. 7. Undefined numbers are numbers in the form 0 k Example 1: Circle all of the words that can be used to describe the number 25. Even, Odd, Positive, Negative, Prime, Composite, Natural, Whole, Rational, Irrational, Real Real Numbers Irrational Rational Undefined Integers Whole Numbers Natural Numbers 3 Example 2: Classify each of the following numbers: 24 _______ -12 ______ 2.5 ______ 3 5 ______ 7 ______ Example 3:       − − − 35 , 4 1 25 , 23 , 11 , 23 . 1 , 5 7 , 0 , 5 , 8 , 2 . 10 Give the list of all Rational numbers: Irrational numbers: Even integers: Odd natural numbers: Whole numbers: Negative real numbers: Prime numbers: Composite numbers: Real numbers: Undefined numbers: 4 Order on a Number Line The real number line: We can graph real numbers on a number line. For each point on the number line there corresponds exactly one real number, and this number is called the coordinate of that point. If a real number x is less than a real number y, we write x < y. On the number line, x is to the left of y. Example 4: For each pair of real numbers, place one of the symbols < , =, or > in the blank. a) 2 2 b) -5 _ - 6 c) 4 1 5 1 d) 2 5 __ 2.1 e) 2 9 _____ 2 1 4
15998	https://dspace.mit.edu/bitstream/handle/1721.1/104426/6-042j-spring-2010/contents/lecture-notes/MIT6_042JS10_lec01_sol.pdf	Massachusetts Institute of Technology 6.042J/18.062J, Fall ’09: Mathematics for Computer Science February 3 Prof. Albert R. Meyer revised January 25, 2010, 1076 minutes Solutions to In-Class Problems Week 1, Wed. Problem 1. Identify exactly where the bugs are in each of the following bogus proofs.1 (a) Bogus Claim: 1/8 > 1/4. Bogus proof. 3 > 2 3 log10(1/2) > 2 log10(1/2) log10(1/2)3 > log10(1/2)2 (1/2)3 > (1/2)2 , and the claim now follows by the rules for multiplying fractions. Solution. log x < 0, for 0 < x < 1, so since both sides of the inequality “3 > 2” are being multiplied by the negative quantity log10(1/2), the “>” in the second line should have been “<.” (b) Bogus proof : 1¢ = $0.01 = ($0.1)2 = (10¢)2 = 100¢ = $1. Solution. $0.01 = $(0.1)2 6= ($0.1)2 because the units $2 and $ don’t match (just as in physics the difference between sec2 and sec indicates the difference between acceleration and velocity). Similarly, (10¢)2 = 100 6 ¢. (c) Bogus Claim: If a and b are two equal real numbers, then a = 0. Bogus proof. a = b a 2 = ab a 2 − b2 = ab − b2 (a − b)(a + b) = (a − b)b a + b = b a = 0. Creative Commons 2010, Prof. Albert R. Meyer. 1From Stueben, Michael and Diane Sandford. Twenty Years Before the Blackboard, Mathematical Association of Amer ica, ©1998. 2 Solutions to In-Class Problems Week 1, Wed. Solution. The bug is at the ﬁfth line: one cannot cancel (a − b) from both sides of the equation on the fourth line because a − b = 0. Problem 2. It’s a fact that the Arithmetic Mean is at least as large the Geometric Mean, namely, a + b √ ab 2 ≥ for all nonnegative real numbers a and b. But there’s something objectionable about the following proof of this fact. What’s the objection, and how would you ﬁx it? Bogus proof. a + b ? √ ab, so 2 ≥ a + b ≥ ? 2 √ ab, so ? a 2 + 2ab + b2 ≥ 4ab, so ? a 2 − 2ab + b2 ≥ 0, so (a − b)2 ≥ 0 which we know is true. The last statement is true because a − b is a real number, and the square of a real number is never negative. This proves the claim. Solution. In this argument, we started with what we wanted to prove and then reasoned until we reached a statement that is surely true. The little question marks presumably are supposed to indicate that we’re not quite certain that the inequalities are valid until we get down to the last step. At that step, the inequality checks out, but that doesn’t prove the claim. All we have proved is that if (a + b)/2 ≥ √ ab, then (a − b)2 ≥ 0, which is not very interesting, since we already knew that the square of any nonnegative number is nonnegative. To be fair, this bogus proof is pretty good: if it was written in reverse order – or if “is implied by” was simply inserted after each line – it would actually prove the Arithmetic-Geometric Mean Inequality: Proof. a + b 2 ≥ √ ab is implied by a + b ≥ 2 √ ab, which is implied by a 2 + 2ab + b2 ≥ 4ab, which is implied by a 2 − 2ab + b2 ≥ 0, which is implied by (a − b)2 ≥ 0. The last statement is true because a − b is a real number, and the square of a real number is never negative. This proves the claim. 3 Solutions to In-Class Problems Week 1, Wed. But the problem with the bogus proof as written is that it reasons backward, beginning with the proposition in question and reasoning to a true conclusion. This kind of backward reasoning can easily “prove” false statements. Here’s an example: Bogus Claim: 0 = 1. Bogus proof. ? 0 = 1, so ? 1 = 0, so ? 0 + 1 = 1 + 0, so 1 = 1 which is trivially true, which proves 0 = 1. We can also come up with very easy “proofs” of true theorems, for example, here’s an easy “proof” of the Arithmetic-Geometric Mean Inequality: Bogus proof. a + b ? √ ab, so 2 ≥ 0 a + b ≥ ? 0 √ ab, so · 2 · 0 ≥ 0 which is trivially true. So watch out for backward proofs! Problem 3. Albert announces that he plans a surprise 6.042 quiz next week. His students wonder if the quiz could be next Friday. The students realize that it obviously cannot, because if it hadn’t been given before Friday, everyone would know that there was only Friday left on which to give it, so it wouldn’t be a surprise any more. So the students ask whether Albert could give the surprise quiz Thursday? They observe that if the quiz wasn’t given before Thursday, it would have to be given on the Thursday, since they already know it can’t be given on Friday. But having ﬁgured that out, it wouldn’t be a surprise if the quiz was on Thursday either. Similarly, the students reason that the quiz can’t be on Wednesday, Tuesday, or Monday. Namely, it’s impossible for Albert to give a surprise quiz next week. All the students now relax, having concluded that Albert must have been blufﬁng. And since no one expects the quiz, that’s why, when Albert gives it on Tuesday next week, it really is a surprise! What do you think is wrong with the students’ reasoning? 4 Solutions to In-Class Problems Week 1, Wed. Solution. The basic problem is that “surprise” is not a mathematical concept, nor is there any generally accepted way to give it a mathematical deﬁnition. The “proof” above assumes some plausible axioms about surprise, without deﬁning it. The paradox is that these axioms are incon sistent. But that’s no surprise :-), since —mathematically speaking —we don’t know what we’re talking about. Mathematicians and philosophers have had a lot more to say about what might be wrong with the students’ reasoning, (see Chow, Timothy Y. The surprise examination or unexpected hanging paradox, American Mathematical Monthly (January 1998), pp.41–51.) MIT OpenCourseWare 6.042J / 18.062J Mathematics for Computer Science Spring 2010 For information about citing these materials or our Terms of Use, visit:
15999	https://www.albert.io/blog/system-of-linear-equations-made-simple-a-sat-math-study-guide/	Skip to content ➜ System of Linear Equations Made Simple: A SAT® Math Study Guide The Albert Team Last Updated On: What We Review Introduction Systems of linear equations show up in nearly every SAT® Math section. Therefore, understanding the system of linear equations is a high-value skill for raising your score. This guide explains how to create, solve, and interpret two-variable systems with ease. Along the way, you will learn how to move among tables, equations, and graphs while spotting whether a system has a unique solution, no solution, or infinitely many solutions. By the end, solving linear systems will feel routine instead of risky. Start practicing SAT® Math (Digital) on Albert now! The Building Blocks: A Single Linear Equation Before handling two equations, let’s review one. Variable – the unknown number Coefficient – the number in front of the variable Constant – the stand-alone number Term – each separate “piece” joined by plus or minus signs Most SAT® items present lines in slope-intercept form, y = mx + b. Here, m is slope and b is the y-intercept. Mini Example Graph y = 2x + 3. Intercept: Plot (0, 3). Slope: rise 2, run 1 to reach (1, 5). Draw the line through both points. A single line is nice, but two lines working together can solve real SAT® problems in seconds. Turning Words into Math: Creating a System from Real-World Context Translating plain language into algebra is a core SAT® skill. Common contexts include mixtures, money, motion, and geometry. Example: Concert Tickets Student tickets cost \$6 each, and adult tickets cost \$10 each. A club sells 40 tickets for \$320. How many of each type were sold? Define variables x = number of student tickets y = number of adult tickets Write the equations Quantity: x + y = 40 Money: 6x + 10y = 320 Note on interpretation Solving will give an ordered pair (x, y). That pair must make sense—negative tickets are impossible. Three Ways to View the Same System Seeing a system three different ways helps you work faster. Tabular – a table of values Algebraic – two equations Graphical – two lines on the coordinate plane Example Table: \| \| \| --- \| \| x \| y \| \| 0 \| 4 \| \| 2 \| 0 \| \| \| \| --- \| \| x \| y \| \| 0 \| –1 \| \| 4 \| 1 \| Convert each row set into slope-intercept form. First line: The slope is \dfrac{0 – 4}{2 – 0} = –2. So y = -2x + 4. Second line: The slope is \dfrac{1 – (–1)}{4 – 0} = 0.5. So y = 0.5x - 1. Graph both lines. They intersect at (2, 0). Move easily from table → equation → graph to confirm answers. Ready to boost your SAT® Math (Digital) scores? Explore our plans and pricing here! Solving Systems Efficiently A. Graphing Method Pros: visual, quick estimate. Cons: exact answers hard if intersection is off-grid. Example: Solve \begin{cases} y = -x + 6 \ y = 2x - 3 \end{cases} by graphing. Plot the y-intercepts (0, 6) and (0, –3). Draw both lines. They intersect at (3, 3). Therefore, the unique solution is (3, 3). B. Substitution Method Best when one equation is solved for a variable. Example: Solve \begin{cases} y = 4x - 1 \ 2x + 3y = 17 \end{cases} \| Step \| Reason \| --- \| \| 2x + 3(4x - 1) = 17 \| Substitute 4x - 1 for y \| \| 2x + 12x - 3 = 17 \| Distribute 3 to both terms \| \| 14x = 20 \| Combine like terms and add 3 to both sides \| \| x = \dfrac{10}{7} \| Divide both sides by 14 \| \| y = 4\left(\dfrac{10}{7}\right) - 1 = \dfrac{40}{7} - 1 = \dfrac{33}{7} \| Substitute x into the equation for y \| \| \left(\dfrac{10}{7}, \dfrac{33}{7}\right) \| Final solution ✅ \| C. Elimination Method Ideal when the coefficients line up. Example: Solve \begin{cases} 3x + 2y = 16 \ 5x - 2y = 4 \end{cases} \| Step \| Reason \| --- \| \| (3x + 2y) + (5x - 2y) = 16 + 4 \| Add both equations to eliminate y \| \| 8x = 20 \| Combine like terms \| \| x = 2.5 \| Divide both sides by 8 \| \| 3(2.5) + 2y = 16 \| Substitute x = 2.5 into the first equation \| \| 7.5 + 2y = 16 \| Multiply \| \| 2y = 8.5 \| Subtract 7.5 from both sides \| \| y = 4.25 \| Divide both sides by 2 \| \| (2.5, 4.25) \| Final solution ✅ \| D. Choosing the Best Method—Decision Checklist If coefficients already match, use elimination. If a variable is isolated, use substitution. If both intercepts are integers and the grid is given, quick-sketch graphing can suffice. Under time pressure, scan for mental-math cancellations first. Solution Conditions: One, None, or Infinitely Many? Unique solution – lines intersect once; slopes different. No solution – lines are parallel; slopes equal but intercepts differ. Infinitely many – lines coincide; both slope and intercept match. Quick slope test: put each equation into slope-intercept form. Compare m and b. Example Classify without solving. a) \begin{cases} 4x - 2y = 8 \ 2x - y = 4 \end{cases} First, divide the first equation by 2: 2x - y = 4. This is the same as the second equation. Therefore, there are infinitely many solutions. b) \begin{cases} y = 3x + 1 \ 6x - 2y = 5 \end{cases} Rearrange the second equation: -2y = -6x + 5 → y = 3x - 2.5. The slopes of both equations are equal, but their y-intercepts differ. So, there are no solutions. c) \begin{cases} y = -0.5x + 7 \ y = 2x - 1 \end{cases} The slopes of both equations are different. Therefore, there is one solution. Bringing the Algebra Back to the Story After computing (x, y), always ask, “What does this pair mean?” Example: Geometry Perimeter A rectangle’s perimeter is 26 cm. Its length exceeds its width by 3 cm. Find its dimensions. Variables: w = width, l = length Equations: \begin{cases} l = w + 3 \ 2l + 2w = 26 \end{cases} One variable is isolated, so we can use substitution to solve. \| Step \| Reason \| --- \| \| 2(w + 3) + 2w = 26 \| Substitute w+3 into the second equation \| \| 2w + 6 + 2w = 26 \| Distribute 2 into parentheses \| \| 4w + 6 = 26 \| Combine like terms \| \| 4w = 20 \| Subtract 6 from both sides \| \| w = 5 \| Divide both sides by 4 \| If the width is 5 cm, the length is 5+3=8 cm. Check for extras: negative lengths would be rejected. Common SAT® Traps & Time-Saving Tips Misaligned variables – double-check that x pairs with x, y with y. Disguised like terms – 0.5x equals \frac{x}{2}; rewrite to match. Off-grid intersections – if slopes differ but intercepts are difficult to read, switch from graphing to elimination. Mental-math elimination – add 7x and –7x mentally to cancel before writing. Bubble strategy – for grid-ins, always reduce fractions like \frac{6}{8}to \frac{3}{4} to avoid scoring machines misreading. IX. Quick-Reference Vocabulary Chart \| \| \| \| --- \| Term \| Simple Definition \| SAT® Tip \| \| Variable \| A symbol (x or y) that stands for an unknown number \| Label units to avoid mix-ups \| \| Coefficient \| Number that multiplies the variable \| Watch the sign (+/–) \| \| Constant \| Fixed number with no variable attached \| Often moves when rearranging \| \| System of Linear Equations \| Two equations using the same variables \| On SAT®, almost always two variables \| \| Solution to a System \| An (x, y) pair that makes BOTH equations true \| The point where the lines meet \| \| Elimination Method \| Add or subtract equations to cancel a variable \| Align like terms first \| Final Takeaways & Next Steps Systems of linear equations power many SAT® Math questions. Master the core steps: create equations from words, represent them in multiple ways, solve efficiently, and interpret answers. Therefore, practice timed drills to lock in speed. For more targeted practice, explore additional SAT® math questions and track accuracy. Confidence grows with each solved system. 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. SAT® Score Calculator See how scores on each section impacts your overall SAT® score ACT® Score Calculator See how scores on each section impacts your overall ACT® score Grammar Review Hub Comprehensive review of grammar skills AP® Posters Download updated posters summarizing the main topics and structure for each AP® exam.

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