id stringlengths 1 6 | url stringlengths 16 1.82k | content stringlengths 37 9.64M |
|---|---|---|
189000 | https://innovation.luskin.ucla.edu/wp-content/uploads/2024/06/Addressing-the-Discriminatory-Impacts-of-Redlining-and-Highway-Development-in-California.pdf | Addressing the Discriminatory Impacts of Redlining and Highway Development in California Informing effective and equitable environmental policy BY JANTZEN HALE, ISAAC BUSHNELL, ELIZABETH FLORES, ITZEL VASQUEZ-RODRIGUEZ, AND MAKENNA CAVANAUGH JUNE 2024 STUDENT REPORT J A N T Z E N H A L E , I S A A C B U S H N E L L , E L I Z A B E T H F L O R E S , I T Z E L VA S Q U E Z- R O D R I G U E Z , A N D M A K E N N A C A V A N A U G H J U N E 2 0 2 4 U C L A L U S K I N S C H O O L O F P U B L I C A F F A I R S in California Addressing the Discriminatory Impacts of REDLINING & HIGHWAY DEVELOPMENT ADDRESSING THE DISCRIMINATORY IMPACTS OF REDLINING AND HIGHWAY DEVELOPMENT IN CALIFORNIA i ABOUT THE AUTHORS ITZEL VASQUEZ-RODRIGUEZ (she/her/ella) focuses on water justice policy. At UCLA, Itzel worked as a Research Fellow for Environmental Justice at the Luskin Center for Innovation and a Policy Fellow with the Latino Policy and Politics Institute. Originally from Lakewood, CA, Itzel is a proud Xicana of Cora descent and is passionate about making sure all Californians have access to safe and affordable drinking water.
JANTZEN HALE focuses on tribal policy. Jantzen works at the National Congress of American Indians (NCAI), communicating stories of impact and need about tribal communities to diverse audiences. Originally from Tucson, AZ, and a tribal citizen of the Navajo Nation, she is excited about the lasting impact she will make in Indian Country as she advocates for the advancement of tribal policies and legislation for Tribal Nations.
MAKENNA CAVANAUGH focuses on environmental and economic justice policy. Throughout her time at UCLA, she has worked at the Environmental Defense Fund, the Container Recycling Institute, and the Water Resources Group at the Institute for Environment and Sustainability. Originally from San Diego, Makenna is pursuing the data analytics certificate and is excited to work towards a just transition after graduation.
ELIZABETH FLORES is an MD/MPP student who focuses on health policy. She currently works on the Structural Racism and Health Equity curriculum at the UCLA medical school. She was also elected as a Co-Director for the Latino Medical Student Association Western Region. As a future family physician, Elizabeth looks forward to combining clinical care with activism to work towards healthy communities.
ISAAC BUSHNELL focuses on racial and economic equity, climate change, and environmental justice. He specializes in conducting data and spatial analysis to inform justice-oriented policymaking. Having previously worked as a middle school teacher and non-profit organizer, his most recent role was with the Integrated Climate Adaptation and Resilience Program at the California Governor’s Office of Planning and Research.
ADDRESSING THE DISCRIMINATORY IMPACTS OF REDLINING AND HIGHWAY DEVELOPMENT IN CALIFORNIA ii ACKNOWLEDGMENTS We would like to thank the following individuals for their support and guidance in the creation of this report: • Client Team: Paula Torrado Plazas, Laura August, Kevin Olp, Jaimie Huynh, and Walker Wieland • Informal Advisors: Tamika L. Butler, Esq., Dr. Lara Cushing, Deputy Executive Officer Chanell Fletcher, Ariana Hernandez, Dr. Michael C. Lens, Deputy Secretary Darwin Moosavi, Dr. Paul Ong, Marina Pérez, Dr. Gregory Pierce, and Dr. Zachary Steinert-Threlkeld We also wish to thank our advisors and peer reviewers for providing valuable insight that strengthened our report: • Advisor: Dr. Michael A. Stoll • Secondary Advisor: Prof. Michael Masserman • Peer Reviewers: Jala Abner, Sebastian Cazares, Reese Howard, Andrew Rock, and Ana Rodriguez Finally, we would like to thank the UCLA Luskin Center for Innovation for providing financial support for this project.
Report designed by Lauren Dunlap and copy edited by Miranda Raney. Cover photo credit: B Negin / Flickr DISCLAIMER This report was prepared in partial fulfillment of the requirements for the Master of Public Policy degree in the Department of Public Policy at the University of California, Los Angeles (UCLA). It was prepared at the direction of the Department and the Office of Environmental Health Hazards Assessment as a policy client. The views expressed herein are those of the authors and may not necessarily reflect those of the Department, the UCLA Luskin School of Public Affairs, UCLA as an institution, or the client.
ADDRESSING THE DISCRIMINATORY IMPACTS OF REDLINING AND HIGHWAY DEVELOPMENT IN CALIFORNIA iii CONTENTS Executive Summary iv 1. Introduction 2 1.1. Project Overview 2 1.2. Our Client 3 1.3. Policy and Research Design Questions 4 2. Methodology 6 2.1. Methods Overview 6 2.2. Establishing the History: Literature Review 7 2.3. Case Studies: Stockton and Los Angeles 7 2.4. Data Analysis: Finding Evidence of Persisting Impact 8 2.5. Generating Policy Options for Evaluation 9 3. Problem Identification 11 3.1. Problem Overview 11 3.2. Redlining 11 3.3. Highway Development in Redlined Communities 16 3.4. Impacts of Highway Development and Redlining in California 22 4. Policy Options 30 4.1. Chapter Overview 30 4.2. Policy Context 30 4.3. Policy Options 36 5. Policy Evaluation 42 5.1. Policy Evaluation Overview 42 5.2. Evaluation Results 45 5.3. Recommendations Summary 51 5.4. Conclusion 53 6. References 55 7. Appendix 62 7.1. Chapter 2 - Methodology 62 7.2. Chapter 3 - Problem Identification 65 7.3. Chapter 4 - Policy Options 73 7.4. Chapter 5 - Policy Evaluation 80 ADDRESSING THE DISCRIMINATORY IMPACTS OF REDLINING AND HIGHWAY DEVELOPMENT IN CALIFORNIA iv EXECUTIVE SUMMARY This report is a collaboration between the California Office of Environmental Health Hazard Assessment and a research team of graduate students in the Master of Public Policy Program at the UCLA Luskin School of Public Affairs. In this report, we conducted research to understand the intertwined history of redlining and highway development in the state of California, the lasting discriminatory legacies of this history throughout the state, and potential policy solutions to rectify these legacies and advance racial and environmental justice for impacted communities of color. To conclude this research, we analyzed a set of policy options to provide specific recommendations for California State agencies to adopt in order to best achieve these goals.
The report begins with an overview of the history of redlining in California, illustrating how decisions in highway development were built on the policy foundation of redlining to disrupt and dispossess marginalized communities and people of color across the state. With predominantly non-white residential areas in California cities officially redlined as “high-risk” areas due to the race of their residents, federal policy formally recommended the use of highways as physical barriers in those neighborhoods to enforce segregation. Additionally, federal policies incentivized local authorities to concentrate disruptive and high-polluting land uses in redlined communities of color so as not to devalue residential neighborhoods that were not already devalued by redlining. Together, these discriminatory policy practices have saddled generations of non-white Californians with socioeconomic, cultural, and environmental burdens not experienced by their white counterparts. These burdens persist today and are linked to diminished outcomes of health, wealth, happiness, and safety for Californians of color.
This report primarily examines two specific impacts: residential racial segregation and PM2.5 pollution concentration. Through a combination of research, mapping, and statistical analysis, striking evidence emerges of the continued heavy segregation of people of color in previously redlined areas and neighborhoods near highways throughout the state. This trend is mirrored and amplified for the segregation of white people in formerly A-rated areas. Furthermore, a positive correlation between highway proximity and PM2.5 levels is amplified in formerly redlined areas and is indicative of the increased burden of overall pollution faced by these communities in California. The research and analysis that underpin these findings utilize data and historical records from all eight California cities subjected to redlining assessments. Throughout the narrative of this history and mapping of its contemporary impacts, Stockton and Los Angeles serve as case studies.
After illustrating the ongoing discriminatory harms stemming from the history of redlining and highway development in California, the report examines the current policy landscape across the state to identify challenges and opportunities relevant to efforts aimed at addressing these harms. Challenges include the continued dependence on highways and the political divisiveness of policies explicitly focused on racial justice, while opportunities include increased funding for environmental initiatives and growing support for pro-housing, integrated communities across California. ADDRESSING THE DISCRIMINATORY IMPACTS OF REDLINING AND HIGHWAY DEVELOPMENT IN CALIFORNIA v Based on this overview of the policy landscape, the report identifies four distinct policy approaches to address the impacts of redlining and highway development and advance environmental and racial justice in California. These approaches are: 1) Explicitly focusing on race and ethnicity in environmental programs and initiatives; 2) Empowering community input and engagement and increase the decision-making authority of impacted communities; 3) Changing zoning and planning goals and requirements; 4) Making data on equity and environmental justice more accessible to impacted communities and available for community use.
Across these four approaches, the report evaluates 13 different state-level policies for California using a Criteria Alternatives Matrix weighted to reflect these goals. Through this evaluation process, we recommend eight policies for adoption by California State agencies: RECOMMENDATIONS CREATE ‘OVERBURDENED’ POLLUTION STANDARD The California Environmental Protection Agency (CalEPA) should create a regulatory standard to identify communities overburdened by pollution and require the implementation of land uses that relieve the pollution burden in such communities.
SEGREGATION ELEMENT Institute a state requirement for formerly redlined cities to include a Segregation Element within their General Plan that establishes strategies to promote integration.
STATEWIDE WAIRE California Air Resources Board (CARB) should use the South Coast Air Quality Management District’s Warehouse Indirect Source Rule as a statewide model to manage emissions and pollution from all high-polluting industrial sources.
HIGHWAY REDESIGN AND TRUCK REROUTING The California Department of Transportation (CalTrans) should prioritize and fund truck rerouting and highway redesign plans across the state in its 2025 California Transportation Plan update.
DECISION-MAKING POWER FOR IMPACTED COMMUNITIES The state should entrust decision-making power to formerly redlined communities or communities of color over policies concerning environmental justice and equity.
INCENTIVIZE COMMUNITY INVOLVEMENT State agencies should systematically incentivize community participation and input around environmental justice policies and initiatives.
40% OF FEDERAL FUNDS INTO COMMUNITIES OF COLOR State agencies should ensure at least 40% of funds from federal environmental and transportation programs are invested into communities of color.
INCREASE GRANT ACCESSIBILITY State agencies providing grant funding for environmental justice and racial equity programs should increase accessibility and support for potential grant recipients.
ADDRESSING THE DISCRIMINATORY IMPACTS OF REDLINING AND HIGHWAY DEVELOPMENT IN CALIFORNIA 1 1 INTRODUCTION Josh Begley / joshbegley.com ADDRESSING THE DISCRIMINATORY IMPACTS OF REDLINING AND HIGHWAY DEVELOPMENT IN CALIFORNIA 2 1. INTRODUCTION 1.1. Project Overview Across California, people of color disproportionately face crippling burdens of environmental injustice and socioeconomic inequality. These ongoing disparities are firmly rooted in the state’s history of redlining, which began in the late 1930s. Redlining was a process by which non-white communities were formally and informally identified as risky, targeted for segregation, and excluded from the benefits of the Federal Housing Authority’s Homeowners Loan Corporation Act (HOLC) that helped white communities build wealth. These racist housing and financial policies were the central nodes of a broad policy network that systematically dispossessed non-white people throughout the United States. A connected and integral node in this discriminatory apparatus was a set of racist land use practices in highway development, which were particularly prevalent in California. In building the country’s most robust highway system, developers and planners explicitly designed infrastructure that cut through redlined areas, irreversibly disrupting many communities of color and sparking lasting socioeconomic and environmental degradation. While significant research and policy-making efforts have been dedicated to understanding and addressing the impacts of redlining, the intertwined story of redlining and highway development necessitates a more detailed examination.
This project originated from our partners at the California Office of Environmental Health Hazard Assessment (OEHHA) and their desire to better inform the public and state-level policymakers about the history of redlining and highway development and its persisting legacies. An extensive literature review and original data analysis were conducted to construct the story of redlining and highway development in California from the 1930s to the present day. Through an examination of historical trends and new analysis, we identified links to inequitable pollution burdens and residential racial segregation still faced by Californians of color today. Based on these findings and a review of potential Misterfarmer / Flickr ADDRESSING THE DISCRIMINATORY IMPACTS OF REDLINING AND HIGHWAY DEVELOPMENT IN CALIFORNIA 3 policy options, we designed a variety of state-level policy options to address these discriminatory legacies. In conclusion, the report recommends eight policies that, based on thorough analysis, will meaningfully address these legacies by promoting environmental and racial justice for all Californians.
1.2. Our Client OEHHA is a state agency within the California Environmental Protection Agency (CalEPA) - see Figure 1 - with a mission to “protect and enhance the health of Californians and the state’s environment through scientific evaluations that inform, support and guide regulatory and other actions.”1 OEHHA is responsible for evaluating health risks associated with environmental contaminants and spearheading research efforts to help shape environmental initiatives in California. OEHHA is committed to equitably protecting and advancing the health and well-being of all Californians.
Figure 1: Flowchart of state agencies that comprise CalEPA.2 We worked with OEHHA staff members and CalEPA Racial Equity Team members, led by OEHHA’s Racial Equity and Environmental Justice Program Manager, Paula Torrado Plazas. This dedicated team consists of members from various departments within CalEPA and is committed to ensuring that the state’s environmental efforts acknowledge and adequately address the disparate environmental challenges faced by different communities in the state.3 Its primary focus involves analyzing environmental racism and advocating for equitable solutions.
1 California Office of Environmental Health Hazard Assessment (OEHHA), “About OEHHA,” accessed April 8, 2024, Link.; California Environmental Protection Agency (CalEPA), “About CalEPA,” accessed April 8, 2024, Link. 2 California Environmental Protection Agency (CalEPA), “About CalEPA,” accessed April 8, 2024, Link. 3 California Environmental Protection Agency (CalEPA), “CalEPA Racial Equity,” accessed April 8, 2024, Link. ADDRESSING THE DISCRIMINATORY IMPACTS OF REDLINING AND HIGHWAY DEVELOPMENT IN CALIFORNIA 4 The Racial Equity Team operates the Pollution and Prejudice Story Map, which details the relationship between racist policy regimes, such as redlining in California and present-day environmental injustice. The team seeks to incorporate new areas of research into the tool to illustrate this relationship more comprehensively, and this project’s research on redlining and highway development will be utilized to further build out this story for the public and decision-makers in state government.
1.3. Policy and Research Design Questions This project aims to answer the following questions for OEHHA: • How have redlining and similar discriminatory policy practices in California shaped regional highway development, and what are the consequential impacts on PM2.5 concentration and segregation levels in these communities?
• What policies and interventions can California State agencies advocate for to address these impacts and promote environmental justice and equity?
OEHHA staff members OEHHA / Twitter ADDRESSING THE DISCRIMINATORY IMPACTS OF REDLINING AND HIGHWAY DEVELOPMENT IN CALIFORNIA 5 2 METHODOLOGY Tony Barnard / Los Angeles Times ADDRESSING THE DISCRIMINATORY IMPACTS OF REDLINING AND HIGHWAY DEVELOPMENT IN CALIFORNIA 6 2. METHODOLOGY 2.1. Methods Overview First, our team conducted a qualitative literature review to explore the historical intersection between highway development and racist practices related to redlining in California. The review included sources on statewide historical policies and trends, highlighting Stockton and Los Angeles to gain a more nuanced understanding of the issue. The literature review was then used to construct a narrative of the critical aspects and interconnection between redlining and highway development in California, and throughout, specific examples from Stockton and Los Angeles were deployed to enhance the story. We then combined evidence from existing literature with original statistical and geospatial analysis to illustrate and quantify the impacts of this history on PM2.5 pollution burdens and residential racial segregation in California today. Next, we researched and compiled policy options to address such impacts to promote equity and environmental justice. Lastly, we evaluated these policy options using a Criteria Alternatives Matrix and made recommendations for relevant California State agencies.4 4 The methods for the policy analysis and recommendations are detailed at the beginning of the Policy Evaluation chapter.
Members of the California Strategic Growth Council and California EPA Secretary Garcia meet with implementing partners of Watts Rising, a Transformative Climate Communities grantee organization. California Strategic Growth Council / Twitter ADDRESSING THE DISCRIMINATORY IMPACTS OF REDLINING AND HIGHWAY DEVELOPMENT IN CALIFORNIA 7 2.2. Establishing the History: Literature Review To begin the literature review, our team leveraged its UCLA and OEHHA connections to study essential resources needed to examine the impacts of highway development within historically redlined communities in California.5 Additional networks we would have liked to engage with in this process, had it not been for time and resource constraints,6 were leaders of community organizations throughout California working to address the legacies of redlining and highway development, as well as community members in Stockton and Los Angeles. This approach would have broadened the perspectives and sources in our literature review, incorporating ideas beyond academia and professional policy circles. Overall, the resources we reviewed included scholarly articles, books, legislation, policy proposals and advocacy, data sets, maps, and other government resources.
Our review of these resources began by focusing broadly on the histories of redlining and highway development. To understand how these racist policies functioned, literature on the nature of racial discrimination and redlining in the State of California from the 1930s-1980s was examined. This time period was selected because the official redlining policies dictated by the Federal Housing Authority (FHA) were in effect from 1936 to the mid-1960s, while policies with similar intent and impact to redlining persisted following the end of redlining as codified law. Additionally, broad trends of highway development throughout the state over time were reviewed. Following this initial review, attention was directed toward sources that specifically addressed the intersection of the two histories. Throughout this research, the links between this history and outcomes of PM2.5 concentration and racial segregation levels in targeted communities were tracked. 2.3. Case Studies: Stockton and Los Angeles Our team included a case study of two formerly redlined cities as a useful method to illuminate the similarities and differences in these histories for areas with disparate socioeconomic, geographic, and demographic characteristics. Specifically, OEHHA was interested in examining the difference between redlined communities in urban and rural areas of California. Los Angeles was selected because it is the largest and most diverse city in California, serving as the nexus of the nation’s largest freeway system and has a wealth of existing research on highway development, redlining, and pollution levels. Stockton was selected because of its rural character, more northern regional location, and demographic diversity. Furthermore, Stockton has been at the forefront of collecting and sharing its pollution and environmental justice data with state agencies, including participation in the state’s Transformative Climate Communities initiative.
5 See Appendix 2.1 for sources utilized within each network.
6 Our team had eight months to complete all work for this report, from our formation in September 2023 to the deadline for completion of the report set by UCLA on April 11, 2024.
ADDRESSING THE DISCRIMINATORY IMPACTS OF REDLINING AND HIGHWAY DEVELOPMENT IN CALIFORNIA 8 Based on the findings of this research, examples of neighborhoods, social movements, highway projects, and policies in these two cities were identified, helping uncover the differential impacts of the intersection between highway development and redlining across California.
2.4. Data Analysis: Finding Evidence of Persisting Impact Following the literature review, data analysis was conducted to augment and address gaps in the existing scholarship on the links between the history of redlining and highway development in California and contemporary issues of racial segregation and PM2.5 pollution concentration. This consisted of spatial analysis to map connections between present-day outcome variables and past discriminatory policies, as well as statistical analysis to isolate significant relationships between outcome variables and relevant historical factors.
ArcGIS Pro was utilized to create maps illustrating the connections among former redlining areas and their proximity to existing highways. Subsequently, contemporary median household income, contemporary racial demographics, and contemporary PM2.5 levels by census tract were layered onto these maps.7 Using ArcGIS Pro and data from the United States Census and CalEnviroscreen, we built a dataset containing key variables for all census tracts in California’s eight formerly redlined cities. Among the variables in this data set, an index was created to chart the overall highway proximity of each census tract.8 Moreover, we constructed two sets of multivariate fixed effects regression models to analyze the impacts of redlining and highway development on PM2.5 levels and levels of racial segregation.9 Models 1 & 2: pctNwhti,c = β0 + β1Redi + β2Greeni + β3HWproxi + ac + ϵi,c pctNwhti,c = β0 + β1Redi + β2Greeni + β3HWproxi + β4RediHWproxi + ac + ϵi,c These models test the links between a census tract’s current non-white population and its highway proximity and past HOLC risk grade.
10 Models 3 & 4: PM2.5i,c = β0 + β1Redi + β2HWproxi + β3Greeni + β4PopDensi + ac + ϵi,c PM2.5i,c = β0 + β1Redi + β2HWproxi + β3Greeni + β4PopDensi + β5Redi HWproxi + ac + ϵi,c 7 See Appendix 2.2 for data used for spatial analysis.
8 See Appendix 2.3 for Highway Proximity Index methods.
9 See Appendix 2.4 for more on the process of compiling the data set and variables for regression analysis. 10 See Appendix 2.5 for a full explanation of the design and variables used in Model 1 and Model 2.
ADDRESSING THE DISCRIMINATORY IMPACTS OF REDLINING AND HIGHWAY DEVELOPMENT IN CALIFORNIA 9 These models test the links between a census tract’s current PM2.5 levels and its highway proximity and past HOLC risk grade.
11 Using these four models as a baseline, our team ran a total of 20 statistical tests to provide more nuanced quantitative evidence of the legacies of redlining and highway development today in California.
12 2.5. Generating Policy Options for Evaluation To finalize the research process, current legislation, initiatives, data, and reporting on the contemporary realities connected to the history of redlining and highway development in California were reviewed. We focused on two central impacts: segregation and PM2.5 pollution. Through this process, our team generated a list of 27 possible policy options to address these ongoing impacts and promote environmental justice. These policy options fit into four general approaches. • Creating an explicit focus on race and ethnicity • Improving community empowerment and decision-making authority • Changing zoning and planning goals • Making data more accessible Due to time constraints, 13 a maximum of four policy options within each approach were selected to evaluate as possible recommendations to California State agencies. The final set of policy approaches and options were determined based on OEHHA’s level of interest in studying the policy, the specificity of the policy design, the amount of evidence available to analyze the policy, and hypotheses about how each policy will address PM2.5 and segregation levels. Thirteen policy options were evaluated based on these criteria.
11 See Appendix 2.6 for a full explanation of the design and variables used in Model 3 and Model 4.
12 See Appendix 2.7 for a full explanation of the design and variables used on the 20 statistical tests run on urban and rural areas as well as Los Angeles and Stockton.
13 Our team had eight months to complete all work for this report, from our formation in September 2023 to the deadline of completion of this report set by UCLA on April 11, 2024.
ADDRESSING THE DISCRIMINATORY IMPACTS OF REDLINING AND HIGHWAY DEVELOPMENT IN CALIFORNIA 10 3 PROBLEM IDENTIFICATION Josh Begley / joshbegley.com ADDRESSING THE DISCRIMINATORY IMPACTS OF REDLINING AND HIGHWAY DEVELOPMENT IN CALIFORNIA 11 3. PROBLEM IDENTIFICATION 3.1. Problem Overview This chapter details the historical injustices of redlining and highway development in California by exploring their connections to ongoing issues of pollution concentration and racial segregation that disproportionately impact Californians of color. Throughout the chapter, the cities of Stockton and Los Angeles serve as case studies to illustrate localized examples of this history and its consequences.
3.2. Redlining What was Redlining?
While the term “redlining” is now broadly used to refer to all systems of discrimination targeting neighborhoods of color, it is crucial to pinpoint the origins of the term in federal housing policy to illuminate its intertwined history with highway development in California. Redlining was first established by the FHA in response to the HOLC. The HOLC aimed to provide security to a fragile national mortgage market amid the Great Depression by offering Americans favorable, low-interest loans to purchase new homes backed by federal government funds. To limit the risk of people defaulting on these loans, the FHA developed criteria to assess relative lending risk on a neighborhood-by-neighborhood basis. Throughout the late 1930s and early 1940s, HOLC task forces visited American cities with populations of more than 40,000 people to create maps that applied this criterion locally. In classifying neighborhood risk on a four-part scale from A (“Best”) to D (“Hazardous”), these A segment of the Great Wall of Los Angeles features a depiction of how highways divided neighborhoods. SPARC Archives / SPARCinLA.org ADDRESSING THE DISCRIMINATORY IMPACTS OF REDLINING AND HIGHWAY DEVELOPMENT IN CALIFORNIA 12 government agents factored in the race and ethnicity of residents, as mandated by official FHA policy.
14 Areas inhabited by “inharmonious racial or nationality groups”15 were deemed riskier and were more likely to be assigned “D” ratings, meaning they were outlined in red – or “redlined” – on official HOLC maps. Assessors, particularly for the eight redlined cities in California, identified almost every racial and ethnic group except for non-Hispanic white individuals as carrying these inharmonious and subversive influences.
Through this explicitly racist policy, the FHA and collaborating local authorities across the United States artificially lowered property values in predominantly non-white neighborhoods and ensured only 2% of the $120 billion in federal home loans went to non-white Americans.
16 This policy systematically deprived people of color in America of wealth and opportunity while explicitly promoting segregation as a goal of federal housing policy. The FHA official manual in 1936 stated: “Areas surrounding a location are investigated to determine whether incompatible racial and social groups are present, for the purpose of making a prediction regarding the probability of the location being invaded by such groups. If a neighborhood is to retain stability, it is necessary that properties shall continue to be occupied by the same social and racial classes. A change in social or racial occupancy generally contributes to instability and a decline in values.”17 This guidance established that integrated neighborhoods were more likely to be redlined, thus incentivizing public and private collaboration to force residents of color out of integrated neighborhoods and into devalued homes in segregated neighborhoods.
Following pushback from civil rights organizers, the FHA removed explicit references to race and ethnicity in its 1947 and 1958 manuals. However, it substituted those references with thinly veiled terms intended to uphold the same discriminatory policies rather than redressing the racial inequality explicitly sowed in federal housing policy; these changes encouraged government officials to continue perpetuating such disparities in practice.
18 Redlining in California Stockton, Los Angeles, San Diego, Fresno, Sacramento, San Jose, Oakland, and San Francisco were the eight cities in California redlined by the federal government. The official HOLC evaluations of Stockton and Los Angeles serve as prime examples of the explicit racial discrimination factored into exclusionary classifications throughout the state.
14 “Pollution and Prejudice - Redlining and Environmental Justice in California,” Pollution and Prejudice, August 16, 2021.
15 “Federal Housing Authority Underwriting Manual” (1938).
16 “Pollution and Prejudice - Redlining and Environmental Justice in California,” Pollution and Prejudice, August 16, 2021.
17 “Federal Housing Authority Underwriting Manual” (1938).
18 Richard Rothstein, The Color of Law: A Forgotten History of How Our Government Segregated America (First edition; New York: Liveright Publishing Corporation, a division of W.W. Norton & Company, 2017).
ADDRESSING THE DISCRIMINATORY IMPACTS OF REDLINING AND HIGHWAY DEVELOPMENT IN CALIFORNIA 13 Figure 2: Map of modern-day Stockton, CA, overlaid with 1939 HOLC risk classifications.
Figure 3: Map of modern-day Los Angeles, CA, overlaid with 1939 HOLC risk classifications.
Figure 2 depicts the 1939 HOLC areas in Stockton. Government officials noted that Homestead (Area D11) would have received a “C” rating were it not for “subversive racial influences in the area.” For Barrio del Chivo (Area D3), officials logged that “subversive races exist [as] there is a concentration of Mexican residents in the area as well as many negroes and orientals.” As a result, the area was deemed “low red” and “an area to develop into a business or industrial section.” In Japantown and Chinatown (Area D7), officials noted the “well-improved streets” and “proximity to recreational centers,” but overrode these positives due to their perception that “the area is infested with subversive racial influences [of] Chinese merchants and Japanese farmers and laborers” and duly redlined it.
19 In Los Angeles, risk assessors redlined large swaths of East and Central Los Angeles due to its concentration of non-white residents. In Elysian Park (Area D35 in Figure 3 and present-day home of Dodger Stadium), officials declared that “racial hazards are so great that higher than ‘medial red’ could not be assigned.” In Boyle Heights (Area D53), assessors praised the community’s “conveniently available schools, churches, trading centers, recreational areas, and transportation.” Y et, the same assessors could not overlook that the area was “literally honeycombed with diverse and subversive racial elements” such as “American Mexicans, Japanese, and … negroes” and redlined it. In Watts (Area D61), the district was noted as “the largest concentration of negroes in Los Angeles County” and flagged with a “low red” grade.20 19 “Home Owners’ Loan Corporation Act Assessment - Stockton, California (1939),” Mapping Inequality: Redlining in New Deal America, 2023.
20 Ibid.
ADDRESSING THE DISCRIMINATORY IMPACTS OF REDLINING AND HIGHWAY DEVELOPMENT IN CALIFORNIA 14 These examples of the language that dictated official government housing policy starkly illustrate that the race and ethnicity of Californians were not just considered in decisions about which neighborhoods were riskier to invest in, but were a leading factor in government decisions to systematically deny benefits to Californians of color, devalue their property, and segregate them into dispossessed communities.
Lasting Impacts of Redlining Examining the racial wealth gap in relation to formerly redlined areas reveals some of the fundamental impacts of redlining. Some of the more elemental tolls of redlining can be charted by looking at the racial wealth gap in relation to formerly redlined areas. Data from the University of Richmond indicates that, across the 200 U.S. cities that received HOLC assessments in the 1930s and 1940s, median household income in formerly redlined areas was only 70% of that in non-redlined areas of the same cities.21 In California, nearly half of the population in redlined neighborhoods (46%) currently live below the poverty line— a rate 40% higher than the average poverty rate for residents of non-redlined parts of the state’s eight HOLC-assessed cities.22 Furthermore, it is evident that more than 80 years after the advent of redlining policies and more than 50 years after their formal abolition, these wide disparities in outcomes are disproportionately experienced by populations of color who still predominantly reside there.23 Just 31% of the 11 million residents of these areas across the country identify as non-Hispanic white, compared to 58% of the city-dwelling U.S. population overall.24 25 In California, non-Hispanic white individuals make up just 18% of residents in formerly redlined areas, compared to 67% of residents in formerly A-rated areas.26 A city-level examination of Los Angeles exemplifies the reality that people of color, both nationwide and in California, are still concentrated in formerly redlined areas, while wealth is concentrated elsewhere.
Figures 4 and 5 depict the striking disparity in wealth distribution within formerly redlined areas in Los Angeles. Census tracts in the city’s lowest quartile for median household earnings strongly correlate with formerly D-rated areas, while wealth is concentrated in 21 Andre M. Perry and David Harshbarger, “America’s formerly redlined neighborhoods have changed, and so must solutions to rectify them,” October 14, 2019, The Brookings Institution 22 See Appendix 3.1.
23 Paul Ong, Ananya Roy Yoon, and Chhandara Pech, “Redlining and Beyond: Development Within and Outside HOLC Spaces in Los Angeles County” (Los Angeles: UCLA Center for Neighborhood Knowledge, 2023), accessed April 8, 2024, Link. 24 Andre M. Perry and David Harshbarger, “America’s formerly redlined neighborhoods have changed, and so must solutions to rectify them,” October 14, 2019, The Brookings Institution 25 United States Department of Agriculture Economic Research Service, “Percent of Urban and Rural Populations by Race/Ethnicity,” October 13, 2020 26 Anthony Nardone, et. al., “Associations between historical residential redlining and current age-adjusted rates of emergency department visits due to asthma across eight cities in California: an ecological study,” The Lancet Planetary Health Journal, Vol. 4, Issue 1, January 2020, Pages e24-e31 ADDRESSING THE DISCRIMINATORY IMPACTS OF REDLINING AND HIGHWAY DEVELOPMENT IN CALIFORNIA 15 Figure 4: Map of Median Household Income by Census Tract in Los Angeles, CA (2021), overlaid with 1939 HOLC risk classifications (redlined areas). Figure 5: Map of Median Household Income by Census Tract in Los Angeles, CA (2021), overlaid with 1939 HOLC risk classifications (A and B-rated areas).
Figure 6: Maps of Segregation Levels by Census Tract in Los Angeles, CA (2021), overlaid with 1939 HOLC risk classifications (redlined areas). Figure 7: Maps of Segregation Levels by Census Tract in Los Angeles, CA (2021), overlaid with 1939 HOLC risk classifications (A-rated areas). ADDRESSING THE DISCRIMINATORY IMPACTS OF REDLINING AND HIGHWAY DEVELOPMENT IN CALIFORNIA 16 formerly A-or-B-rated areas. Similarly, Figures 6 and 7 show a strong connection between redlining and present-day racial segregation.27 Formerly A-rated areas overwhelmingly overlap with census tracts where white individuals are segregated, whereas formerly D-rated areas overlap with areas where non-white populations are segregated. Another notable trend in Los Angeles is the demographic shift occurring in many of these redlined areas – from predominantly Black to predominantly Latine and vice versa – since the 1940s. However, this shift has not altered the reality that dispossessed communities of color are still highly segregated in these underserved areas.28 While redlining has served as the policy foundation for lasting discrimination and dispossession of people of color in California and nationwide, the state’s history of robust highway development generated further discriminatory impacts for non-white communities.
3.3. Highway Development in Redlined Communities Regional Highway Development in California The establishment of the California State Highway System and The Bureau of Highways in 1895 marked the beginning of the State’s unprecedented commitment to highway infrastructure. The development of highways and roads was slow and underfunded until the Collier Burns Act of 1947 increased gas and diesel taxes to help fund highway construction. Initiated the same year, the California Master Plan designed a comprehensive system of interconnected freeways to meet the State’s evolving transit needs. The Collier Burns Act influenced President Eisenhower’s 1958 Federal-Aid Highway Act, creating a gas tax and trust fund that accelerated highway projects. This shift to state and federal funding of highways transformed California’s highway and freeway development.29 However, this windfall of funding for the expansion of highway development triggered backlash. The following decade was characterized by the freeway revolts, a period of widespread protests across the U.S. to force planners to consider the impacts of development projects on communities. Revolts were predominantly successful in white communities, but bulldozed in communities of color. In the 1960s and 1970s, highways contributed to suburbanization trends, leading to lower land values and population loss in urban neighborhoods.30 From the 1980s onward, the State focused on modifying and expanding existing roads to meet its changing transportation needs and challenges.31 Today, California’s expansive road network continues to be a vital component of its identity and economic vitality. Highways remain integral to land use and transportation 27 Data Source for segregation levels: Stephan Menendian, Samir Gambhir, and Arthur Gailes, “The Roots of Structural Racism Project,” The Othering and Belonging Institute, June 21, 2021.
28 Stephan Menendian, Samir Gambhir, and Arthur Gailes, “The Roots of Structural Racism Project,” The Othering and Belonging Institute, June 21, 2021.
29 California, State Of. “Caltrans History | Caltrans,” n.d. 30 Mohl RA. The Interstates and the Cities: The U.S. Department of Transportation and the Freeway Revolt, 1966–1973. Journal of Policy History. 2008;20(2):193-226.
31 California, State Of. “Caltrans History | Caltrans,” n.d. ADDRESSING THE DISCRIMINATORY IMPACTS OF REDLINING AND HIGHWAY DEVELOPMENT IN CALIFORNIA 17 systems in the movement of people and goods.32 The legacy of these highways extends beyond physical infrastructure: they shaped urban landscapes, influenced demographic patterns, and contributed to conversations about equitable development. As California grapples with evolving transportation challenges and considers sustainable options for the future, the history of its highway development becomes a basis for conversations about the future of highways and addressing past racial injustices.
Targeting Redlined Communities for Highway Development To develop vast highway networks in California, state and local governments needed to identify suitable land for efficient route construction while mitigating negative impacts on land use patterns. With the help of federal directives from the FHA, one of the main strategies they resorted to was constructing highways in redlined communities when traversing residential areas.33 Today, 77% of redlined areas across California have highways bordering or bisecting them, compared to just 49% of all other HOLC-assessed areas.34 At the census tract level, tracts in formerly redlined areas across California have roughly twice as many highways within 0.75 miles of them as formerly A-rated tracts and are roughly twice as close to the nearest highway on average.35 These statistics are blatantly reinforced by maps of California’s redlined cities that overlay highways with historic HOLC risk classifications. Figures 8 and 9 depict the phenomenon in Stockton and Los Angeles, while Figures 10 and 11 provide further examples of Fresno and San Francisco to reinforce the pervasiveness of this practice in redlined areas across the state.
These policy decisions to route highways through redlined areas were motivated by two central factors rooted in the FHA’s Underwriting Manual. The first was the desire of local authorities to confine land use or development associated with pollution or poor air quality to areas of cities and regions that had already been redlined. The criteria used in the HOLC assessment to categorize neighborhood risk incentivized compounding risk in redlined areas. In addition to considering the race and ethnicity of an area’s residents, the HOLC assessment also decreed that the “presence of smoke, odors, or fog” in the area should also inform a neighborhood’s risk categorization.36 With predominantly non-white neighborhoods in cities already redlined as “hazardous” on account of their racial and ethnic makeup, local authorities were motivated to locate any land uses associated with negative environmental outputs in or near redlined communities for years to come in order to avoid the devaluation of A and B-rated areas. Due to the noise, smells, and smog associated with them, highways fell into this category of land uses and were consequently built through redlined areas.37 32 California Department of Transportation, 2022 Caltrans Facts, accessed April 8, 2024, Link 33 “Pollution and Prejudice - Redlining and Environmental Justice in California,” Pollution and Prejudice, August 16, 2021.
34 See Appendix 3.2.
35 See Appendix 3.3 and 3.4 36 “Federal Housing Authority Underwriting Manual” (1938).
37 “Pollution and Prejudice - Redlining and Environmental Justice in California,” Pollution and Prejudice, August 16, 2021.
ADDRESSING THE DISCRIMINATORY IMPACTS OF REDLINING AND HIGHWAY DEVELOPMENT IN CALIFORNIA 18 Figure 8: Map of modern-day Stockton, CA (highways in brown), overlaid with 1939 HOLC risk classifications.
Figure 9: Map of modern-day Los Angeles, CA (highways in brown), overlaid with 1939 HOLC risk classifications.
Figure 10: Map of modern-day Fresno, CA (highways in brown), overlaid with 1939 HOLC risk classifications.
Figure 11: Map of modern-day San Francisco, CA (highways in brown), overlaid with 1939 HOLC risk classifications.
ADDRESSING THE DISCRIMINATORY IMPACTS OF REDLINING AND HIGHWAY DEVELOPMENT IN CALIFORNIA 19 The second factor was the explicit recommendation of the FHA to use “high-speed traffic arteries” as a land use tool to “prevent the expansion of inharmonious uses to the other side of the street.”38 Having already established that they viewed the coexistence of different racial groups in the same residential area as an “inharmonious use” of land, this section of FHA guidance created a precedent for local authorities to use highways as a barrier to enforce segregation between white communities and redlined communities of color. Since the interstate highway system began to take shape after the first Federal Highway Act of 1956 and after the Supreme Court struck down traditional tools of segregation, highways held the capacity to continue the de facto enforcement of racial zoning lines to keep racial groups segregated. Throughout California, officials leveraged this capacity and routed highways to curb the physical and socioeconomic mobility of people of color and preserve wealth for white spaces.
Discrimination and Displacement in Los Angeles & Stockton Case studies of Los Angeles and Stockton illustrate the profound consequences and destruction of communities of color wrought by the development of highways in redlined areas. Throughout Los Angeles, highways “worked to hide the brutal violence of racial segregation and also helped to maintain it.”39 In 1968, the Century Freeway displaced 3,550 families, 117 businesses, and numerous parks, schools, and churches, primarily in Black Watts and Willowbrook.40 CalTrans encouraged those who could do so to leave, forced residents out by eminent domain, or trapped remaining residents in bisected communities that were maintained through racial covenants.41 In Boyle Heights, a thriving Mexican-American and Mexican immigrant community, residents were displaced by six freeways despite their activism. Some residents with the ability to leave left the area, while the acceptance of relocation assistance by poorer residents was misinterpreted by planners as an endorsement for the highways.42 Sugar Hill, formerly known as West Adams Heights, was once the wealthiest Black neighborhood in the city of Los Angeles before CalTrans utilized eminent domain to construct the I-10 freeway and reinforce segregation.43 Despite protests from Sugar Hill residents, the choice between the freeway cutting through the University of Southern California’s (USC) fraternity and sorority row or destroying Sugar Hill led to the destruction of Sugar Hill, while USC’s Greek row remained intact as shown in Figure 12.
38 “Federal Housing Authority Underwriting Manual” (1938).
39 Archer, D. N. (2020). “White Men’s Roads Through Black Men’s Homes”: Advancing Racial Equity Through Highway Reconstruction. Vanderbilt Law Review, 73(5), 1259–1330.” 40 Mohl, RA supra note 27.
41 Lutenski, E. (2019). Dickens Disappeared: Black Los Angeles and the Borderlands of Racial Memory. American Studies (Lawrence), 58(3), 15–35. Link 42 Other non-white neighborhoods impacted by freeways in Los Angeles include Lincoln Heights, Watts, and Wilmington. See Jaffe, Eric. “The Forgotten History of l.a.’s Failed Freeway Revolt.” Bloomberg.
Com, Bloomberg, 23 July 2014. 43 This came after an unsuccessful attempt to evict Black families through a court case focused on enforcing racial covenants, see Susaneck, Adam Paul. “Sugar Hill.” SEGREGATION BY DESIGN, Link. Accessed 2 Jan. 2024. ADDRESSING THE DISCRIMINATORY IMPACTS OF REDLINING AND HIGHWAY DEVELOPMENT IN CALIFORNIA 20 SUGAR HILL Figure 12: Map of Sugar Hill, South Los Angeles, illustrated in red before and after highway construction.
Figure 13: Map of East Los Angeles neighborhoods affected by highway development. Figure 14: Map of Highway 4 cutting through Stockton’s Asian enclaves.44 44 Ong, P. M., Pech, C., Do, C.-H., Yoon, A., & Wasserman, J. L. (2023). Stockton’s Crosstown Freeway, Urban Renewal, and Asian Americans: Systemic Causes and Impacts.
ADDRESSING THE DISCRIMINATORY IMPACTS OF REDLINING AND HIGHWAY DEVELOPMENT IN CALIFORNIA 21 Los Angeles’s 1958 Master Plan is marked by the absence of freeways slicing through predominantly white neighborhoods.45 Freeway revolts in Beverly Hills successfully resisted the proposed Beverly Hills Freeway, while similar efforts in Boyle Heights failed to prevent the construction of six freeways.46 Compared to the 2.4% of the land used for freeways in Los Angeles, East Los Angeles saw 9.3% and Boyle Heights 12% of its land taken by freeways as seen in Figure 13.47 While the freeway revolts eventually spurred federal policy changes in 1965 that prioritized local input, these changes arrived too late for most or were implemented despite opposition from communities of color.48 In Stockton, the scenario was no different— low-income communities were subject to extensive demolition. Stockton’s 1958 Master Plan for Highway 4 routed through “blighted” areas of racial enclaves for “slum clearance,” destroying neighborhoods like Little Manila and Barrio del Chivo,49 and impacting Chinatown and Japantown, as seen in Figure 14. Little Manila was once the largest Filipino community outside the Philippines.50 Barrio del Chivo was a mainly Mexican and Black neighborhood and government officials determined that “the best that can be hoped for in this area, is that it [is] develop[ed] into a business or industrial section” and graded it as a low red area optimal for highway development.51 Neighborhoods utilized a range of responses to evade negative community impacts but were dismissed or ignored by city officials. Residents from Barrio del Chivo tried to secede from the City of Stockton to avoid demolition. The appeal failed, leading to its residents relocating and the community being destroyed. Organized resistance in Little Manila challenged displacement and redevelopment but faced two unsuccessful court challenges, one of which was predetermined by the city government. Ultimately, the Crosstown Freeway’s construction “displaced more than 1,000 people and destroyed nearly 800 housing units,” impacting mainly people of color.52 45 “Hidden Long-Term Effects on the Latino Community of East Los Angeles.” Foundations of Law and Society (blog). Accessed April 8, 2024. Link.
46 Other wealthier areas like Westwood and West Hollywood that also resisted the thwarted Beverly Hills Freeway reaped the benefits. Gamboa et al., 2021 47 Levine, Jonathan. “Justice and the Interstates: The Racist Truth about Urban Highways, Edited by Ryan Reft, Amanda K. Phillips de Lucas, and Rebecca C. Retzlaff: Washington, DC, Island Press, 2023.” Journal of Urban Affairs, (2023), 1–2. doi:10.1080/07352166.2023.2217045.
48 Communities of color in LA were also further impacted by projects outside of Los Angeles, as the 69 highway displaced over 4,000 Black and Mexican-American residents from Pasadena into inner-city Los Angeles communities. Mohl, RA supra note 27.
49 Ong, P. M., Pech, C., Do, C.-H., Yoon, A., & Wasserman, J. L. (2023). Stockton’s Crosstown Freeway, Urban Renewal, and Asian Americans: Systemic Causes and Impacts.
50 Planners used highway development in tandem with the urban renewal West End Redevelopment Project to achieve this. Through parcel land grabs, eventually Little Manila was completely destroyed. Mabalon pg. 271 51 This designation and later “slum” identification would lead to a comprehensive land grab and destruction of the neighborhood. Madrigal-Lauchland, Vanessa. “The Shifting Meanings of Stockton’s Barrio del Chivo.” California History 96 (2019): 97-100. Link.
52 Ong, P. M., Pech, C., Do, C.-H., Yoon, A., & Wasserman, J. L. (2023). Stockton’s Crosstown Freeway, Urban Renewal, and Asian Americans: Systemic Causes and Impacts.; Ibid.
ADDRESSING THE DISCRIMINATORY IMPACTS OF REDLINING AND HIGHWAY DEVELOPMENT IN CALIFORNIA 22 In both cities, communities experienced displacement at the hands of local governments through eminent domain, either through comprehensive land grabs or through parcel land grabs. Parcel land grabs caused lasting damage as homes, community centers, and businesses were destroyed, notably in Chinatown in Stockton and Sugar Hill in Los Angeles. In Los Angeles, city officials justified building through redlined neighborhoods by citing low construction costs and the need to preserve industrial sites. City government officials in Stockton openly used highway development to “wipe clean” areas with Asian racial enclaves.53 Through eminent domain, city governments provided relocation incentives, offering alternative housing options in temporary camps along with compensation for lost housing. However, these alternatives fell short, providing inferior living conditions to residents’ original homes and inadequate compensation for new housing costs.54 Following pushback from civil rights organizers, the FHA removed explicit references to race and ethnicity in its 1947 and 1958 manuals. However, it substituted those references with thinly veiled terms intended to uphold the same discriminatory policies.
3.4. Impacts of Highway Development and Redlining in California The impacts of highway development history reach far beyond physical infrastructure, encompassing vast environmental, socioeconomic, and sociocultural implications. Due to this project’s time constraints and OEHHA’s priorities, our team focused its analysis on two main impacts: racial segregation and particulate matter (PM) pollution above 2.5.
Racial Segregation Racial segregation is the degree to which individuals are concentrated into distinct areas, institutions, or groups based on their different racial identities. In the context of historical redlining, however, racial segregation focuses on housing segregation and the degree to which individuals outside the non-Hispanic white racial group are concentrated in distinct residential areas. While acknowledging the existence and importance of segregation between different racial groups of color, prioritizing the binary of white and non-white mirrors the designations of “inharmonious racial or ethnic groups” used by the FHA. Understanding the importance of studying racial segregation levels requires highlighting the inevitable negative impacts of segregation on people of color and society as a whole. From an institutional perspective, racial segregation allows for the continued implementation of policies that can be race-neutral in language but racially discriminatory in practice. When people of color are concentrated in specific residential areas with particular levels of wealth and political capital, policymakers can enact measures that 53 This divide is further demonstrated by officials and white neighborhoods labeling Asian neighborhoods as Skid Row while residents referred to the area as the West End. Ong et al. supra note 46.
54 In rural Stockton, homes were on average estimated to be around $8,500 in Little Manila and Chinatown, but residents received only $3,000 for their homes. In some urban areas like Sugar Hill with higher land costs residents were paid at around $12,000 each, still $5,000 less than the actual values of their homes. Ibid. ADDRESSING THE DISCRIMINATORY IMPACTS OF REDLINING AND HIGHWAY DEVELOPMENT IN CALIFORNIA 23 disadvantage such areas and, consequently, people of color, without explicitly considering racial factors. Today, such policies, which produce racist outcomes in the absence of overt racist intent, are more politically and legally viable than openly racist policies like those codified by the FHA in 1936, and they are perpetuated by persisting racial segregation. Such policies contribute to negative outcomes in health, wealth, and freedom for segregated communities of color in California and across the country. There are many stark examples of this process in practice today. Segregated non-white communities have less access to green space and public parks compared to white communities, which causes residents to face increased levels of pollution, extreme heat, and worsened physical and mental health.55 Segregated non-white communities are targeted with a higher level of police presence and arrest rates than white communities, which leads to the drastic overrepresentation of people of color in the California and U.S. carceral system.56 Schools in segregated non-white areas receive significantly less funding than predominantly white schools, which contributes to worse educational outcomes for youth of color and limits their future career opportunities.57 A more integrated society would render such racially discriminatory policies tactically and politically infeasible.
The demographic statistics of current residents in formerly-redlined areas highlight the effectiveness of federal housing policy from the 1930s-1960s in enforcing and perpetuating racial segregation in residential areas, which continues to strongly influence the concentration of non-white racial groups in these areas today. According to a 2020 segregation index built by researchers at UC Berkeley, 73% of census tracts in formerly redlined areas in California exhibit high levels of segregation among people of color, while 93% of census tracts in formerly A-rated areas are characterized by heavy or moderate segregation among white residents.58 However, it is crucial to further analyze and isolate the relationships between the histories of redlining and highway development and their impact on contemporary outcomes of racial segregation. This examination will help to understand whether the federal government’s strategies to enforce segregation in the mid-20th century continue to influence current discriminatory practices. Models 1 and 2 provide data-based evidence of the magnitude and significance of these relationships and the many nuanced factors still driving racial segregation in California today. 55 Lathan, Nadia. “50 years after being outlawed, redlining still drives neighborhood health inequities.” University of California, Berkeley Institute of Public Health. Accessed September 28, 2023.; Hsu, Angel, Glenn Sheriff, Tirthankar Chakraborty, and Diego Manya. 2021. “Disproportionate Exposure to Urban Heat Island Intensity across Major US Cities.” Nature Communications 12 (1): 2721–11.
56 Public Policy Institute of California. “Racial Disparities in California Arrests.” PPIC. October 2019.
57 Fuller, Bruce, Yoonjeon Kim, Claudia Galindo, Shruti Bathia, Margaret Bridges, Greg J. Duncan, and Isabel García Valdivia. 2019. “Worsening School Segregation for Latino Children?” Educational Researcher 48 (7): 407–20.
58 See Appendix 3.5 ADDRESSING THE DISCRIMINATORY IMPACTS OF REDLINING AND HIGHWAY DEVELOPMENT IN CALIFORNIA 24 Model 1 & 2 Results: There are many significant correlations between redlining, highway development, and segregation. Because the interaction term in Model 2 is insignificant, the relationships predicted by Model 1 are the most important to analyze. The model associates census tracts in formerly redlined areas with a non-white population 12.1% greater than their city’s average. Conversely, census tracts in formerly A-rated are associated with a 28% smaller non-white population. In addition to this relationship with HOLC risk classifications, a census tract scoring 10/10 on the highway proximity index is correlated with a 5.73% greater non-white population than a census tract scoring 5/10. In comparing these relationships in urban and rural areas, the relationship between highway proximity and segregation is significantly stronger in rural areas than in urban areas. A five-point increase in the highway proximity index is associated with an 11% increase in non-white population in rural census tracts, compared to just a 3.8% increase in urban census tracts. This disparity is inverted when comparing Stockton and Los Angeles, however. The same increase in highway proximity in Los Angeles is associated with a 7.35% increase in the non-white population, while it is linked to a 3.7% increase in Stockton. This is likely explained by Stockton’s unique history of displacement by highway construction as well as the rebuilding of more expensive housing near highways in its urban renewal projects that relocated some non-white residents. These differences in urban versus rural areas and Los Angeles versus Stockton are both statistically significant at a 95% confidence level.59 In considering the implications of these model results, it is critical to note that, at 99.9% confidence, the margin by which a select census tract in California’s eight redlined cities could deviate from its city’s overall percentage of non-white population due to random chance is at most +/- 2.41%. Given that margin, these relationships all indicate links to high levels of segregation throughout California.60 59 See Appendix 3.7.
60 See Appendix 3.6.
ADDRESSING THE DISCRIMINATORY IMPACTS OF REDLINING AND HIGHWAY DEVELOPMENT IN CALIFORNIA 25 Particulate Matter 2.5 Pollution PM2.5 levels refer to the concentration in the air of particulate matter between 0.1 and 2.5 micrometers. Much of the PM2.5 in the air is naturally produced from environmental sources such as the interaction of wind with dust, sea salt, and soil particles. But, human activities like traffic exhaust, fossil fuel burning, and construction have been adding unhealthy amounts of PM2.5 to the air for decades.61 Traffic exhaust and traffic density, in particular, are leading contributors to elevated PM2.5 levels in cities across California, with specific nexuses of PM2.5 near busy roads and highways.62 The latest guidelines from the World Health Organization identify an annual average of 5 µg/m3 as a healthy standard of average human exposure to PM2.5,63 64 but the average concentration of PM2.5 across California in the latest data collected by the state was as high as 10.18 µg/m3. In Los Angeles, in particular, the median census tract has an annual average concentration of 11.88 µg/m3, which ranks in the 99th percentile for cities nationwide.65 66 Stockton, which has an annual average concentration of 11.11 µg/m3, ranks in the 91st percentile nationally.67 These staggering totals are cause for concern due to the serious health risks associated with high levels of PM2.5 exposure. These particles can trigger reactionary inflammation in the lungs when inhaled, which is correlated with severe health issues such as low birth rate, heart disease, stroke, lung cancer, and premature death generally.68 Consistently high PM2.5 exposure is also heavily correlated with higher prevalence and more severe episodes of asthma and can target the developing lungs in unborn fetuses and young children to devastating effect.69 From a public policy perspective, an im portant finding about the health risks associated with PM2.5 is that the relationship between PM2.5 levels and life expectancy is not linear. The closer initial PM2.5 levels are to zero, the effect of a change in PM2.5 levels on health outcomes will be greater,70 suggesting that interventions in areas with heavier concentrations would need to deliver major reductions in PM2.5 to improve community health.
61 Environmental Protection Agency, “Particulate Matter Pollution,” EPA Website, Accessed January 21, 2024.
62 Chen Y, Gu P, Schulte N, Zhou X, Mara S, Croes BE, Herner JD, Vijayan A. A new mobile monitoring approach to characterize community-scale air pollution patterns and identify local high pollution zones, Atmospheric Environment, Volume 272, 2022. Link.
63 World Health Organization, “2021 Global Air Quality Guidelines,” Accessed January 24, 2024.
64 California Air Resources Board, “Inhalable Particulate Matter and Health (PM2.5 and PM10),” CARB Website, Accessed February 2, 2024 65 California Office of Environmental Health and Hazard Assessment, CalEnviroScreen 4.0 Online Database, accessed Winter, 2024.
66 Environmental Defense Fund, U.S. Climate Vulnerability Index, accessed February 2, 2024.
67 Ibid.
68 Correia, A. W., Pope, C. A. III, Dockery, D. W., Wang, Y., Ezzati, M., & Dominici, F. “Effect of air pollution control on life expectancy in the United States: an analysis of 545 U.S. counties for the period from 2000 to 2007.” Epidemiology (Cambridge, Mass.) 24, no. 1 (2013): 23–31. Link.
69 Xing, Y. F., Xu, Y. H., Shi, M. H., & Lian, Y. X. “The impact of PM2.5 on the human respiratory system.” Journal of Thoracic Disease 8, no. 1 (2016): E69–E74. Link.
70 Colin D Mathers, Dejan Loncar, “Projections of global mortality and burden of disease from 2002 to 2030,” The Lancet 369, no. 9573 (2007): 243-250, accessed February 6, 2024 ADDRESSING THE DISCRIMINATORY IMPACTS OF REDLINING AND HIGHWAY DEVELOPMENT IN CALIFORNIA 26 The past tenants of redlining that incentivized authorities and city planners to concentrate industrial and highway development in redlined areas have had the lasting legacy of disproportionately burdening those same predominantly non-white areas with disproportionate PM2.5 pollution today. Nationally, 55% of formerly redlined neighborhoods face PM2.5 levels significantly above the mean in their cities, while 68% of A-rated areas experience levels significantly below their city-level mean.71 In California, the mean concentration of PM2.5 in D-rated areas is greater than the citywide mean in all eight formerly redlined cities.72 Figures 15 and 16 spatially illustrate this correlation between past redlining and current PM2.5 pollution in Los Angeles and Stockton, and further reinforce the connection between PM2.5, the proximity of highways, and former HOLC ratings. This disparity in PM2.5 pollution burden in the state is linked to disparities in the health outcomes connected to PM2.5 exposure. For example, residents of formerly redlined areas in California experience a rate of asthma-induced emergency room visits connected to elevated PM2.5 exposure that is 240% higher than residents of formerly A-rated areas.73 71 Lane, HM., Morello-Frosch, R., Marshall, JD., Apte JS. Historical Redlining is Associated with Present-Day Air Pollution Disparities in U.S. Cities. Environ. Sci. Technol. Lett. 2022, 9, 4, 345–350. Link.
72 See Appendix 3.8.
73 Nardone A, Casey JA, Morello-Frosch R, Mujahid M, Balmes JR, Thakur N. Associations between historical residential redlining and current age-adjusted rates of emergency department visits due to asthma across eight cities in California: an ecological study. Lancet Planet Health. 2020 Jan;4(1):e24-e31.
Figure 15: Map of PM2.5 pollution concentration levels in modern-day Los Angeles, CA, overlaid with 1939 HOLC risk classifications for A and D-rated areas and highway routes.
Figure 16: Map of PM2.5 pollution concentration levels in Stockton, CA (2021) overlaid with 1939 HOLC risk classifications for A and D-rated areas and highway routes.
ADDRESSING THE DISCRIMINATORY IMPACTS OF REDLINING AND HIGHWAY DEVELOPMENT IN CALIFORNIA 27 Existing research has established correlations between PM2.5 pollution burdens and redlined areas and between PM2.5 pollution burdens and highway proximity. However, it is critical to parse the specific connections between the intertwined history of redlining and highway development and present-day PM2.5 levels. Models 3 and 4 help isolate the significant effects of highway development and redlining on PM2.5 concentration, as well as study the interaction between all three variables.
Model 3 & 4 Results: The results of Model 3 predict significant baseline relationships between both redlining and PM2.5 levels and highway proximity and PM2.5 levels. The previous redlining of a census tract is associated with a .175 µg/m3 increase in annual average PM2.5 today, while a tract at the high end of the highway proximity index is correlated with a .337 µg/m3 increase compared to an identical tract at the low end. Additionally, urban areas and Los Angeles have a statistically significant increase in PM2.5 levels associated with highway proximity when compared to rural areas and Stockton, respectively.74 However, the interaction between redlining, highway proximity, and PM2.5 concentration in Model 4 illuminates the impact on communities affected by both these variables, as is the case for many California communities today. The interaction suggests that increases in PM2.5 exposure associated with increased highway proximity are much stronger in formerly redlined communities than in non-redlined areas (.505 µg/m3 compared to .337 µg/m3 predicted by Model 3). However, Model 4 also predicts a weaker direct relationship between redlining and PM2.5 when an interaction term is included. This suggests that much of the positive correlation expressed in Model 3 can be attributed to this interaction with highway proximity, rather than past redlining in isolation. This association with greater PM2.5 levels for the interaction between highway proximity and former redlining is likely produced by other factors that are positively correlated with all three variables. For example, industrial development that produces elevated PM2.5 74 See Appendix 3.7.
ADDRESSING THE DISCRIMINATORY IMPACTS OF REDLINING AND HIGHWAY DEVELOPMENT IN CALIFORNIA 28 could have also been disproportionately concentrated in redlined areas and areas near highways. Additionally, the lack of open green spaces in redlined communities may be particularly prevalent in highway-adjacent areas, exacerbating residents’ exposure to PM2.5. Further research that incorporates data on these new variables would be necessary to provide more concrete support for these claims.
It is important to frame these predicted relationships within the context of established correlations between PM2.5 levels and different health impacts. The strongest relationship the models predict is that past redlining and maximum highway proximity in a California community is linked to an increased PM2.5 burden of .505 µg/m3. Existing research links that difference with just a 0.1% increase in the prevalence of respiratory diseases, a 0.25% increase in overall mortality by 0.25%, and an eight-day decrease in life expectancy, amongst other health impacts.75 There are 1.28 million people in California who live in formerly-redlined census tracts that score above halfway on the highway proximity scale, meaning this intersected history can be linked to about 1,280 respiratory diseases and 3,200 causes of death annually in this disproportionately non-white population statewide.
Further important context for examining this problem of PM2.5 burdens in a policy context is that PM2.5 particles are widely dispersed in the environment. Although sources of pollution might predominate near certain types of census tracts, the airborne particles are not contained within a census tract-sized area. In all eight of California’s formerly redlined cities, the average difference between a census tract in the 25th percentile of PM2.5 concentration and a census tract in the 75th percentile was just 0.48 µg/m3. This relative uniformity within cities at the census tract level stands in contrast to other harmful pollutants like diesel particulate matter, which remains more concentrated in the areas where it is produced.76 Despite the variable’s significant negative correlation with highway proximity, the positive relationship predicted by both models for formerly A-rated tracts emphasizes the dispersed nature of PM2.5.
Overall, evidence shows California’s intersecting history of redlining and highway development has placed a disproportionate burden of PM2.5 pollution on communities of color that still overwhelmingly inhabit formerly redlined areas and areas near highways. This burden is especially amplified in communities that fit both of these descriptions. While the magnitude of increased PM2.5 pollution in such communities is relatively small compared to other communities within the same city, it still produces measurable negative impacts on health outcomes that target non-white Californians.
75 Xing, Y. F., Xu, Y. H., Shi, M. H., & Lian, Y. X., supra note 68. Correia, A. W., Pope, C. A. III, Dockery, D. W., Wang, Y., Ezzati, M., & Dominici, F. supra note 79.
76 California Air Resources Board, “Overview of Diesel Exhaust and Health,” California Air Resources Board website, accessed March 6, 2024.
ADDRESSING THE DISCRIMINATORY IMPACTS OF REDLINING AND HIGHWAY DEVELOPMENT IN CALIFORNIA 29 4 POLICY OPTIONS CA Department of Water Resources ADDRESSING THE DISCRIMINATORY IMPACTS OF REDLINING AND HIGHWAY DEVELOPMENT IN CALIFORNIA 30 4. POLICY OPTIONS 4.1. Chapter Overview This chapter reviews the current policy context relevant to addressing the historical injustices of redlining and highway development in California. This review identifies four policy approaches for achieving this goal and advancing racial and environmental justice and concludes by introducing a set of potential state-level policies for further evaluation.
4.2. Policy Context Institutions and actors across all levels of government are responsible for the persistence of racist legacies of redlining and highway infrastructure in California. As part of its mission, OEHHA has a vested interest in exploring options for State agencies to address the impacts of these legacies. In addition to OEHHA, other relevant State agencies mentioned include the California State Transportation Agency (CalSTA), which houses CalTrans, tasked with managing all highways and interstate highways within California; CARB, responsible for overseeing air pollution for the State; and the California Transportation Commission (CTC), entrusted with approving state transportation funding.77 77 California Department of Transportation, “About Caltrans,” accessed April 5, 2024, Link.; California Air Resources Board, “About ARB,” accessed April 5, 2024, Link.; California Transportation Commission, “Home,” accessed April 6, 2024, Link.; California State Transportation Agency, accessed April 8, 2024, Link. Traveljunction / Flickr ADDRESSING THE DISCRIMINATORY IMPACTS OF REDLINING AND HIGHWAY DEVELOPMENT IN CALIFORNIA 31 Key Challenges While actors in California have taken strides to address the adverse effects of highway development and redlining while promoting equity and environmental justice, there are many remaining obstacles. The three most prominent challenges include: Dependence on Highway-Based Transportation Infrastructure The majority of Californians rely on car-based transportation, given the State’s 248 highways, more than 27 million licensed drivers, and 36 million vehicles.78 Highways are permanent transportation infrastructure and the transition away from highway dependence requires substantial investments of both time and money. However, decision-makers continue to invest in highway-based transportation infrastructure despite opposition from Tribal Nations and other communities.79 Even justice-oriented solutions like freeway caps (building green space on top of freeways) reinforce highway dependence; despite their benefits to neighborhood connection and pollution reduction, they have been weaponized to counter efforts to decommission highways.80 Additionally, current and future freeway cap projects around the country do not adequately prioritize communities of color.81 Despite the magnitude of the challenge, grassroots movements have demonstrated that it is possible to stop freeway expansions and remove stretches of freeways altogether. After years of fierce community opposition, activists successfully blocked a $6-billion plan to widen the 710 Freeway, which would have uprooted hundreds of Black and Latine families.82 In San Francisco, the notorious Embarcadero Freeway was removed in 1991, and removal is currently being considered for the I-980 freeway in Oakland.83 Political Resistance to Investing in Reparations for People of Color The California Reparations Report in 2023 proposed a powerful set of reparative policy measures to seek justice for past harms against Black Californians, including a detailed 78 Statista. “Total Number of U.S. Licensed Drivers by State.” Accessed April 8, 2024, Link.; California Department of Transportation, 2022 Caltrans Facts, accessed April 8, 2024, Link. 79 For an example of California tribes opposing transportation infrastructure, see: ABC7 News, “Native American Tribes Protest Willits Bypass,” accessed April 5, 2024, Link.
80 Houston, Douglas, and Michelle E. Zuñiga. “Put a park on it: How freeway caps are reconnecting and greening divided cities.” Cities 85 (2019): 98-109. Accessed January 2022. Link.
81 People of color are only 1.1 more likely to be within 0.5 miles to a freeway cap than people of color throughout the entire region. Ibid.
82 Rong-Gong Lin II (Uranga), “710 Freeway expansion dropped after decades of planning, marking a milestone for L.A.,” Los Angeles Times, March 15, 2022, Link.; We use the term “Latine” as a gender-neutral variation of the words Latino/a/@ and Hispanic.
83 Julian Mark, “‘Sky’s the limit’: Caltrans is getting serious about replacing I-980,” The Oaklandside, October 30, 2023, Link.; KRON4 News, “Why Was San Francisco’s Waterfront a Freeway?” accessed April 5, 2024, Link.; California Department of Transportation (Caltrans), “Vision 980,” accessed April 5, 2024, Link.; There have also been moves to block freeway expansions in areas with high levels of pollution and poverty altogether, see Streetsblog California, “No Freeway Expansion Bill Dies in Senate Committee,” accessed April 5, 2024, Link.; California Legislature, “Analysis of Assembly Bill 1778 (2021-2022),” accessed April 5, 2024, Link.
ADDRESSING THE DISCRIMINATORY IMPACTS OF REDLINING AND HIGHWAY DEVELOPMENT IN CALIFORNIA 32 plan to address redlining and highway development.84 However, discussions leading up to and following the publication of the report revealed fear, prejudice, and misunderstanding around the feasibility and necessity of reparations in California.85 Furthermore, Title VI of the Civil Rights Act continues to be used as a legal tool to challenge policies or programs aimed at supporting communities of color.86 This challenge with California’s multi-billion dollar budget shortfall for 2022-2025, which is likely to impact decision-makers’ ability and willingness to invest in new programs and fully fund current programs.87 Lack of Political and Economic Influence of Impacted Communities Throughout California’s history, many policies, including redlining, have limited the political and economic influence of non-white populations. Beyond being segregated into areas with the least amount of resources and the highest pollution levels, legal restrictions have prevented minority groups from voting or holding office, while labor exploitation extracted wealth from these communities into white-dominated areas.88 The state legislature has failed to ensure that infrastructure investments adequately target these most vulnerable communities. For example, Assembly Bill 2419 (2022) failed to advance beyond committee. The bill mandated that “a minimum of 40 percent of federal infrastructure funds coming to California from the Bipartisan Infrastructure Law prioritize communities of color that have been overlooked or harmed by past infrastructure choices.”89 Federal legislators have also failed to funnel investment into impacted communities. For instance, the U.S. Investing in Opportunity Act of 2017 established tax incentives for investors in designated “Opportunity Zones,” but research suggests that these zones do not lift low-income communities out of poverty and instead exacerbate wealth inequality, displacement, and gentrification.90 This has systematically diminished the capacity of non-white Californians to influence decision-making processes or hold direct power over decisions themselves, thus presenting an obstacle to passing policies that specifically benefit these communities at the expense of wealthy ones.
84 California Department of Justice, “AB 3121 Task Force Report,” accessed April 6, 2024, Link.; Similar conversations surround a 2024 reparations bill in California, see New York Times, “Chavez Ravine Reparations for Dodger Stadium,” March 26, 2024, accessed April 6, 2024, Link. 85 “Reparations in California: What They Are and What’s Happening,” CalMatters, accessed April 6, 2024, Link. 86 The US EPA has published guidance on how to navigate Title VI and Environmental Justice programs, see: U.S. Environmental Protection Agency, “Title VI and Environmental Justice,” accessed April 6, 2024, Link. But, legal challenges still arise, see a 2010 example here: Federal Transit Administration, BART Title VI Final Report (Washington, D.C.: U.S. Department of Transportation, April 16, 2010), accessed April 5, 2024, Link.
87 “Q&A: What Does the Budget Shortfall Mean for California?,” California Budget and Policy Center, accessed April 5, 2024, Link.
88 Shaffer, Ralph E. “California Reluctantly Implements the Fifteenth Amendment: White Californians Respond to Black Suffrage, March - June, 1870.” Presentation at Cal Poly Pomona, 2020.
89 California Budget & Policy Center, “Q&A: What Does the Budget Shortfall Mean for California?” (accessed April 5, 2024), Link.
90 Opportunity Zones are tools aimed at assisting low-income, economically distressed communities. Researchers find that this policy does not have its intended effect. Gabriel Zucman and Emmanuel Saez, “The False Promise of Opportunity Zones,” Boston Review, accessed April 5, 2024, Link.; Urban Institute, Early Assessment of Opportunity Zones for Equitable Development Projects (Washington, D.C.: Urban Institute, n.d.), accessed April 5, 2024, Link. ADDRESSING THE DISCRIMINATORY IMPACTS OF REDLINING AND HIGHWAY DEVELOPMENT IN CALIFORNIA 33 Key Opportunities The climate crisis and justice-oriented movements have propelled significant opportunities to implement comprehensive reforms aimed at addressing the impacts of redlining and highway development. Initiatives Beyond California The federal government has enacted several pieces of legislation and programs to offer funds for green investment and combating climate change. These initiatives could be used to address the environmental impacts of highway development on redlined communities by protecting communities from pollution and transitioning away from highways. For instance, the Inflation Reduction Act of 2022 emphasizes clean energy economies and established targeted incentives for investing in low-income communities, particularly those disproportionately affected by pollution. Through incentivizing investment in underserved communities, including those affected by redlining, this bill could improve environmental conditions in redlined communities.91 The White House also announced the Justice40 Initiative in 2021, which directs 40% of benefits from billion-dollar federal environmental investments towards “disadvantaged communities.”92 It further created the Climate and Economic Justice Screening tool to help administer Justice40 funding. However, it’s worth noting that Justice40 funding may not effectively mitigate racial and ethnic disparities, as the tool does not include race or ethnicity as a factor.93 Programs explicitly targeting the impacts of redlining are gaining significance across the country. The U.S. Department of Justice began a Combating Redlining Initiative in 2021 to target racist practices in the banking sector. The initiative has secured millions of dollars by winning settlement agreements with banks and mortgage lenders, including in Los Angeles.94 Furthermore, some state and local initiatives have offered reparations to non-white citizens through down-payment assistance. Washington was the first state to approve a program of this nature and is set to begin implementation in July 2024. Local governments in redlined cities across the U.S. have also begun exploring similar initiatives.
California State Initiatives California is adopting and amending several new regulatory tools that have the potential to address the impacts of highway development on redlined communities. The Transportation Development Act, amended in 2023, could impact funding for equity 91 White House, Inflation Reduction Act Guidebook (Washington, D.C.: White House, 2022), accessed April 5, 2024, Link.
92 Delger Erdenesanaa, “Signature Biden Program Won’t Fix Racial Gap in Air Quality, Study Suggests,” The New York Times, July 20, 2023, sec. Climate, Link. 93 Yuzhou Wang et al., “Air Quality Policy Should Quantify Effects on Disparities,” Science 381, no. 6655 (2023): 272–74, Link.
94 U.S. Department of Justice, “Justice Department Reaches Significant Milestone in Combating Redlining Initiative,” accessed April 5, 2024, Link. ADDRESSING THE DISCRIMINATORY IMPACTS OF REDLINING AND HIGHWAY DEVELOPMENT IN CALIFORNIA 34 in regional transportation planning.95 California has been at the forefront of adopting indirect source rules (ISRs) under the Clean Air Act. ISRs target and regulate emissions that indirectly come from pollution sites (like vehicle emissions from trucks that deliver goods to and from the facility). In 2021, The South Coast Air Quality Management District enacted an ISR to regulate indirect warehouse emissions, which may soon become federally enforceable.96 CARB has also considered roadside vegetative barriers as a strategy to combat the impacts of air pollution from freeways.97 In 2013, CalEPA created its Environmental Justice Task Force to coordinate the compliance and enforcement work of the agency in “areas of California . . . burdened by multiple sources of pollution and are disproportionately vulnerable to its effects,” including Stockton and Los Angeles.98 The Task Force has demonstrated early success. The Community Air Protection Program is a seminal program established in 2017 to protect communities disproportionately impacted by air pollution.99 An example of efforts within the program is a locally driven study in Fresno to reroute trucks out of overburdened residential areas.
100 The Transformative Climate Communities program (TCC), established in 2016, seeks to fight redlining and climate change by awarding grants for developing and implementing neighborhood-level plans that reduce greenhouse gas emissions and provide multiple benefits.
101 TCC puts frontline communities in charge and requires “all projects to develop a collaborative governance structure between local government, community-based organizations, and residents.”102 State legislation has also created more equitable approaches to local and regional planning. Senate Bill 1000 (2016) officially incorporated environmental justice into local and regional planning, while Senate Bill 375 (2008) established a “bottom-up” approach to these processes.
103 Senate Bill 1137 (2022) mandates a 3,200-foot health and safety buffer zone between new and reworked oil and gas wells and “sensitive land uses” such as schools and hospitals.
104 The establishment of these buffer zones would limit exposure to PM2.5 in 95 California Legislature, “Bill Comparison: Senate Bill 125 (2023-2024 Session),” accessed April 5, 2024, Link.; The National Law Review, “Warehouse and Logistics Operations Targeted in Regulatory Push for Indirect Source Rules,” accessed April 5, 2024, Link. 96 The National Law Review, “Warehouse and Logistics Operations Targeted in Regulatory Push for Indirect Source Rules,” accessed April 5, 2024, Link. 97 California Air Resources Board, “Strategy Snapshots,” accessed April 5, 2024, Link. 98 California Environmental Protection Agency (CalEPA), “Environmental Justice Compliance and Enforcement Task Force,” accessed April 5, 2024, Link.
99 California Legislature, Assembly Bill 617, 2017-2018 Reg. Sess., accessed April 5, 2024, Link.; The Program, however, has faced criticism, which prompted improvement efforts. See California Air Resources Board, “About the California Cap-and-Trade Program,” accessed April 5, 2024, Link. 100 California Air Resources Board, “Community Identified Project Approval Notice: San Joaquin Fresno Truck Study,” accessed April 5, 2024, Link. 101 UCLA Luskin Center for Innovation, “Transformative Climate Communities,” accessed April 5, 2024, Link. 102 Greenlining Institute, “Transformative Climate Communities,” accessed April 5, 2024, Link. 103 California Legislature, Senate Bill 1000, 2015-2016 Reg. Sess., accessed April 5, 2024, Link.; California Institute for Local Government (CA ILG), “The Basics of SB 375,” accessed April 5, 2024, Link. 104 California Legislature, Senate Bill 1137, 2021-2022 Reg. Sess., accessed April 5, 2024, Link. ADDRESSING THE DISCRIMINATORY IMPACTS OF REDLINING AND HIGHWAY DEVELOPMENT IN CALIFORNIA 35 overburdened areas if it survives the oil industry-backed veto referendum on the upcoming ballot.
105 State legislators also formed a Select Committee on Reconnecting Communities in 2023, aiming to “explore ways the state can reconnect neighborhoods that decades ago were torn apart by interstates and highways,” and take advantage of the federal funds through the Reconnecting Communities and Neighborhoods Grant Program.
106 Culture Shift in State Agencies California State agencies are initiating cultural changes aimed at prioritizing racial equity. A 2022 Executive Order mandates state entities to “embed and institutionalize racial equity strategies across their policies, programs, and initiatives,” which has led to the creation of the Racial Equity Commission to support that effort.
107 Some leaders within agencies like CalTrans profess that the agency now places equity and engagement at its core.
108 To reflect this commitment, CalSTA created a Climate Action Plan for Transportation Infrastructure and is in the process of developing a Transportation Equity Index that will help CalTrans prioritize and evaluate plans and projects.
109 But, this comes at a time when CalTrans is facing increased scrutiny from community groups and the federal government over its handling of proposed freeway expansion projects.
110 Housing Nexus Exploring the intersection between transit and California’s housing crisis holds the potential to capitalize on funding opportunities and create more integrated communities. In response to long-standing housing inequities, California has passed bills to address the consequences of redlining and similar discriminatory practices. Of note are Assembly Bill 1466, which required counties across California to remove any unlawful restrictive covenant language on historical public records, and Assembly Bill 686, which required all 105 California Legislature, “Analysis of Senate Bill 1137 (2021-2022 Session),” accessed April 5, 2024, Link.; California Geologic Energy Management Division (CalGEM), NTO 2023-03, accessed April 5, 2024, Link. 106 “Reconnecting Communities,” CalMatters, February 2023, accessed April 5, 2024, Link.; Select Committee on Reconnecting Communities, “Reconnecting Communities: CA Freeways - Past and Present,” YouTube video live stream, 2:15:27, Select Committee on Reconnecting Communities, December 8, 2023, Link.; “Select Committee on Reconnecting Communities,” California State Assembly, accessed April 5, 2024, Link. 107 California Office of Planning and Research, “Racial Equity Action Plan,” accessed April 6, 2024, Link. 108 “Undoing the Past: Lawmakers Seek to Mend California Neighborhoods Sliced by Highways,” KQED, February 7, 2024, accessed April 5, 2024, Link.; Select Committee on Reconnecting Communities, “Reconnecting Communities: CA Freeways - Past and Present,” YouTube video live stream, 2:15:27, Select Committee on Reconnecting Communities, December 8, 2023, Link.
109 CalTrans also outlines this commitment in its guidance documents, including its Equity statement, the California Transportation Plan 2050, and the CalTrans 2020-2024 Strategic Plan. California Department of Transportation (Caltrans), “Caltrans Equity, Quality, and Inclusion (EQI) Program,” accessed April 5, 2024, Link. 110 For an example, see details on the Highway 99 Project at “Highway 99 Fresno,” Fresno Land, January 12, 2024, accessed April 5, 2024, Link.; “Showdown on the I-15,” Politico California Climate Newsletter, January 25, 2024, accessed April 6, 2024, Link. ADDRESSING THE DISCRIMINATORY IMPACTS OF REDLINING AND HIGHWAY DEVELOPMENT IN CALIFORNIA 36 public agencies to affirmatively further fair housing in California.
111Additionally, leadership changes in both houses of the state legislature suggest future laws will be more pro-housing overall.
112 In anticipation of the 2028 Olympics, Los Angeles is implementing infrastructure projects to enhance its transportation capacity. If leveraged well, these projects could benefit formerly redlined communities. The city and local transit organizations are relying on partnerships with academics and private industry to complete transportation projects quickly and sustainably such as TRACtion.
113 Additionally, the Los Angeles CleanTech Incubator established the Transportation Electrification Partnership with the express goal of “accelerating transportation electrification and zero emissions goods movement throughout the Los Angeles region in advance of the 2028 Olympic and Paralympic Games by pursuing bold targets, pilots, initiatives, and policies that are equity-driven, create quality jobs, and grow the economy.”114 4.3. Policy Options Based on a review of the context surrounding the legacies of redlining and highway development in California, the following four policy approaches have been identified and developed alongside 13 distinct policy options for further evaluation.
115 We use these four approaches to better organize and categorize our 13 policy options. Focus Explicitly on Race and Ethnicity The review suggests that policies addressing the lasting impacts of redlining and highway development should codify an explicit focus on communities of color that were directly affected by these discriminatory decisions into law. We have identified the following as potential policy options within this approach: 1. 40% of Federal Funds into Communities of Color: In line with the Justice40 Initiative, California State agencies should ensure at least 40% of funds from major federal sources are invested into communities of color to advance efforts in environmental 111 Los Angeles County Registrar-Recorder/County Clerk, “Restrictive Covenant Modifications,” accessed April 5, 2024, Link.; “L.A. County Will Remove Racist Restrictive Covenant Language from Millions of Documents,” Los Angeles Times, February 6, 2024, accessed April 5, 2024, Link.; “Restrictive Covenant Modification Implementation Plan,” Los Angeles County Registrar-Recorder/County Clerk (RR/CC), accessed April 5, 2024, Link.; California Department of Housing and Community Development (HCD), “Affirmatively Furthering Fair Housing,” accessed April 5, 2024, Link. 112 CalMatters, “California Legislature Makes Unprecedented Investments in Housing,” November 14, 2023, accessed April 5, 2024, Link. ; “Oil’s No Good, Very Bad Week,” Politico California Climate newsletter, February 2, 2024, accessed April 5, 2024, Link.; “Mike McGuire Takes Reins of California Senate,” CalMatters, February 5, 2024, accessed April 5, 2024, Link. 113 UCLA Sustainable LA Grand Challenge, “Traction: Transportation,” accessed April 5, 2024, Link. 114 Los Angeles Cleantech Incubator, “Transportation,” accessed April 5, 2024, Link. 115 Due to space and time constraints, we chose to only evaluate 13 policy options for this report. For our full list of policy options, including context on Tribal Nations, see Appendix 4.2 and 4.3.
ADDRESSING THE DISCRIMINATORY IMPACTS OF REDLINING AND HIGHWAY DEVELOPMENT IN CALIFORNIA 37 justice.
116 All funds should be tracked and publicly posted to help ensure funds are being allocated as such. This includes 1) CalEPA using Inflation Reduction Act funds and 2) CalTrans using Department of Transportation Re-Connecting Communities Plan funds to redress harms and address environmental hazards.
2. Statewide WAIRE: CARB should use the South Coast Air Quality Management District’s (AQMD) Warehouse Indirect Source Rule (WAIRE) as a statewide model to manage emissions and pollution from power plants, waste sites, ports, refineries, and vehicles. Fines collected from violators could be utilized to fund compliance activities and be directly redistributed to formerly redlined communities. As an incentive, fines collected could also go back to facilities that meet the required standards in formerly redlined communities.
3. Homeownership Funds for Communities of Color: The California Department of Housing & Community Development (HCD) should provide hyper-local grants or contracts that focus on homeownership assistance to communities of color and formerly redlined communities. This will help address housing discrimination and promote the racial integration of neighborhoods throughout California.
4. Cash Reparations for Black Californians: Budget allocations for state agencies should follow the reparations Task Force recommendations to provide reparations payments to descendants of enslaved people. Because the U.S. Census does not currently identify the number of such descendants in the state, the report uses the number of census respondents who identified as Black or African American alone as a rough estimate.
Increase and Improve Community Empowerment and Decision-Making Authority The literature suggests that although many policies aimed at addressing redlining and highway development have environmental justice principles and community empowerment provisions, in reality, these principles have often been overridden, thereby preventing communities from fully participating or leading projects and decisions impacting them. This approach gives historically redlined communities more power to bring about just solutions that are rooted in their own community needs. We will examine the following policy options within this approach: 5. Decision-Making Power for Impacted Communities: Give the power to formerly redlined communities or communities of color to make decisions on policies related to environmental justice and equity by: a. Mandating approval from impacted communities for (1) any new industrial development plans or re-zoning that would increase pollution in formerly 116 CalEnviroscreen can be used to identify these communities, although there are criticisms and shortcomings of the tool. California law defines environmental justice as the “fair treatment of people of all races, cultures, and incomes with respect to the development, adoption, implementation, and enforcement of environmental laws and policies” (Government Code section 65040.12).
ADDRESSING THE DISCRIMINATORY IMPACTS OF REDLINING AND HIGHWAY DEVELOPMENT IN CALIFORNIA 38 redlined areas and (2) the development of the Community Air Monitoring Plans (CAMPs), Community Emissions Reduction Programs (CERPs), and substantive actions by the Air Districts supported by California Air Resources Board (CARB) to significantly reduce emissions.
b. Create substantial and institutionalized avenues for residents of formerly redlined communities to exercise leadership (including serving on boards, commissions, and advisory councils) within programs, grant allocations, and regulatory decision-making processes. These bodies should be allocated a portion of the city budget to pursue initiatives that address the impacts of redlining.
6. Incentivize Community Involvement: The state should systematically incentivize community involvement in shaping environmental justice policies and initiatives.
a. Enhance public notice and scoping meeting requirements under the California Environmental Quality Act (CEQA) for projects that propose the siting and expansion of polluting land uses in overburdened communities.
117 b. Increase community involvement in public forums by (1) facilitating CBO participation through accessible government stipends or honorariums and (2) providing childcare, food, and language access services at public meetings to accommodate all income levels and family dynamics. This may require creating carve-out funding or incentives in programs.
c. Allow agency staff more time and an increased mandate to plan meaningful public opportunities to gather community input and partner with community members on key initiatives.
7. Flexible Budgets and Timelines: In line with the Clean Mobility Equity Playbook, “allow more flexibility for programs to manage their own budgets and timelines. Equity programs often require greater resources, capacity, and longer timelines to [achieve] their goals.”118 Use Zoning and Planning Goals The literature review suggests that incorporating policies to address the impacts of redlining and highway development by altering zoning and planning goals can facilitate more equitable and inclusive communities. We will examine the following policy options within this approach: 8. Highway Redesign and Truck Rerouting: In their 2025 update to the California Transportation Plan (CTP), CalTrans should: a. Use AB 617 funding to mandate that local governments develop plans to reroute trucks away from pollution-burdened communities 117 Some strategies include establishing a proportion of the population that has to provide feedback on a project, creating advisory groups, and requiring community members be included in earlier parts of the process.
118 Greenlining Institute, Clean Mobility Equity: A Playbook (Oakland, CA: Greenlining Institute, 2021), accessed April 8, 2024, Link. ADDRESSING THE DISCRIMINATORY IMPACTS OF REDLINING AND HIGHWAY DEVELOPMENT IN CALIFORNIA 39 b. Designate state funds to require Regional Transportation Planning Associations and Metropolitan Planning Organizations to conduct studies into possible freeway re-design projects – such as freeway capping to removal – and grant the CalTrans Racial Equity and Transportation Advisory Committee with ultimate authority to fund the implementation of worthy projects, and c. Strengthen measures to protect community members from displacement due to transportation development projects. d. Appoint the Racial Equity and Transportation Advisory Committee established in Recommendation 4.3 of the 2021 CTP as the main liaison for community input on these initiatives and as partners in all plan development and funding decisions.
9. Segregation Element: Using data from CalEPA and HCD tools as evidence, the state should require all formerly redlined cities in California to include a specific Segregation Element within their General Plan that addresses current levels of segregation in the city and establishes strategies to promote integration, such as inclusionary zoning or investment in fair housing projects. It should also establish and maintain a public database of integration planning resources and examples, managed by HCD, as well as a standing advisory board of representatives specifically from formerly-redlined communities to advise and participate in the planning process of this element. The public database should be available to all jurisdictions that want to pursue a segregation element in its General Plan, not just the mandated eight cities.
10. Create ‘Overburdened’ Pollution Standard: CalEPA should create a regulatory standard that indexes pollution levels in a community and sets a threshold for what constitutes an overburdened community in California. This standard should further require city plans to establish which communities are overburdened in their Environmental Justice Elements and amend land use policies to prevent further development of high-polluting land uses (highway expansion, industrial development) and promote land uses that relieve pollution (green space development, traffic reduction) in those communities. The state should offer subsidies for jurisdictions that implement these requirements and fines for those that do not.
Make Data More Accessible The literature suggests that although data is available, it is hard to access and understand, which prevents it from being a useful resource to best pinpoint where and how solutions should be implemented. This approach does not directly reduce the lasting impacts of redlining and highway development, but provides tools for communities to use data to identify inequities and advocate for solutions. We have selected the following policy options within this approach for further evaluation: 11. Increase Grant Accessibility: All state agencies providing grant funding for environmental justice and racial equity programs should increase the level of accessibility and support for potential grant recipients to 1) improve the California ADDRESSING THE DISCRIMINATORY IMPACTS OF REDLINING AND HIGHWAY DEVELOPMENT IN CALIFORNIA 40 Grants Portal to make programs more accessible through use of plain language and multiple languages 2) provide increased technical assistance for grant applications to relieve the need for grant staff at organizations.
12. Add Segregation and Redlining Data to CalEnviroscreen: OEHHA should incorporate the following data into the publicly accessible CalEnviroscreen tool so that communities may use it to understand the connections between past redlining and present environmental injustice and use it as evidence to help advocate to eradicate such injustices: a. Data on segregation levels by census tract from UC Berkeley’s Roots of Structural Racism Project. As this data was gathered in 2020, OEHHA should continue to periodically update such data using the latest census tract information and the methods used by UC Berkeley researchers to create this segregation index.
b. Past HOLC areas, using data from the University of Richmond’s Mapping Inequality Project. This data could also be applied at the census tract level to allow users to view the former risk grade of each census tract where applicable.
13. Expand CalTrans Equity Index Data: OEHHA and CARB should work with CalTrans to develop an Equity Index with a wider range of data to include UC Berkeley’s segregation index, University of Richmond’s HOLC Areas, and pollution levels from CES 4.0 to consider pollution in overexposed communities. These efforts should include funding to update, improve, and maintain this data.
ADDRESSING THE DISCRIMINATORY IMPACTS OF REDLINING AND HIGHWAY DEVELOPMENT IN CALIFORNIA 41 5 POLICY EVALUATION US EPA / Facebook ADDRESSING THE DISCRIMINATORY IMPACTS OF REDLINING AND HIGHWAY DEVELOPMENT IN CALIFORNIA 42 5. POLICY EVALUATION 5.1. Policy Evaluation Overview This chapter breaks down the evaluation process of the 13 identified policy options and provides recommendations for which should be implemented by state agencies.
Key Criteria To evaluate the final list of policy options and provide recommendations, we compiled five key criteria: • Environmental justice principles, particularly those centered around the themes of justice, autonomy, and “policy, politics, and economic processes,” and how well the policy adheres to those principles.
119 • Address segregation and PM2.5 levels, which we define as the extent to which these policy options are designed to alleviate the disproportionate socioeconomic and environmental burdens caused by redlining and highway development on impacted communities in California.
• Socio-political feasibility, which considers the levels of support for the policy and the relative ease or difficulty of getting it into law.
120 • Efficacy, which we define as the ability of the policy option to produce its intended result. This includes considerations of the financial, social, and administrative costs of the policy.
121 • Generalizability, which we define as the extent to which research findings can be applied to other settings or contexts.
119 Dorceta Taylor, “The Rise of the Environmental Justice Paradigm: Injustice Framing and the Social Construction of Environmental Discourses,” American Behavioral Scientist 43 (2000): 508-580, Link. 120 H. Lawford-Smith, “Understanding Political Feasibility,” Journal of Political Philosophy 21 (2013): 243-259, Link. 121 Collins Dictionary, s.v. “Efficacy,” accessed April 5, 2024, Link. Josh Begley / joshbegley.com ADDRESSING THE DISCRIMINATORY IMPACTS OF REDLINING AND HIGHWAY DEVELOPMENT IN CALIFORNIA 43 These criteria were chosen for the policy evaluation because they are project-specific, equity-focused, and policy-oriented. The criteria will target segregation and PM2.5 levels, which are emblematic of broader socio-economic and environmental impacts. Our team and client are interested in advancing equitable solutions, which promote the use of environmental justice principles as key criteria. This decision stems from acknowledging that “environmental justice” is cited and operationalized in current policies and programs in this field. Lastly, political feasibility and efficacy serve as policy-specific metrics that will gauge the function and use of our ultimate recommendations.
Evaluation Criteria Breakdown To analyze the 13 policy options, we utilized a Criteria Alternatives Matrix (CAM). CAM allows for a systematic and transparent evaluation of different policies across a range of criteria to assess each of their strengths and weaknesses and assign each a composite score to reflect its fit with our policy goals. Each of the five evaluation criteria was assigned a possible maximum score to weigh its importance to the overall quality of the policy in question. The maximum composite score for any given policy is 100. A composite score above 50 indicates a policy that has benefits that outweigh its flaws and effectively addresses the discriminatory legacies of redlining and highway development in California. To ensure the robustness of our recommendation threshold based on the sensitivity tests we conducted, 122 we applied a margin of error of two points to this threshold, raising the target score for recommendations to 52.
For each of the criteria, our evaluation is guided by a series of sub-questions about the nature of the policy, detailed in Figures 17-21. Together, the sub-questions are designed to cover all the important components of the wider criterion.
Figure 17: Promotion of Environmental Justice Principles sub-questions and points.
123 Promotion of Environmental Justice Principles 30 total points Does the policy help enforce the right of communities to be free from ecological destruction?
6 points Is the policy based on mutual respect and justice for all peoples, free from any form of discrimination or bias?
6 points Does the policy protect or establish the right of communities to participate as equal partners at every level of decision making (needs assessment, planning, implementation, enforcement, and evaluation)?
6 points Does the policy protect or promote the right of victims of environmental injustice to receive full compensation and reparations for damages?
6 points Does the policy help mandate the right to ethical, balanced, and responsible uses of land and renewable resources in the interest of a sustainable planet for humans and other living things?
6 points 122 For more information on our sensitivity tests, see Appendix 5.8.
123 See Appendix 5.1 for further explanations of the sub-questions and weights assigned to them.
ADDRESSING THE DISCRIMINATORY IMPACTS OF REDLINING AND HIGHWAY DEVELOPMENT IN CALIFORNIA 44 Figure 18: Promotion of Environmental Justice Principles sub-questions and points.
124 Reduction of PM 2.5 Pollution and/or Racial Segregation 30 total points Does the policy help reduce the burden of PM 2.5 pollution on formerly redlined communities/communities of color in California?
10 points Does the policy help reduce the high levels of residential racial segregation in California connected to its histories of redlining and highway development?
7 points Does the policy help address other harms to communities of color that stem from residential racial segregation?
5 points Does the policy help meaningfully reduce overall levels of PM2.5 pollution levels?
4 points Does the policy help reduce the burden of related pollutants (diesel particulate matter, PM 10, more) on formerly redlined communities/communities of color in California?
4 points Figure 19: Socio-Political Feasibility sub-questions and points.
125 Socio-Political Feasibility 20 total points How much political support is there for the policy?
3 points How much political opposition is there to the policy?
3 points if no opposition Is there a reason to think support for the policy might increase in the future?
2 points Could the policy be officially adopted without the legislature or a ballot measure? 4 points What is the level of support for the policy amongst the communities it seeks to support? 4 points Did communities facing environmental injustice in California participate in developing and advocating for the policy?
4 points Figure 20: Efficacy sub-questions and points.
126 Efficacy 14 total points What are the monetary and social costs of the policy, relative to its designed impacts?
6 points if low How manageable would it be to implement and administer the policy?
4 points How long would it take for the policy to begin producing meaningful impacts?
2 points Are there foreseeable scenarios that would cause the policy to become ineffective or irrelevant in the near future?
2 points if no Figure 21: Generalizability sub-questions and points.
127 Generalizability 6 total points Can the policy be effective across jurisdictions with different socioeconomic, environmental, and cultural characteristics in California?
3 points Can the policy be an effective model for jurisdictions outside California?
3 points 124 See appendix 5.2 for further explanations of the sub-questions and weights assigned to them.
125 See appendix 5.3 for further explanations of the sub-questions and weights assigned to them.
126 See appendix 5.4 for further explanations of the sub-questions and weights assigned to them.
127 See appendix 5.5 for further explanations of the sub-questions and weights assigned to them.
ADDRESSING THE DISCRIMINATORY IMPACTS OF REDLINING AND HIGHWAY DEVELOPMENT IN CALIFORNIA 45 5.2. Evaluation Results128 CAM analysis results for each four policy approaches. Total scores and criteria-specific scores for each approach reflect the mean score of all the proposed policy options within that approach. Figure 22 shows the composite scores of the four different policy approaches, generated by averaging the scores for each individual policy within each approach across the five evaluation criteria. Figures 23-26 show the scores assigned to each individual policy. We discuss the relative strengths and weaknesses of each policy and each wider approach in the next pages.
Legend 1: for Figures 22-26 Total Score Coding: Criteria Score Coding: Strongly Recommend (>60 points) # Extreme Strength Recommend (56-60 points) # Strength Tentatively Recommend (53-55 points) # Neither Strength nor Weakness Neither Recommend nor Reject (49-52 points) # Weakness Tentatively Reject (46-48 points) # Extreme Weakness Reject (40-45 points) relative to scores of other policy options See Appendix 5.6 for breakdown of ranges for each evaluation criteria Strongly Reject (<40 points) 128 See appendix 5.7 for a full matrix of the scores assigned to each of the 13 policies we reviewed for every sub-question within our criteria.
Figure 22: CAM analysis results for each four policy approaches. Total scores and criteria-specific scores for each approach reflect the mean score of all the proposed policy options within that approach.
Policy Approach Focus Explicitly on Race and Ethnicity Community Empowerment and Decision-Making Zoning and Planning Goals Make Data More Accessible Criteria Weight (points out of 100) Average Score of Policy Approach Promotion of environmental justice principles 30 12.75 15 17.2 11.67 Reduction of PM 2.5 pollution and/or racial segregation 30 11.25 10.33 17.3 5.67 Socio-political feasibility 20 13 10.67 13 13.83 Efficacy 14 7.25 6.5 8.7 11.67 Generalizability 6 5 6 4.8 6 TOTAL SCORE 100 49.25 48.5 61 48.84 ADDRESSING THE DISCRIMINATORY IMPACTS OF REDLINING AND HIGHWAY DEVELOPMENT IN CALIFORNIA 46 Focus Explicitly on Race and Ethnicity Figure 23: CAM analysis results for the three proposed policy options within the Race and Ethnicity approach. Refer to Legend 1 for color codes.
Focus Explicitly on Race and Ethnicity #1: 40% of Federal Funds into Communities of Color #2: Expanded Statewide Warehouse Indirect Source Rule #3: Homeownership Grant Fund for Communities of Color #4: Cash Reparations for Black Californians Criteria Weight (points out of 100) Average Score of Policy Approach Promotion of environmental justice principles 30 16 15 9 11 Reduction of PM 2.5 pollution and/or racial segregation 30 12 16 14 3 Socio-political feasibility 20 14 14 11 13 Efficacy 14 8 8 8 5 Generalizability 6 5 5 4 6 TOTAL SCORE 100 55 58 46 38 We recommend Options #1 and #2 within this policy approach. Both policies are well-rounded and score above 50 for each criterion. Both do particularly well in promoting environmental justice principles and garnering community support. Moreover, they can be implemented without legislative action and can utilize existing regulations as implementation models, with Option #2 using AQMD’s WAIRE program and Option #1 using the Justice40 initiative, respectively. However, widespread adoption may take more time without legislative action. Option #1 has the hurdle of attaining implementation in each California State agency. Even with environmental justice groups’ support, the process may take considerable time to implement across agencies and face opposition. Legislative or ballot measures could expedite the process but would necessitate advocacy, campaigning, and grassroots organizing efforts. While Option #2 does not address residential racial segregation, it most effectively reduces PM2.5 pollution and other environmental pollutants in communities of color facing disproportionate environmental burdens, making it a top policy recommendation. Options #3 and #4 fall short in promoting environmental justice principles compared to other policy options. They also face political opposition and legislative challenges that render them less politically feasible. Option #4, in particular, offers minimal direct reduction of PM2.5 pollution levels or racial segregation; cash reparations, without community empowerment, may not foster the necessary collaboration needed to create systemic change. Option #4 also faces high costs with no direct funding source present, which makes it the lowest-scoring option in the category.
ADDRESSING THE DISCRIMINATORY IMPACTS OF REDLINING AND HIGHWAY DEVELOPMENT IN CALIFORNIA 47 Increase and Improve Community Empowerment and Decision-Making Authority Figure 24: CAM analysis results for the three proposed policy options within the Community Empowerment approach. Refer to Legend 1 for color codes.
Community Empowerment and Decision-Making #5: Decision-Making Power for Impacted Communities #6: Incentivize Community Involvement #7: Flexible Budgets/ Timelines for Programs Criteria Weight (points out of 100) Average Score of Policy Approach Promotion of environmental justice principles 30 18 18 9 Reduction of PM 2.5 pollution and/or racial segregation 30 16 15 0 Socio-political feasibility 20 10.5 10 11.5 Efficacy 14 6.5 6.5 6.5 Generalizability 6 6 6 6 TOTAL SCORE 100 57 55.5 33 We recommend Option #5 and Option #6 within the community empowerment approach. Both options excel in promoting environmental justice principles and reducing the burden of PM2.5 pollution on formerly redlined communities and communities of color. Leveraging existing regulations like the community benefits policies in Option #5 and CEQA in Option #6 effectively addresses environmental justice concerns and fosters community engagement. These options can also be implemented without legislative action, although state involvement helps provide comprehensive frameworks and support to ensure successful implementation. Options #5 and #6 could serve as effective models for jurisdictions outside of California seeking to implement robust environmental review processes and ensure meaningful community involvement in development decisions. The primary challenge for both options is the feasibility of implementing and administering the policies. For Option #5, municipal governments may face resource constraints and compliance issues between developers and the community. For Option #6, the lengthy and costly processes of incentivizing and facilitating community participation on all relevant local issues constitute a major challenge to the efficacy of the policy.
Option #7 would have no significant impact in helping relevant programs reduce pollution burdens and promote integration, even on more flexible timelines, and therefore would not help achieve our central policy goals.
ADDRESSING THE DISCRIMINATORY IMPACTS OF REDLINING AND HIGHWAY DEVELOPMENT IN CALIFORNIA 48 Use Zoning and Planning Goals Figure 25: CAM analysis results for the three proposed policy options within the Zoning and Planning approach. Refer to Legend 1 for color codes.
Zoning and Planning Policy Options #8: Highway Redesign and Truck Rerouting #9: Segregation Element #10: Create 'Overburdened' Pollution Standard Criteria Weight (points out of 100) Average Score of Policy Approach Promotion of environmental justice principles 30 17 16.5 18 Reduction of PM 2.5 pollution and/or racial segregation 30 18 19.5 14.5 Socio-political feasibility 20 14 9.5 15.5 Efficacy 14 4 9.5 12.5 Generalizability 6 5 3.5 6 TOTAL SCORE 100 58 58.5 66.5 We recommend all three of our proposed policy options in this approach. As Figure 22 details, the CAM analysis finds the most utility in using zoning and planning goals to address the ongoing impacts of redlining and highway development in California. The three zoning and planning policies averaged an overall score well above the 52 threshold for recommendation, significantly above any of the other policy approaches that were considered. Furthermore, the approach averaged scores greater than 52 for each evaluation criterion, indicating its widespread effectiveness across all key aspects of policy-making. Zoning and planning policies were especially well-equipped to promote environmental justice principles and address the impacts of racial segregation and PM2.5 pollution burden – the two most important evaluation criteria. The minor weaknesses of this approach include the amount of time it takes for policies to begin producing intended effects and the amount of participation and decision-making authority they afford to impacted communities.
Option #10 scored the best of the 13 options analyzed. This proposal – for CalEPA to prohibit further high-polluting land uses in overburdened communities above a defined pollution threshold – scored 85% of possible points across the feasibility, efficacy, and generalizability criteria. This was due to the unique combination of cost-effectiveness, political support, and community involvement woven into the policy. The model for such pollution-reducing zoning regulations comes from Stockton’s General Plan, which was developed with robust community engagement on environmental justice components. The policy also uses data-driven standards to add teeth to the statewide environmental justice planning requirements established by SB 1000, a bill that community-based organizations played a large role in developing back in 2016. Overall, the set of ADDRESSING THE DISCRIMINATORY IMPACTS OF REDLINING AND HIGHWAY DEVELOPMENT IN CALIFORNIA 49 regulations in Option #10 would instigate meaningful change to disrupt the continuing patterns of over-pollution in formerly redlined communities and would prompt cities to craft innovative solutions alongside communities to reduce the disproportionate pollution burden faced by communities of color moving forward.
Options #8 and #9 should also be adopted by the State. The facilitation of truck rerouting plans and freeway redesign projects like freeway capping in Option #8 promotes increased social and financial mobility for Californians of color alongside its main function of radically improving air quality and reducing pollution in their communities. The policy is also modeled on community-driven initiatives in California, such as Fresno’s Truck Reroute Study and proposals for the Park 101 freeway cap in Los Angeles. However, the immense disruption and monetary and time costs associated with freeway reduction and freeway capping are drawbacks to its efficacy.
Option #9 requires formerly redlined cities to implement a segregation element in their general plan. The binding requirements for pro-integration planning solutions provide enough unique opportunities to advance environmental justice and integration to overcome the feasibility and efficacy constraints of having to pass such requirements into law through the legislature. Its eventual implementation, however, could be fraught with local turmoil in cities like San Francisco and Los Angeles, as community leaders from NIMBY groups, as well as from communities of color that fear gentrification and displacement, could resist integration planning initiatives.
Make Data More Accessible Figure 26: CAM analysis results for the three proposed policy options within the Data Accessibility approach. Refer to Legend 1 for color codes.
Making Data More Accessible Policy Options #11: Increase Support for EJ and Equity Grant-Seekers #12: Add Segregation and Redlining Data to CalEnviroscreen #13: Expand CalTrans Equity Index Criteria Weight (points out of 100) Average Score of Policy Approach Promotion of environmental justice principles 30 13 12 10 Reduction of PM 2.5 pollution and/or racial segregation 30 10 5 2 Socio-political feasibility 20 13.5 14 14 Efficacy 14 10 13 12 Generalizability 6 6 6 6 TOTAL SCORE 100 52.5 50 44 ADDRESSING THE DISCRIMINATORY IMPACTS OF REDLINING AND HIGHWAY DEVELOPMENT IN CALIFORNIA 50 We recommend Option #11 in this approach. The overall strength of the policies from this category is that they all are relatively easy to implement, have little costs attached to them, and increase communities’ understanding of local issues. This promotes equity and autonomy for communities to use information in support of their collective local interests. Its main weakness is that increased data accessibility does not directly improve PM2.5 pollution or segregation on its own. Option #11 is the best option from the category because it includes direct help with grant writing for state funds. Grant writing is a niche skill that is a barrier for many grassroots organizations. This policy alleviates that burden and makes state funding more attainable for impacted communities to push forward work that addresses local concerns of environmental and racial justice. The weakness of this policy is that it does not provide increased funding or availability for grants that address PM2.5 pollution or segregation. Option #12 reached the threshold for recommendation and would supplement the current CalEnviroScreen tool by adding segregation data and outlining historically redlined communities. This is relatively simple to implement because data sources for this already exist and can be combined into CalEnviroScreen to make the tool more robust in its visualization of where pollution is concentrated and the broader factors linked to that pollution. This would be a great tool for advocacy and would also point to where California agencies will need to allocate their resources for true environmental justice. The main weakness of this policy is that it provides no funding or direct action to address the central issues and could have an overall limited reach. Additionally, this option did not pass the margin of error, preventing its recommendation.
Option #13 prioritizes mobility using the CalTrans Equity Index. Although incorporating data on pollution and segregation into this index is important, it may not be as effective in advocating for these issues compared to the CalEnviroScreen.
ADDRESSING THE DISCRIMINATORY IMPACTS OF REDLINING AND HIGHWAY DEVELOPMENT IN CALIFORNIA 51 5.3. Recommendations Summary In summary, we recommend eight of our proposed policy options for the state to adopt.
Strongly recommend: • Option #10 (66.5): CalEPA should create a regulatory standard to identify communities overburdened by pollution and require the implementation of land uses in such communities that relieve said pollution burden.
• Option #10 is cost-effective and has strong political support and community involvement components.
Further recommend: • Option #9 (58.5): Institute a State requirement for formerly redlined cities to include a Segregation Element within their General Plan that establishes strategies to promote integration.
• Option #9’s binding requirements provide enough unique opportunities to advance environmental justice and integration.
• Option #2 (58): CARB should use the South Coast AQMD’s Warehouse Indirect Source Rule as a statewide model to manage emissions and pollution from all high-polluting industrial sources.
• Option #2 effectively reduces PM2.5 pollution and other environmental pollutants in communities of color facing disproportionate environmental burdens.
• Option #8 (58): CalTrans should prioritize and fund truck rerouting and highway redesign plans across the state in their 2025 CTP update.
• Option #8 is especially well-equipped to promote environmental justice principles and address the impacts of racial segregation and PM2.5 pollution burden.
• Option #5 (57): The state should entrust decision-making power to formerly redlined communities or communities of color over policies concerning environmental justice and equity. • Option #5 excels in promoting environmental justice principles and reducing the burden of PM2.5 pollution on formerly redlined communities and communities of color. ADDRESSING THE DISCRIMINATORY IMPACTS OF REDLINING AND HIGHWAY DEVELOPMENT IN CALIFORNIA 52 Relying on the status quo of policies is not enough to dismantle the systems leading to racial segregation and disproportionate pollution burdens for non-white communities around California. The unequal outcomes in health, income, safety, and more that stem from these systemic inequalities throughout the state are only getting worse.
129 Adopting this set of policy recommendations, especially those that score above 55 points in the CAM analysis, is an urgently necessary set of actions that would help address gaps in statewide policy related to the environmental justice needs of these communities.
129 “Segregation Is Getting Worse in the U.S. The Bay Area Is No Exception,” KQED, accessed April 5, 2024, Link.
Tentatively recommend: • Option #6 (55.5): State agencies should systematically incentivize community participation and input around environmental justice policies and initiatives.
• Option #6 effectively addresses environmental justice concerns and fosters community engagement.
• Option #1 (55): State agencies should ensure at least 40% of funds from federal environmental and transportation programs are invested in communities of color.
• Option #1 promotes environmental justice principles and garners community support.
• Option #11 (52.5): All state agencies providing grant funding for environmental justice and racial equity programs should increase the level of accessibility and support for potential grant recipients.
• Option #11 is relatively easy to implement, has low costs, and increases community understanding of local issues. ADDRESSING THE DISCRIMINATORY IMPACTS OF REDLINING AND HIGHWAY DEVELOPMENT IN CALIFORNIA 53 5.4. Conclusion This report examined California’s intertwined history of discriminatory redlining and highway development, illuminating troubling connections between that history and present-day deficits in environmental and socioeconomic outcomes for non-white Californians. In this examination, we specifically focused on the disparities in PM2.5 pollution exposure between redlined and non-redlined communities today, as well as the current levels of residential racial segregation that correspond to past HOLC risk grades. Employing a combination of existing research and original spatial and data analysis, we uncovered significant findings. We demonstrated a staggering association between previously redlined areas and current non-white segregation levels, and an even stronger association between previously A-rated HOLC areas and current levels of white segregation. Moreover, we showed that the targeting of highway development in redlined areas is connected to higher present-day concentrations of people of color near highways, and that highway proximity has a significant positive correlation with PM2.5 exposure, especially in formerly redlined communities. Overall, we delineated how people of color in California experience health issues and discrimination due to segregation and pollution exposure that stem from the state’s racist history of redlining and highway development, and that these ongoing harms have yet to be fully addressed.
Building on these findings, research was conducted into policies that the State of California could pursue to advance environmentally just and equitable solutions to these problems. Four potentially fruitful policy approaches were identified – designing policies that focus explicitly on race and ethnicity, changing zoning and planning goals, increasing community empowerment and decision-making authority, and increasing the accessibility of data – and specific policy proposals within those approaches were analyzed. The analysis yielded eight policies that the state should implement to better address the ongoing impacts of its discriminatory history of redlining and highway development, with zoning and planning policies being the most effective overall policy approach.
Although this project represents a comprehensive effort, given our time and capacity constraints, this topic warrants future analysis and exploration. There are many lasting impacts of redlining and highway development beyond PM2.5 pollution and segregation that researchers could study in detail. Additionally, there is room to expand our case studies of Stockton and Los Angeles to include the other six formerly redlined cities in California. Research into the legacies of redlining and transportation development in other states, as well as the policy efforts that have been made to address them, would also be valuable.
We envision this project serving as a valuable reference for decision-makers at the state, regional, and local levels. Our maps and findings from this report can be explored visually on OEHHA’s Pollution and Prejudice story map.
130 It is crucial that California addresses the detrimental impacts of redlining and highway development to foster an environment that brings about true justice and opportunity for communities of color.
130 Available in June 2024 at Link. ADDRESSING THE DISCRIMINATORY IMPACTS OF REDLINING AND HIGHWAY DEVELOPMENT IN CALIFORNIA 54 6 REFERENCES Dicklyon / Wikimedia Commons ADDRESSING THE DISCRIMINATORY IMPACTS OF REDLINING AND HIGHWAY DEVELOPMENT IN CALIFORNIA 55 6. REFERENCES ABC7 News. “Native American Tribes Protest Willits Bypass.” Accessed April 5, 2024. https:// abc7news.com/willits-bypass-highway-101-native-american-tribes-caltrans/1060842/.
Archer, D. N. (2020). “White Men’s Roads Through Black Men’s Homes”: Advancing Racial Equity Through Highway Reconstruction. Vanderbilt Law Review, 73(5), 1259–1330.” CalEPA. “Pollution and Prejudice Map.” ArcGIS StoryMaps. Published August 16, 2021. Accessed April 5, 2024.
California Air Resources Board. “About the California Cap-and-Trade Program.” Accessed April 5, 2024.
California Air Resources Board. “About ARB.” Accessed April 5, 2024. about.
California Air Resources Board. “Community Identified Project Approval Notice: San Joaquin Fresno Truck Study.” Accessed April 5, 2024. community-identified-project-approval-notice-san-joaquin-fresno-truck-study.
California Air Resources Board. “Strategy Snapshots.” Accessed April 6, 2024.
ca.gov/our-work/programs/resource-center/strategy-development/strategy-snapshots.
California Budget & Policy Center. “Q&A: What Does the Budget Shortfall Mean for California?” Accessed April 5, 2024.
California Department of Housing and Community Development (HCD). “Affirmatively Furthering Fair Housing.” Accessed April 5, 2024.
California Department of Justice. “AB 3121 Task Force Report.” Accessed April 6, 2024. https:// oag.ca.gov/ab3121/report. California Department of Transportation. 2022 Caltrans Facts. Accessed April 8, 2024. https:// dot.ca.gov/-/media/dot-media/programs/research-innovation-system-information/documents/ caltrans-fact-booklets/2022-caltrans-facts-a11y.pdf. California Department of Transportation. “About Caltrans.” Accessed April 5, 2024.
ca.gov/about-caltrans.
California Department of Transportation (Caltrans). “Caltrans Equity, Quality, and Inclusion (EQI) Program.” Accessed April 5, 2024.
California Department of Transportation (Caltrans). “Vision 980.” Accessed April 5, 2024. https:// dot.ca.gov/caltrans-near-me/district-4/d4-projects/vision-980.
California Environmental Protection Agency (CalEPA). “About CalEPA.” Accessed April 8, 2024. California Environmental Protection Agency (CalEPA). “CalEPA Racial Equity.” Accessed April 8, 2024. California Environmental Protection Agency (CalEPA). “Environmental Justice Compliance and Enforcement Task Force.” Accessed April 5, 2024. environmental-justice-compliance-and-enforcement-task-force/.
ADDRESSING THE DISCRIMINATORY IMPACTS OF REDLINING AND HIGHWAY DEVELOPMENT IN CALIFORNIA 56 California Geologic Energy Management Division (CalGEM). NTO 2023-03. Accessed April 5, 2024.
pdf.
California Institute for Local Government (CA ILG). “The Basics of SB 375.” Accessed April 5, 2024.
California Legislature. “Analysis of Assembly Bill 1778 (2021-2022).” Accessed April 5, 2024.
California Legislature. “Analysis of Senate Bill 1137 (2021-2022 Session).” Accessed April 5, 2024.
California Legislature. Assembly Bill 617. 2017-2018 Reg. Sess. Accessed April 5, 2024. https:// leginfo.legislature.ca.gov/faces/billTextClient.xhtml?bill_id=201720180AB617.
California Legislature. “Bill Comparison: Senate Bill 125 (2023-2024 Session).” Accessed April 5, 2024. id=202320240SB125&showamends=false.
California Legislature. Senate Bill 1000. 2015-2016 Reg. Sess. Accessed April 5, 2024. https:// leginfo.legislature.ca.gov/faces/billNavClient.xhtml?bill_id=201520160SB1000.
California Legislature. Senate Bill 1137. 2021-2022 Reg. Sess. Accessed April 5, 2024. https:// leginfo.legislature.ca.gov/faces/billNavClient.xhtml?bill_id=202120220SB1137.
California Office of Environmental Health Hazard Assessment (OEHHA). “About OEHHA.” Accessed April 8, 2024. California Office of Planning and Research. “Racial Equity Action Plan.” Accessed April 6, 2024. California State Assembly. “Select Committee on Reconnecting Communities.” Accessed April 5, 2024.
California State Transportation Agency. Accessed April 8, 2024.
California Transportation Commission. “Home.” Accessed April 6, 2024. CalMatters. “California Legislature Makes Unprecedented Investments in Housing.” November 14, 2023. Accessed April 5, 2024.
CalMatters. “Mike McGuire Takes Reins of California Senate.” February 5, 2024. Accessed April 5, 2024.
CalMatters. “Reconnecting Communities.” February 2023. Accessed April 5, 2024. https:// calmatters.org/california-divide/2023/02/reconnecting-communities/.
California, State Of. “Caltrans History | Caltrans,” n.d. caltrans-history.
Chen Y, Gu P , Schulte N, Zhou X, Mara S, Croes BE, Herner JD, Vijayan A. A new mobile monitoring approach to characterize community-scale air pollution patterns and identify local high pollution zones, Atmospheric Environment, Volume 272, 2022.
atmosenv.2022.118936.
ADDRESSING THE DISCRIMINATORY IMPACTS OF REDLINING AND HIGHWAY DEVELOPMENT IN CALIFORNIA 57 Collins Dictionary. s.v. “Efficacy.” Accessed April 5, 2024. dictionary/english/efficacy.
Correia, A. W., Pope, C. A. III, Dockery, D. W., Wang, Y., Ezzati, M., & Dominici, F. “Effect of air pollution control on life expectancy in the United States: an analysis of 545 U.S. counties for the period from 2000 to 2007.” Epidemiology (Cambridge, Mass.) 24, no. 1 (2013): 23–31.
Douds, K., & Wu, J. (2018). Trust in the Bayou City: Do Racial Segregation and Discrimination Matter for Generalized Trust? Sociology of Race and Ethnicity, 4(4), 567-584.
org/10.1177/2332649217717741 Environmental Justice Network. “Principles of Environmental Justice.” Accessed April 6, 2024. Environmental Protection Agency. (n.d.). Particulate matter (PM) basics. Retrieved January 31, 2024, from Erdenesanaa, Delger. “Signature Biden Program Won’t Fix Racial Gap in Air Quality, Study Suggests.” The New York Times, July 20, 2023, sec. Climate.
com/2023/07/20/climate/justice40-pollution-environmental-justice.html.
Federal Transit Administration. BART Title VI Final Report. Washington, D.C.: U.S. Department of Transportation, April 16, 2010. Accessed April 5, 2024.
dot.gov/files/docs/BART_Title_VI_Final_4_16_10.doc.
“Fresno Land.” “Highway 99 Fresno.” Accessed April 6, 2024. highway-99-fresno/.
Gamboa, Suzanne, et al. “Bulldozed and Bisected: Highway Construction Built a Legacy of Inequality.” NBCNews.Com, NBCUniversal News Group, www.nbcnews.com/specials/ america-highways-inequality/. Accessed 2 Jan. 2024. Greenlining Institute. Clean Mobility Equity: A Playbook. Oakland, CA: Greenlining Institute, 2021. Accessed April 8, 2024. Greenlining Institute. “Transformative Climate Communities.” Accessed April 5, 2024. https:// greenlining.org/work/transformative-communities/transformative-climate-communities/.
“Hidden Long-Term Effects on the Latino Community of East Los Angeles.” Foundations of Law and Society (blog). Accessed April 8, 2024.
com/2017/12/11/eisenhowers-1956-federal-aid-highway-act-and-the-hidden-long-term-effects-on-the-latino-community-of-east-los-angeles/.
Houston, Douglas, and Michelle E. Zuñiga. “Put a park on it: How freeway caps are reconnecting and greening divided cities.” Cities 85 (2019): 98-109. Accessed January 2022.
org/10.1016/j.cities.2018.08.007 Jaffe, Eric. “The Forgotten History of l.a.’s Failed Freeway Revolt.” Bloomberg.Com, Bloomberg, 23 July 2014, www.bloomberg.com/news/articles/2014-07-23/the-forgotten-history-of-l-a-s-failed-freeway-revolt. KQED. “Segregation Is Getting Worse in the U.S. The Bay Area Is No Exception.” Accessed April 5, 2024.
ADDRESSING THE DISCRIMINATORY IMPACTS OF REDLINING AND HIGHWAY DEVELOPMENT IN CALIFORNIA 58 KQED. “Undoing the Past: Lawmakers Seek to Mend California Neighborhoods Sliced by Highways.” February 7, 2024. Accessed April 5, 2024. undoing-the-past-lawmakers-seek-to-mend-california-neighborhoods-sliced-by-highways.
KRON4 News. “Why Was San Francisco’s Waterfront a Freeway?” Accessed April 5, 2024. https:// www.kron4.com/news/bay-area/why-was-sfs-waterfront-a-freeway/.
“L.A. County Will Remove Racist Restrictive Covenant Language from Millions of Documents.” Los Angeles Times, February 6, 2024. Accessed April 5, 2024. story/2024-02-06/l-a-county-will-remove-racist-restrictive-covenant-language-from-millions-of-documents.
Lane, HM., Morello-Frosch, R., Marshall, JD., Apte JS. Historical Redlining is Associated with Present-Day Air Pollution Disparities in U.S. Cities. Environ. Sci. Technol. Lett. 2022, 9, 4, 345–350 Lawford-Smith, H. “Understanding Political Feasibility.” Journal of Political Philosophy 21 (2013): 243-259.
Levine, Jonathan. “Justice and the Interstates: The Racist Truth about Urban Highways, Edited by Ryan Reft, Amanda K. Phillips de Lucas, and Rebecca C. Retzlaff: Washington, DC, Island Press, 2023.” Journal of Urban Affairs, (2023), 1–2. doi:10.1080/07352166.2023.2217045.
Lin II, Rong-Gong (Uranga). “710 Freeway expansion dropped after decades of planning, marking a milestone for L.A.” Los Angeles Times, March 15, 2022. story/2022-03-15/710-freeway-expansion-dropped-after-decades-of-planning. Los Angeles Cleantech Incubator. “Transportation.” Accessed April 5, 2024.
org/transportation/.
Los Angeles County Registrar-Recorder/County Clerk. “Restrictive Covenant Modifications.” Accessed April 5, 2024. restrictive-covenant-modifications.
Los Angeles County Registrar-Recorder/County Clerk (RR/CC). “Restrictive Covenant Modification Implementation Plan.” Accessed April 5, 2024. Restrictive-Covenant-Modification-Implementation-Plan.pdf?v=5.
Lutenski, E. (2019). Dickens Disappeared: Black Los Angeles and the Borderlands of Racial Memory. American Studies (Lawrence), 58(3), 15–35. Madrigal-Lauchland, V. (2019). The Shifting Meanings of Stockton’s Barrio del Chivo. California History (San Francisco), 96(3), 97–100. Mark, Julian. “‘Sky’s the limit’: Caltrans is getting serious about replacing I-980.” The Oaklandside, October 30, 2023. Mohl RA. The Interstates and the Cities: The U.S. Department of Transportation and the Freeway Revolt, 1966–1973. Journal of Policy History. 2008;20(2):193-226. doi:10.1353/jph.0.0014 Nardone A, Casey JA, Morello-Frosch R, Mujahid M, Balmes JR, Thakur N. Associations between historical residential redlining and current age-adjusted rates of emergency department visits due to asthma across eight cities in California: an ecological study. Lancet Planet Health. 2020 Jan;4(1):e24-e31. doi: 10.1016/S2542-5196(19)30241-4. PMID: 31999951; PMCID: PMC10018700.
ADDRESSING THE DISCRIMINATORY IMPACTS OF REDLINING AND HIGHWAY DEVELOPMENT IN CALIFORNIA 59 New York Times. “Chavez Ravine Reparations for Dodger Stadium.” March 26, 2024. Accessed April 6, 2024. NRDC. “Making Good on California’s Clean Transportation Promises.” Accessed April 5, 2024.
Ong, Paul, Ananya Roy Yoon, and Chhandara Pech. “Redlining and Beyond: Development Within and Outside HOLC Spaces in Los Angeles County.” Los Angeles: UCLA Center for Neighborhood Knowledge, 2023. Accessed April 8, 2024. item/2tj3n9s0. Ong, P . M., Pech, C., Do, C.-H., Yoon, A., & Wasserman, J. L. (2023). Stockton’s Crosstown Freeway, Urban Renewal, and Asian Americans: Systemic Causes and Impacts.
OpenAI. “ChatGPT.” Accessed April 3, 2024. Pohl, D. J., Seblova, D., Avila, J. F., Dorsman, K. A., Kulick, E. R., Casey, J. A., & Manly, J. (2021). Relationship between Residential Segregation, Later-Life Cognition, and Incident Dementia across Race/Ethnicity. International Journal of Environmental Research and Public Health, 18(21), 11233-. Politico California Climate Newsletter. “Oil’s No Good, Very Bad Week.” February 2, 2024. Accessed April 5, 2024. Politico California Climate Newsletter. “Showdown on the I-15.” January 25, 2024. Accessed April 6, 2024. showdown-on-the-i-15-00138004?nname=california-climate&nid=00000189-315c-d8dd-a1ed-797dc9f10000&nrid=6c553c21-3dcc-448a-a23e-8aa141adcad9&nlid=2745178.
Select Committee on Reconnecting Communities. “Reconnecting Communities: CA Freeways - Past and Present.” YouTube video live stream. 2:15:27, Select Committee on Reconnecting Communities. December 8, 2023.
Shaffer, Ralph E. “California Reluctantly Implements the Fifteenth Amendment: White Californians Respond to Black Suffrage, March - June, 1870.” Presentation at Cal Poly Pomona, 2020.
Statista. “Total Number of U.S. Licensed Drivers by State.” Accessed April 8, 2024,
statista.com/statistics/198029/total-number-of-us-licensed-drivers-by-state/.
Streetsblog California. “No Freeway Expansion Bill Dies in Senate Committee.” Accessed April 5, 2024.
Stermon, M., & Lukinbeal, C. (2021). Institutionalized Racism: Redlined Districts Then and Now in Boston, Detroit, and Los Angeles. Yearbook of the Association of Pacific Coast Geographers 83, 81-97.
Susaneck, Adam Paul. “Sugar Hill.” SEGREGATION BY DESIGN, www.segregationbydesign.com/ los-angeles/sugar-hill. Accessed 2 Jan. 2024. Taylor, Dorceta. “The Rise of the Environmental Justice Paradigm: Injustice Framing and the Social Construction of Environmental Discourses.” American Behavioral Scientist 43 (2000): 508-580.
ADDRESSING THE DISCRIMINATORY IMPACTS OF REDLINING AND HIGHWAY DEVELOPMENT IN CALIFORNIA 60 The National Law Review. “Warehouse and Logistics Operations Targeted in Regulatory Push for Indirect Source Rules.” Accessed April 5, 2024. warehouse-and-logistics-operations-targeted-regulatory-push-indirect-source-rules.
“The Warehouse and Logistics Operations Targeted in Regulatory Push for Indirect Source Rules.” The National Law Review. Accessed April 5, 2024. warehouse-and-logistics-operations-targeted-regulatory-push-indirect-source-rules.
Tiotiu AI, Novakova P , Nedeva D, Chong-Neto HJ, Novakova S, Steiropoulos P , Kowal K. Impact of Air Pollution on Asthma Outcomes. Int J Environ Res Public Health. 2020 Aug 27;17(17):6212. doi: 10.3390/ijerph17176212. PMID: 32867076; PMCID: PMC7503605.
UCLA Luskin Center for Innovation. “Transformative Climate Communities.” Accessed April 5, 2024.
UCLA Sustainable LA Grand Challenge. “Traction: Transportation.” Accessed April 5, 2024. https:// sustainablela.ucla.edu/traction-transportation.
Urban Institute. Early Assessment of Opportunity Zones for Equitable Development Projects. Washington, D.C.: Urban Institute, n.d. Accessed April 5, 2024. default/files/publication/102348/early-assessment-of-opportunity-zones-for-equitable-development-projects.pdf.
U.S. Department of Justice. “Justice Department Reaches Significant Milestone in Combating Redlining Initiative.” Accessed April 5, 2024.
U.S. Environmental Protection Agency. “Title VI and Environmental Justice.” Accessed April 6, 2024.
White House. Inflation Reduction Act Guidebook. Washington, D.C.: White House, 2022. Accessed April 5, 2024.
Wang, Yuzhou, Joshua S. Apte, Jason D. Hill, Cesunica E. Ivey, Dana Johnson, Esther Min, Rachel Morello-Frosch, et al. “Air Quality Policy Should Quantify Effects on Disparities.” Science 381, no. 6655 (2023): 272–74.
Xing, Y. F., Xu, Y. H., Shi, M. H., & Lian, Y. X. “The impact of PM2.5 on the human respiratory system.” Journal of Thoracic Disease 8, no. 1 (2016): E69–E74.
issn.2072-1439.2016.01.19.
Zucman, Gabriel, and Emmanuel Saez. “The False Promise of Opportunity Zones.” Boston Review. Accessed April 5, 2024.
ADDRESSING THE DISCRIMINATORY IMPACTS OF REDLINING AND HIGHWAY DEVELOPMENT IN CALIFORNIA 61 7 APPENDIX Alfred Twu / Wikimedia Commons ADDRESSING THE DISCRIMINATORY IMPACTS OF REDLINING AND HIGHWAY DEVELOPMENT IN CALIFORNIA 62 7. APPENDIX 7.1. Chapter 2 - Methodology 2.1 To identify sources from our UCLA network, we explored the UC Library system and similar scholarly databases and solicited recommendations from professors and scholars – principally our UCLA Project Advisor, Dr. Michael Stoll – with relevant expertise related to the research topic. Through OEHHA, we not only considered resource recommendations from the policy professionals we worked with directly on the project, but also reached out to contacts in other state agencies, including CalTrans and California Air Resources Board (CARB), who held unique insights into our research topic.
2.2 To conduct the spatial analysis, the study used data on historical redlining areas from the University of Richmond’s Mapping Inequality Project, data on racial demographics and median household income from the U.S. Census Bureau’s 2021 American Community Survey 5-year estimates, data on California Highways from CalTRANS, data on PM2.5 levels collected in 2021 from CalEnviroscreen 4.0, and data on segregation levels from UC-Berkeley’s The Roots of Structural Racism Project.
2.3 The tables below detail the methods behind the creation of the census tract-level Highway Proximity index variable used as an independent variable in Model 1 and Model 2: Criteria Weight # of Highways within 1 mi.
4 Distance of nearest highway 4 Bisecting Highway (Y/N) 2 Total 10 # of Highways within 1 mi.
Distance of nearest highway Bisecting Highway (Y/N) Range Score Range Score Range Score 0 0 3.09–5.54 mi.
0 No (0) 0 1 1 1.92–3.09 mi.
1 Yes (1) 2 2–3 2 1.15–1.92 mi.
2 4 3 0.57–1.15 mi.
3 5–6 4 0.0006–0.57 mi.
4 ADDRESSING THE DISCRIMINATORY IMPACTS OF REDLINING AND HIGHWAY DEVELOPMENT IN CALIFORNIA 63 We used tools in ArcGIS Pro to calculate all three components of this index. To identify the number of highways near a census tract, we instructed ArcGIS to select the number of observations in the CalTrans highway data set that were within 1 mile of the census tract’s border. To calculate the exact proximity of the closest highway to every census tract, we had ArcGIS Pro locate the center point of every census tract and then calculate the distance from that center point to the nearest highway in the CalTrans data set. To identify whether a census tract was bisected by a highway, we had ArcGIS Pro select all tracts that contained a CalTrans highway within their borders. We then ran sensitivity tests by altering the weights of each of the three components by 5% (0.5 points) in either direction. None of these tests changed the direction or statistical significance of the relationships with the independent variable highway proximity in any of our regression models, nor did they change the magnitude of the relationships to a statistically significant level. Thus, the sensitivity tests ensure the robustness of the index.
2.4 The regressions in the study used census tract-level data from CalEnviroscreen 4.0 published in 2021 and spatial analysis tools in ArcGIS Pro to add new key variables for its regression models. The study created Redlined and Green dummy variables, for which census tracts were assigned a value of 1 if their center point was within a quarter-mile of a formerly-red or green HOLC area respectively, and a value of 0 if not. It also created a Highway Proximity variable which indexed three elements of highway proximity for each census tract and assigned a value of 0-10 to each census tract based on these elements described below. The study further added a Non-white Population variable for each census tract by inverting the percentage of the non-Hispanic white population living in that census tract. Other control variables the study created were Population Density for each census tract and a Citywide Non-White Population statistic assigned to each census tract based on the city it was located in. A full list of the variables already included in the CalEnviroscreen 4.0 data can be found by viewing the tool online. After creating these new variables, analysts used the resulting data to run the following two sets of fixed effects regression models to obtain results that would help answer the study’s research questions.
2.5 Model 1 and Model 2 first use census tract-level data from California’s eight formerly-redlined cities to test the effects of Redlined classifications, Green classifications, and Highway proximity on 2021 levels of Non-white population. The two models are designed identically except for the addition of an interaction term between Redlined and Highway proximity in Model 2, which captures the additional impact of highway proximity on a census tract’s racial demographics when that census tract was previously redlined. Both models are included so that the study can analyze the baseline relationship of the dependent variable to all independent variables in the absence of an interaction term, as well as the predicted changes that occur when accounting for the interaction between the study’s two critical independent variables. The model controls for the fixed effects of ADDRESSING THE DISCRIMINATORY IMPACTS OF REDLINING AND HIGHWAY DEVELOPMENT IN CALIFORNIA 64 each of the eight different cities in the data, which most notably includes the overall non-white population of the city. This means the regression controls for all differences in data between cities that may bias the relationships of interest.
2.6 Model 3 and Model 4 use the same data to test the effects of Redlined classifications, Green classifications, and Highway proximity on 2021 levels of PM2.5 concentration. Following the same logic as the prior set of models, the two models are identical except for the addition of an interaction term between Redlined and Highway proximity in Model 3, which captures the additional impact of highway proximity on a census tract’s PM2.5 exposure when that census tract was previously redlined. The model controls for the population density of each census tract, as well as city-level fixed effects, which most notably includes overall citywide PM2.5 levels and other levels of related types of pollution such as Diesel Particulate Matter.
2.7 For more specific analysis, we applied the formulas from Model 1 and Model 2 to different subsections of our data set. These four different subsets represented urban areas, rural areas, Los Angeles, and Stockton. When testing the data sets of Los Angeles and Stockton, we get rid of the city fixed effects element and employ a simple multivariate regression. Models 5-20 in Appendix section 3.7 display these regressions. Subsequently, we use a Wald test to examine statistically significant differences between urban and rural areas, as well as between Los Angeles and Stockton. Wald tests were used to test for statistical significant differences between two regressions. The formula used for the Wald test were: Comparing urban and rural: Test statistic = Comparing Los Angeles and Stockton: Test statistic = At a 95% confidence interval, the test statistic was compared against the critical value z value of 1.96.
ADDRESSING THE DISCRIMINATORY IMPACTS OF REDLINING AND HIGHWAY DEVELOPMENT IN CALIFORNIA 65 7.2. Chapter 3 - Problem Identification 3.1 Using the dataset described in Appendix 2.4, we calculated the mean poverty in formerly-redlined census tracts and weighted it by population, then calculated the mean poverty rate in all other census tracts and weighted it by population.
Results: μRedlinedPovertyRate = 46.003%, μNonRedlinedPovertyRate = 32.078% 3.2 Overlaying the CalTrans highway data with the University of Richmond data on all HOLC areas in ArcGIS Pro, we calculated the percentage of redlined areas in California that currently intersect with a highway, as well as the percentage of A, B, or C-rated areas that currently intersect with a highway. We then ran a two-sample Z-test on this difference in proportions of 77% compared to 49%, which yielded a p-value of .057e-8, meaning we could be more than 99.99% confident that this difference did not occur due to random chance.
3.3 Using the dataset from Appendix 2.4, we calculated the mean number of highways within 0.75 miles of all redlined census tracts, as well as the mean number of highways within 0.75 miles of all A-rated census tracts.
Results: • μred = 2.25 highways • μgreen = 1.26 highways We then conducted a two-sample T-test on these sample means to determine if the difference was statistically significant, which it was at a p-value = .013e-12 (> 99.999% confidence).
3.4 Using the dataset from Appendix 2.4, we calculated the mean distance of the nearest highway to the geographical center of all redlined census tracts, as well as the mean distance of the nearest highway to the geographical center of all A-rated census tracts.
Results: • μred = 0.62 miles • μgreen = 1.1 miles We then conducted a two-sample T-test on these sample means to determine if the difference was statistically significant, which it was at a p-value = .092e-8 (> 99.999% confidence).
ADDRESSING THE DISCRIMINATORY IMPACTS OF REDLINING AND HIGHWAY DEVELOPMENT IN CALIFORNIA 66 3.5 Data on segregation levels comes from the Othering and Belonging Institute at UC Berkeley and their Roots of Structural Racism Project, which was published in 2021. This study paired their census tract-level segregation data with the University of Richmond’s Mapping Inequality project data on former HOLC risk classification areas to assign each census tract in the eight formerly-redlined cities of California both a HOLC risk grade and a segregation level. It then used these two variables to calculate the statistics on segregation for redlined and A-rated areas.
3.6 99.9% confidence intervals were created for each of the eight formerly redlined cities in California by calculating a proportion confidence interval using the following parameters: • Sample size (N) = μCensus Tract Population • Population probability (P) = Citywide % of non-white population • x = N P Here were the exact confidence intervals for non-white population by census tracts for each of the eight cities: • Los Angeles: +/- 2.33% • Stockton: +/- 1.9% • San Diego: +/- 2.34% • San Francisco: +/- 2.41% • San Jose: +/- 1.98% • Sacramento: +/- 2.2% • Fresno: +/- 2.11% • Oakland: +/- 2.38% ADDRESSING THE DISCRIMINATORY IMPACTS OF REDLINING AND HIGHWAY DEVELOPMENT IN CALIFORNIA 67 Model 5 & Model 6 Model 7 & Model 8 Model 9 & Model 10 Model 11 & Model 12 3.7 Models 5-12 ADDRESSING THE DISCRIMINATORY IMPACTS OF REDLINING AND HIGHWAY DEVELOPMENT IN CALIFORNIA 68 Null hypothesis: There is no statistically significant difference between the 8 urban and rural cities in California.
Wald Test values: • Redlined: | 3.62487553 | > |1.96|, Can reject the null hypothesis • Green: |-0.05081071| < |1.96|, Fail to reject the null hypothesis • HWprox: |-3.36682480| > |1.96|, Can reject the null hypothesis Null hypothesis: There is no statistically significant difference between Los Angeles and Stockton.
Wald Test values • Redlined: | 1.148772 |< |1.96|, Fail to reject the null hypothesis • Green: |1.038601|< |1.96|, Fail to reject the null hypothesis • HWprox: |2.865382|> |1.96|, Can reject the null hypothesis Null hypothesis: There is no statistically significant difference between the 8 urban and rural cities in California.
Wald Test values • Redlined: |0.69619795|< |1.96|, Fail to reject the null hypothesis • Green: |0.05395146|< |1.96|, Fail to reject the null hypothesis • HW prox: |3.32260686|> |1.96|, Can reject the null hypothesis • RedHWprox: |1.81678505|< |1.96|, Fail to reject the null hypothesis Null hypothesis: There is no statistically significant difference between Los Angeles and Stockton.
Wald Test values • Redlined:| -0.4729896 | < |1.96|, Fail to reject the null hypothesis • Green: | -1.0331063 | < |1.96|, Fail to reject the null hypothesis • HWprox: |2.4864457| > |1.96|, Can reject the null hypothesis • RedHWprox: |0.2441454|< |1.96|, Fail to reject the null hypothesis ADDRESSING THE DISCRIMINATORY IMPACTS OF REDLINING AND HIGHWAY DEVELOPMENT IN CALIFORNIA 69 Model 13 & Model 14 Model 17 & Model 18 Model 15 & Model 16 Model 19 & Model 20 Models 13-20 ADDRESSING THE DISCRIMINATORY IMPACTS OF REDLINING AND HIGHWAY DEVELOPMENT IN CALIFORNIA 70 Null hypothesis: There is no statistically significant difference between the 8 urban and rural cities in California.
Wald Test values • Redlined: |-1.488367|. Fail to reject the null hypothesis.
• HWprox: |-5.031918|> |1.96|, Can reject the null hypothesis.
• Green: |-1.047843|. Fail to reject the null hypothesis.
Null hypothesis: There is no statistically significant difference between Los Angeles and Stockton.
Fail to reject the null hypothesis for all.
Wald Test values • Redlined: |-1.6672988| • HWprox: |-0.8337777| • Green: |-1.1685054| Null hypothesis: There is no statistically significant difference between the 8 urban and rural cities in California.
Wald Test values • Redlined: |-3.367458 |> |1.96|, Can reject the null hypothesis. • HWprox: |-6.985343|> |1.96|, Can reject the null hypothesis.
• Green: | -1.072796 |. Fail to reject the null hypothesis.
• RedHWprox: |-2.254027| > |1.96|, Can reject the null hypothesis.
Null hypothesis: There is no statistically significant difference between Los Angeles and Stockton.
Fail to reject the null hypothesis for all.
Wald Test values • Redlined: |-0.5219282| • HWprox: |-1.5232856| • Green: |-1.1806447| • RedHWprox: |0.3286962| We also conducted a series of regressions to investigate the effects of independent ADDRESSING THE DISCRIMINATORY IMPACTS OF REDLINING AND HIGHWAY DEVELOPMENT IN CALIFORNIA 71 variables on non-white population levels and PM2.5 levels between urban and rural areas, as well as between Los Angeles and Stockton. Beginning with a naive regression, each model progressively introduced independent variables Redlined, HwProx, Green, and an interaction term of RedlinedHwProx were added one regression at the time to isolate their resulting effects on either the Urban or LAvStock variable. To calculate the impact of each variable on the Urban or LAvStock coefficient, the marginal percent change was calculated using the formula below. Overall, the results of these models aimed to illustrate the impact of redlining and highway development on present day racial segregation levels and PM2.5 levels in California.
The tables below show naive regressions and the resulting outputs of the differences between rural and urban as well as Los Angeles and Stockton on racial segregation and PM2.5 levels. The coefficient for the urban variable, displayed in the Urban and Rural comparison column, slightly changes as independent variables are added to the model. The first regression shows that rural areas have a higher level of non-white population compared to urban areas. This coefficient increases by 201.42% for rural areas when controlling for redlining. Again, the coefficient increases by 7.8% when accounting for highway proximity. Similar trends are seen in the Los Angeles and Stockton regressions. Redlining accounts for a 35.08% increase in the Los Angeles and Stockton coefficient, the lavstock variable, as it increases from -7.948 to -10.736. Highway proximity increases this coefficient again by 3.62%. In both models, controlling for greenlining and the interaction term reduces the magnitude of the urban and lavstock variable coefficients. This is attributed to the influence of greenlining status and interaction effects explaining part of the observed effects.
ADDRESSING THE DISCRIMINATORY IMPACTS OF REDLINING AND HIGHWAY DEVELOPMENT IN CALIFORNIA 72 These tables reflect higher PM2.5 levels in urban areas and Los Angeles than rural areas and Stockton. In both, the comparison coefficient changes slightly when accounting for redlining, highway proximity, and greenlining. With the final inclusion of the interaction term, both coefficients increase slightly when capturing the combined effects of redlining and highway proximity.
3.8 Here are the respective mean annual averages of PM2.5 concentration in California’s eight formerly-redlined cities – for both the formerly redlined census tracts and the city overall: Los Angeles: Redlined = 12.05 µg/m3, Overall = 11.86 µg/m3 Stockton: Redlined = 11.68 µg/m3, Overall = 11.11 µg/m3 Sacramento: Redlined = 9.14 µg/m3, Overall = 8.92 µg/m3 San Diego: Redlined = 10.27 µg/m3, Overall = 9.96 µg/m3 San Francisco: Redlined = 8.64 µg/m3, Overall = 8.62 µg/m3 San Jose: Redlined = 8.92 µg/m3, Overall = 8.49 µg/m3 Fresno: Redlined = 13.79 µg/m3, Overall = 13.59 µg/m3 Oakland: Redlined = 9.29 µg/m3, Overall = 9.18 µg/m3 These statistics were calculated using data from CalEnviroscreen 4.0 for census tract-level PM2.5 concentration, as well as data from the University of Richmond’s Mapping Inequality project to select the historical HOLC classifications of each census tract using tools in ArcGIS Pro.
ADDRESSING THE DISCRIMINATORY IMPACTS OF REDLINING AND HIGHWAY DEVELOPMENT IN CALIFORNIA 73 7.3. Chapter 4 - Policy Options 4.1: Additional Policy Approaches Our team conducted a content review of over 200 pieces of literature related to redlining and highway development in California. We coded emergent themes in each reading that provided policy options, recommendations, or analysis. To determine which readings provided policy options or analysis, we used the search terms “policy”, “recommend”, and “solution,” and also looked at the conclusion section or chapter of each reading, where appropriate.
The following points are additional policy approaches that emerged from our literature review. Though they are important to acknowledge, our team has determined these additional approaches do not warrant further evaluation in this project because of time and word count restraints for this report and their more tangential connections to our key criteria.
• Funding/resources/investment: there is more funding needed to fully address the impacts of redlining and highway development, and more investments are needed to complete projects that are alternatives to highways.
• Research: there is a need to produce more nuanced research that integrates spatial analysis and social sciences and humanities more broadly on this subject to better understand what solutions will be most effective.
• Housing: more affordable and higher-density housing is needed to mitigate the overarching housing crisis in California to, therefore, address segregation levels that have come as a result of redlining and highway development.
4.2: Additional Policy Options Our team originally considered 27 policy options that emerged from our literature review. Collectively, the team performed an initial policy analysis and reorganization process to reduce the list to 14 final policy options to evaluate. If a policy option could not be consolidated elsewhere, the team voted on which policy options to eliminate using our key criteria as a guide. Policy options that received a majority vote (three or more votes out of a team of five) were eliminated.
Below is our list of additional policy options and the brief reason they were eliminated from the final list.
ADDRESSING THE DISCRIMINATORY IMPACTS OF REDLINING AND HIGHWAY DEVELOPMENT IN CALIFORNIA 74 Policy Approach: Focus Explicitly on Race and Ethnicity 1. CalEPA should model policy behaviors that help the Justice40 Initiative and the Climate and Economic Justice Screening tool to better identify and prioritize communities by explicitly including race and ethnicity in its calculations of who can benefit a. Ensure any California legislation inspired by Justice40 includes an explicit recognition of race and ethnicity and/or formerly redlined communities b. Ensure federal agencies issuing funds to California prioritize formerly redlined communities and/or race and ethnicity c. Cultivate and highlight data that demonstrates that more just environmental outcomes can be produced by explicitly prioritizing race and ethnicity in program tenets.
d. Justice40 originated from a New York law. The New York law has since changed the word “benefit” to “investment” in order to more accurately direct funding to “disadvantaged communities.” CalEPA can advocate for changes of the word “benefit” in Justice40 and Justice40-like programs at the state level, to something like “investment” or the dollar amount of the investment. This would specifically direct funds and programs to communities, rather than this vague idea of “benefits.” e. Require a racial and ethnicity equity analysis (like NOAA) on all Justice40 initiatives/projects/plans that score how well it addresses equity to mandate race and ethnicity considerations. ELIMINATION RATIONALE: This is similar to Policy Option #1 that was included in our final list. We consolidated other pieces of this option to our final list.
2. CARB should use Stockton’s local diesel pollution regulations as a model to build a statewide diesel pollution policy.
ELIMINATION RATIONALE: This did not appear to be a standalone policy option we could evaluate fairly. We felt other policy options could encapsulate a similar idea.
3. California Counties should use the fact that they’re required to inventory and remove racial covenant language from historical documents as an opportunity to flag for areas in which to increase funding or target funding.
ELIMINATION RATIONALE: We believed this policy option was too vague, and instead decided to include this information as context for key opportunities in our Policy Options Chapter.
ADDRESSING THE DISCRIMINATORY IMPACTS OF REDLINING AND HIGHWAY DEVELOPMENT IN CALIFORNIA 75 4. Require jurisdictions to complete comprehensive racial equity impact studies prior to any construction of new or expanded transportation infrastructure, housing development, or industrial development. Racial equity impact studies can help ensure development is tailored toward dismantling the discriminatory legacies of redlining rather and highway development rather than further entrenching them.
ELIMINATION RATIONALE: We believed this policy option was better suited under the “Make Data More Accessible” policy approach. We consolidated this policy option into our final list. 5. The California DOJ should mirror the federal government Combatting Redlining Initiative and should pursue cases against the eight formerly-redlined municipalities to compel them to provide investment to address the impacts of past redlining.
ELIMINATION RATIONALE: We consolidated this policy option into our final list as a sub point of another option.
6. Leverage funding and program examples that were explicitly mentioned at this California State Assembly Select Committee on Reconnecting Communities’ committee hearing and ensure they are race-specific. The committee is producing a report to the legislature.
ELIMINATION RATIONALE: We felt this policy option may be covered in the Committee’s eventual report and would require too much time to fully flesh out and vet as a separate policy option.
Policy Approach: Increase & Improve Community Empowerment and Decision-Making Authority 7. CalTrans should enforce and provide staff with adequate time and resources to follow the Reconnecting Communities handbook.
ELIMINATION RATIONALE: We consolidated this idea into another policy option that was included in the final list.
ADDRESSING THE DISCRIMINATORY IMPACTS OF REDLINING AND HIGHWAY DEVELOPMENT IN CALIFORNIA 76 8. For funds awarded to federally recognized tribal governments, state agencies should remove the requirement to submit a limited waiver of their sovereign immunity for purposes of contracting in cases where a waiver is not explicitly required by statute. This creates an unnecessary administrative burden and barrier.
ELIMINATION RATIONALE: Up to this point in the report, we had not thoroughly discussed California Native American tribes and did not feel comfortable leaving in this recommendation without providing more background information, context, and policy options that focus on tribes.
Policy Approach: Use Zoning and Planning Goals 9. The State should create and fund new environmental justice planning requirements that counties and cities must include in their updated plans a. Institute requirements for jurisdictions to include Environmental Justice Impact Assessments in their Hazard Mitigation Plans that especially study racial inequities in environmental burdens, and for environmental justice goals to be included in their general plans based off of such assessments. State agencies should provide public resources and guidance to help jurisdictions implement these requirements in their updated plans.
b. Provide funds to the eight formerly-redlined cities to be allocated to impact studies of redlining and implementation of plans that help redlined communities address its lasting impact.
c. Integrate the history of redlining and highway development and its lasting impacts into the training of transportation planners and decision-makers d. Pass a state bill modeled after SB 379 (2017; climate adaptation planning standards) that provides requirements for cities to include certain EJ assessments in their hazard mitigation plan and EJ goals in their general plan. State agencies could provide public resources and guidance to help local jurisdictions in their efforts.
i. Create requirements in CalEPA by revising the California Code of Regulations (ex. Title 27) ELIMINATION RATIONALE: We conducted further research on this policy option and consolidated parts of this into our final Policy Option #9.
ADDRESSING THE DISCRIMINATORY IMPACTS OF REDLINING AND HIGHWAY DEVELOPMENT IN CALIFORNIA 77 10. CalEPA should ban types of development that would place further environmental stress on already overburdened communities.
a. Ban highway expansion in areas that are subject to pollution levels above a certain threshold. Also require cities to review and amend land use designations so that they prevent further development of all high-polluting land uses in communities above said threshold. The Stockton General Plan and Stockton Port Clean Air Plan are prime models for these policies, and there are anti displacement and highway widening memos from CAlSTA we could reference. b. Prohibit by-right approval of the siting or expansion of polluting land uses near disadvantaged communities, and require project proponents to obtain a discretionary permit, such as a conditional use permit, in order to ensure that projects undergo individualized environmental review. Require local governments to make special findings that a project will not exacerbate environmental degradation or worsen public health outcomes when approving the project.
c. Include strong enforcement measures of these and other guidelines i. Ex: Enforce laws about Urban Development fund use to ensure cities allocate 2-4% to affirmatively further fair housing, moratorium on foreclosures. Focusing environmental enforcement and compliance activity in communities that are the most vulnerable and the most burdened by multiple sources of pollution is a priority of the CalEPA Working Group and its partner agencies.
ELIMINATION RATIONALE: We incorporated parts of this policy option into Policy Options #5 and #10. We were unsure which state agency(ies) this option would target specifically.
11. Address ongoing displacement in communities of color from transportation investments by incorporating ideas from the Climate Action Plan for Transportation Infrastructure (CAPTI) and these anti-displacement memos.
ELIMINATION RATIONALE: We incorporated parts of this policy option into Policy Option #8 and reference the CAPTI in our policy context section.
12. Consider policies that focus on the benefits and needs of reconnecting communities.
ELIMINATION RATIONALE: We believe the California State Assembly Select Committee on Reconnecting Communities would better evaluate this specific policy option. They are tasked with submitting a report to the Legislature in the coming year. We reference this committee in our Policy Options chapter. We also reference the Department of Transportation Re-Connecting Communities Plan in Policy Option #1.
ADDRESSING THE DISCRIMINATORY IMPACTS OF REDLINING AND HIGHWAY DEVELOPMENT IN CALIFORNIA 78 Policy Approach: Make Data More Accessible 13. Ensure that coordinated outreach and application processes such as those used by Access Clean California72 are a formal program requirement for individual climate and transportation equity incentive programs going forward, particularly for programs that have clear overlap with existing programs in terms of geography, technology and proposed beneficiaries. A lack of coordination has been shown to cause inefficiencies and consumer confusion.
ELIMINATION RATIONALE: We opted to make this policy option more general and include certain programs as examples of how to improve outreach and application processes.
14. CalEPA continue to work with USEPA to prioritize making enforcement engagement more transparent, solution-oriented, responsive to community needs, and sustained. ELIMINATION RATIONALE: We did not believe this policy option fit well under this policy approach. We instead incorporated part of this option into our policy context section.
4.3: Background on California American Indian Tribes Tribal Policies and Programs Tribal Nations in California represent a vital and vibrant part of the state’s cultural fabric, history, and contemporary socio-political landscape. Despite the challenges and transformations they have faced over centuries, tribal communities remain active contributors to California’s diversity and governance. With more than 100 tribes, these groups operate as sovereign governments, a status that affords them certain rights and responsibilities independent of state law, underscoring their importance in policy-making and research endeavors within the state.
1 Sovereignty means that any policy development or research within or affecting these lands must involve the tribes as equal partners. Acknowledging their sovereignty not only respects their rights but also ensures that policies are more effective, culturally sensitive, and inclusive. This is especially critical in areas such as environmental management, where Indigenous knowledge can play a key role and tailored approaches are necessary to address community-specific needs.
The concept of sovereignty is central to understanding the role of American Indian tribes in California today. As sovereign entities, these tribes possess the right to govern themselves, make and enforce laws, tax, establish membership criteria, and manage 1 California Tribal Communities - Tribal_projects,” n.d. Link.
ADDRESSING THE DISCRIMINATORY IMPACTS OF REDLINING AND HIGHWAY DEVELOPMENT IN CALIFORNIA 79 their lands and resources. This status is recognized at the federal level and entails a government-to-government relationship between tribes and the United States.
The inclusion of Tribal Nations in community engagement and policy discussions in California has been growing, yet it remains an area requiring continuous effort and improvement. Initiatives such as consultation policies, where state agencies are required to consult with tribes on actions that may affect them, represent steps in the right direction. However, the effectiveness of such measures depends on the genuine commitment to listening to and incorporating the input of Native communities.
Inviting tribes to the conversation not only acknowledges their sovereignty but also leverages their unique perspectives and wisdom, enriching policy-making and community initiatives. There are instances of successful collaboration in environmental stewardship, cultural heritage preservation, and economic development that showcase the potential of such engagements as demonstrated in the Inflation Reduction Act. Its inclusion of funding for tribal climate resilience and energy programs represents a significant step in bolstering support for Tribal Nations. It is the expectation for both the Administration and Congress to continue supporting tribal climate change initiatives as a means to uphold and fulfill the commitments made to tribal communities.2 While our team was unable to deliberately incorporate intentional research, analysis, and evaluation of policies with a focus on tribes, we remain hopeful that future research on this topic can be explored. 2 The White House. “Inflation Reduction Act Guidebook | Clean Energy | The White House,” December 5, 2023. Link.
ADDRESSING THE DISCRIMINATORY IMPACTS OF REDLINING AND HIGHWAY DEVELOPMENT IN CALIFORNIA 80 7.4. Chapter 5 - Policy Evaluation 5.1: Promotion of environmental justice principles weight explanations These five questions are based on the environmental justice principles established by the People of Color Environmental Leadership Summit in 1991. Out of the 17 principles established by the Summit, the questions focus on five particular principles that best encapsulate the desired goals of environmental justice policies and have particular relevance to the history and impacts of redlining and highway development in California. The full list of environmental justice principles can be found online.3 All of the sub-questions have equal weight, as they are all core components of environmental justice. Within the overall policy evaluation, we weighted the environmental justice criterion highly because the history of redlining and highway development in the state has entrenched environmental injustices that persist throughout the state today. Any policy seeking to address that history should thoroughly adhere to these core environmental justice principles.
5.2: Reduction of PM2.5 Pollution and/or Racial Segregation weight explanations The impacts of PM2.5 pollution and residential racial segregation into the same criterion because they are both discriminatory legacies of the same history of redlining and highway development in California. Thus, policies oriented towards promoting justice for the communities who suffer from that history may touch on both impacts. The three sub-questions focusing on PM2.5 pollution burden are worth 15 points in total. We weighted them to ensure this criterion elevates the need to alleviate the disproportionate pollution burden on non-white communities in California, while also recognizing the benefits of policies that reduce PM2.5 concentration generally. The two sub-questions on segregation are also worth 15 points in total. We weighted them to prioritize policies that break down the systemic foundation of residential segregation, rather than just providing means to mitigate its harms without confronting the root cause.
5.3: Socio-Political Feasibility weight explanations These sub-questions form a broader evaluation of how likely a policy is to be officially adopted. We included the final sub-questions to ensure the criterion pays special attention to the voices of the communities most impacted by a given policy. The weights reinforce the environmental justice tenants about the need to empower communities to make decisions about their futures. We gave a lower weight to the sub-question considering future support because it is both harder to measure with certainty and less 3 Environmental Justice Network, “Principles of Environmental Justice,” accessed April 6, 2024, Link. ADDRESSING THE DISCRIMINATORY IMPACTS OF REDLINING AND HIGHWAY DEVELOPMENT IN CALIFORNIA 81 relevant to the task of recommending policies for immediate implementation. We included the sub-question addressing a policy’s capacity to be adopted without a law or a ballot measure and weighted it highly because bypassing the need to rely on votes greatly increases the political feasibility of that policy being adopted.
5.4: Efficacy weight explanations Together, these sub-questions evaluate the likelihood of a policy producing its intended effect. Cost-effectiveness is the most important consideration for this criterion, as a policy cannot be holistically evaluated while ignoring its costs. We also recognized the importance of evaluating the realistic capacity of agencies to achieve the full potential of a policy when putting it into practice, and thus weighted these two sub-questions the highest within the efficacy category. This criterion also considers implementation speed and the potential for roadblocks not otherwise considered to stymie the desired outcomes of a proposed policy.
5.5: Generalizability weight explanations This criterion is important to consider how applicable a given policy may be across widely differing jurisdictions within the diverse State of California. Additionally, while this study and its policy recommendations are unique to California, the federal policy program of redlining existed across the United States and likely produced similar environmental injustices throughout the more than 400 cities assessed by HOLC. Therefore, it is equally important for this criterion to consider the potential added benefits of a policy that could also be exported beyond California to advance environmental and racial justice nationally.
ADDRESSING THE DISCRIMINATORY IMPACTS OF REDLINING AND HIGHWAY DEVELOPMENT IN CALIFORNIA 82 5.6: Criteria Score Coding Breakdown We calculated ranges for what scores constituted a strength, extreme strength, weakness, or extreme weakness within each policy evaluation criterion. These ranges were based on a weighted average of the natural breaks of all the scores we assigned across that evaluation criterion and the even breaks across the full range of possible scores. The exact ranges for each category are listed below: Promotion of Environmental Justice Principles: • Extreme strength: > 19 points • Strength: 17-19 points • Weakness: 10-12.5 points • Extreme weakness: < 10 points Reduction of PM2.5 and/or racial segregation: • Extreme strength: > 17.5 points • Strength: 17.5-15.1 points • Weakness: 11-8 points • Extreme weakness: < 8 points Socio-political feasibility: • Extreme strength: > 16 points • Strength: 14.5-16 points • Weakness: 9-10.4 points • Extreme weakness: < 9 points Efficacy: • Extreme strength: > 12.5 points • Strength: 10.1-12.5 points • Weakness: 6-7.9 points • Extreme weakness: < 6 points Generalizability: • Extreme strength: N/A • Strength: 6 points • Weakness: 3.5-2 points • Extreme weakness: < 2 points 5.7: Full CAM matrixes ADDRESSING THE DISCRIMINATORY IMPACTS OF REDLINING AND HIGHWAY DEVELOPMENT IN CALIFORNIA 83 “Focus Explicitly on Race and Ethnicity” Policy Option Scores ADDRESSING THE DISCRIMINATORY IMPACTS OF REDLINING AND HIGHWAY DEVELOPMENT IN CALIFORNIA 84 “Increase and Improve Community Empowerment and Decision-Making Authority” Policy Option Scores ADDRESSING THE DISCRIMINATORY IMPACTS OF REDLINING AND HIGHWAY DEVELOPMENT IN CALIFORNIA 85 “Use Zoning and Planning Goals” Policy Option Scores ADDRESSING THE DISCRIMINATORY IMPACTS OF REDLINING AND HIGHWAY DEVELOPMENT IN CALIFORNIA 86 “Make Data More Accessible” Policy Option Scores ADDRESSING THE DISCRIMINATORY IMPACTS OF REDLINING AND HIGHWAY DEVELOPMENT IN CALIFORNIA 87 5.8 Sensitivity Tests To conduct our sensitivity test, we converted our category points to weights (ex: 30 points = .30 category weight). We then multiplied the points received in each category by the category weight to get its total score (ex: if a policy option received 15 out of 30 points in a category, its total score is .50). We increased and decreased each category’s weight by .04 and .08 to see if that changed our recommendation outcome. A change of .04 was selected to easily offset the change of one weight across the remaining criteria. A change of .08 was selected to investigate if a larger impact was needed to change our policy recommendations. As a note, criteria 5 had an original weight of .06 so we only subtracted .06 for that test and redistributed .15 to the other criteria instead of .20. Remember - we recommended policy options with a total score of .50 or higher.
Our top scoring policy options in each category remained the same after all sensitivity tests. The policy options we did not recommend never made it above a .50 threshold, and were never recommended on any of our sensitivity tests.
Policy Option #11 failed to reach the threshold twice: when the weight for criteria 2 was increased by .08 and when the weight for criteria 5 was decreased by .06.
Policy Option #12 failed to reach the threshold a total of ten times: when the weights for criteria 1 and 4 were increased by .04 and when the weights for criteria 3, 4, and 5 were decreased by .04. Similarly, it failed to reach the threshold for recommendation when the weights for criteria 1 and 4 were increased by .08, when the weights for criteria 3 and 4 were decreased by .08, and when the weight for criteria 5 was decreased by .06.
Although Policy Option #12 has a total score of .50 (50 points) on our initial evaluation, it failed the sensitivity test and was therefore not included in our final recommendation. Our sensitivity test results are in the table below. ADDRESSING THE DISCRIMINATORY IMPACTS OF REDLINING AND HIGHWAY DEVELOPMENT IN CALIFORNIA 88 How many times was the recommendation reversed?
Policy Option # Original Total Score First Round Sensitivity Test (+ or - .04 pts) Second Round Sensitivity Test (+ or - .08 pts) 1 0.55 0 0 2 0.58 0 0 3 0.46 0 0 4 0.38 0 0 5 0.57 0 0 6 0.555 0 0 7 0.315 0 0 8 0.58 0 0 9 0.585 0 0 10 0.665 0 0 11 0.525 0 2 12 0.5 5 5 13 0.44 0 0 Total 8 recommended 10 tests 10 tests The table above illustrates the number of times a Policy Option recommendation status was changed in response to a sensitivity test. Column 1 is the Policy Option number. Column 2 is the original score of the Policy Option. Column 3 designates the number of times a Policy Option did not reach the policy option cutoff when the weights were changed by .04. Column 4 designates the number of times a Policy option did not reach the policy option cutoff when the weights were changed by 0.80.
Criteria 5 was subtracted by .06 To ensure that our results were significant, we conducted a binomial probability calculation using ChatGPT for Policy Options #11 and #12: P(X = 18) = 190 × (0.5)18 × (1 – 0.5)2 ≈ 0.000181 We found that there is a .018% chance that Policy Option #11 would fail two of our 20 significance tests if the true outcomes of recommending and rejecting the policy were equally likely. This means we can be >99.9% confident in recommending it.
P(X = 10) = 184,756 × (0.5)10 × (1 – 0.5)10 ≈ 0.176 There is a 17.6% chance that Policy Option #12 would fail 10 of the 20 significance if the true outcomes of recommending and rejecting the policy were equally likely. Because of this, we are not confident enough to recommend Policy Option #12 as a priority at this time.
ADDRESSING THE DISCRIMINATORY IMPACTS OF REDLINING AND HIGHWAY DEVELOPMENT IN CALIFORNIA 89 |
189001 | https://artofproblemsolving.com/wiki/index.php/Titu%27s_Lemma?srsltid=AfmBOopKXEJTq0tamUhDVzDCII8cfPpz5pzwLNPT5TIagw4eIx5eBc5e | Art of Problem Solving
Titu's Lemma - 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 Titu's Lemma
Page
ArticleDiscussionView sourceHistory
Toolbox
Recent changesRandom pageHelpWhat links hereSpecial pages
Search
Titu's Lemma
Titu's lemma states that:
It is a direct consequence of Cauchy-Schwarz inequality.
Equality holds when for .
Titu's lemma is named after Titu Andreescu and is also known as T2 lemma, Engel's form, or Sedrakyan's inequality.
Contents
[hide]
1 Examples
1.1 Example 1
1.1.1 Solution
1.2 Example 2
1.2.1 Solution
1.3 Example 3
1.3.1 Solution
2 Problems
2.1 Introductory
2.2 Intermediate
2.3 Olympiad
Examples
Example 1
Given that positive reals , , and are subject to , find the minimum value of . (Source: cxsmi)
Solution
This is a somewhat standard application of Titu's lemma. Notice that When solving problems with Titu's lemma, the goal is to get perfect squares in the numerator. Now, we can apply the lemma.
Example 2
Prove Nesbitt's Inequality.
Solution
For reference, Nesbitt's Inequality states that for positive reals , , and , We rewrite as follows. This is the application of Titu's lemma. This step follows from .
Example 3
Let , , , , , , , be positive real numbers such that . Show that (Source)
Solution
By Titu's Lemma,
Problems
Introductory
There exists a smallest possible integer such that for all real sequences . Find the sum of the digits of . (Source)
Intermediate
Prove that, for all positive real numbers (Source)
Olympiad
Let be positive real numbers such that . Prove that (Source)
Let be positive real numbers such that . Prove that
(Source)
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. |
189002 | https://mathmistakes.info/mistakes/calculus/Examples/2/ccm.html | Common Calculus Mistakes Example: Tangent Lines
mathmistakes.infoReal Mistakes from Real Student Work and Math Facts Students Should KnowYour Resource for Stronger Math Skills
Home
Calculus Mistakes
This Mistake
Plain Web Page
PDF File
Other Mistakes
Algebra Mistakes
Trigonometry Mistakes
Facts
Algebra Facts
Trigonometry Facts
Calculus Facts
Community/Services
Blog
Discussion
Newsletters
Multimedia
Products
About this Site
Links
Browser Settings
Terms of Use
Privacy Policy
Help this Site
Special Offers
Common Calculus Mistakes
Example: Tangent Lines
The Goal
Find the equation of the tangent line to
at x = 2.
The Mistake
Find the mistake:
(Roll the mouse over the math to see a hint in red)
The Correction
(Roll the mouse over the area above to see the correction in blue)
An Explanation
The derivative of a function f(x) provides a formula f'(x) from which the slope of a tangent line can be computed at a point x=a by evaluating f'(x) at x=a, that is, by computing m=f'(a). The result is a number, which then can be used to find the equation of the tangent line at x=a using the point-slope form for a line, in this situation: y - f(a) = m(x-a).
(Note that in the mistaken solution the proposed tangent line equation is not the equation of a line.)
Home Page | Common Calculus Mistakes | Privacy Policy
Copyright © 2006-2019 Russell Blyth. All rights reserved. |
189003 | https://ocw.mit.edu/courses/18-01sc-single-variable-calculus-fall-2010/resources/mit18_01scf10_ses98a/ | �
�
Introduction to Taylor Series
Why are we looking at power series? If we reverse the equation for the geometric
series: 1
1 + x + x2 + x3 + =· · · 1 − x
we get a description of 1−
1
x
in terms of a series. In fact, we can represent all of
the functions we’ve encountered in this course in terms of series.
The technique is similar to the use of a decimal expansion to represent 1/3
x
or √2. When we describe a function like e or arctan x in terms of a series we
can approximate and manipulate those functions as easily as we do polynomial
functions.
Rules for convergent power series
What sorts of manipulations might we want to perform? Addition, multiplica tion, division, substitution (composition), integration and differentiation. The
rules for manipulation of power series are essentially the same as those for ma nipulating polynomials!
d
f (x) + g(x), f (x) g(x), f (g(x)) , f (x)/g (x), f (x), f (x)dx · dx
We can do all of these with power series; in this class integration and differen tiation will be the most interesting manipulations.
We take the derivative of a power series just as we do for polynomials:
d (a0 + a1x + a2x2 + a3x3 + ) = a1 + 2 a2x + 3 a3x2 +
dx · · · · · ·
Similarly, the formula for the integral of a power series is:
234
(a0 + a1x + a2x2 + a3x3 + )dx = c + a0x + a1x + a2x + a3x +· · · 2 3 4 · · ·
Here the arbitrary constant c takes the place of the constant term in the new
series.
Question: Is that a series or a polynomial?
Answer: It’s a polynomial if it ends; if it goes on infinitely far then it’s a
series.
Question: You can add up terms of x in a series?
Answer: When we introduced series we described them as infinite sums of
numbers . At the start of this class we rewrote the geometric series using the
variable x in place of the “constant value” a. When we plug in a value for x we
get a sum of infinitely many numbers, so as long as we remember that x is a
placeholder for a numerical value there’s no problem.
In other words, what we’re working with here are functions of x. These func tions are defined for values of x inside the radius of convergence and undefined
1for values of x that are too large, just as the function f (x) = √x is defined for
positive values of x and undefined for x < 0.
Power series are infinite sums of powers of x, with coefficients. People also
study and use series that are infinite sums of sines and cosines and lots of other
series, but we’re only going to study power series here.
2MIT OpenCourseWare
18.01SC Single Variable Calculus�� Fall 2010 ��
For information about citing these materials or our Terms of Use, visit: . |
189004 | https://artofproblemsolving.com/wiki/index.php/Quadratic_formula?srsltid=AfmBOoqRd0eAlFTYf-190wtT_Fzw9CQhz2t2HY26BYGAck9M8PeZu_PN | Art of Problem Solving
Quadratic formula - 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 Quadratic formula
Page
ArticleDiscussionView sourceHistory
Toolbox
Recent changesRandom pageHelpWhat links hereSpecial pages
Search
Quadratic formula
The quadratic formula is a general expression for the solutions to a quadratic equation. It is used when other methods, such as completing the square, factoring, and square root property do not work or are too tedious.
Contents
[hide]
1 Statement
1.1 Proof
1.2 Variation
2 See Also
Statement
For any quadratic equation , the following equation holds.
Proof
We start with
Dividing by , we get
Add to both sides in order to complete the square:
Complete the square:
Move to the other side:
Take the square root of both sides:
Finally, move the to the other side:
This is the quadratic formula, and we are done.
Variation
In some situations, it is preferable to use this variation of the quadratic formula:
See Also
Quadratic equation
This article is a stub. Help us out by expanding it.
Retrieved from "
Categories:
Algebra
Quadratic equations
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. |
189005 | https://www.ck12.org/flexi/geometry/classify-polygons/how-do-you-find-a-missing-angle-in-a-polygon/ | How do you find a missing angle in a polygon?
Flexi Says:
To find a missing angle in a polygon, you can use the following steps:
Step 1: Determine the total sum of the interior angles of the polygon.
Step 2: Subtract the measures of the known angles from the total sum. The result will be the measure of the missing angle.
Example:
Calculate the value of @$\begin{align}x\end{align}@$ in the figure.
@$$\begin{align}\text{Sum of interior angles of a pentagon} & = (n-2) \cdot 180^\circ \ x + 106^\circ + 93^\circ + 128^\circ + 76^\circ & = (5-2) \cdot 180^\circ \ x + 403^\circ & = 540^\circ \ x & = 540^\circ - 403^\circ \ x & = 137^\circ\end{align}@$$
Click here to learn more about polygons!
Try Asking:
How many sides does a circle have?What are the twelve different types of polygons?Name a solid shape with 6 equal faces.
By messaging Flexi, you agree to our Terms and Privacy Policy
Image Attribution
| |
| Credit: |
| Source: |
| License: | |
189006 | https://course.ccs.neu.edu/cs5002f18-seattle/lects/cs5002_lect9_fall18_notes.pdf | CS 5002: Discrete Structures Fall 2018 Lecture 9: November 8, 2018 1 Instructors: Adrienne Slaughter, Tamara Bonaci Disclaimer: These notes have not been subjected to the usual scrutiny reserved for formal publications.
They may be distributed outside this class only with the permission of the Instructor.
Introduction to Algorithms Readings for this week: Rosen, Chapter 2.1, 2.2, 2.5, 2.6 Sets, Set Operations, Cardinality of Sets, Matrices 9.1 Overview Objective: Introduce algorithms.
1. Review logarithms 2. Asymptotic analysis 3. Define Algorithm 4. How to express or describe an algorithm 5. Run time, space (Resource usage) 6. Determining Correctness 7. Introduce representative problems 1. foo 9.2 Asymptotic Analysis The goal with asymptotic analysis is to try to find a bound, or an asymptote, of a function. This allows us to come up with an “ordering” of functions, such that one function is definitely bigger than another, in order to compare two functions. We do this by considering the value of the functions as n goes to infinity, so for very large values of n, as opposed to small values of n that may be easy to calculate.
Once we have this ordering, we can introduce some terminology to describe the relationship of two functions.
9-1 9-2 Lecture 9: November 8, 2018 2 3 4 5 6 7 8 1 2 4 8 16 32 64 128 256 512 1024 2048 4096 n!
nn 2n n2 n log(n) log(n) n √n 1 Growth of Functions From this chart, we see: 1 ≪log n ≪√n ≪n ≪n log(n) ≪n2 ≪2n ≪n! ≪nn (9.1) Complexity Terminology Θ(1) Constant Θ(log n) Logarithmic Θ(n) Linear Θ(n log n) Linearithmic Θ nb Polynomial Θ(bn) (where b > 1) Exponential Θ(n!) Factorial 9.2.1 Big-O: Upper Bound Definition 9.1 (Big-O: Upper Bound) f(n) = O(g(n)) means there exists some constant c such that f(n) ≤c · g(n), for large enough n (that is, as n →∞).
We say f(n) = O(g(n)) Example: I claim 3n2 −100n + 6 = O(n2). I can prove this using the definition of big-O: Lecture 9: November 8, 2018 9-3 f(n) = 3n2 −100n + 6 (9.2) g(n) = n2 (9.3) ⇒3n2 −100n + 6 ≤c · n2 for some c (9.4) If c = 3 : 3n2 −100n + 6 ≤3n2 (9.5) To prove using Big-O: • Determine f(n) and g(n) • Write the equation based on the definition • Choose a c such that the equation is true.
– If you can find a d, then f(n) = O(g(n)). If not, then f(n) ̸= O(g(n)).
These statements are all true: 3n2 −100n + 6 = O(n2) (9.6) 3n2 −100n + 6 = O(n3) (9.7) 3n2 −100n + 6 ̸= O(n) (9.8) Proving 9.7: f(n) = 3n2 −100n + 6 (9.9) g(n) = n3 (9.10) ⇒3n2 −100n + 6 = c · n3 (for some c) (9.11) If c = 1 : 3n2 −100n + 6 ≤n3 (when n > 3) (9.12) We also know this to be true because order is transitive: if f(n) = O(g(n)), and g(n) = O(h(n)), then f(n) = O(h(n)). Since n2 = O(n3), then any f(n) = O(n2) is also O(n3).
Proving 9.8: f(n) = 3n2 −100n + 6 (9.13) g(n) = n (9.14) For any c : cn < 3n2 (when n > c) (9.15) 9.2.2 Big-Omega: Lower Bound Definition 9.2 (Big-Omega: Lower Bound) f(n) = Ω(g(n)) means there exists some constant c such that f(n) ≥c · g(n), for large enough n (that is, as n →∞).
We say f(n) = Ω(g(n)) or “f of n is Big Omega of g of n” 9-4 Lecture 9: November 8, 2018 Example: I claim 3n2 −100n + 6 = Ω(n2). I can prove this using the definition of big-Omega: f(n) = 3n2 −100n + 6 (9.16) g(n) = n2 (9.17) ⇒3n2 −100n + 6 ≥c · n2 for some c (9.18) If c = 2 : 3n2 −100n + 6 ≤2n2 (9.19) We show Big-Omega the same way we show Big-O.
These statements are all true: 3n2 −100n + 6 = Ω(n2) (9.20) 3n2 −100n + 6 ̸= Ω(n3) (9.21) 3n2 −100n + 6 = Ω(n) (9.22) Proving 9.21: f(n) = 3n2 −100n + 6 (9.23) g(n) = n3 (9.24) ⇒3n2 −100n + 6 ≥c · n3 (for some c) (9.25) If c = 1 : 3n2 −100n + 6 ≥n3 (when n > 3) (9.26) Proving 9.22: f(n) = 3n2 −100n + 6 (9.27) g(n) = n (9.28) For any c : cn < 3n2 (when n > 100c) (9.29) 9.2.3 Big-Theta: “Tight” Bound Definition 9.3 (Big-Theta: “Tight” Bound) f(n) = Θ(g(n)) means there exists some constants c1 and c2 such that f(n) ≤c1g(n) and f(n) ≥c2g(n).
We say f(n) = Θ(g(n)) or “f of n is Big-Theta of g of n”.
Definition 9.4 (Theta and “order of”) When f(x) = Θ(g(x)), it is the same as saying f(x) is the order of g(x), or that f(x) and g(x) are the same order.
3n2 −100n + 6 = Θ(n2) Both O and Ωapply Lecture 9: November 8, 2018 9-5 3n2 −100n + 6 ̸= Θ(n3) Only O applies 3n2 −100n + 6 ̸= Θ(n) Only Ωapplies Interesting Aside Donald Knuth popularized the use of Big-O notation. It was originally inspired by the use of “ell” numbers, written as L(5), which indicates a number that we don’t know the exact value of, but is less than 5. That allows us to reason about the value without knowing the exact value: we know L(5) < 100, for example.
Theorem 9.5 If f(x) = anxn + an−1xn−1 + · · · + a1x + a0, then f(x) = O(n) a) f(x) = 17x + 11 b) f(x) = x2 + 1000 c) f(x) = x log x d) f(x) = x4/2 e) f(x) = 2x f) f(x) = ⌊x⌋· ⌈x⌉ 9.2.4 Logs, Powers, Exponents We’ve seen f(n) = O(nd). If d > c > 1, then nc = O(nc). nc is O nd , but nd is not O (nc).
logb n is O(n) whenever b > 1. Whenever b > 1, c and d are positive: (logb n)c is O nd , but nd is not (O (logb n)c) (9.30) This tells us that every positive power of the logarithm of n to the base b, where b ¿ 1, is big-O of every positive power of n, but the reverse relationship never holds. In Example 7, we also showed that n is O(2n). More generally, whenever d is positive and b ¿ 1, we have nd is O (bn) , but bn is not O nd (9.31) This tells us that every power of n is big-O of every exponential function of n with a base that is greater than one, but the reverse relationship never holds. Furthermore, we have when c ¿ b ¿ 1, bn is O (cn) but cn is not O (bn) (9.32) This tells us that if we have two exponential functions with different bases greater than one, one of these functions is big-O of the other if and only if its base is smaller or equal.
9.2.5 Adding Functions There are a set of rules that govern combining functions together.
O(f(n)) + O(g(n)) →O(max(f(n), g(n))) (9.33) Ω(f(n)) + Ω(g(n)) →Ω(max(f(n), g(n))) (9.34) Θ(f(n)) + Θ(g(n)) →Θ(max(f(n), g(n)) (9.35) These statements express the notion that the largest term of the statement is the dominant one. For example, n3 + 2n2 + 3 = O(n3).
Example: Prove that n2 = O(2n).
9-6 Lecture 9: November 8, 2018 Example: Prove that if f1(n) = O(g1(n)) and f2(n) = O(g2(n)), then f1(n)+f(n) = O(g1(n)+g2(n)).
Example: f(n) = n + log n (9.36) g(n) = p (n) (9.37) Is f(n) = O(g(n)), g(n) = O(f(n)), or both?
Example: If f(n) = n + log n + √n, find a simple function g such that f(n) = Θ(g(n)).
Lecture 9: November 8, 2018 9-7 Summary • f(n) = O(g(n)) means c · g(n) is an upper bound on f(n). Thus there exists some constant c such that f(n) is always ≤c · g(n), for large enough n (i.e., n ≥n0 for some constant n0).
• f(n) = Ω(g(n)) means c · g(n) is a lower bound on f(n). Thus there exists some constant c such that f(n) is always ≥c · g(n), for all n ≥n0.
• f(n) = Θ(g(n)) means c1 · g(n) is an upper bound on f(n) and c2 · g(n) is a lower bound on f(n), for all n ≥n0. Thus there exist constants c1 and c2 such that f(n) ≤c1 · g(n) and f(n) ≥c2 · g(n).
This means that g(n) provides a nice, tight bound on f(n).
9.2.6 Introduction to Algorithms • An algorithm is a set of instructions for accomplishing a task.
• Technically, any program is an algorithm • We talk about algorithms as general approaches to specific problems • An algorithm is general, but is implemented in code to make it specific Algorithms are like Recipes • If I were to use a simile, I’d say algorithms are like recipes.
• People have been cooking and baking for a looong time – Let’s take advantage of solved problems and use them as starting blocks • There are general approaches to different kinds of foods • Each recipe for a chocolate chip cookie is a little different, but follows the same general structure.
• I can adapt a recipe for chocolate chip cookies to a different kind of cookie if I want.
• I might modify my recipe depending on the context I’m cooking in: cooking for a 200 person formal dinner versus playing around on a Saturday afternoon.
What is an algorithm?
• An algorithm is the part of the “recipe” that stays the same no matter what it’s implemented in or what hardware it’s running on.
• An algorithm solves a general, specified problem • An algorithmic problem is specified by describing: – The set of instances it works on – Desired properties of the output Example: Sorting Input: A sequence of N numbers: n1, n2, n3, . . . , nn Output: The permutation of the input sequence such as n1 ≤n2 ≤n3 . . . ≤nn We look to ensure that an algorithm is: 9-8 Lecture 9: November 8, 2018 • Correct • Efficient in time • Efficient in space The rest of today: • Example algorithms – Binary Search – Selection Sort • Algorithm Analysis – Proving Correctness (briefly) – Run time: How long does it take for an algorithm to run?
– Run space: How much extra memory/storage does an algorithm require?
• Asymptotic Analysis and Growth of Functions 9.3 Some Algorithms 9.3.1 Expressing Algorithms Expressing Algorithms We need some way to express the sequence of steps in an algorithm.
In order of increasing precision: • English • Graphically • Pseudocode • real programming languages (C, Java, Python, etc) Unfortunately, ease of expression moves in the reverse order.
An algorithm is an idea. If the idea is not clear when you express the algorithm, then you are using a too low-level way to express it.
9.3.2 Binary search Searching Input: A set of N values: n1, n2, n3, . . . , nn and a target value t Output: Whether the set contains t Imagine...
A (sub) roster of athletes on the USA Olympic Ski & Snowboard team for 2018: Lecture 9: November 8, 2018 9-9 1 2 3 4 5 6 7 8 Andy Newell Bryan Fletcher Chloe Kim Jessie Diggins Lindsey Vonn Sadie Bjornsen Sophie Caldwell Taylor Fletcher Is Chloe Kim on the US Ski & Snowboard team for 2018?
Is Chloe Kim on the US Ski-Snowboard team for 2018?
Let’s make this a little more complicated...
Assume I have 1,000,000 athletes. How do I answer this question?
OR: Maybe I can’t actually see the list here in its entirety.
How do I search?
A slight aside...
Consider a dictionary (the book kind!) You want to look up a word. First, you open up to the middle.
If you’ve gone too far, you split the first half of the dictionary; if you haven’t gone far enough, you split the second half of the dictionary.
Binary Search A: Array to search I: Item to find min: starting point max: end point Binary-Search(A, I, min, max) 1 if (min == max) 2 return false 3 mid = min + ⌊(max −min)/2⌋ 4 if (A[mid] == I) 5 return true 6 if (A[mid] < I) 7 Binary-Search(A, I, mid, max) 8 else 9 Binary-Search(A, I, min, mid) 1 Binary-Search(A, I, 1, numElems) What’s interesting about Binary Search?
• It’s recursive – It’s defined in terms of itself • Each time we call Binary-Search, we are searching on only half the size of the input – Since we know what the middle element is, we know whether our final element is before or after that one, so can discard half the array with each comparison!
9-10 Lecture 9: November 8, 2018 Quick Review 1. I stated the problem 2. I described a solution in English (using a metaphor) 3. I described the solution with psuedocode 4. I provided a graphical solution 9.3.3 Selection sort Selection Sort Input: A sequence A of n numbers: n1, n2, n3, . . . , nn, and empty sequence B Output: The input sequence such that B contains the elements of A ordered such that n1 ≤n2 ≤ n3 . . . ≤nn Selection-Sort(A, B) 1 for i = 1 to A.length 2 min ind = 0 3 for j = 1 to A.length 4 if A[j] < A[min ind] 5 min ind = j 6 B[i] = A[min ind] 7 A[min ind] = Inf The C code 1 void selection_sort(int a[], int b[], int len){ 2 3 for (int i=0; i<len; i++){ 4 int min_ind = 0; 5 for (int j=0; j<len; j++){ 6 if (a[j] < a[min_ind]){ 7 min_ind = j; 8 } 9 } 10 b[i] = a[min_ind]; 11 a[min_ind] = 10000; //sentinel val 12 } 13 } whew What did we just do?
• Expressing Algorithms – English – Psuedocode – Programming Language (C) – Graphically!
• Binary Search– a first search algorithm • Selection sort– a first sort algorithm • Next up: Analyzing the algorithms 9.4 Analysis What is Algorithm Analysis?
When we analyze algorithms, we are analyzing for 3 things: Lecture 9: November 8, 2018 9-11 1. Correctness 2. Run time 3. Run space 9.4.1 Correctness Proving Correctness How to prove that an algorithm is correct?
For any algorithm, we must prove that it always returns the desired output for all legal instances of the problem.
For sorting, this means even if the input is already sorted or it contains repeated elements.
Proof by: • Induction • Counterexample • Loop Invariant Proof by Counterexample Searching for counterexamples is the best way to disprove the cor-rectness of a heuristic.
• Think about small examples • Think about examples on or around your decision points • Think about extreme examples (big or small) Proof by Induction Failure to find a counterexample to a given algorithm does not mean “it is obvious” that the algorithm is correct.
Mathematical induction is a very useful method for proving the correctness of recursive algorithms.
1. Prove base case 2. Assume true for arbitrary value n 3. Prove true for case n + 1 Proof by Loop Invariant Built offproof by induction.
Useful for algorithms that loop.
1. Find p, a loop invariant 2. Show the base case for p 3. Use induction to show the rest.
9.4.2 Run time What is run time?
The amount of time it takes for an algorithm to run, in terms of the size of the input n.
A faster algorithm running on a slower computer will always win for sufficiently large instances.
Usually, problems don’t have to get that large before the faster algorithm wins.
This is where Big-O comes in.
Essentially, the number of lines of code that are run.
What is run time?
9-12 Lecture 9: November 8, 2018 • Best Case – Given an optimal input and all the best decisions that can be made, how many steps until the algorithm terminates?
– In a sort problem, it’s usually a sorted input.
• Worst Case – Given the worst possible input and all the worst decisions that can be made, how many steps until the algorithm terminates?
– In a search problem, it’s usually the last item looked at.
• Average Case – Somewhere between the two; frequently an averaging of best & worst.
Run Time of Selection Sort Selection-Sort(A, B) 1 for i = 1 to A.length.
/ / n 2 min ind = −1 / / 1 3 for j = 1 to A.length / / n 4 if A[j] < A[min ind] / / 1 5 min ind = j / / 1 6 B[i] = A[min ind] / / 1 7 A[min ind] = Inf / / 1 TODO: Show run time in terms of a sum?
Pn i=1(3 + Pn j=1 2) Run time: n · (n + 1 + 1 + 1 + 1 + 1) ⇒O(n2) Is this best, worst or average?
Run Time of Selection Sort Best case: {1, 2, 3, 4, 5, 6} (It’s already sorted) Worst case: {6, 5, 4, 3, 2, 1 } (It’s reverse sorted) Average case: {1, 6, 4, 5, 2, 3} (It’s a little of this and that) Actually, for Selection Sort, there’s no difference in run time for Best/Worst/Average case.
In all cases, we still iterate through all the elements.
⇒O(n2) Run Time of Binary Search Lecture 9: November 8, 2018 9-13 Binary-Search(A, I, min, max) 1 if (min == max) 2 return false 3 mid = min + ⌊(max −min)/2⌋ 4 if (A[mid] == I) 5 return true 6 if (A[mid] < I) 7 Binary-Search(A, I, mid, max) 8 else 9 Binary-Search(A, I, min, mid) Best case: Binary-Search({1, 2, 3}, 2, 1, 3) Worst case: Binary-Search({1, 2, 3, 4, 5, 6, 7, 9, 10}, 2, 1, 10) Run Time of Binary Search Binary-Search(A, I, min, max) 1 if (min == max) 2 return false / / 1 3 mid = min + ⌊(max −min)/2⌋ / / 1 4 if (A[mid] == I) 5 return true / / 1 6 if (A[mid] < I) 7 Binary-Search(A, I, mid, max) / / R(n/2) 8 else 9 Binary-Search(A, I, min, mid) / / R(n/2) R(n) = 1 + 1 + 1 + R(n/2) R(n) = O(lg n) Run Time of Binary Search Best Case: O(1) Worst Case: O(lg n) Average Case: O(lg n) Example: Given the following algorithm: PrintFoobar(n) 1 for i = 1 to n/2 2 for j = i to n −1 3 for k = 1 to j 4 Print(’foobar’) Assume n is even. Let T(n) denote the number of times ‘foobar’ is printed as a function of n.
9-14 Lecture 9: November 8, 2018 • Express T(n) as three nested summations.
• Simplify the summation.
Run time as Clock time So far we’ve focused on counting “number of instructions” as a proxy for measuring the “clock time” (that is, number of seconds) that an algorithm will run. However, we can use the number of instructions as a tool to help us figure out clock time, when we have some specifics.
Number of Instructions Instructions Per Second = Number of Seconds (9.38) The number of instructions is measured in terms of n, the size of the input to the algorithm. While (for the most part) the number of instructions is about the same from machine to machine, what varies is the number of instructions per second that are run. This gives us 3 variables: number of instructions, instructions per second and number of seconds to run the algorithm. Therefore, if we know 2 we can calculate the third.
Here’s an example: Let’s say I’ve implemented Selection-Sort on my MacBook Pro. I know Selection-Sort takes n2 instructions to run. I choose to run it on an input length of 10,000 items. It takes 2 clock seconds to run (this is a number I’m choosing for illustration purposes; that’s way too long!).
Number of Instructions Instructions per Second = Number of Seconds (9.39) n2 Instructions Per Second = 2 seconds (9.40) n = 10, 000 : 10, 0002 x Instructions per Second = 2 seconds (9.41) 10, 0002 instructions 2 seconds = x = 50 MIPS2 (9.42) 2Millions of Instructions per Second. For reference, the iPhone 6 was probably around 25,000 MIPS (in 2014).
Lecture 9: November 8, 2018 9-15 Thus, if we know our algorithm has n2 instructions, and we measure that it takes 2 seconds to run on our machine with an input of 10,000 items, then our machine runs at about 50 MIPS.
Further, now that we know our machine runs at 50 MIPS, we can use that to estimate how long it will take to run a different algorithm (that is, different run time) or different input size.
Let’s say we have one million items as input to the same algorithm: n = 1, 000, 000 : (1, 000, 000)2 50 MIPS =?
(9.43) (1, 000, 000)2 50, 000, 000 = 20, 000 seconds (9.44) 9.4.2.1 Runtime, Clocktime, and Efficiency Let’s take two relatively recent machines. One is powered by the Intel Core i7 500U which runs at 49,360 (roughly 50K MIPS), and the other is an Intel Core i7 2600K at 117,160 MIPS.
Runtime Size of Input Intel i7 A Intel i7 B n 1,000 1000 50,000 MIPS = 0.02µsec3 1000 117,000 MIPS = 0.009µsec linear 10,000 10000 50,000 MIPS = 0.2µsecs 1,000,000 1,000,000 50,000 MIPS = 20µsecs 1,000,000 117,000 MIPS = 8.5µsecs n log n 1000 1000 log 1000 50,000 MIPS = 3000 50,000 MIPS = 0.06µsec 3000 117,000 MIPS = 0.03µsec logarithmic 10000 1,000,000 1000000 log 1000000 50,000 MIPS = 6,000,000 50,000 MIPS = 120µsec 6,000,000 117,000 MIPS = 51.3µsec n2 1000 10002 50,000 MIPS = 20µsecs 10002 117,000 MIPS = 8.5µsecs quadratic/ 10000 1,000,000 cn 50 250 50,000 MIPS = 1,125,899,906,842,624 50,000,000,000 = 22518µsecs 250 117,000 MIPS = 9623µsecs4 exponentia 10000 1,000,000 Run Time, Summary • We count up the number of statements that are run.
• Consider whether there’s a difference in how long it takes a function to run given different inputs.
• Selection sort is O(n2), pretty much all the time.
• Binary search can be either O(1) or O(lg n), depending on what the input looks like.
9.4.3 Memory Use How much memory?
Another resource that we sometimes care about is the amount of memory it takes to run an algorithm.
Sometimes this is total, sometimes it’s just the amount in addition to the input.
Binary Search: Needs nothing, so memory is O(1) (constant— nothing other than the input.
41µsec = 10−6 seconds 4= 1,125,899,906,842,624 117,000,000,000 9-16 Lecture 9: November 8, 2018 Selection sort: Needs a whole other array! O(n) Note: this is just how it was implemented here, in this discussion. If we chose not to use that other array, it wouldn’t be O(n).
9.5 Representative Problems Stable Matching • Gale-Shapley • The problem of: matching residents with med schools • Each resident has a prioritized list of schools they want to go to • Each school has a prioritized list of students they want to accept • A stable match is one where if either the school or the student is offered another match, they won’t change.
Interval Scheduling The Problem: We have a resource r, such as a classroom, and a bunch of requests q : {start, finish}.
How can we schedule the requests to use the resource?
• We want to identify a set S of requests such that no requests overlap.
• Ideally, the S that we find contains the maximum number of requests.
In this diagram, we see three sets of requests.
Which set of requests is the preferred choice for the interval scheduling problem as defined?
time Solution: A simple heuristic that is an example of a greedy algorithm.
Weighted Interval Scheduling The Problem: Same as interval scheduling, but this time, the request has a weight.
• The weight may be how much we’ll earn by satisfying this request • Find a subset that maximizes the weights • This is very similar to the problem we just saw, but these weights cause a problem.
• Consider: If all requests except one have weight = 1, and one has weight greater than the sum of all the others.
Solution: An approach called dynamic programming, where we calculate the weight of each subset and use that to find the best set overall.
Bipartite Matching The Problem: We have two groups of objects that we need to match, or assign to another object.
• An example: matching residents with med schools • Each resident has a prioritized list of schools they want to go to • Each school has a prioritized list of students they want to accept • A stable match is one where if either the school or the student is offered another match, they won’t change.
An Aside: Bipartite Graph Lecture 9: November 8, 2018 9-17 • A graph G = (V, E) is bipartite if the nodes V can be partitioned into sets X and Y in such a way that every edge has one end in X and the other in Y .
• It’s just a graph, but we tend to depict bipartite graphs in two columns to emphasize the bipartite-ness.
1 2 3 4 5 6 1 2 3 4 5 6 Bipartite Matching • Bipartite Matching is relevant when we want to match one set of things to another set of things.
– Nodes could be Jobs and Machines; Edges indicate that a given machine can do the job.
– Nodes could be men and women; Edges indicate that a given man is married to a given woman.
(Okay, in the real world it’s more complex, but this is a classic “problem” I feel required to present...) Solution: use backtracking and augmentation to solve the problem, which contributes to network flow problems Independent Set Independent Set is a very general problem: • Given a graph G = (V, E), a set of nodes S ⊆V is independent if no two nodes in S are joined by an edge.
• Goal: Find an independent set that is as large as possible.
• Applicable to any problem where you are choosing a collection of objects and there are pairwise conflicts.
1 2 3 4 5 6 7 In this graph, the largest independent set is {1, 4, 5, 6} Independent Set 9-18 Lecture 9: November 8, 2018 • Example: Each node is a friend, and each edge indicates a conflict between those two friends. Use Independent Set to find the largest group of people you can invite to a party with no conflicts.
• Interval Scheduling is a special case of Independent Set: – Define graph G = (V, E) where V is the set of requests or intervals, and E is the set of edges that indicate conflicts between two requests.
• Bipartite Matching is a special case of Independent Set: – A little more complex than I want to explain in class; see the book and we’ll cover it later.
• Solution: No efficient algorithm is known to solve this problem.
• However: If we’re given an independent set for a given graph G, we can easily check that it’s a correct answer.
Competitive Facility Location This time, we have a two-player game.
• Dunkin Donuts puts a cafe at one location.
• Then Starbucks does.
• BUT! Caf es can’t be too close (zoning requirement) • Goal: Make your shops in the most convenient locations as possible.
• Model the problem: – Consider each location as a zone (rather than a point) that has an estimated value or revenue.
– Model the problem as a graph: G = (V, E) where V is the set zones as noted above, and E represents whether two zones are adjacent.
– The zoning requirement says that the set of cafes is an independent set in G.
10 1 15 5 1 15 • Can’t put a caf e in adjacent zones • Can’t put two cafes in one zone • Edges indicate two zones are adjacent • The set of caf es opened must be an independent set.
Competitive Facility Location • Another question: Can we find a strategy such that Starbucks, no matter where Dunkin Donuts opens a cafe, can open caf es in locations with a total value of at least B?
• Even if I could give you a strategy, you’d have a hard time believing that the strategy is correct.
• This is in contrast to Independent Set!
• This problem is what we call a PSPACE-complete problem • Independent Set is a NP complete problem Representative Problems: Summary Interval Scheduling Solved easily with a greedy algorithm Weighted Interval Scheduling Solved with dynamic programming Bipartite Matching Solved with backtracking and augmentation Independent Set No efficient approach to generate a solution, but it’s easy to check a given solution Competitive Facility Location No easy way to generate a solution, and NO EASY WAY to check a given solution What is efficient?
• In general, it’s pretty easy to come up with a brute force solution to a problem.
• For example, we can generate all possible solutions for a problem, and then check which one is correct (or acceptable).
• Definition attempt 1: An algorithm is efficient if it achieves qualitatively better worst-case perfor-mance, at an analytical level, than brute-force search.
Lecture 9: November 8, 2018 9-19 • What’s “qualitatively better”?
• Final definition: An algorithm is efficient if it has a polynomial running time.
Why does this matter?
• Algorithms are important – Many performance gains outstrip Moore’s law: We can’t always just throw hardware at the problem.
• Simple problems can be hard – Factoring, TSP • Simple ideas don’t always work – Nearest neighbor, closest pair heuristics • Simple algorithms can be very slow – Brute-force factoring, TSP • Changing your objective can be good – Guaranteed approximation for TSP • And: for some problems, even the best algorithms are slow Readings for NEXT week: Rosen, Chapter 4.1, 4.2, 4.3, 4.4 Divisibility and Modular Arithmetic, Integer Representations and Algorithms, Primes and Greatest Common Divisors, Solving Congruences Solving Congruences 9-20 Lecture 9: November 8, 2018 9.6 Appendix: Logarithms 9.7 Logarithms 9.7.1 Definition What is a logarithm?
bx = y ⇕ logb(y) = x We say “log base b of y equals x” Once again, b is called the base x is called the exponent Some Practice log7(49) =?
⇒7? = 49 ⇒72 = 49 ⇒log7(49) = 2 Let’s re-write this using the formula we have.
That let’s us change the question to “49 is the 7 raised to what power?” Or, “What is the exponent?” 9.7.2 Properties Special Logs • Base b = 2: binary logarithm, also referred to as lg x • Base b = e: natural logarithm, also referred to ln x, where e = 2.718.....
– The inverse of ln x is exp(x) = ex ⇒exp(ln x) = x • Base b = 10: The common logarithm, also referred to as log x.
• If it’s not one of these, the base is specified.
Restrictions logb(a) is only defined when b > 1 and a > 0.
Practice: Use what you know about exponents to convince yourself why this is true.
The Product Rule loga(xy) = loga(x) + loga(y) Lecture 9: November 8, 2018 9-21 The logarithm of a product is the sum of the logs of its factors.
The Quotient Rule loga x y = loga(x) −loga(y) The logarithm of a quotient is the difference of the logs of its factors.
The Power Rule loga(xy) = y loga(x) When the term of a logarithm has an exponent, it can be pulled out in front of the log.
Change of Base Rule loga b = logc b logc a 9.7.3 Logarithm Exercises 2. solve these.
(a) 3+log7 x 2−log7 x = 4 x > 0 3 + log7 x = 8 −4 log7 x 5 log7 x = 5 log7 x = 1 x = 71 = 7 k = {7} (9.45) (b) 5 + log x 3 −log x x > 0 2= 8 x > 0 5+log x 3−log x = 3 5 + log x = 9 −3 log x 4 log x = 4 log x = 1 x = 101 = 10 K = {10} (9.46) 9-22 Lecture 9: November 8, 2018 (c) log4 x2 −9 −log4(x + 3) = 3 llog4 x2 −9 −log4(x + 3) = 3 x > 3 ∧x > −3 ⇒x ∈(3; 00) log4 x2 −9 −log4(x + 3) = log4 64 log4 x2 −9 x + 3 = log4 64 x2 −9 x + 3 = 64 (x −3)(x + 3) x + 3 = 64 x = 67 ∈(3, 00) x = {67} Summary The Product Rule loga(xy) = loga(x) + loga(y) The Quotient Rule loga x y = loga(x) −loga(y) The Power Rule loga(xy) = y loga(x) The Change of Base Rule loga b = logc b logc a 9.7.4 Relevance of Logs 9.7.4.1 Logs and Binary Search Logs and Binary Search TODO: Something about in asymptotic analysis log base 10 is equivalent to log base 3 or whatever.
Binary search is O(log n).
Given a telephone book of n names: • Looking for person p • Compare p to the person in the middle, or the n 2 nd name • After one comparison, you discard 1 2 of the names in this book.
• The number of steps the algorithm takes = the number of times we can halve n until only one name is left.
= ⇒log2 n comparisons • In this case, x =? , y = n, and b = 2 9.7.4.2 Logs and Trees Logs and Trees A binary tree of height 2 can have up to 4 leaves: Lecture 9: November 8, 2018 9-23 What is the height h of a binary tree with n leaf nodes?
For n leaves, n = 2h ⇒h = log2 n 9.7.4.3 Logs and Bits Logs and Bits Let’s say we have 2 bit patterns of length 1 (0 and 1), and 4 bit patterns of length 2 (00, 01, 10, 11).
How many bits w do we need to represent any one of n different possibilities, either one of n items, or integers from 1 to n?
• There are at least n different bit patterns of length w • We need at least w bits where 2w = n ⇒w = log2 n bits Takeaway Logs arise whenever things are repeatedly halved or doubled.
9-24 Lecture 9: November 8, 2018 9.8 Appendix: Selection Sort, graphically 2 3 7 9 2 4 8 5 1 6 A B |
189007 | https://artofproblemsolving.com/wiki/index.php/Shoelace_Theorem?srsltid=AfmBOoo5HfltIprvdENiMQOajJiqLH4FWDIXP-p8bpiX1za6RJFBI06- | Art of Problem Solving
Shoelace 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 Shoelace Theorem
Page
ArticleDiscussionView sourceHistory
Toolbox
Recent changesRandom pageHelpWhat links hereSpecial pages
Search
Shoelace Theorem
The Shoelace Theorem is a nifty formula for finding the area of a simple polygon given the coordinates of its vertices.
Contents
1 Theorem
2 Other Forms
3 Proof 1
3.1 Proof of claim 1:
3.2 Proof:
4 Proof 2
5 Proof 3
6 Problems
6.1 Introductory
6.2 Exploratory
7 External Links
Theorem
Suppose the polygon has vertices , , ... , , listed in clockwise order. Then the area () of is
You can also go counterclockwise order, as long as you find the absolute value of the answer.
The Shoelace Theorem gets its name because if one lists the coordinates in a column, and marks the pairs of coordinates to be multiplied, the resulting image looks like laced-up shoes.
Other Forms
This can also be written in form of a summation or in terms of determinants as which is useful in the variant of the Shoelace theorem. Note here that and .
The formula may also be considered a special case of Green's Theorem
where and so .
Proof 1
Claim 1: The area of a triangle with coordinates , , and is .
Proof of claim 1:
Writing the coordinates in 3D and translating so that we get the new coordinates , , and . Now if we let and then by definition of the cross product .
)
Proof:
We will proceed with induction.
By claim 1, the shoelace theorem holds for any triangle. We will show that if it is true for some polygon then it is also true for .
We cut into two polygons, and . Let the coordinates of point be . Then, applying the shoelace theorem on and we get
Hence
as claimed.
~ShreyJ
Proof 2
Let be the set of points belonging to the polygon. We have that where . The volume form is an exact form since , where Using this substitution, we have Next, we use the Theorem of Stokes to obtain We can write , where is the line segment from to . With this notation, we may write If we substitute for , we obtain If we parameterize, we get Performing the integration, we get More algebra yields the result
Proof 3
This is a very nice approach that directly helps in understanding the sum as terms which are areas of trapezoids.
The proof is in this book:
(The only thing that needs to be slightly modified is that one must shift the entire polygon up by k, until all the y coordinates are positive, but this term gets canceled in the resulting sum.)
Problems
Introductory
In right triangle , we have , , and . Medians and are drawn to sides and , respectively. and intersect at point . Find the area of .
Exploratory
Observe that is the area of a triangle with vertices and is the volume of a tetrahedron with vertices . Does a similar formula hold for Dimensional triangles for any ? If so how can we use this to derive the D Shoelace Formula?
External Links
A good explanation and exploration into why the theorem works by James Tanton:
Nice geometric approach and discussion for proving the 3D Shoelace Theorem by Nicholas Patrick and Nadya Pramita:
Nice integral approach for proving the 3D Shoelace Theorem (ignoring sign of volume) by @george2079: AOPS
Retrieved from "
Categories:
Geometry
Theorems
Mathematics
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. |
189008 | https://mathoverflow.net/questions/288751/union-of-pairwise-almost-disjoint-sets | co.combinatorics - Union of pairwise almost disjoint sets - MathOverflow
Join MathOverflow
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
MathOverflow helpchat
MathOverflow 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
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
Union of pairwise almost disjoint sets
Ask Question
Asked 7 years, 9 months ago
Modified7 years, 8 months ago
Viewed 676 times
This question shows research effort; it is useful and clear
5
Save this question.
Show activity on this post.
I asked this question on stackexchange :
but I got no answer after 24 hours, so I ask it here.
Let r and n be two natural numbers, with n≥2. What is known about the least possible cardinality of the union of r sets with cardinality n, such that any two of these sets have at most one element in common ?
Let f(r,n) denote this least possible cardinality.
If n=2, the condition that two of the sets have at most one common element amounts to say that these sets are distinct, thus f(r,2) is the least natural number k such that (k 2)≥r.
In the general case, each of the r sets contains (n 2) pairs and two sets never contain a same pair, thus the union of the r sets contains at least r(n 2) pairs, thus
(1) f(r,n)≥k(r,n), where k(r,n) denotes the least k such that (k 2)≥r(n 2).
This is not optimal, in the sense that f(r,n)>k(r,n) happens. For example, f(2,3)=5 and k(2,3)=4.
I have two questions :
1° do you know a better minoration of f(r,n) than (1) ?
2° (1) gives f(30,4)≥20; can it be proved that f(30,4)≥21 ?
Thanks in advance.
co.combinatorics
Share
Share a link to this question
Copy linkCC BY-SA 3.0
Cite
Improve this question
Follow
Follow this question to receive notifications
edited Jan 12, 2018 at 9:55
PanurgePanurge
asked Dec 18, 2017 at 12:17
PanurgePanurge
1,225 7 7 silver badges 16 16 bronze badges
0
Add a comment|
4 Answers 4
Sorted by: Reset to default
This answer is useful
5
Save this answer.
Show activity on this post.
Your question is a special case of the set packing problem. My reference for this stuff (which I know nothing about) is A. E. Brouwer, Packing and covering of (k t) sets, in: A. Schrijver (ed.), Packing and Covering in Combinatorics, Mathematical Centre Tracts 106 (1979), 89–97. (This is obviously not the last word on the subject, but for some reason I happen to have a copy.) Paraphrasing the definition from Brouwer:
Let 0≤t≤k≤v, and define D(t,k,v)=max{|B|:B⊂P k(v)and no two elements of B have t points in common}
where P k(v)is the collection of k-subsets of a fixed v-set.
In this notation, your f(r,n) is the least v such that D(2,n,v)≥r.
According to Brouwer's 1979 survey, the exact values of D(2,k,v) are known for k=3 and k=4. For k=4 the results are attributed to A. E. Brouwer, Optimal packings of K 4's into a K n, Math. Centre Report no. zw 92/97, Math. Centre, Amsterdam, 1977; J. Combinatorial Theory (A) 26 (1979), 278–297, and are as follows (paraphrased):
For v≠8,9,10,11,17,19, we have D(2,4,v)=⌊v 4⌊v−1 3⌋⌋−ε,
where ε=1for v≡7or 10(mod 12)and ε=0otherwise;
and for v=8,9,10,11,17,19we have D(2,3,v)=2,3,5,6,20,25.
In particular, D(2,4,19)=25 and D(2,4,20)=30, so f(30,4)=20.
Share
Share a link to this answer
Copy linkCC BY-SA 3.0
Cite
Improve this answer
Follow
Follow this answer to receive notifications
edited Dec 19, 2017 at 5:09
answered Dec 19, 2017 at 0:45
bofbof
15k 2 2 gold badges 48 48 silver badges 72 72 bronze badges
Add a comment|
This answer is useful
4
Save this answer.
Show activity on this post.
You won't get an exact bound, as we do not even know for which values do finite projective planes exist. I think that you should find the least q+1≤n for which (q 2+q+1 2)≥r(q+1 2), build a projective plane on this q (if you can!) and just let the remaining parts of the sets be singletons.
Share
Share a link to this answer
Copy linkCC BY-SA 3.0
Cite
Improve this answer
Follow
Follow this answer to receive notifications
edited Dec 18, 2017 at 12:41
Tony Huynh
32.5k 11 11 gold badges 114 114 silver badges 189 189 bronze badges
answered Dec 18, 2017 at 12:34
domotorpdomotorp
19.6k 3 3 gold badges 62 62 silver badges 134 134 bronze badges
Add a comment|
This answer is useful
4
Save this answer.
Show activity on this post.
As it happens, f(30,4)=20. It is a fairly unique construction based on some almost magic seeming coincidences. You can check that f(4,3)=6. Given a base set S={a,b,c,d,e,f} there are (6 3)=20 triples. I'll write a b c (in any order) for the triple {a,b,c}.
Two systems showing f(4,3)=6 are {a c e,a d f,b c f,b d e} and {a c f,a d e,b c e,b d f}. They come in a certain way from the partition a b|c d|e f into disjoint pairs. In all there are 15 such partitions and 30 such systems. The details are an easy exercise.
So for f(30,4)=20 I take the 20 "points" to be the triples from S and the 30 "lines" to be the quadruples of points forming a system as above for f(4,3)=6. Again, this construction does not generalize to other cases.
I use the language above to help point to the literature. I'll be a little loose with the definitions here and let you check out all the conditions. Say that a design (another name is hypergraph) is a family B of b subsets called blocks from a base set V with v points. I've replaces r with b. I will also replace n with k and call the design k-uniform if all the blocks have size k.
A design, which might or might not be uniform, is called a linear space if every pair of points belongs to a unique block. In that case the blocks are often called lines since two points determine a unique line. A k-uniform linear space is called a Balanced Incomplete Block Design BIBD-(v,b,k,r,1) or a Steiner k system. The r is the constant (in this case) number of blocks containing each point. The parameters are related by v r=b k and v(v−1)=b k(k−1). So given b,k both v and r are determined. I think that for fixed k there is such a system provided b is large enough and the values of v and r are integers.
If there is such a system then f(b,k)=v. There can't quite be one with with b=30,k=4 since that gives v(v−1)=360 but 19⋅18=360−19 and 20⋅19=360+20. However v=20 might be just right if we allow each point to have a unique complement not in any block with it. And that works.
A partial linear space (which might be uniform) is like a linear space except that two points determine at most one line. So you have defined f(b,k) to be the smallest v for which there is a (b,v,k)−partial linear space. Since we can always throw away some blocks, f(b−1,k)≤f(b,k) so I would expect that f(b−j,k)=f(b,k)=v in the event that there is a BIBD-(b,v,k,r,1) and j is small enough.
The example at the top was a rather nice partially balanced incomplete block design with λ 1+λ 2=1. That paper shows a few nice constructions. It is from 1955 so more is known now then then, but it is a good place to look.
Share
Share a link to this answer
Copy linkCC BY-SA 3.0
Cite
Improve this answer
Follow
Follow this answer to receive notifications
edited Jan 12, 2018 at 19:44
answered Dec 18, 2017 at 19:41
Aaron MeyerowitzAaron Meyerowitz
30.3k 2 2 gold badges 50 50 silver badges 105 105 bronze badges
1
It seems that we must avoid comments like "Thanks", but I think I must thank you. Sorry I forgot it.Panurge –Panurge 2018-01-12 10:12:23 +00:00 Commented Jan 12, 2018 at 10:12
Add a comment|
This answer is useful
2
Save this answer.
Show activity on this post.
edit: It seems that you are more interested in the case r>n. I hope that answer for r≤n still helps.
Why not simply construct sets such that the union is minimal?
W.l.o.g., the first set is X 1={1,2,…,n}.
The second set should intersect this set in at most one element. Now, if it doesn't intersect at all, the union will surely be not minimal, so we should have one element in the intersection, wlog this element is 1. Then X 2={1,n+1,n+2,…,2 n−1}. With the same argumentation, X 3 should intersect both X 1 and X 2. But it should not intersect them both in the same element n, as in this case the union will no longer be minimal. Therefore, wlog we can say that X 3={2,n+1,2 n,…3 n−3}.
Continuing like that, it should not be hard to show that f(r,n)=r∑i=1 n−i=n r−(n 2)
as long as r≤n.
Share
Share a link to this answer
Copy linkCC BY-SA 3.0
Cite
Improve this answer
Follow
Follow this answer to receive notifications
edited Dec 18, 2017 at 13:01
answered Dec 18, 2017 at 12:41
DirkDirk
819 4 4 silver badges 9 9 bronze badges
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
co.combinatorics
See similar questions with these tags.
Featured on Meta
Spevacus has joined us as a Community Manager
Introducing a new proactive anti-spam measure
Linked
4Maximum number of subsets of the same size intersecting in at most one point
Related
24Pairwise intersecting sets of fixed size
6Literature about a property of union closed families?
7One more strengthening of Frankl's conjecture
6Union-closed family generated by n 2-sets
2Number of members of a separating union-closed family whose universe has given cardinality
1Intersection of members in a separating union-closed family of sets
1Existence of a family of sets with some properties
2Number of disjoint set triplets in a union-closed family
7Counterexample to a generalization of Frankl's union-closed sets conjecture
Question feed
Subscribe to RSS
Question feed
To subscribe to this RSS feed, copy and paste this URL into your RSS reader.
Why are you flagging this comment?
It contains harassment, bigotry or abuse.
This comment attacks a person or group. Learn more in our Code of Conduct.
It's unfriendly or unkind.
This comment is rude or condescending. Learn more in our Code of Conduct.
Not needed.
This comment is not relevant to the post.
Enter at least 6 characters
Something else.
A problem not listed above. Try to be as specific as possible.
Enter at least 6 characters
Flag comment Cancel
You have 0 flags left today
MathOverflow
Tour
Help
Chat
Contact
Feedback
Company
Stack Overflow
Teams
Advertising
Talent
About
Press
Legal
Privacy Policy
Terms of Service
Your Privacy Choices
Cookie Policy
Stack Exchange Network
Technology
Culture & recreation
Life & arts
Science
Professional
Business
API
Data
Blog
Facebook
Twitter
LinkedIn
Instagram
Site design / logo © 2025 Stack Exchange Inc; user contributions licensed under CC BY-SA. rev 2025.9.26.34547
By clicking “Accept all cookies”, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy.
Accept all cookies Necessary cookies only
Customize settings |
189009 | https://www.bbc.co.uk/bitesize/articles/zvxpscw | KS2
Mental addition and subtraction methods
Part of MathsAdding and subtracting
Save to My Bitesize
Addition and subtraction methods
You can add or subtract numbers in lots of different ways.
You can use number lines, number bonds, fact families, 10 squares and counters, or even just writing the numbers down.
Sometimes it is faster and easier to do calculations in your head. This is called mental calculation.
Back to top
Activity: Addition and subtraction methods
Complete this interactive activity to explore some different methods for addition and subtraction. Then put your knowledge to the test.
Back to top
Making calculations simpler
It can be quicker and easier to do some calculations in your head.
An easy number to work with could be a single digit number, or a multiple of 10, or a multiple of 100.
This is what these types of numbers look like in a place value chart.
Take a look at this slideshow to see how you can use your knowledge of place value to quickly do this calculation:
4384 + 200 = ?
Image gallery
Skip image gallery
1 of 3
Slide 1 of 3, The number sentence 4384 + 200. In a place value chart in the first row there are the values 4 in the thousands column, 3 in the hundreds column, 8 in the tens column and 4 in the ones column. In the second row the thousands column is blank, there is 2 in the hundreds column, 0 in the tens column and 0 in the ones column. A plus symbol is to the left of the chart between the two rows., To solve this problem which place value column will you need to add?
End of image gallery
Back to top
Using the compensation method
Another way to calculate is to use a compensation method. This method helps if a number is close to a multiple of 10 or 100.
Look at this calculation:
4384 + 190 = ?
It would be easier to add 200 to 4384 than 190.
But if you add 200, you need to compensate by taking 10 away at the end.
The calculations to find the answer would look like this:
4384 + 200 = 4584
4584 - 10 = 4574
This tells you that:
4384 + 190 = 4574
This is what the calculation looks like on a number line:
Image gallery
Skip image gallery
1 of 2
Slide 1 of 2, The number sentence 4384 + 190. Below is a number line starting at 4384. Marked with an arrow a short distance before 4584 is 4574. There is one jump above the number line that goes from 4384 to the end of the number line and is labelled plus 200., It is easier to add 200 to a number than it is to add 190.
End of image gallery
Back to top
Counting up to solve subtraction problems
Counting up is a useful method, if you are subtracting two numbers that are close in value to each other.
Look at this calculation:
5125 - 4980 = ?
To solve this you can count up from 4980 to 5000 (which is 20).
Then you can count on from 5000 to 5125 (which is 125).
In total, you will have have counted up 145.
So:
5125 - 4980 = 145
Image gallery
Skip image gallery
1 of 4
Slide 1 of 4, The number sentence 5125 - 4980. Below this is a number line with 4980, 5000 and 5125 marked. There is a jump marked with an arrow going from 4980 to 5000. It is labelled + 20., You can count up from 4980 to 5000 (which is 20).
End of image gallery
Back to top
Example 1
To find out the answer to this calculation you can just add the hundreds, the other values will not change.
What is 8276 + 300?
✓ 8276 + 300 = 8576
To find the answer you just need to add the 300 and not change the other columns.
Back to top
Example 2
In this example you could use a compensation strategy to find the answer.
Which number could you compensate to make it easier?
✓ A good strategy here is to add 300 first.
This number line shows the result if you add 300:
As 300 is 10 too many, you need to subtract 10 at the end.
This is what that calculation looks like on a number line:
The calculations are:
6426 + 300 = 6726
6726 - 10 = 6716
So the answer is 6716.
Back to top
Example 3
The two numbers in this calculation are close together.
What strategy could you use to find the answer?
✓ You could count up from 5993 to 6012.
First, jump from 5993 to 6000. This is a jump of 7.
Next, jump from 6000 to 6012. This is a jump of 12.
Adding those jumps of 7 and 12 together gives you 19.
So:
6012 - 5993 = 19
Back to top
NEW! Play Guardians: Defenders of Mathematica - the Halloween update. game
NEW! Play Guardians: Defenders of Mathematica - the Halloween update
Experience Mathematica as you’ve never seen it before, with all-new backgrounds and costumes for Halloween. Available for a limited time only. Use your maths skills to save the day before it's too late!
Back to top
More on Adding and subtracting
Find out more by working through a topic |
189010 | https://rarediseases.org/rare-diseases/post-polio-syndrome/ | Post Polio Syndrome - Symptoms, Causes, Treatment | NORD
We value your privacy
We use cookies to enhance your browsing experience, serve personalized ads or content, and analyze our traffic. By clicking "Accept All", you consent to our use of cookies.
By continuing to use this website, you agree to the Terms & Conditions and Privacy Policy.
Customize Reject All Accept All
Customize Consent Preferences
We use cookies to help you navigate efficiently and perform certain functions. You will find detailed information about all cookies under each consent category below.
The cookies that are categorized as "Necessary" are stored on your browser as they are essential for enabling the basic functionalities of the site. ...Show more
Necessary Always Active
Necessary cookies are required to enable the basic features of this site, such as providing secure log-in or adjusting your consent preferences. These cookies do not store any personally identifiable data.
Cookie __cf_bm
Duration 1 hour
Description This cookie, set by Cloudflare, is used to support Cloudflare Bot Management.
Cookie AWSALBCORS
Duration 7 days
Description Amazon Web Services set this cookie for load balancing.
Cookie cookielawinfo-checkbox-necessary
Duration 1 year
Description Set by the GDPR Cookie Consent plugin, this cookie records the user consent for the cookies in the "Necessary" category.
Cookie cookielawinfo-checkbox-functional
Duration 1 year
Description The GDPR Cookie Consent plugin sets the cookie to record the user consent for the cookies in the category "Functional".
Cookie cookielawinfo-checkbox-performance
Duration 1 year
Description Set by the GDPR Cookie Consent plugin, this cookie stores the user consent for cookies in the category "Performance".
Cookie cookielawinfo-checkbox-analytics
Duration 1 year
Description Set by the GDPR Cookie Consent plugin, this cookie records the user consent for the cookies in the "Analytics" category.
Cookie cookielawinfo-checkbox-advertisement
Duration 1 year
Description Set by the GDPR Cookie Consent plugin, this cookie records the user consent for the cookies in the "Advertisement" category.
Cookie cookielawinfo-checkbox-others
Duration 1 year
Description Set by the GDPR Cookie Consent plugin, this cookie stores user consent for cookies in the category "Others".
Cookie wpEmojiSettingsSupports
Duration session
Description WordPress sets this cookie when a user interacts with emojis on a WordPress site. It helps determine if the user's browser can display emojis properly.
Cookie __cfruid
Duration session
Description Cloudflare sets this cookie to identify trusted web traffic.
Cookie connect.sid
Duration 1 year 1 month 4 days
Description This cookie is used for authentication and for secure log-in. It registers the log-in information.
Cookie XSRF-TOKEN
Duration 15 minutes
Description Wix set this cookie for security purposes.
Cookie __stripe_mid
Duration 1 year
Description Stripe sets this cookie to process payments.
Cookie __stripe_sid
Duration 1 hour
Description Stripe sets this cookie to process payments.
Cookie _GRECAPTCHA
Duration 3 years
Description Gravity Forms and ReCAPTCHA
Cookie _grecaptcha
Duration 3 years
Description Google ReCAPTCHA
Cookie rc::a
Duration Never Expires
Description This cookie is set by the Google recaptcha service to identify bots to protect the website against malicious spam attacks.
Cookie rc::c
Duration session
Description This cookie is set by the Google recaptcha service to identify bots to protect the website against malicious spam attacks.
Cookie cookieyes-consent
Duration 1 year
Description CookieYes sets this cookie to remember users' consent preferences so that their preferences are respected on subsequent visits to this site. It does not collect or store any personal information about the site visitors.
Cookie PHPSESSID
Duration session
Description This cookie is native to PHP applications. The cookie stores and identifies a user's unique session ID to manage user sessions on the website. The cookie is a session cookie and will be deleted when all the browser windows are closed.
Cookie wordpress_test_cookie
Duration session
Description WordPress sets this cookie to determine whether cookies are enabled on the users' browsers.
Cookie wordpress_
Duration past
Description WordPress sets this cookie when a user logs in to the WordPress administration. It stores the user's authentication details, allowing them to access the backend.
Cookie wordpress_sec_
Duration past
Description WordPress sets this cookie to provide protection against hackers and store account details.
Cookie wordpress_logged_in_
Duration past
Description WordPress sets this to indicate that you are logged in and who you are, for most interface use.
Cookie wp-settings-
Duration past
Description WordPress sets this cookie to preserve the user’s wp-admin settings.
Cookie wp-settings-time-
Duration past
Description WordPress sets this cookie to preserve the user’s wp-admin settings.
Cookie wordpressuser_
Duration past
Description WordPress sets this cookie to authenticate users and maintain their login session. It ensures that logged-in users can navigate the website securely and access authorised areas such as their account dashboard or the admin panel.
Cookie wordpresspass_
Duration past
Description WordPress sets this cookie to authenticate logged-in users securely. It stores an encrypted version of the password to verify user credentials while ensuring that the actual password is never exposed.
Functional
[x]
Functional cookies help perform certain functionalities like sharing the content of the website on social media platforms, collecting feedback, and other third-party features.
Cookie li_gc
Duration 6 months
Description Linkedin set this cookie for storing visitor's consent regarding using cookies for non-essential purposes.
Cookie lidc
Duration 1 day
Description LinkedIn sets the lidc cookie to facilitate data center selection.
Cookie wp-wpml_current_language
Duration session
Description WordPress multilingual plugin sets this cookie to store the current language/language settings.
Cookie yt-remote-device-id
Duration Never Expires
Description YouTube sets this cookie to store the user's video preferences using embedded YouTube videos.
Cookie ytidb::LAST_RESULT_ENTRY_KEY
Duration Never Expires
Description The cookie ytidb::LAST_RESULT_ENTRY_KEY is used by YouTube to store the last search result entry that was clicked by the user. This information is used to improve the user experience by providing more relevant search results in the future.
Cookie yt-remote-connected-devices
Duration Never Expires
Description YouTube sets this cookie to store the user's video preferences using embedded YouTube videos.
Cookie yt-remote-session-app
Duration session
Description The yt-remote-session-app cookie is used by YouTube to store user preferences and information about the interface of the embedded YouTube video player.
Cookie yt-remote-cast-installed
Duration session
Description The yt-remote-cast-installed cookie is used to store the user's video player preferences using embedded YouTube video.
Cookie yt-remote-session-name
Duration session
Description The yt-remote-session-name cookie is used by YouTube to store the user's video player preferences using embedded YouTube video.
Cookie yt-remote-fast-check-period
Duration session
Description The yt-remote-fast-check-period cookie is used by YouTube to store the user's video player preferences for embedded YouTube videos.
Cookie FORMASSEMBLY
Duration session
Description FormAssembly sets this cookie to offer online forms to our users.
Cookie yt-remote-cast-available
Duration session
Description The yt-remote-cast-available cookie is used to store the user's preferences regarding whether casting is available on their YouTube video player.
Cookie dextp
Duration 6 months
Description Audience Manager sets this cookie to record when the last data synchronization call was made.
Cookie _uid
Duration 6 months
Description AddThis sets this cookie to enable the visitors to share the web pages on social platforms via sharing buttons on the website.
Analytics
[x]
Analytical cookies are used to understand how visitors interact with the website. These cookies help provide information on metrics such as the number of visitors, bounce rate, traffic source, etc.
Cookie _gcl_au
Duration 3 months
Description Google Tag Manager sets the cookie to experiment advertisement efficiency of websites using their services.
Cookie _ga
Duration 1 year 1 month 4 days
Description Google Analytics sets this cookie to calculate visitor, session and campaign data and track site usage for the site's analytics report. The cookie stores information anonymously and assigns a randomly generated number to recognise unique visitors.
Cookie _gid
Duration 1 day
Description Google Analytics sets this cookie to store information on how visitors use a website while also creating an analytics report of the website's performance. Some of the collected data includes the number of visitors, their source, and the pages they visit anonymously.
Cookie _gat_UA-
Duration 1 minute
Description Google Analytics sets this cookie for user behaviour tracking.
Cookie ga
Duration 1 year 1 month 4 days
Description Google Analytics sets this cookie to store and count page views.
Cookie prism_
Duration 1 month
Description Active Campaign sets this cookie to track and store interactions.
Cookie vuid
Duration 1 year 1 month 4 days
Description Vimeo installs this cookie to collect tracking information by setting a unique ID to embed videos on the website.
Cookie sid
Duration session
Description The sid cookie contains digitally signed and encrypted records of a user’s Google account ID and most recent sign-in time.
Cookie CFID
Duration session
Description CFID cookie, set by Adobe, is used in conjunction with the "CFTOKEN" cookie. It stores a unique ID for the visitor, the visitor's device and browser.
Cookie CFTOKEN
Duration session
Description The CFTOKEN cookie, set by Adobe, denotes the type of device used by the visitor for the website to be properly formatted. These details are stored in the CFID cookie.
Cookie demdex
Duration 6 months
Description The demdex cookie, set under the domain demdex.net, is used by Adobe Audience Manager to help identify a unique visitor across domains.
Cookie s_ecid
Duration 2 years
Description Adobe installs this cookie to allow persistent ID tracking in the 1st party state and is used as a reference ID if the AMCV cookie has expired.
Cookie _fbp
Duration 3 months
Description Facebook sets this cookie to display advertisements when either on Facebook or on a digital platform powered by Facebook advertising after visiting the website.
Cookie CLID
Duration 1 year
Description Microsoft Clarity set this cookie to store information about how visitors interact with the website. The cookie helps to provide an analysis report. The data collection includes the number of visitors, where they visit the website, and the pages visited.
Cookie MR
Duration 7 days
Description This cookie, set by Bing, is used to collect user information for analytics purposes.
Cookie SM
Duration session
Description Microsoft Clarity cookie set this cookie for synchronizing the MUID across Microsoft domains.
Cookie pardot
Duration past
Description The pardot cookie is set while the visitor is logged in as a Pardot user. The cookie indicates an active session and is not used for tracking.
Cookie bugsnag-anonymous-id
Duration Never Expires
Description BugSnag/Juicer sets this cookie for bug reporting and other analytical purposes.
Cookie DST
Duration 6 months
Description Adobe Audience Manager sets this cookie when an error occurs while sending data to a destination.
Cookie s_vi
Duration 2 years
Description An Adobe Analytics cookie that uses a unique visitor ID time/date stamp to identify a unique vistor to the website.
Performance
[x]
Performance cookies are used to understand and analyze the key performance indexes of the website which helps in delivering a better user experience for the visitors.
Cookie AWSALB
Duration 7 days
Description AWSALB is an application load balancer cookie set by Amazon Web Services to map the session to the target.
Cookie AMCV_AdobeOrg
Duration 1 year 1 month 4 days
Description Adobe-Dtm sets this cookie to find the unique user ID that recognises the user on returning visits.
Cookie AMCVS_AdobeOrg
Duration session
Description Adobe-Dtm sets this cookie to store a unique ID to identify a unique visitor.
Cookie SRM_B
Duration 1 year 24 days
Description Used by Microsoft Advertising as a unique ID for visitors.
Advertisement
[x]
Advertisement cookies are used to provide visitors with customized advertisements based on the pages you visited previously and to analyze the effectiveness of the ad campaigns.
Cookie bcookie
Duration 1 year
Description LinkedIn sets this cookie from LinkedIn share buttons and ad tags to recognize browser IDs.
Cookie test_cookie
Duration 16 minutes
Description doubleclick.net sets this cookie to determine if the user's browser supports cookies.
Cookie YSC
Duration session
Description Youtube sets this cookie to track the views of embedded videos on Youtube pages.
Cookie VISITOR_INFO1_LIVE
Duration 6 months
Description YouTube sets this cookie to measure bandwidth, determining whether the user gets the new or old player interface.
Cookie VISITOR_PRIVACY_METADATA
Duration 6 months
Description YouTube sets this cookie to store the user's cookie consent state for the current domain.
Cookie NID
Duration 6 months
Description Google sets the cookie for advertising purposes; to limit the number of times the user sees an ad, to unwanted mute ads, and to measure the effectiveness of ads.
Cookie everest_g_v2
Duration 1 year
Description The cookie is set under the everesttech.net domain to map clicks to other events on the client's website.
Cookie dpm
Duration 6 months
Description The dpm cookie, set under the Demdex domain, assigns a unique ID to each visiting user, hence allowing third-party advertisers to target these users with relevant ads.
Cookie everest_session_v2
Duration session
Description This cookie, set by Everesttech, is used for targeted ads and to document efficacy of each individual ad.
Cookie yt.innertube::requests
Duration Never Expires
Description YouTube sets this cookie to register a unique ID to store data on what videos from YouTube the user has seen.
Cookie yt.innertube::nextId
Duration Never Expires
Description YouTube sets this cookie to register a unique ID to store data on what videos from YouTube the user has seen.
Cookie MUID
Duration 1 year 24 days
Description Bing sets this cookie to recognise unique web browsers visiting Microsoft sites. This cookie is used for advertising, site analytics, and other operations.
Cookie ANONCHK
Duration 10 minutes
Description The ANONCHK cookie, set by Bing, is used to store a user's session ID and verify ads' clicks on the Bing search engine. The cookie helps in reporting and personalization as well.
Cookie IDE
Duration 1 year 24 days
Description Google DoubleClick IDE cookies store information about how the user uses the website to present them with relevant ads according to the user profile.
Cookie rlas3
Duration 1 year
Description RLCDN sets this cookie to provide users with relevant advertisements and limit the number of ads displayed.
Cookie pxrc
Duration 2 months
Description This cookie is set by pippio to provide users with relevant advertisements and limit the number of ads displayed.
Cookie uid
Duration 1 year 1 month 4 days
Description This is a Google UserID cookie that tracks users across various website segments.
Uncategorized
[x]
Other uncategorized cookies are those that are being analyzed and have not been classified into a category as yet.
Cookie _cfuvid
Duration session
Description Calendly sets this cookie to track users across sessions to optimize user experience by maintaining session consistency and providing personalized services
Cookie CSRF-TOKEN
Duration session
Description Description is currently not available.
Cookie m
Duration 1 year 1 month 4 days
Description No description available.
Cookie acceptCookies
Duration 1 year
Description No description available.
Cookie CWEB2ID
Duration 1 year
Description Description is currently not available.
Cookie sprdConsent
Duration 1 year
Description Description is currently not available.
Cookie direct_affiliate
Duration 1 hour
Description No description available.
Cookie affiliate
Duration 15 days
Description No description available.
Cookie any_affiliate
Duration 3 months
Description No description available.
Cookie PP
Duration session
Description Description is currently not available.
Cookie p_url
Duration session
Description Description is currently not available.
Cookie raygun4js-userid
Duration Never Expires
Description Description unavailable.
Cookie classy-session-id
Duration session
Description Description is currently not available.
Cookie __Secure-ROLLOUT_TOKEN
Duration 6 months
Description Description is currently not available.
Cookie AWSALBTGCORS
Duration 7 days
Description No description available.
Cookie AWSALBTG
Duration 7 days
Description No description available.
Cookie _dd_s
Duration 4 hours
Description No description available.
Cookie i18next
Duration session
Description No description available.
Cookie wp-postpass_7a45526c2bbf3827a311ad10171fe873
Duration past
Description Description is currently not available.
Cookie gfmUserId
Duration session
Description Description is currently not available.
Cookie pgdid
Duration 1 year 1 month 4 days
Description Description is currently not available.
Cookie temp_ee
Duration Less than a minute
Description Description is currently not available.
Cookie adEdition
Duration 1 day
Description No description available.
Cookie geoEdition
Duration 1 day
Description No description available.
Cookie nbcu
Duration 6 months
Description No description available.
Cookie debugmode
Duration session
Description Description is currently not available.
Cookie __Secure-YEC
Duration past
Description Description is currently not available.
Accept All Save My Preferences Reject All
Powered by
Patients & Caregivers
Patient Organizations
Clinicians & Researchers
NORD en Español
Contact NORD
Donate
Rare Disease News
Resource Library
About Us
Events
Search for:
Donate
Understanding Rare Disease
Where to start
Rare Disease Facts and Statistics
NORD’s Rare Disease Database
Rare Disease Video Library
What It Means To Be Undiagnosed
Find A Rare Disease Organization
A Parent’s Guide to Rare Disease Diagnosis in Children
A PCP Primer to Diagnosing Rare Disease in Children
Stay informed
Stories That Inspire
RareEdu® – Online Learning Platform
Rare Disease Day
Resource Library
State Resource Center – Find Local Resources
Publications On Rare Disease
NORD® Living Rare, Living Stronger® Patient & Family Meeting
Learn More
Living with a Rare Disease
Manage your care
Getting Help & Support
Managing Your Disease
We can help
How NORD Can Help
Speak To Someone at NORD
Rare Disease Center Of Excellence
Patient Assistance Programs
Explore Clinical Trials
Find A Patient Organization
Caregiver Resources
State Resource Center – Discover Local Resources
Community Support
Where to start
Rare Diseases Defined
Financial & Medical Assistance
Call Center & Information Services
Bringing Together Your Community
Mentoring organizations
NORD Member List
Start a Rare Disease Organization
Membership Program
Becoming Research Ready
Launching Registries & Natural History Studies
Patient-Focused Drug Development
Improving clinical care
Rare Disease Centers of Excellence
Continuing Medical Education (CME)
Partnering with the community
Corporate Council
National Partnerships
Global Parnerships
NORD® Living Rare, Living Stronger® Patient & Family Meeting
Learn More
Advancing Research
For patients
List of Rare Diseases
Gene Therapy for Rare Disease
Find Clinical Trials & Research Studies
For researchers
Request for Proposals
Research Grant Programs
Data Standards for Rare Diseases
For clinicians
Resources for Patients
Find a Rare Disease Care Center
Continuing Medical Education (CME)
Connect
IAMRARE® Program Powered by NORD
Rare Disease Cures Accelerator (RDCA-DAP)
Add Your Expertise
NORD Rare Disease Scientific Symposium
Learn More
Driving Policy
NORD policy and you
Today’s Policy Issues
NORD’s Policy Statements
Rare Disease Advisory Councils
NORD State Report Card
Taking action
Join the Rare Action Network®
Policy & Advocacy Taskforce
Contact your Representative
Take Action on Key Issues
Advocate for Rare Disease Policies as a Health Care Professional
Join A Current Campaign.
Learn about our current policy goals
Get Involved
Raising awareness and funds with NORD
Do-It-Yourself NORD Fundraiser
NORD Students for Rare
Sports & Fitness Fundraisers
Media Inquiries
In your community
Attend An Upcoming Event
Find a Rare Disease Patient Organization
Stay Informed With NORD’s Email Newsletter
Rare Disease Day®
Patient stories
Share Your Story
Careers At NORD
Intern At NORD
Jobs At Patient Disease Organizations
Show your support
Donate to NORD
Volunteer with NORD
Visit the NORD Store
Rare Disease News
Resource Library
About Us
For Clinicians & Researchers
For Patient Organizations
Home/Rare Diseases/Post Polio Syndrome
Disease Overview
Synonyms
Signs & Symptoms
Causes
Affected Populations
Disorders with Similar Symptoms
Diagnosis
Standard Therapies
Clinical Trials and Studies
References
Programs & Resources
Complete Report
Get Information about a Rare Disease
Use the form below to explore NORD's comprehensive rare disease database. Search for detailed information on rare diseases, including NORD's authoritative Rare Disease Reports and data from other reliable sources.
= NORD Rare Disease Report
_Interested in exploring the database in detail?
Click here to explore the database in alphabetical order._
Disease Overview
Synonyms
Signs & Symptoms
Causes
Affected Populations
Disorders with Similar Symptoms
Diagnosis
Standard Therapies
Clinical Trials and Studies
References
Programs & Resources
Complete Report
Post Polio Syndrome
Print
Last updated: April 08, 2009
Years published: 1987, 1989, 1991, 1992, 1998, 2005, 2009
Advertisement
Disease Overview
Post-polio syndrome (PPS) is a syndrome that affects some people who have had polio (poliomyelitis) and occurs many years (typically from 10 to 40 years) after recovery from the initial infection. It is characterized by the development of progressive weakness in muscles that were affected by the original polio infection. In addition, those affected may experience extreme fatigue and joint pain. Skeletal deformities, such as scoliosis, may occur as a result of this syndrome. There is variation in the severity of symptoms. In severe cases, symptoms may mimic those of the rare disorder known as Lou Gehrig’s disease (amyotrophic lateral sclerosis). The degree of muscle atrophy during the post-polio period appears to reflect the severity of the impact of the initial polio infection. People who were significantly affected by polio are more likely to experience severe symptoms from post-polio syndrome. The cause of this syndrome is not known. Although exact numbers are not available, it has been estimated that there are 300,000 polio survivors in the United States and that from one-fourth to one-half of them may ultimately develop some degree of post-polio syndrome.
View Full ReportShow Less
Print / Download as PDF
Next section >
< Previous section
Next section >
Synonyms
Polio, Late Effects
Post-Polio Muscular Atrophy
Postpoliomyelitis syndrome
Post-Polio Sequelae
< Previous section
Next section >
< Previous section
Next section >
Signs & Symptoms
Post-polio syndrome occurs at least 10 years after a person has been stricken by polio. It is characterized by gradual deterioration of muscle function and increased weakness that usually occurs in the limbs that were most severely affected by polio. Sometimes, the disorder involves muscles that appeared to be fully recovered or that were not involved in the original polio attack, including muscles necessary for respiration. Other symptoms may include fatigue, muscle pain and twitching (fasciculations).
The fatigue associated with PPS comes on unexpectedly, sometimes as a feeling of total exhaustion throughout the body. Pain in the muscles and joints is not uncommon. Also, the affected person may experience problems related to swallowing, breathing, and sleeping. People with PPS are less able to tolerate the cold, and they are less able to carry on with day-to-day activities such bathing, dressing, and walking.
< Previous section
Next section >
< Previous section
Next section >
Causes
The cause of post-polio syndrome is not known at this time. One theory is that it may be related to the recovery from the original polio. During recovery from polio, nerve cells in affected muscles may regrow many smaller branches (dendrites) from the message-transmitting large branches (axons) of nerve cells. These sprouts take over the function of neurons killed by the polio virus. It is thought that, after years of functioning beyond capacity, the nerve cells weaken and lose their ability to maintain these tiny dendrites, which then begin to shrink, and the whole muscle becomes weaker. Additional research is needed to determine whether this may, in fact, be the cause of post-polio syndrome.
Although this syndrome sometimes resembles amyotrophic lateral sclerosis (ALS), it is not considered to be a form of that disease.
< Previous section
Next section >
< Previous section
Next section >
Affected populations
Post-polio syndrome affects people who have had acute episodes of poliomyelitis. It occurs 10 years or more after the original illness, and can occur as long as 40 years afterward. According to one estimate, 25% to 50% of the 300,000 polio survivors in the United States may develop the syndrome. At the present time, there is no known way to prevent the syndrome.
< Previous section
Next section >
< Previous section
Next section >
Disorders with Similar Symptoms
Symptoms of the following disorders can be similar to those of post-polio syndrome. Comparisons may be useful for differential diagnosis:
Amyotrophic lateral sclerosis (ALS) is one of a group of disorders known as motor neuron diseases. It is characterized by the progressive degeneration and eventual death of nerve cells (motor neurons) in the brain, brainstem and spinal cord that facilitate communication between the nervous system and voluntary muscles of the body. Ordinarily, motor neurons in the brain (upper motor neurons) sent messages to motor neurons in the spinal cord (lower motor neurons), and then to various muscles. ALS affects both the upper and lower motor neurons, so that the transmission of messages is interrupted, and muscles gradually weaken and waste away. As a result, the ability to initiate and control voluntary movement is lost. Ultimately, ALS leads to respiratory failure because affected individuals lose the ability to control muscles in the chest and diaphragm. ALS is often called Lou Gehrig’s disease. (For more information on this disorder, choose “amyotrophic lateral sclerosis” as your search term in the Rare Disease Database.)
Multiple sclerosis is a neuroimmunologic (both the nervous system and immunological system are involved) disorder involving the brain, spinal chord and optic nerves. By means of a mechanism that is not clearly understood, the protective insulating sheath (myelin sheath) that covers the nerve is destroyed. The inflammatory attacks that produce the characteristic scarring (plaques or patches) of the myelin sheath occur randomly, in varying intensity, and at multiple sites. The course of the disease may advance, relapse, remit, or stabilize. The randomness of the location of plaques or patches affects the nerve’s ability to transmit information (neurotransmission) and causes a wide range of neurological symptoms, which may vary from person to person. (For more information on this disorder, choose “multiple sclerosis” as your search term in the Rare Disease Database.)
< Previous section
Next section >
< Previous section
Next section >
Diagnosis
Diagnosis of post-polio syndrome is made on the basis of a thorough history, a neurological examination, and the process of excluding other possible diseases through various tests. In making the diagnosis, physicians will be aware of three factors. a prior diagnosis of polio, an interval of one or more decades since the original acute episode, and slow, steady, progressive deterioration.
< Previous section
Next section >
< Previous section
Next section >
Standard Therapies
Treatment
There are no specific treatments for PPS at this time. The goal of management of this disorder is to make the patient as comfortable as possible.
Affected individuals are taught to conserve energy by pacing their activities and combining these with periods of rest. Mechanical assists such as canes, walkers, and scooters may be helpful.
Moderate exercise is seen by many physicians as beneficial for those affected. Swimming is one type of exercise that is sometimes recommended. Speech therapy may be helpful for individuals whose swallowing has been affected.
Also, occupational therapy can lead to adjustments in the home environment that may allow those affected to carry out common activities in ways that are less energy-consuming.
< Previous section
Next section >
< Previous section
Next section >
Clinical Trials and Studies
Two clinical trials sponsored by the National Institute of Neurological Disorders and Stroke (NINDS) are being carried out (2005) at the Clinical Center on the NIH campus in Bethesda. For additional information, contact the NIH Clinical Center at the phone numbers or email address listed above.
One is designed to determine whether electromyography (a test of the communication between nerves and muscle cells) is a reliable diagnostic tool for post-polio syndrome. The other is designed to determine whether the drug Modafinil can decrease fatigue in patients with post-polio syndrome.
Information on current clinical trials is posted on the Internet at www.clinicaltrials.gov. All studies receiving U.S. government funding, and some supported by private industry, are posted on this government web site.
For information about clinical trials being conducted at the NIH Clinical Center in Bethesda, MD, contact the NIH Patient Recruitment Office:
Tollfree: (800) 411-1222
TTY: (866) 411-1010
Email: prpl@cc.nih.gov
For information about clinical trials sponsored by private sources, contact:
www.centerwatch.com
< Previous section
Next section >
< Previous section
Next section >
References
TEXTBOOKS
Beers MH, Berkow R., eds. The Merck Manual, 17th ed. Whitehouse Station, NJ: Merck Research Laboratories; 1999:1486, 2342.
Berkow R., ed. The Merck Manual-Home Edition.2nd ed. Whitehouse Station, NJ: Merck Research Laboratories; 2003:576, 1578.
Mandell GL, Bennett JE, Dolan R, eds. Mandell, Douglas and Bennett’s Principles and Practice of Infectious Diseases. 4th ed. Churchill Livingstone Inc. New York, NY; 1995:1617-18.
Rowland LP. Ed. Merritt’s Neurology. 10th ed. Lippincott Williams & Wilkins. Philadelphia, PA. 2000:137, 712.
REVIEW ARTICLES
Trojan DA, Cashman NR. Post-poliomyelitis syndrome. Muscle Nerve. 2005;31:6-19.
Bartels MN, Omura A. Aging in polio. Phys Med Rehabil Clin N Am. 2005;16:197-218.
Khan F. Rehabilitation for postpolio sequelae. Aust Fam Physician. 2004;33:621-24.
Jubelt B, Agre JC. Characteristics and management of postpolio syndrome. JAMA. 2000;284:412-14.
JOURNAL ARTICLES
Sorenson EJ, Daube JR, Windebank AJ. A 15-year follow-up of neuromuscular function in patients with prior poliomyelitis. Neurology. 2005;22:1070-72.
Sliwa J. Postpolio syndrome and rehabilitation. Am J Phys Med Rehabil. 2004;83:909.
Sandberg A, Stalberg E. How to interpret normal electromyographic findings in patients with an alleged history of polio. J Rehabil Med. 2004;36:169-76.
Finch LE, Venturini A, Mayo NE, et al. Effort-limited treadmill walk test: reliability and validity in subjects with postpolio syndrome. Am J Phys Med Rehabil. 2004;613-23.
Sanberg A, Stalberg E. changes in macro electromyography over time in patients with a history of polio: A comparison of 2 muscles. Arch Phys Med Rehabil. 2004;85:1174-82.
Gordon T, Hegedus J, Tam SL. Adaptive and maladaptive motor axonal sprouting in aging and motoneuron disease. Neurol. Res. 2004;26:174-85.
Jubelt B. Post-Polio Syndrome. Curr Treat Options Neurol. 2004;6:87-93.
Markstrom A, Sundell K, Lysdahl M, et al. Quality-of-life evaluation of patients with neuromuscular and skeletal diseases treated with noninvasive and invasive home mechanical ventilation. Chest. 2002;122:1695-700.
FROM THE INTERNET
NINDS Post-Polio Syndrome Information Page. NINDS. Last updated March 29, 2005. 2pp.
www.ninds.nih.gov/disorders/post_polio/post_polio_pr.htm
Post-polio syndrome. MayoClinic.com. March 04, 2005. 6pp.
www.mayoclinic.com/invoke.cfm?id=DS00494
Post-Polio Syndrome. Quick Reference and Fact Sheets. March of Dimes. 2005. 2pp.
www.marchofdimes.com/printableArticles/681_1284.asp?printable=true
< Previous section
Next section >
< Previous section
Programs & Resources
Assistance Programs
Patient Organizations
More Information
RareCare® Assistance Programs
NORD strives to open new assistance programs as funding allows. If we don’t have a program for you now, please continue to check back with us.
Additional Assistance Programs
MedicAlert Assistance Program
NORD and MedicAlert Foundation have teamed up on a new program to provide protection to rare disease patients in emergency situations.
Learn more
Rare Disease Educational Support Program
Ensuring that patients and caregivers are armed with the tools they need to live their best lives while managing their rare condition is a vital part of NORD’s mission.
Learn more
Rare Caregiver Respite Program
This first-of-its-kind assistance program is designed for caregivers of a child or adult diagnosed with a rare disorder.
Learn more
Learn more about Patient Assistance Programs >
Patient Organizations
Post-Polio Health International
Phone:314-534-0475Email:info@post-polio.orgFax: 314-534-5070
Related Rare Diseases:Post Polio Syndrome
View Profile >
British Polio Fellowship
Email:info@britishpolio.org.uk
Related Rare Diseases:Post Polio Syndrome
View Profile >
Centers for Disease Control and Prevention
Phone:404-639-3534Email:cdcinfo@cdc.gov
Related Rare Diseases:Recurrent Pericarditis, Mucormycosis, Clostridial Myonecrosis, ...
View Profile >
New Horizons Un-Limited, Inc.
Phone:414-299-0124Email:horizons@new-horizons.orgFax: 414-347-1977
Related Rare Diseases:MECP2 Duplication Syndrome, Collagen Type VI-Related Disorders, Spinal Muscular Atrophy, ...
View Profile >
Learn more about Patient Organization & Membership >
More Information
The information provided on this page is for informational purposes only. The National Organization for Rare Disorders (NORD) does not endorse the information presented. The content has been gathered in partnership with the MONDO Disease Ontology. Please consult with a healthcare professional for medical advice and treatment.
GARD Disease Summary
The Genetic and Rare Diseases Information Center (GARD) has information and resources for patients, caregivers, and families that may be helpful before and after diagnosis of this condition. GARD is a program of the National Center for Advancing Translational Sciences (NCATS), part of the National Institutes of Health (NIH).
View report
Orphanet
Orphanet has a summary about this condition that may include information on the diagnosis, care, and treatment as well as other resources. Some of the information and resources are available in languages other than English. The summary may include medical terms, so we encourage you to share and discuss this information with your doctor. Orphanet is the French National Institute for Health and Medical Research and the Health Programme of the European Union.
View report
< Previous section
For Patients & Caregivers
For Organizations
For Clinicians & Researchers
Sign Up for NORD News
National Organization for Rare Disorders (NORD)
1900 Crown Colony Drive
Suite 310
Quincy, MA 02169
Phone: 617-249-7300
Other Locations:
Danbury, CT Office
7 Kenosia Avenue
Danbury, CT 06810
Phone: 203-744-0100
Fax: 203-263-9938
Newport Coast, CA Office
21163 Newport Coast Drive
Suite 254
Newport Coast, CA 92657
Washington, DC Office
1779 Massachusetts Avenue
Suite 500
Washington, DC 20036
Phone: 202-588-5700
Understanding Rare Disease
Where to Start
Stay Informed
Patient Stories
Living with a Rare Disease
Manage Your Care
We Can Help
Community Support
Where to Start
Mentoring Organizations
Improving Clinical Care
Partnering With the Community
Community Support
Advancing Research
For Patients
For Researchers
For Clinicians
Connect
Publications
Driving Policy
NORD Policy Priorities
Taking Action
Join the Rare Action Network®
Donate
Donate to NORD
Get Involved
Legal
TradeMark & Brand Guidelines
Terms & Conditions
Privacy Policy
Rare Disease News
Resource Library
About Us
Media Inquiries
Get Involved
Careers
Sitemap
Contact
Copyright ©2025 NORD – National Organization for Rare Disorders, Inc. All rights reserved.
NORD is a registered 501(c)(3) charity organization.
Please note that NORD provides this information for the benefit of the rare disease community. NORD is not a medical provider or health care facility and thus can neither diagnose any disease or disorder nor endorse or recommend any specific medical treatments. Patients must rely on the personal and individualized medical advice of their qualified health care professionals before seeking any information related to their particular diagnosis, cure or treatment of a condition or disorder.
Your Name(Required)
Your Email(Required)
I show my stripes for...(Required)
Max. 5 words
Photo(Required)
Max file size 5MB. Min. image size 260px X 260px
Drop files here or Select files
Max. file size: 5 MB.
Your Message(Required)
Max. 15 words
Consent(Required)
[x] I agree to the terms and conditions
In these Standard Terms and Conditions, “Your Content” refers to any audio, video, text, images, or other material you provide or display. By providing “Your Content”, you grant NORD a license to read, use, reproduce, adapt, modify, publish, translate, and distribute the content in our marketing materials and to all media. “Your Content” must be your own and must not invade any third‐party’s rights. When you submit content or information to NORD, you are allowing the public to access and use that information, and to associate it with you. NORD reserves the right to remove “Your Content” at any time, without notice. We take your privacy very seriously. Subject to the Terms and Conditions above, NORD will never sell or disclose your personal information.
Phone
This field is for validation purposes and should be left unchanged.
×
Name(Required)
First Last
Email(Required)
Enter Email Confirm Email
Photos
Drop files here or Select files
Accepted file types: jpeg, png, jpg, Max. file size: 3 MB.
Comments(Required)
Please let us know what's on your mind. Have a question for us? Ask away.
Email
This field is for validation purposes and should be left unchanged.
×
Name(Required)
First Last
Email(Required)
Enter Email Confirm Email
Comments(Required)
Please let us know what's on your mind. Have a question for us? Ask away.
URL
This field is for validation purposes and should be left unchanged.
×
Notifications
Understanding Rare Disease
Where to start
Rare Disease Facts and Statistics
NORD’s Rare Disease Database
Rare Disease Video Library
Help for People with Undiagnosed Medical Conditions
Find A Rare Disease Organization
A Parent’s Guide to Rare Disease Diagnosis in Children
A PCP Primer to Diagnosing Rare Disease in Children
Stay informed
Stories That Inspire
RareEdu® – Online Learning Platform
Rare Disease Day
Resource Library
State Resource Center
Publications On Rare Disease
Featured NORD Event
NORD® Living Rare, Living Stronger® Patient & Family Meeting
Learn More
Living with a Rare Disease
Manage your care
Getting Help & Support
Managing Your Disease
Telemedicine Resources
We can help
How NORD Can Help
Speak To Someone at NORD
Rare Disease Centers Of Excellence
Patient Assistance Programs
Explore Clinical Trials
Find A Patient Organization
Caregiver Resources
Discover Local Resources
Community Support
Where to start
Rare Diseases Defined
Financial & Medical Assistance
Call Center & Information Services
Bringing Together Your Community
Mentoring organizations
NORD Member List
Start a Rare Disease Organization
Membership Program
Becoming Research Ready
Patient-Focused Drug Development
Improving clinical care
Claim Your Care
Rare Disease Centers of Excellence
Continuing Medical Education (CME)
Partnering with the community
Corporate Council
National Partnerships
Global Partnerships
Rare Cancer Coalition
NORD Rare Impact Awards®
Featured NORD Event
NORD® Living Rare, Living Stronger® Patient & Family Meeting
Learn More
Advancing Research
For patients
List of Rare Diseases
Gene Therapy for Rare Disease
Find Clinical Trials & Research Studies
For researchers
Request for Proposals
Research Grant Programs
Data Standards for Rare Diseases
For clinicians
Resources for Patients
Find a Rare Disease Care Center
Continuing Medical Education (CME)
Connect
IAMRARE® Program Powered by NORD
Rare Disease Cures Accelerator (RDCA-DAP)
Add Your Expertise
Featured NORD Event
NORD Rare Disease Scientific Symposium
Learn More
Driving Policy
NORD policy and you
Public Policy Positions
Policy Statements & Letters to Policymakers
Rare Disease Advisory Councils
NORD State Report Card®
Taking action
Join the Rare Action Network®
Policy & Advocacy Taskforce
Take Action on Key Issues
Advocate for Rare Disease Policies as a Health Care Professional
Join A Current Campaign.
Learn about our current policy goals
Take Action
Featured NORD Event
NORD® Rare Diseases & Orphan Products Breakthrough Summit®
Learn More
Get Involved
Raising Awareness & Funds with NORD
Do-It-Yourself NORD Fundraiser
NORD Students for Rare
Sports & Fitness Fundraisers
Media Inquiries
In your community
Attend An Upcoming Event
Find a Rare Disease Patient Organization
Stay Informed With NORD’s Email Newsletter
Rare Disease Day®
Patient stories
Share Your Story
Careers At NORD
Intern At NORD
Jobs At Patient Disease Organizations
Corporate Council
About the Corporate Council
Corporate Council Members
Join Corporate Council
Code of Conduct
Show Your Support
Donate to NORD
Volunteer with NORD
Visit the NORD Store
Close |
189011 | https://store.macmillanlearning.com/us/product/Quantitative-Chemical-Analysis/p/131948770X?srsltid=AfmBOop8CqosKi1F-lT1ScF5mwtwpBlnvA4IfAoVz-M4XZCsBkA2TV9H | Quantitative Chemical Analysis 11th Edition | Daniel C. Harris | Macmillan Learning
Skip to Main Content
Instructor Site
Student Store
United States
Canada
Student StoreStudent Store
Expand navigation
Sign in // Register
I'M AN INSTRUCTOR
I'M A STUDENT
#### Find what you need to succeed.
0
United States
Canada
Who We Are
Who We Are
back
Who We Are
Student Benefits
Student Benefits
back
Special Offers
Rent and Save
Flexible Formats
College Quest Blog
Discipline
Discipline
back
AstronomyBiochemistryBiologyChemistryCollege SuccessCommunicationEconomicsElectrical EngineeringEnglishEnvironmental ScienceGeographyGeologyHistoryMathematicsMusic & TheaterNutrition and HealthPhilosophy & ReligionPhysicsPsychologySociologyStatisticsValue
Digital Products
Digital Products
back
Achieve
E-books
iClicker Student App (Student Response System)
Support
Support
back
Get Help
Rental and Returns
Support Community
Student Options Explained
×
Rental FAQs
GET FREE SHIPPING!
Use Promo Code SHIPFREE at Step 4 of checkout.
Free Shipping only applicable to US orders. Restrictions apply.
Quantitative Chemical Analysis
Instant Access
Eleventh Edition|©2026 Daniel C. Harris; Charles A. Lucy
Format
Packages
Study Extras
Achieve $124.99
ISBN:9781319490294
Online course materials that will help you in this class. Includes access to e-book and iClicker Student.
Access ( 1 term )
$124.99
Subscribe until 02/28/2026
You will need to find your course in order to purchase Achieve.
Find Your Course
A grace period may be available for this course.
Visit Achieve to find out.
E-book from $89.99
ISBN:9781319576899
Take notes, add highlights, and download our mobile-friendly e-books.
Access ( 6 months )
$89.99
Subscribe until 03/28/2026
Add to Cart
Buy
$129.99
Add to Cart
Loose-Leaf $227.99
ISBN:9781319591380
Save money with our hole-punched, loose-leaf textbook.
Buy
$227.99
Add to Cart
Paperback from $70.00
ISBN:9781319487706
Read and study old-school with our bound texts.
Rent ( 6 months )
$70.00
Rent until 03/31/2026
Includes eBook Trial Access
(14-day)
Rent ( 1 year )
$140.00
Rent until 09/27/2026
Includes eBook Trial Access
(14-day)
Loose-Leaf + Achieve $174.99
ISBN:9781319605636
This package includes Achieve and Loose-Leaf.
Buy Loose-Leaf +
Access Achieve ( 1 term )
$174.99
You will need to find your course in order to purchase Achieve.
Find Your Course
Paperback + Achieve from $149.99
ISBN:9781319605612
This package includes Achieve and Paperback.
Rent Paperback ( 4 months ) +
Access Achieve ( 1 term )
$149.99
Rent until 02/09/2026
Includes eBook Trial Access
(14-day)
You will need to find your course in order to purchase Achieve.
Find Your Course
Solutions Manual for Quantitative Chemical Analysis from $41.99
ISBN:9781319619343
Take notes, add highlights, and download our mobile-friendly e-books.
Access ( 6 months )
$41.99
Rent until 03/31/2026
Add to Cart
Buy
$62.99
Add to Cart
Loose-leaf Version for Solutions Manual Quantitative Chemical Analysis $64.99
ISBN:9781319618612
Save money with our hole-punched, loose-leaf textbook.
Buy
$64.99
Includes eBook Trial Access
(14-day)
Add to Cart
About
Digital Options
Contents
Authors
About
Master Analytical Chemistry with Confidence
Harris/Lucy, Quantitative Chemical Analysis, Eleventh Edition, is your essential guide to mastering analytical chemistry and boosting your grades. This digital textbook offers the latest insights and practical applications, ensuring you have the most reliable tools at your fingertips. With expertly crafted content and innovative resources, it’s designed to support your academic success and help you excel in every chemistry challenge.
Digital Options
E-book
Read online (or offline) with all the highlighting and notetaking tools you need to be successful in this course.
Learn More
E-book
Read online (or offline) with all the highlighting and notetaking tools you need to be successful in this course.
Learn More
Achieve
Achieve is a single, easy-to-use platform proven to engage students for better course outcomes
Learn More
Contents
Table of Contents
Chapter 0 The Analytical Process
Chapter 1 Chemical Measurements
Chapter 2 Tools of the Trade
Chapter 3 Experimental Error and Uncertainty
Chapter 4 Statistics
Chapter 5 Quality Assurance and Calibration Methods
Chapter 6 Chemical Equilibrium
Chapter 7 Let the Titrations Begin
Chapter 8 Activity and the Systematic Treatment of Equilibrium
Chapter 9 Monoprotic Acid-Base Equilibria
Chapter 10 Polyprotic Acid-Base Equilibria
Chapter 11 Acid-Base Titrations
Chapter 12 Analytical Complexation
Chapter 13 Sampling and Sample Preparation
Chapter 14 Fundamentals of Electrochemistry
Chapter 15 Electrodes and Potentiometry
Chapter 16 Redox Titrations
Chapter 17 Electroanalytical Techniques
Chapter 18 Fundamentals of Spectrophotometry
Chapter 19 Applications of Spectrophotometry
Chapter 20 Spectrophotometers
Chapter 21 Atomic Spectroscopy
Chapter 22 Mass Spectrometry
Chapter 23 Introduction to Chromatography
Chapter 24 Gas Chromatography
Chapter 25 High-Performance Liquid Chromatography
Chapter 26 Chromatographic Methods and Capillary Electrophoresis
Chapter 27 Gravimetric and Combustion Analysis
Authors
Daniel C. Harris
Dan Harris was born in Brooklyn, NY in 1948. He earned degrees in Chemistry from MIT in 1968 and Caltech 1973 and was a postdoc at Albert Einstein College of Medicine in New York. After teaching at the University of California at Davis from 1975-1980 and at Franklin and Marshall College from 1980-1983, he moved to the Naval Air Systems Command at China Lake, California, where he is now a Senior Scientist and Esteemed Fellow. While teaching analytical chemistry at Davis, he wrote his lectures in bound form for his students. This volume caught the attention of publishers representatives wandering through the college bookstore. The first edition of Quantitative Chemical Analysis was published in 1982. The first edition of Exploring Chemical Analysis came out in 1996. Both have undergone regular revision. Dan is also co-author of Symmetry and Spectroscopy published in 1978 by Oxford University Press and now available from Dover Press. His book Materials for Infrared Windows and Domes was published by SPIE press in 1999. Dan and his wife Sally were married in 1970. They have two children and four grandchildren. Sallys work on every edition of the books is essential to their quality and accuracy.
Charles A. Lucy
Chuck Lucy is Professor Emeritus and 3M National Teaching Fellow at the University of Alberta. He has published over 160 papers (26 with undergraduate researchers) and been on the editorial advisory boards of seven analytical journals, including Analytical Chemistry and Analyst. He is a passionate teacher who has taught classes ranging from large first-year courses to discovery-based graduate lectures. Chuck has organized teaching workshops and chemistry education symposia across North America and has received numerous awards, including the Chemical Institute of Canada Award for Chemistry Education and the American Chemical Society Division of Analytical Chemistry’s J. Calvin Giddings Award for Excellence in Education. After contributing content to several chapters in the ninth edition of Quantitative Chemical Analysis, Chuck has broad responsibilities as a co-author of this tenth edition. The statistics, spectrophotometry, and chromatography chapters have benefited most from Chuck’s authorship. He highlights contributions of undergraduate student researchers and his new end-of-chapter problems emphasize graphical examination of data. Many of his problems incorporate primary sources and research publications, or make use of online tools such as a chromatography simulator with which you can observe the effects of varying experimental conditions.
Elevating Practical Applications with Cutting-Edge Updates and Enhanced Problem-Solving Practice
Master Analytical Chemistry with Confidence
Harris/Lucy, Quantitative Chemical Analysis, Eleventh Edition, is your essential guide to mastering analytical chemistry and boosting your grades. This digital textbook offers the latest insights and practical applications, ensuring you have the most reliable tools at your fingertips. With expertly crafted content and innovative resources, it’s designed to support your academic success and help you excel in every chemistry challenge.
E-book
Read online (or offline) with all the highlighting and notetaking tools you need to be successful in this course.
Learn More
E-book
Read online (or offline) with all the highlighting and notetaking tools you need to be successful in this course.
Learn More
Achieve
Achieve is a single, easy-to-use platform proven to engage students for better course outcomes
Learn More
Table of Contents
Chapter 0 The Analytical Process
Chapter 1 Chemical Measurements
Chapter 2 Tools of the Trade
Chapter 3 Experimental Error and Uncertainty
Chapter 4 Statistics
Chapter 5 Quality Assurance and Calibration Methods
Chapter 6 Chemical Equilibrium
Chapter 7 Let the Titrations Begin
Chapter 8 Activity and the Systematic Treatment of Equilibrium
Chapter 9 Monoprotic Acid-Base Equilibria
Chapter 10 Polyprotic Acid-Base Equilibria
Chapter 11 Acid-Base Titrations
Chapter 12 Analytical Complexation
Chapter 13 Sampling and Sample Preparation
Chapter 14 Fundamentals of Electrochemistry
Chapter 15 Electrodes and Potentiometry
Chapter 16 Redox Titrations
Chapter 17 Electroanalytical Techniques
Chapter 18 Fundamentals of Spectrophotometry
Chapter 19 Applications of Spectrophotometry
Chapter 20 Spectrophotometers
Chapter 21 Atomic Spectroscopy
Chapter 22 Mass Spectrometry
Chapter 23 Introduction to Chromatography
Chapter 24 Gas Chromatography
Chapter 25 High-Performance Liquid Chromatography
Chapter 26 Chromatographic Methods and Capillary Electrophoresis
Chapter 27 Gravimetric and Combustion Analysis
Daniel C. Harris
Dan Harris was born in Brooklyn, NY in 1948. He earned degrees in Chemistry from MIT in 1968 and Caltech 1973 and was a postdoc at Albert Einstein College of Medicine in New York. After teaching at the University of California at Davis from 1975-1980 and at Franklin and Marshall College from 1980-1983, he moved to the Naval Air Systems Command at China Lake, California, where he is now a Senior Scientist and Esteemed Fellow. While teaching analytical chemistry at Davis, he wrote his lectures in bound form for his students. This volume caught the attention of publishers representatives wandering through the college bookstore. The first edition of Quantitative Chemical Analysis was published in 1982. The first edition of Exploring Chemical Analysis came out in 1996. Both have undergone regular revision. Dan is also co-author of Symmetry and Spectroscopy published in 1978 by Oxford University Press and now available from Dover Press. His book Materials for Infrared Windows and Domes was published by SPIE press in 1999. Dan and his wife Sally were married in 1970. They have two children and four grandchildren. Sallys work on every edition of the books is essential to their quality and accuracy.
Charles A. Lucy
Chuck Lucy is Professor Emeritus and 3M National Teaching Fellow at the University of Alberta. He has published over 160 papers (26 with undergraduate researchers) and been on the editorial advisory boards of seven analytical journals, including Analytical Chemistry and Analyst. He is a passionate teacher who has taught classes ranging from large first-year courses to discovery-based graduate lectures. Chuck has organized teaching workshops and chemistry education symposia across North America and has received numerous awards, including the Chemical Institute of Canada Award for Chemistry Education and the American Chemical Society Division of Analytical Chemistry’s J. Calvin Giddings Award for Excellence in Education. After contributing content to several chapters in the ninth edition of Quantitative Chemical Analysis, Chuck has broad responsibilities as a co-author of this tenth edition. The statistics, spectrophotometry, and chromatography chapters have benefited most from Chuck’s authorship. He highlights contributions of undergraduate student researchers and his new end-of-chapter problems emphasize graphical examination of data. Many of his problems incorporate primary sources and research publications, or make use of online tools such as a chromatography simulator with which you can observe the effects of varying experimental conditions.
Related Titles
Find Your School
Select Your Discipline
Select Your Course
Find Your School
No schools matching your search criteria were found !
No active courses are available for this school.
No active courses are available for this discipline.
Can't find your course? Back Continue Back Purchase Access
Find Your Course
Confirm Your Course
Enter the course ID provided by your instructor
Find Your Course
We found the following course. Does this look correct?
We found the following course. To properly enroll in your course, please use the link provided in your school's course system (LMS Example: Canvas, Blackboard, D2L, Etc).
Your Achieve account needs to be linked with your school's account.
Back Purchase Access
Back Add to Cart
Back Add to Cart
Find Your School
Select Your Course
School Name
No schools matching your search criteria were found.
Instructor Name (Optional)
Academic Term
Search
Select Your Course
No Courses found for your selection.
Need an instructorless course?Back Purchase Access
macmillanlearning.com
//Privacy Notice
//Ads & Cookies
//Terms of Purchase/Rental
//Terms of Use
//Piracy
//Products
//Site Map
//Customer Support
We have updated the terms of our Privacy Notice. Click here for more details.
We are processing your request. Please wait...
×
Meet Tivity by Macmillan Learning!
Stay organized, manage assignments, and find balance with this free AI-powered app.
Get Tivity Today |
189012 | https://ieltsliz.com/ielts-line-graph-vocabulary/ | IELTS Line Graph Vocabulary
IELTS Liz
IELTS Preparation with Liz: Free IELTS Tips and Lessons, 2025
Welcome! I'm Liz. Develop your IELTS exam skills to maximise your score.
Home
Test Info
Test Information FAQ
Band Scores
IELTS Candidate Success Tips
GT IELTS
Computer IELTS: Pros & Cons
How to Prepare
Remarking
Useful Links & Resources
Recommended Books
Listening
Reading
Writing
TASK 1
TASK 2
Speaking
Vocabulary
Topics
Speaking Part 1 Topics
Speaking Part 2 Topics
Speaking Part 3 Topics
100 Essay Questions
On The Day Tips
Top Results
Advanced IELTS
IELTS Line Graph Vocabulary
by Liz162 Comments
Essential vocabulary to describe an IELTS writing task 1 line graph. To get a good band score you must show the examiner a range of different words to show upward and downward trends as well as key features. IELTS line graphs show change over a period of time and you must vary your language and vocabulary when you write your report.
Important: This page is about line graphs. But it is possible to use some of this language with bar charts that show change over time as well as tables or pie charts that show change over time.
This page contains:
VERBS & NOUNS WITH PRACTICE LESSON
2. ADVERBS & ADJECTIVES with PRACTICE LESSONS
3. TIME PHRASES
4. LINK TO MODEL LINE GRAPH
1. IELTS Line Graph Verbs & Nouns
Below are lists of verbs and nouns that can be used to describe a line graph in IELTS writing task 1. You should alter your sentences to use the noun form of the word and the verb form of the word.
Upward Trend Words
to rise / a rise
to increase / an increase
to climb / a climb
to grow / a growth
to go up
Past Tense Verbs:
to rise / rose
to increase / increased
to climb / climbed
to grow / grew
to go up / went up
Downward Trend Words
to decrease / a decrease
to drop / a drop
to fall / a fall
to decline / a decline
to go down
Past Tense Verbs:
to decrease / decreased
to drop / dropped
to fall / fell
to decline / declined
to go down / went down
Question: Can you use the verb “increase” and the noun “an increase” in the same writing or is it considered repetition of words?
Answer: You can definitely use the verb and noun form of the same word. This shows great flexibility that you can change word form. Your task is to show the examiner this flexibility. You will use these verbs and nouns in all line graphs because all line graphs show upward and downward trends.
Other Vocabulary
to fluctuate / a fluctuation
to dip / a dip
to remain steady
to remain stable
to remain unchanged
to level off
to rocket / to soar
Peaks & Lows
to peak at
(reach) a peak of
hit a high of
hit a low of
bottom out
plateau
Tips: You will use the above words according to what is shown in the line graph you are given. Fluctuations are when a line goes up and down repeatedly. A dip is when the line goes down but then returns to its previous point. Nearly all line graphs will have a peak, which means the highest point. Each line only has one peak, which is good to highlight in your report. To rocket and soar are both verbs that describe extremely dramatic increases. Never use these two verbs unless the increase truly is dramatic.
Mistakes when using words will affect your band score. This includes spelling.
Band 6 = some vocabulary errors.
Band 7 = few vocabulary errors.
2. Practice with Nouns and Verbs
Fill in the gaps using the vocabulary above. Look at the graph below and then fill in the gaps in the sentences 1-5 below.
IELTS Line graphs never have one line, except occasionally in a multiple task. So, this isn’t a line graph to practice with for a full model answer. For sample charts, click here: Sample Practice Charts
QUESTIONS: 1-5
Fill in the gaps with the correct nouns and verbs. You should use no more than two words for each gap.
The number of reported cases of influenza began at 40 in 1985 after which the figure (1). …………………… steadily to reach about 55 in 1987. In 1988, there was a (2) …………. in the number to below 50 before (3) ……………. to reach a (4) ………… (5) ……….. approximately 75 in 1991. After that date, the figure dropped to about 55 in the final year.
ANSWERS
Click below to reveal the answers:
Answers
rose / increased / climbed / grew / went up
a fall / a decline / a drop / a dip / a decrease
climbing / increasing / rising / going up / growing
peak
of
NOTICE: to peak at / a peak of (the preposition changes when you use the verb or noun)
.
3. Line Graph Adverbs & Adjectives
Below are useful adverbs and adjectives to describe change.
Adverbs for Line Graphs
steadily / gradually /eventually
slightly / marginally
rapidly / dramatically
significantly / considerably
relatively
Adjectives for Line Graphs
steady / gradual
slight / marginal
rapid / steep / dramatic
considerable / significant
Adjective Definition Exercise:
What is the real meaning of these adjectives and adverbs? Below you will find a table with adjectives and meanings. Match the adjectives with the correct meaning.
1) rapid / steep / dramatic A) the change took a long time
2) marginal / minimal B) a change that went from very low to very high
3) steady / gradual C) a change that went from about 50 to 100 out of 1,000
4) considerable / significant D) a big change that happened over a very short period of time
ANSWERS
Click below to reveal answers:
Answers
D
C
A
B
.
4. Time Expressions for Periods of Change
IELTS line graphs show change over a period of time. Below are some expressions to help you describe the passing of time without too much repetition.
Time phrases:
steady / gradual = these words do not show how much change, but rather that it took a long time for the change
over the next three days
three days later
after three days
over the following three days
the next three days show
from…to… / between … and…
the last year / the final year
the first year / initially
over the period / given period / period given
at the beginning of the period
at the end of the period
over a ten-year period
you must have an article (a/the)
the word year has no s
ten-year has a hyphen
RECOMMENDED FOR YOU:
IELTS LINE GRAPH MODEL ANSWER
Multiple Task: Line Graph & Bar Chart Model
ALL MODEL ANSWERS & TIPS FOR WRITING TASK 1
I hope you found this page useful. All the best, Liz
FREE SUBSCRIBE
Subscribe to get new lessons & tips by email.
Email Address
Subscribe
.
Filed Under: IELTS Writing Task 1
Comments
Favour says July 9, 2025 at The line graph gives a review of the reported number of incidences of flu in a specific village in the UK over a period spanning from 1985 to 1995.
Overall, the data reveals the significant fluctuations in the number of recounted issues of influenza throughout the ten-year period and the notable peaks and declines in some years.
In 1985, the aggregate of the stated occasions of the virus stood at 40. This figure grew steadily to 45 in 1986 and hit a high of 55 in 1987. The following year experienced a steady drop to approximately 47, representing a decrease of 8 cases. This fall continued until 1990 when a resurgence occurred with the figure of narrated events of influenza at 70. The most significant increase was observed in 1991, when the issues escalated further to 75 before witnessing a decline.
Following the peak in 1991, the subsequent years experienced fluctuations. In 1992, the aggregate of reported incidences stood at roughly 62, and the downward trend persisted into 1993, where cases decreased to 60. By 1995, the stated cases rounded off to approximately 60, suggesting a gradual tapering off in flu occurrences among the elderly population. Reply
2. Trang says May 9, 2025 at Hi,
There is a bar chart showing the percentage of men holding driving licenses. According to the chart, the figure for men was 70% in both 1976 and 1981, which are the first two time points provided.
So, can I say: “The percentage of men holding driving licenses remained stable at 70% between 1976 and 1981”?
I’m not sure if I’m being too strict, but I was confused. What if the figure in 1980 was different, even though the chart doesn’t show that year? Reply
Liz says May 9, 2025 at In that case, you could write “the percentage remained relatively stable at around 70%”. This means is was stable, but not totally stable. It allows for some flexibility with data. Also, adding “around 70%” allows for further flexibility if needed. Reply
Trang says May 10, 2025 at thank you so much!!!!!!!!! Reply
Grace says February 26, 2025 at Hi Liz, do you have any advice what section to review first? I’m preparing for my IELTS but I’m not sure where to start. MUCH THANKS–your online student. Reply
Liz says February 27, 2025 at There is no particular place to start. Start wherever you want. Choose something you enjoy and that way you can have a positive first start. Reply
Genet says January 13, 2025 at Thank you Liz for the wonderful contribution Reply
Liz says January 13, 2025 at You’re welcome 🙂 Reply
Dorothy says October 28, 2024 at Hi Liz,
Can “to leap” or “to jump” be used to indicate an upward trend? Are they informal for academic writing? Reply
Liz says October 28, 2024 at They are not suitable to use. You already have enough vocabulary to indicate increases as you can see on the page above. You have both a range of nouns, verbs, adjectives and adverbs. Use those that are suitable and it is enough for band 9. Reply
Dorothy says October 28, 2024 at Thank you! You’re truly my online tutor. Reply
Nkuli Chilomba says July 27, 2025 at Thank you for providing us with such valuable information on this website. You’re truly the best and kindest IELTS teacher online. we appreciate every tip and feedback you give. Reply
Margaret says October 20, 2024 at Hi Liz,
Greetings to you : )
Does this sound ok?
‘The line graph gives information about the reported influenza cases in people above 65 years old in a particular village in the United Kingdom over a 10 years period (from 1985 to 1995).’ Reply
Liz says October 20, 2024 at Spot on! Reply
Margaret says October 20, 2024 at Thank you Liz : )! Reply
Ayesha says September 18, 2024 at This line graph shows in some village of UK, the incidence of influenza virus in people elder than age of 65 year from 1985 to 1995.
initially in 1984 the 1St case of influenza seen in age of 40,and cause went up till 1987 in age above 40 year, however a fluctuation in cases occur from 1986 to 1989.Moreover the the reported cases of influenza climbed up in 1990 in age group of 50-70 year and it hit a high of in 1995 in age slightly above 70 year.
from then a dip of cases sewn in age of 60 in 1992 and attain a plataeu from 1992-1993.
after that a variability in cases observed in later years.
Dear life please go through from my report, I will be very thankful to you for this act of kindness Reply
Liz says September 18, 2024 at Although I don’t offer a feedback or marking service, I will just point out that this graph doesn’t show ages. It shows how many people got flu each year. Also, this is a single line graph and I’ve put a note under the graph to say that you won’t get one line on it’s own as a complete task in writing task 1. See other model line graphs on this page: Reply
Ayesha says September 18, 2024 at Thank-you liz Reply
Rajveer kaur says November 27, 2024 at thanku so much mam for giving wonderful information Reply
Dian says September 18, 2024 at Hi, Liz. I would like to kindly ask for your permission to use some of your materials in my lessons. I find your materials really easy to understand and I believe by adding them to my lessons will help my student a lot. Thank you. Reply
Liz says September 18, 2024 at Unfortunately, all my materials on this website are copyrighted and cannot be used or distributed in public or to other people. All my materials are to be used directly on my website. My recommendation is to use the IELTS Cambridge test books as sources for your lessons and direct your students to my website as homework so that they can come and learn directly on my website as a way to consolidate what you have taught. I really hope that one day I’ll have materials that teachers can download for use in the classroom. As soon as my health recovers and I’ve caught up with my massive work load, I’ll turn my eye to doing that. Reply
Ola says September 11, 2024 at Hi liz
I am trying to work on paraprasing my introduction , for the example given : The line graph illustrates the accumulated cases of influenza diagnosed in the elderly in a village in UK between 1985 to 1995 .
I thinks the repetitive use of in is bad? Reply
Liz says September 11, 2024 at Very good. Repetition of words is about vocabulary. In this case, the word “in” is a preposition which forms part of essential grammar in the English language and is 100% crucial for the sentence to be grammatically correct. So, it is 100% fine and perfectly good. Even a band score 9 will need to use this preposition in this way. When you look at repeated vocabulary, it is about repeating words like “celebrity” which could easily be altered to “famous person” or “people in the limelight” or “people often in the media” or “stars”. Some words, often nouns or verbs, can easily be paraphrased, other nouns and verbs are not often paraphrased. But prepositions form part of grammar, not vocabulary.
If you ever find a sentence that does feel a bit uncomfortable, it is possible to alter the grammar structure of the sentence to avoid problems with the structure and wording. However, in this case, it is actually acceptable. Reply
Ola says September 14, 2024 at Noted Liz, thank you. Reply
Ali says September 9, 2024 at thank you so much, Liz! I think they removed the words to grow and to go up in acad writing usage and to go down and something. Right? Reply
Liz says September 10, 2024 at On that is not true at all. It is 100% acceptable and in fact very appropriate to use the words – grow, go up, go down and fall in relation to numbers. Reply
Abdallah El-Kaiod says September 4, 2024 at hi liz greeting from Egypt , does this work
The graph illustrates the number of the reported influenza cases for people over 65 from period 1985 to 1995 in a certain village of the UK. Reply
Liz says September 4, 2024 at Greetings 🙂 All good, except check your prepositions – “in the UK”. Reply
Charisse says August 17, 2024 at Hi Liz, is this ok for an introduction?
The graph gives information about a group of people, more than 65 years of age who had flu in a particular village in the UK over the course of 10 years (1985-1995). Reply
Liz says August 17, 2024 at Well done, but these are reported cases rather than just a group of people. So, you can either use the same words “reported cases” or write “cases that were reported”. Reply
upreallll says May 11, 2024 at Hi Liz, can I say add adver/adjective in (fluctuate) term, like it fluctuated continously or it experienced a steady fluctuation for the remainder of the period?
Thank uuu Lizz.. Reply
Liz says May 11, 2024 at Yes, it is possible to use an adjective before the word “fluctuation” or an adverb with the verb “fluctuate”. However, you need to be careful which ones you use. To say it “fluctuated continuously” isn’t very meaningful when fluctuations are movements that are continuous anyway. But you can highlight if the fluctuation is minimal or significant – highlighting the degree of fluctuation. Reply
Sunal says July 15, 2024 at Peace be upon you
Felt like an avatar came down for us.
Take love Reply
Salman says February 28, 2024 at Is this a good introduction?
The graph underneath illustrates the number of elderly, influenced by influenza in a specific village in the UK between 1985 and 1965 Reply
Liz says February 28, 2024 at “underneath” = you will be writing on paper or on a blank screen. There will be nothing underneath your writing. Reply
Foysal Ahmed says April 25, 2024 at I am very poor student. I don’t have enough money to enroll. help me Reply
Liz says April 25, 2024 at This website contains a lot of pages of model answers, tips, topics, practice lessons etc – all free. Go to the HOME page to learn how to use it. Reply
Jenny says July 7, 2023 at The line graph describes data about those aged 65 and above who had flu over a 10-year period between 1985 and 1995 in a British village.
Overall, it can be clearly seen that the quantity of cases had grown from 1985 to the final year with the highest number of sick people reported in 1991.
During the first 5-year period, the number of cases climbed by around 80%: from 40 in 1985 to 70 in 1990. Over the first 2 years, the figure was constantly growing to its first peak at nearly 55 in 1987, which was followed by a gradual drop by approximately 10 cases one year later. Further marginal growth to roughly 48 by 1988 ended up with a dramatic jump to 70 cases in 1990.
For the following 5-year period, the figure experienced some fluctuation. Between 1990 and 1991, the trajectory was inclining to its highest point at nearly 75 cases, which was almost 2 times more than in 1985. Followed by a decline to about 61 cases in 1992, there was a last peak at about 64 reported cases in 1994. By the final year, 60 people had influenza, which was by 40 cases more than 10 years before. Reply
Sandeep Regmi says April 3, 2024 at Good one! Reply
Swaranjit kaur says June 13, 2023 at I’m very glad to read your lessons
Thanks alot for this
God bless you 😇🙏🏻 Reply
Liz says June 20, 2023 at You’re welcome 🙂 Reply
Manisha says June 23, 2023 at mam why you don’t post videos on YouTube now ?😞 Reply
Liz says June 23, 2023 at Due to a serious long-term health problem. Reply
Merin says June 26, 2023 at Get well soon. All the prayers for you. Reply
Waqas Yousaf says September 17, 2023 at May you recover soon, your blog really helps me a lot. Reply
OLUWAFEMI RILWAN AKINSOLA says October 13, 2023 at I’m really sorry to hear about this. You’re in my thoughts and prayers Reply
Image 97Maryam says August 14, 2023 at Hy liz! can we write our test with lead pencil so if we make any mistake we erase it or we need to use pens or ballpoints?
And secondly in writing can we write task 2 first and then task 1. or we need to go in sequence means first write task 1 and then task 2? Reply
Liz says August 14, 2023 at See this page for your answers: Always take time to read all pages on this website to learn about IELTS. Reply
Image 101Muhammad Abdullah says January 21, 2024 at you are Great mam Reply
Diao Yue says November 7, 2022 at Could you explain straightly the meanings of “dip “&”hit a high of “&”hit a low of”? I search these phrases on the Internet, but I cannot find their exact meanings. Hope to receive your response soon:) Reply
Liz says November 7, 2022 at “To hit a high of x” means to reach a peak and this is the highest point on the graph. “To hit a low of x” is the opposite. It means to reach the lowest point on the graph. “To dip” means the number drops and then returns back to its original number (or close to) – this is similar to a dip in the road when you are driving. Reply
Mr Dan says September 10, 2022 at Wonderful insight. I am new here n it’s been so helpful. Keep the good work. Thank you. Reply
Liz says September 11, 2022 at I’m glad it’s useful. Reply
Mahzuna Safara says October 16, 2022 at hello Liz. my name is Oysha. I’m uzbek 🇺🇿 Thank you very much for your useful information about line graph. I have a question for you. Is it possible to increase from 3.5 degrees to 7.5 degrees in 2 months? Reply
Liz says October 21, 2022 at Are you cold? Try putting on a sweater. Just kidding. You mean to increase from 3.5 band score to 7.5 band score. It would depend on the reason you only got 3.5 in your test. If the reason is that you didn’t understand the test, ran out of time, got confused by the types of questions and didn’t prepare the topics, then it’s possible to quick address all those issues and increase your score. But if the reason you got band 3.5 is because your English language is very limited, you make frequent errors, you can’t understand words etc, then it would take a lot of time to improve your English to a level which is similar to 7.5. Reply
Harsh says September 5, 2022 at The line chart illustrates the number of cases of influenza among people above 65 years in the countryside in the United Kingdom between 1985 to 1995. Units measured in years and number of people.
Looking from the overall perspective, during the decade, the graph fluctuated. In the beginning year, it had the least amount of cases and after 6 years it was at its peak.
In 1985, 40 people were infected and then it gradually increased till 1987 with nearly 55 persons. For the next 2 years, 5 cases were decreased compared to 1987. For the next couple of years, it was rapidly raised to one-quarter of humans and it was the highest number of cases registered in a decade.
From 1991 to 1993, cases dropped by approximately 15. For the last 2 years, there were ups and downs with 10 cases in 1994 and 1995 respectively and ending with 60 cases in the year 1995. Reply
20. Daya Mathew says August 15, 2022 at How is your health now?
Come back soon 🙂 Reply
Liz says August 16, 2022 at Thanks. I’m just in the middle of moving house which shatters my health. Hopefully once I move and rest, I can start gaining strength again. Reply
JessicaMary says August 5, 2022 at I love you Liz, keep being strong. Reply
Liz says August 5, 2022 at Thank you. I appreciate your comment Reply
Muthukumar says June 28, 2022 at Liz are you live?
Why don’t you make new YouTube videos? Reply
Liz says June 28, 2022 at Of course I’m alive. See this page: Reply
Ja Nu says June 15, 2022 at Hi, Liz
Please help me for below my task,
The graph illustrates the number of people over 65 who were infected by influenza, in a particular village. Th data described the survey period start from 1985 to till the end of 1995.
Overall, the graph shows how, within a decade many people are suffered by flue in every years.
In the beginning of the year 1985, total 40 people have infected by the flu. The disease was steeply increased to 55 in 1987. In 1988 the virous spread rate slightly dope to 48 but not last long till 1989, the flu hit to 50 people. Since than, the virous spread rate sharply increase to 70. The report from the graph, in the 1991, 75 people have been infected by influenza which is the highest cause rate in this decade.(1985 to 1995)
The effected rate was gradually drop to 62 in 1992. The following year of 1993 to end of 1995, the number of people are slightly up and down. The flu rate was end up as over 65 people are suffered in this ten years. Reply
24. Silpa says June 10, 2022 at The graph demonstrates the reported number of people over the age of 65 who were infected by influenza in a particular village in the period from 1985 to 1995. In particular, the graph shows how many people are affected by the flu every year.
Firstly, in the year 1985, about 40 people over the age of 65 were attacked by the virus. The number went up gradually to around 45 in 1986 and reached a peak at 55 in 1987. Then, a significant decrease in the number of cases was observed in the consecutive year. Again, there was a slight increment in cases identified between 1988 and 1989. From 1989 onwards, there was a drastic improvement in cases touching 70 in 1990 and reaching around 75 in 1991. From then onwards, the number of people suffering from the flu decreased gradually in between the years 1991 and 1992 and then a slight drop off till 1993.
60 people were infected in the years 1993 and 1995 and around 65 in the middle year. Although 40 is not a small number, it was the least number which was recorded in the 10 years and it was in 1985. In 1991, the highest number of cases was reported as 75. Reply
Silpa says June 10, 2022 at Thanks a lot Liz. Biggest support and inspiration 🙂 Reply
najma says June 9, 2022 at Each word i read i ask God to bless you. You are doing an incredible job by helping poor students gain knowledge for free . Reply
Liz says June 9, 2022 at Such a lovely comment. I’m so glad you like my lessons 🙂 Reply
Abdullajon says June 19, 2022 at Hi I got useful insights
Thank you 😊 💓 Reply
Diyora says March 9, 2024 at hi do you have a telegraph channels Reply
Liz says March 9, 2024 at No, I don’t. I have this website, my Youtube channel, my online store and one Facebook page only. Anything else you see online is fake. I’m glad you contacted me to confirm this. Reply
Mueksh says June 2, 2022 at Great thanks mam 💖 Reply
Shima says April 8, 2022 at Hi dear Liz.
No word to appreciate your free help. Your expressions are wonderful and informative.
You are amazing Liz.
Thanks alot
Shima Reply
Liz says April 8, 2022 at You’re welcome 🙂 Reply
sukh says September 10, 2022 at you are a wonderful lady ….. thanks to you for providing best material Reply
Rupi says September 27, 2020 at Thanks mam ,ur teaching method is great Reply
abhay says August 9, 2020 at Hi Liz,
Can I write “in the penultimate year”? I mean is it only used in graduation years or can I use it in a graph too? Reply
Liz says August 9, 2020 at It’s fine to use. Reply
Manjeet Kaur says September 21, 2021 at Thnk uh very much mam Reply
Azimbek says July 23, 2020 at Thank you for all of your lessons they are staggering! Just I can say wonderful! I found it very interesting. Reply
Liz says July 24, 2020 at You’re welcome 🙂 Reply
Oliver says June 1, 2022 at You are the best of the best Reply
Julius says July 16, 2020 at I really want to appreciate your method of teachings. They have been very insightful and easy to understand. Keep up your good job madam! . I have now developed confidence for the exam. May God bless you richly. Reply
Liz says July 16, 2020 at Glad the lessons are helping you 🙂 Reply
Mina says June 5, 2020 at I think I’m determined not to leave a work unread in your fascinating website for it seems you have prepared everything appropriately for your guests. I wonder if you could provide actual written answers for writing task 1 and 2 so that we can estimate our writings.
Thanks Reply
Liz says June 6, 2020 at You can find everything on the main pages of this website. The model answers for task 1 are in the writing task 1 section of this site. You can find each section accessed either through the RED MENU BAR at the top of the site or on the HOME PAGE 🙂 Reply
abror says July 13, 2020 at thank you very much. your lessons are enriching my knowledge. they are incomparable. Reply
Atinder Singh Toor says May 30, 2020 at This website is a blessing especially in times like these. Thank you so much for providing such amazing information and that too for free. You are such an amazing teacher. God bless you. 🙂 Reply
Liz says May 30, 2020 at I’m glad you are finding my site useful. Good luck with your test 🙂 Reply
Rajwinder kaur says May 15, 2020 at Exactly! Reply
Md Ataur Rahman says April 28, 2020 at Dear Liz,
Thank you so much for your excellent support for my IELTS preparation. From the couple of days, I am following your tips and lectures which are available here.
I am so much pleased as I have found my confidence to sit for the exam recently.
Wish you all the best.
Thank you again. Reply
Liz says April 28, 2020 at I’m glad you’ve found my site useful 🙂 Reply
cardy says May 28, 2020 at I see lot of foreinger use icon ” 🙂 ” when replied someone’s comment. Idk, if it’s normal to your country when responed like that. But in VietNam, that icon deemed not serious about the reply. Btw, I love your video. It’s free but more qualified than several courses in my country.
Again thank youuu <3 Reply
Liz says May 28, 2020 at To give someone a smile is normal online 🙂 Reply
md najmul says February 23, 2020 at Hi mam..!
I’m new here and my elder brother given your link.Because Once he followed..!
And now i following you..!
thanks for everything..!
and you pray for me so that i can improve myself.. “ Reply
Liz says February 23, 2020 at Good luck!! 🙂 Reply
Prabash says February 1, 2020 at Dear madam! You are doing great job.thank you very much for that
In the answer sheet task 1,,,there is a word rose,i think it should be rise
Good bless you Reply
Liz says February 1, 2020 at I don’t understand. There is no answer sheet on the page above. I don’t know which word you are referring to. Can you explain more? Reply
Peter Shuai says February 14, 2020 at I guess what Prabash said was the word “rose”. he thought it should be “rise”, however in my opinion, he is wrong, “rose” is the correct one. Reply
Liz says February 14, 2020 at Is he referring to question 1? It is passed tense: rose / increased etc. You can’t use a present tense when past tense is required. You are correct, Peter 🙂 Reply
Image 203Preet says February 3, 2022 at You are so kindhearted Liz ma’am 😊 and thanks for this Reply
Image 205Erfana Hanifa says February 2, 2020 at In writing task 1,you should maintain the use of one tense throughout.For example,if your explanation is in ‘past tense’ you should only use past tense throughout.
Thatswhy,Madam Liz used ‘rose’ instead of ‘rise’ in her graph explanation. Reply
Liz says February 14, 2020 at Sorry I didn’t see your comment before. Exactly correct 🙂 Reply
Bahare Mohammadkhani says November 29, 2019 at Thanks Liz. Reply
Bahare Mohammadkhani says November 12, 2019 at Hi Liz,
I was wondering whether all line graphs have trend or not
thanks Reply
Liz says November 12, 2019 at Yes, they all have trends. They show change over a period of time and that is the trend you are reporting as well as any other significant key feature. Reply
Shamwil says October 8, 2019 at That is great Reply
Rajesh Bhosale says September 16, 2019 at This helps a lot. Reply
Shi-Ming Chen says August 23, 2019 at Thanks. It is helpful for me. Reply
Kamalpreet says November 20, 2019 at Yes definitely Reply
Syed Mohsin says August 17, 2019 at Thank u Mam God bless u Reply
Miranda says August 14, 2019 at Thank you Liz, I have my IELTS on the 30th of August and your website is very helpful and useful.
Mimi Reply
Liz says August 14, 2019 at Good luck 🙂 Reply
Maitha says July 22, 2019 at Interesting website, lots of helpful materials in an organized manner, and the videos are very clear and informative.
Thanks a lot! Reply
46. Mary says July 21, 2019 at Thank you Liz.your help and useful advises . Reply
47. KVN Raju says May 22, 2019 at Hi Liz,
Your teaching lesson skills and tips are outstanding that’s helping me for IELTS preparation well.
Thank you and appreciate for your kind support.
Regards,
Raju Reply
manoj parmar says June 22, 2019 at dear liz
tremandous job you are doing, carving us towards acheiving higher band scores mile stone.
i appreciate for your great work Reply
Mani says May 18, 2019 at Hi dear Master. Well done. I really appreciate it. Many thxs Reply
Jass grewal says May 10, 2019 at Nice and greate work ….ieltsliz.com
.
.
From ludhiana-machhiwara 😊 Reply
Roop kaur says May 11, 2019 at Actually you have a great experience mam i also agree with jass ..so really thanks a lot. Reply
jass grewal says August 6, 2019 at thanks roop 🙂 Reply
Dilpreet kaur says May 3, 2019 at Dear ,you are doing a good job….👍👍 Reply
Karona Sutradhar says April 30, 2019 at It is very helpful to me. Reply
kulwinderkaur says May 19, 2019 at It’s really very fruitfull for me
Thanks mam your website very propitious for us Reply
Baljit says April 19, 2019 at Thank u mam you helped me a lot. Reply
Aziz says April 18, 2019 at Hii Liz I’ve just started IELTS . Thank you for all information given! I think there is a wrong word in the other vocabulary that is a drip. It should be a dip. Have a nice day 🙂 Reply
Liz says April 18, 2019 at Well spotted! Good proof reading skills 🙂 Reply
abeselom says December 11, 2019 at Hey Liz i just want to ask you if i scored 5 on my ielts are there any college that would accept me? Reply
Liz says December 11, 2019 at You would have to ask each individual college if they would accept you. Reply
Nurse Mujaheed says April 2, 2019 at Thank you very much ma,Nigerian nurses are also proud of you. Reply
Liz says April 4, 2019 at Thanks 🙂 Reply
gel says March 31, 2019 at Hi liz if by chance you read this please do write a comment, I am still not confident in doing task 1. Anyway, I see this as a great help to me , thank you.
Here is my sample report for the line graph above , please correct me especially on my grammar since I am mostly doing a self review and I am thankful I found this because you freely give suggestions.
The line graph gives information about the number of people above 65 with reported cases of influenza in some village of UK starting 1985 up to 1995. The data is measured according to the number of people affected by the flu in the given time frame.
Overall, there is a fluctuation in the statistics throughout the period. Observably, its lowest cases is in the beginning while the highest occurs in the middle years (1991 to 1992) of the decade.
For over 3 years , starting from 1985 until 1987 there is a gradual increase of influenza cases, specifically from 40 going up to 50 respectively. In the following years on the other hand, it starts to fluctuate , it moves downward from the stated latter period above , afterwards the number of people with reported cases continuously rise reaching its peak data of more than 70 in 1991.
Meanwhile , in the next 4 years including the end of the period the number of people with influenza steadily decrease . Consequently, from above 70 it then decline to exactly 60 (1991 to 1995). Reply
Ahmed Abdullah says September 5, 2019 at This is well written! But do check the last line where, I think you have got a grammatical mistake. Instead of using ‘declined’, you used the word ‘decline’. Reply
Mr Hadi says March 22, 2019 at Really appreciate your help ,May God bless you with good health and blessings.Thanks for being so generous. Reply
amin says March 3, 2019 at thanks for your lesson, my teacher. it is very useful Reply
Baljit Singh says February 23, 2019 at Excellent tech mam
Thanks mam 😊 😊 😊 😊 Reply
59. Md Sohan Uddin says February 15, 2019 at I wish I could talk with Liz.🙄 Reply
60. Marija says February 3, 2019 at Rubie expressed what i am feeling just right now. Dear Liz, you are so generous and I would like you to understand how much do I appreciate your free help.
Right words – God bless you:)
hugs and kisses,
Marija Reply
Liz says February 4, 2019 at Thanks 🙂 Reply
Shiva says March 1, 2019 at Yes, off course.. Their is no doubt 😊 Reply
Rubie Fabia says February 3, 2019 at You are very generous to share these ideas, tips, examples and teach those struggling examiners for their IELTS test. I am one of those people who’s very thankful when I found your website. It was very helpful for me.
Godbless your good heart Ms. Liz!
love lots,
Rubie Reply
62. Nencilia Botha says February 2, 2019 at Thanks liz that was very informative Reply
63. taqi adeel says January 17, 2019 at as regard of my experience ietlsliz website has such a good materials for preparing ielts exam . Reply
64. Mary says November 28, 2018 at Hi Liz.
I’ve started studying for IELTS for two months. My teacher taught me the way to practice for writing that I want to ask your opinion about that. She told us that, for beginners of course, we should spend for example one or two days on one sample of writing, study about topic and learn new words or collocations then try to write a good essay. What do you think? Is it a good way? Reply
Liz says November 28, 2018 at Why don’t you try it and see if it works. There is no one way to prepare for IELTS or to learn English. You need to decide which way works most effectively for yourself. Reply
mati says April 7, 2019 at It worked for me! Reply
Said Belmeddah says November 27, 2018 at Thank you a lot teacher for such fruitful initiative. Reply
Mohamed says November 25, 2018 at Thank you so much for your helpful material and tips. Reply
Inderjit Singh says November 24, 2018 at Very good speak 👍 Reply
Kharistini says November 23, 2018 at Thank you miss, this is really help me to study about english and ielts.. good job miss, hope you keep health Reply
KARIMA LANDOULSI says November 22, 2018 at Thank you so much Reply
Abdul says November 22, 2018 at Good job. Reply
D. B. Baniya says November 22, 2018 at Thank you so much for such a useful materials. No words to appreciate your hardworking. Reply
Aminat says November 22, 2018 at Thanks so much ma’am, your lesson has always being wonderful. Reply
Maryann says November 21, 2018 at This is good Reply
Shakawath Hussain says November 21, 2018 at Thank you very much Liz for a nice organised writing about task1 . Reply
Marykutty Mathew says November 21, 2018 at Thank you so much, I think it would help me to get good score next time. Thank you Mam. Reply
Marykutty Mathew says November 21, 2018 at Thank you so much, I think it would help me to get good score next time, thank you Mam Reply
Shupikai muchada says November 20, 2018 at Thanks Liz that was very informative Reply
aseel says November 20, 2018 at thankyou teacher , that was really helpful , so if just used these phrases in my exam i will det a high score or at least 5.5 ? Reply
Liz says November 20, 2018 at Vocabulary counts for 25% of your marks. Reply
Myo Aung says November 20, 2018 at ThankQ so much for your wonderful work you have given us. Reply
Thank you very much for your boundless generosity and kindness in helping the IELTS examinees. says November 20, 2018 at Wow awesome tips. Many thanks for your boundless generosity and kindness. Reply
Speak Your Mind Cancel reply
Name
Email
Website
[x] Notify me of follow-up comments by email.
[x] Notify me of new posts by email.
Search for ...
Search
IELTS LIZ STORE
Recent New Lessons
Thank you for your support, from Liz
I’m struggling and reaching out to you
Happy New Year to You All in 2025!!
Questions about Sport for IELTS Speaking Part 3
50% DISCOUNT Advanced IELTS – Ends TODAY !
LIZ'S PERSONAL STORY
Lovely! Thanks from Liz
About me
Hi, my name is Elizabeth (Liz). I am your teacher and the author of this website. I am an experienced, qualified English teacher specialising in IELTS test preparation since 2008. Please enjoy the 300 plus pages of free lessons, tips, model answers, topics etc.
Click Below to Learn:
IELTS Test Information
Listening
Reading
Writing Task 1
Writing Task 2
Speaking
Vocabulary
Archives
Archives
Copyright Notice
Copyright © Elizabeth Ferguson, 2014 – 2025
All rights reserved.
Privacy Policy & Disclaimer
Click here:Privacy Policy
Click here: Disclaimer
Return to top of page
Copyright ©2025 · Prose on Genesis Framework · WordPress · Log in
error: Content is protected !! |
189013 | https://www.fishersci.no/shop/products/dichloromethane-99-8-spectroscopy-stabilized-amylene-thermo-scientific/p-217075 | Dichloromethane, 99.8%, for spectroscopy, stabilized with amylene 500 mL | Buy Online | Thermo Scientific Acros | Fisher Scientific
Skip to main content Enable accessibility for low vision Open the accessibility menu
UserName
Pipetters and Dispensers! | Save Up to 23% Get Started
Search
Order Status
Quick Order
Support
Offers
Special Offers
Spotlight
Product of the Month
Outlet Corner
View All
Sign In
Shopping Tools
My Lists
Quick Order
Chemical Search
Order Oligo & Assay Products
Orders
Order Status
Account
Account Dashboard
My Profile
Password & Security
Create Web Profile
Register for an Account
Sign In
Sign In
Don't have an account?Create an Account
Account
Main Menu
Account
Account Dashboard
My Profile
Shipping & Billing Addresses
Add New Delivery Address
Password and Security
Create Web Profile
Register for an Account
Orders
Main Menu
Orders
Order Status
Quotes
Invoices
Returns
Multi-Tier requisitions
Shopping Tools
Main Menu
Shopping Tools
My Lists
Quick Order
Chemical Search
Search Third Party Products
Request for Non-Catalogue Product
Order Non-Catalogue Products
Order Oligo Products / TaqMan® Assays
Products
Antibodies
Products
Antibodies
Antibody Panels and Kits
Antibody Production and Purification Reagents and Kits
Bovine Secondary Antibodies
Donkey Secondary Antibodies
Goat Secondary Antibodies
Buffers and Standards
Primary Antibodies for ELISA
Primary Antibodies for Immunohistochemistry
Primary Antibodies for Western Blotting
Secondary Antibodies
Antibodies Advanced Search
Shop All Antibodies
Chemicals
Products
Chemicals
Acids
Bases
Biochemical Reagents
Buffers and Standards
Organic Compounds
Organometallic Compounds
Salts and Inorganics
Solvents
Water
Search by Chemical Structure
Shop All Chemicals
Lab Consumables
Products
Lab Consumables
Beakers and Lids
Bottles, Jars, and Jugs
Dishes
Dispensers
Filters and Filtration
Flasks
Microplates
Pipettes
Tubes
Vials
Shop All Lab Consumables
Lab Equipment and Instruments
Products
Lab Equipment and Instruments
Balances and Scales
Centrifuges
Cold Storage Products
Fume Hoods
Hotplates and Stirrers
Incubators
Microscopes
pH and Electrochemistry
Pumps and Tubing
Shakers
Shop All Lab Equipment and Instruments
Lab Furniture and Storage
Products
Lab Furniture and Storage
Desks and Tables
Fume Hoods
Laboratory Benches and Tables
Laboratory Carts and Accessories
Safety Cabinets
Seating
Shelving
Sinks and Drains
Storage Cabinets
Task Lighting
Shop All Lab Furniture and Storage
Life Sciences Products
Products
Life Sciences Products
Antibodies
Biochemical Reagents
Cell Analysis Products
Cell Culture Media
Cellular Imaging
Gel Electrophoresis Equipment and Supplies
Life Sciences Buffers
Molecular Biology Reagents and Kits
PCR Equipment and Supplies
Protein Analysis Reagents
Recombinant Proteins
Shop All Life Sciences Products
Safety Products
Products
Safety Products
Chemical Detection
Cleaners and Disinfectants
Eye and Face Protection
Fall Protection and Confined Space Entry
Hand Protection
Lockout Tagout Products
Radiation Monitoring Instrumentation
Respiratory Protection
Safety Clothing
Signs and Tags
Spill Control and Containment
Wipes and Wipe Accessories
Shop All Safety Products
Shop All Products
Browse Featured Brands
Tools
Search by Chemical Structure
Chem dex Chemical Search
Custom Assays, Antibodies, Oligos
eMolecules
Applications
Cell Biology
Applications
Cell Biology
Cell Culture and Modification
Cell Analysis
Chemistry
Applications
Chemistry
Analytical Testing Methods
Organic Synthesis
Chromatography
Genomics
Applications
Genomics
Molecular Biology Methods
PCR and qPCR
Microbiology
Production & Bioprocessing
Applications
Production & Bioprocessing
Cell & Gene Therapy
Proteomics
Applications
Proteomics
Protein Biology Methods
Safety
Applications
Safety
Contamination Control and Controlled Environments
Browse All Applications
Business Solutions
Biopharma
Biotech
Green Solutions
Food and Beverage
Testing Lab and Environmental
Chemical and Cosmetic
Clean Energy and Battery
Programs and Services
Edge Program
Encompass Procurement Services
Every Space
New Lab Start-Up Program
News Corner
SureTRACE Program
Sustainability Program
Services
Programs and Services
Services
Scan2Order
SureTRACE+Program
Unity Lab Services
Browse All Programs
Documents and Certificates
Main Menu
Documents and Certificates
Certificates
Safety Data Sheets
SureTRACE
Offers
Main Menu
Special Offers
Spotlight
Product of the Month
Outlet Corner
View All
Main Menu
Special Offers
Spotlight Save with our Spotlight Offers throughout the year on a variety of products and categories.
Product of the month Discover Fisherbrand's Product of the Month and upgrade your lab now.
Outlet Corner Take advantage of exceptional discounts on unsold products.
View All Special Offers
Help and Support
How to order
Order History
My Lists
Request a Quote
Business Account
Chemical Structure Search
Finding Safety Data Sheets
Finding Product Certificates
Account Creation
Help & Support Center
Register for a Business AccountMy ListsQuick Order
missing translation for 'onlineSavingsMsg'
Learn More
Dichloromethane, 99.8%, for spectroscopy, stabilized with amylene Product Code.: 10092190
819.00 NOK
814.57 NOK/ 500mL
Save 4.43 NOK
0.54% Off
Home
Products
Chemicals
Solvents
Methylene Chloride
Dichloromethane, 99.8%, for spectroscopy, stabilized with amylene
Structure Search
Print
Favorite item to keep tabs on itYou have reached the limit of 350 favorited items.Manage Your Favorites
SDS
Dichloromethane, 99.8%, for spectroscopy, stabilized with amylene
Product Code. p-217075Shop All Thermo Scientific Chemicals Products
500 mL, Glass Bottle, 500mL
Click to view available options
Quantity:
500 mL
2.5 L
Packaging:
Glass Bottle
Unit Size:
2.5 L
500mL
This item is not returnable. View return policy
Product Code. 10092190
Brand: Thermo Scientific Acros167775000
819.00 NOK
814.57 NOK/ 500mL
Save 4.43 NOK
0.54% Off
Request Bulk or Custom Format
Please sign in to purchase this item. Need a web account? Register with us today!
This item is not returnable. View return policy
Description
Specifications
Compare Specifications
Product Suggestions
Videos
Documents
Certificates
Safety and Handling
Safety Services
Promotions
Description
CAS: 75-09-2 | CH2Cl2 | 84.93 g/mol
This Thermo Scientific Chemicals brand product was originally part of the Acros Organics product portfolio. Some documentation and label information may refer to the legacy brand. The original Acros Organics product / item code or SKU reference has not changed as a part of the brand transition to Thermo Scientific Chemicals.
For UV/VIS spectroscopy
Specifications
Chemical Identifiers
CAS 75-09-2, 513-35-9
Molecular Formula CH2Cl2
Molecular Weight (g/mol)84.93
MDL Number MFCD00000881
InChI Key YMWUJEATGCHHMB-UHFFFAOYSA-NShow MoreShow Less
Synonym methylene chloride, methylene dichloride, methane, dichloro, methylene bichloride, methane dichloride, solaesthin, solmethine, freon 30, narkotil, aerothene mmShow MoreShow Less
PubChem CID6344
ChEBICHEBI:15767
IUPAC Name dichloromethane
SMILES ClCCl
Specifications
Melting Point-97.0°C
Absorbance(1 cm cell vs HPLC−grade water), (1 cm cell vs HPLC−grade water) at 0nm, 0.01 max. at 270nm, 0.01 max. at 280nm, 0.02 max. at 260nm, 0.05 max. at 250nm, 0.2 max. at 240nm, 1 max. at 233nm
Density 1.3250g/mL
Boiling Point 39.0°C to 40.0°C
Quantity 500 mL
Packaging Glass Bottle
Physical Form Liquid
Assay Percent Range 99.8%
Residue after Evaporation 0.0003% max.
Specific Gravity 1.325
Linear Formula CH 2 Cl 2
Viscosity 0.43 mPa.s (20°C)
Beilstein 01, 60
Water 0.02% max.
Fieser 01,676; 02,273; 04,337; 07,239; 13,106
Merck Index 15, 6135
Solubility Information Solubility in water: 20g/L (20°C). Other solubilities: soluble in alcohol, ether and dmf
Formula Weight 84.93
Percent Purity 99.8%
Grade Spectroscopy
Chemical Name or Material Dichloromethane, Stabilized
Show MoreShow Less
Product Suggestions
You may be interested in these alternatives
Product Code. 10082190 Dichloromethane, 99.8%, for spectroscopy, stabilized with amylene
2045.77 NOK
2035.59 NOK/ 2.50 L
Save 10.18 NOK
0.50% Off
Product Code. 11308377 Dichloromethane, Spectrophotometric Grade, 99.7+%, stab. with amylene
930.00 NOK / 1 L
Customers who viewed this item also viewed This information does not imply a recommendation or representation of any kind and any action taken upon the information provided is strictly at your own risk. The products shown have not been qualified by Fisher Scientific for any particular purpose, application or use and it is the responsibility of the purchaser to ensure that the performance of the product is suitable for the purchasers specific use or intended application.
Viewing 1-3 of 7
Catalog No. 10001334 Acetonitrile, Optima ™ LC/MS Grade, Fisher CheAcetonitrile, Optima ™ LC/MS Grade, Fisher Che kr 6 504,52 / 2.50 L
Catalog No. 10244202 Acetone, Extra Pure, SLR, Fisher Chemical ™Acetone, Extra Pure, SLR, Fisher Chemical ™ kr 501,65 / 5 L
Catalog No. 10162180 Acetone, Certified AR for analysis, meets analyAcetone, Certified AR for analysis, meets analy kr 417,65 / 2.50 L
Catalog No. 10080130 Ethyl Acetate, Extra Pure, SLR, Fisher ChemicalEthyl Acetate, Extra Pure, SLR, Fisher Chemical kr 810,00 / 2.50 L
Catalog No. 10131560 Acetone, for HPLCAcetone, for HPLC kr 849,65 / 2.50 L
Catalog No. 10458210 Dichloromethane, 99+%, Extra Pure, Stabilised wDichloromethane, 99+%, Extra Pure, Stabilised w kr 914,84 / 2.50 L
Catalog No. 10429252 Dichloromethane, 99.9%, Extra Dry, stabilized,Dichloromethane, 99.9%, Extra Dry, stabilized, kr 691,61 / 100mL
Documents
Certificates
Certificates
A lot number is required to show results for certificates. To find your lot number on previous orders use our order status area.
Lot Number Certificate Type Search
5 results found
| Lot Number | Certificate Type | Date | Product Code |
--- --- |
| Lot NumberA0459396 | Certificate Type Certificate of Analysis | Date 17/07/2025 | Product Code 167770025,167775000 |
| Lot NumberA0459396 | Certificate Type Certificate of Analysis | Date 17/01/2024 | Product Code 167770025,167775000 |
| Lot NumberA0440639 | Certificate Type Certificate of Analysis | Date 29/09/2023 | Product Code 167770025,167775000 |
| Lot NumberA0401081 | Certificate Type Certificate of Analysis | Date 14/12/2022 | Product Code 167770025,167775000 |
| Lot NumberA0409896 | Certificate Type Certificate of Analysis | Date 13/12/2022 | Product Code 167770025,167775000 |
1
Safety and Handling
Product Identifier
Dichloromethane
Signal Word
Warning
Hazard Category
Carcinogenicity Category 2
Serious eye damage/eye irritation Category 2
Skin corrosion/irritation Category 2
Specific target organ toxicity Category 3
Hazard Statement
H315-Causes skin irritation.
H319-Causes serious eye irritation.
H336-May cause drowsiness or dizziness.
H351-Suspected of causing cancer.
Precautionary Statement
P280-Wear protective gloves/protective clothing/eye protection/face protection.
P302+P352-IF ON SKIN: Wash with plenty of water/soap.
P304+P340-IF INHALED: Remove person to fresh air and keep comfortable for breathing.
P312-Call a POISON CENTER/doctor/if you feel unwell.
P337+P313-If eye irritation persists: Get medical advice/attention.
Supplemental information
MIXTURE LIST-Contains: Dichloromethane
SDS
RUO – Research Use Only
Product Content Correction
Your input is important to us. Please complete this form to provide feedback related to the content on this product.
Product Title
Dichloromethane, 99.8%, for spectroscopy, stabilized with amylene >
Your Name
Email Address
Type of Issue
Detailed Description of Issue/Request
By clicking Submit, you acknowledge that you may be contacted by Fisher Scientific in regards to the feedback you have provided in this form. We will not share your information for any other purposes. All contact information provided shall also be maintained in accordance with our Privacy Policy.
CancelSubmit
Thank You! Your feedback has been submitted. Fisher Scientific is always working to improve our content for you. We appreciate your feedback.
Spot an opportunity for improvement?Share a Content Correction
About Us
Fisher Scientific
Quality Management
Work at Thermo Fisher Scientific
Policies and Notices
Customer Support
Help & Support Center
Returns
Account Dashboard
Report a Site Issue
Business Solutions
Biopharma
Biotech
Industry
Green Solutions
Programs & Services
Edge Program
New Lab Start-Up Program
SureTRACE Program
SureTRACE+ Program
Sustainability Program
Unity Lab Services
Connect with Us
Lab Reporter
News Corner
Subscribe to News
Webinars
We keep science moving forward by offering over 2.6 million products and extensive support services to the research, production, testing laboratories, healthcare, and science education markets. Count on us for an unrivaled selection of lab, life sciences, safety, and facility management supplies—including chemicals, consumables, equipment, instruments, diagnostics, and much more—along with exceptional customer care from an industry-leading team that’s proud to be part of Thermo Fisher Scientific.
Cookie Preferences
Website Terms & Conditions
Sales Terms & Conditions
Privacy Notice
Cancellation and Returns Policy
How Cookies are Used
Transparency Act Statement
Thermo Fisher Scientific - Fisher Scientific AS - Postboks 114, Smestad - 0309 Oslo - Norway © 2025 Thermo Fisher Scientific Inc. All rights reserved.
All trademarks are the property of Thermo Fisher Scientific and its subsidiaries unless otherwise specified.
Your choices regarding cookies on this site
We and our affiliates and vendors use cookies and similar technologies to operate our sites, recognize visitors to our sites, provide secure log-in, collect statistics to optimize site functionality, and deliver content tailored to your interests. Click Accept All to accept all cookies and go directly to the site, or click on Manage Settings to see detailed descriptions of the types of cookies and choose whether to accept certain cookies while on the site.
Manage Settings
Accept All
Feedback |
189014 | https://twitter.com/RAEinforma/status/1475749192987467784 | RAE on X: "@stevenuvlogs #RAEconsultas Las dos formas son correctas. En muchas áreas hispanohablantes es normal el uso del pretérito perfecto simple en contextos en que en otras zonas se usa el compuesto. V. las secciones 23.7 y 23.8 de la «NGLE»: / X
Don’t miss what’s happening
People on X are the first to know.
Log in
Sign up
Post
See new posts
Conversation
RAE
@RAEinforma
#RAEconsultas Las dos formas son correctas. En muchas áreas hispanohablantes es normal el uso del pretérito perfecto simple en contextos en que en otras zonas se usa el compuesto. V. las secciones 23.7 y 23.8 de la «NGLE»:
Translate post
8:43 AM · Dec 28, 2021
2
7
New to X?
Sign up now to get your own personalized timeline!
Sign up with Apple
Create account
By signing up, you agree to the Terms of Service and Privacy Policy, including Cookie Use.
Trending now
What’s happening
Paris Fashion Week 2025 Womenswear SS26
LIVE
Sports · Trending
Rays
10.9K posts
Politics · Trending
Grand Blanc
155K posts
San Diego Padres · Trending
Juan Soto
2,809 posts
Politics · Trending
Stop Trump
45.8K posts
Show more
Terms of Service
|
Privacy Policy
|
Cookie Policy
|
Accessibility
|
Ads info
|
More
© 2025 X Corp. |
189015 | https://es.khanacademy.org/math/ap-calculus-bc/bc-series-new/bc-10-1/v/partial-sum-notation | Introducción a las sumas parciales (video) | Khan Academy
Salta al contenido principal
Si estás viendo este mensaje, significa que estamos teniendo problemas para cargar materiales externos en nuestro sitio.
Si estás detrás de un filtro de páginas web, por favor asegúrate de que los dominios .kastatic.org y .kasandbox.org estén desbloqueados.
Explorar
Matemáticas: De preescolar a 8.º
Matemáticas: Cursos de preparación
Matemáticas: De 9.° a la universidad
Ciencia
Computación
Economía y finanzas
Selecciona una categoría para ver sus cursos.
Buscar
Haz una donaciónInicia sesiónRegístrate
Busca cursos, habilidades y videos
Saltar al contenido de la lección
Cálculo avanzado 2 (AP Calculus BC)
Curso: Cálculo avanzado 2 (AP Calculus BC)>Unidad 10
Lección 1: Definir series infinitas convergentes y divergentes
Sucesiones convergentes y divergentes
Ejemplo resuelto: convergencia o divergencia de una sucesión
Convergencia y divergencia de sucesiones
Introducción a las sumas parciales
Sumas parciales: fórmula para el enésimo término de la suma parcial
Sumas parciales: el valor de un término a partir de la suma parcial
Introducción a las sumas parciales
Series infinitas como el límite de sumas parciales
Sumas parciales y series
Matemáticas por area>
Cálculo avanzado 2 (AP Calculus BC)>
Series y sucesiones infinitas>
Definir series infinitas convergentes y divergentes
© 2025 Khan Academy
Términos de usoPolítica de privacidadAviso de cookiesDeclaración de accesibilidad
Introducción a las sumas parciales
Google Classroom
Microsoft Teams
Acerca de Acerca de este video Transcripción
La suma parcial de una serie infinita como la suma de los primeros términos (y por lo tanto "suma parcial"). Este concepto aparentemente sencillo es muy útil al pensar en series infinitas.
Saltar al final de las discusiones
Preguntas Sugerencias y agradecimientos
¿Quieres unirte a la conversación?
Inicia sesión
Ordenar por:
Más votados
federico-nicolas hace 2 años Publicado hace hace 2 años. Enlace directo a la publicación “este vídeo está mal, al f...” de federico-nicolas mayor este vídeo está mal, al final toma el 6 término de la sucesión, no está evaluando la sumatoria de 1 a 6, que seria sumar todos los términos de la sucesión desde el primero hasta el sexto
Respuesta Botón que navega a la página de registro •Comentar Botón que navega a la página de registro
(3 votos)
Votar a favor Botón que navega a la página de registro
Dar voto negativo Botón que navega a la página de registro
Marca Botón que navega a la página de registro
mayor
Respuesta
Mostrar vista previa Mostrar opciones de formato
Publicar respuesta
¿Sabes inglés? Haz clic aquí para ver más discusiones en el sitio en inglés de Khan Academy.
Transcripción del video
digamos que tienes una serie s muy bien que no es otra cosa más que una suma infinita verdad es la suma infinita desde n igual a 1 hasta infinito de los términos a n y esto sólo es una forma digamos compacta de escribir que estamos sumando a uno con a dos más a tres y así sumamos todos estos términos por siempre verdad es bueno porque son una infinidad de términos muy bien entonces lo que quiero hacer en este vídeo es introducirte a la idea de lo que es una suma parcial ok y lo que tenemos aquí es una serie mientras que una forma parcial por ejemplo sería ésta la sexta suma parcial la sexta suma parcial sería sumar los primeros seis términos de esta serie entonces tendríamos a uno más a dos más a tres 4 + a 5 + a 6 así de fácil esto es una suma parcial y a lo mejor esto está muy abstracto porque no sabe uno quién son los agenes entonces vamos a hacer un caso más concreto digamos que nuestra serie es la suma desde n igual a 1 hasta infinito y nuestros genes estarán dados por 1 entre n cuadrada muy bien entonces esto me está diciendo bueno por ejemplo cuando n vale 1 tenemos 1 entre 1 al cuadrado y eso vale 1 cuando en el vale 2 tenemos 1 entre 2 al cuadrado que es un cuarto y cuando n vale 3 tenemos 1 entre 3 al cuadrado que es un noveno y así seguimos sumando todos estos términos verdad todos estos términos entonces la pregunta sería por ejemplo cuál es el valor de la tercera suma parcial y como siempre te invito a que hagas una pausa y trates de resolver este problema por tu propia cuenta entonces ahora vamos a hacerlo juntos muy bien entonces el primer término habíamos visto que era 1 el segundo término era 1 entre 2 al cuadrado que es un cuarto y el tercer término como ya vimos es un noveno entonces cuánto vale esta suma bueno tenemos que encontrar un denominador común que puede ser 36 y un entero son 36 sobre 36 verdad un cuarto son 9 sobre 36 simplemente hacemos 36 entre 4 que son 9 y luego 36 entre 9 son 4 por 1 son 4 guardar un noveno son 4 entre 36 entonces si sumamos esto tenemos 36 más 4 que son 40 más 9 son 49 sobre 36 muy bien entonces este es el valor explícito de la tercera suma parcial que surge a partir de esta serie muy bien entonces el objetivo de este vídeo en realidad es apreciar la idea de lo que es una suma parcial y vamos a hacer otro ejercicio para seguir practicando que íbamos a quitar todo esto y digamos que nuevamente tenemos una serie que es la suma desde n igual a 1 hasta infinito de los términos a n pero ahora digamos que nuestro nuestra enésima suma parcial tiene una fórmula y digamos que es n cuadrada menos 3 sobre n cúbica más 4 muy bien y recordemos quién es ese en nuestra enésima suma parcial es sumar todos los términos de esta serie a 1 + a 2 hasta llegar al enésimo término al n y esto nos está diciendo que es n cuadrada menos 3 sobre n cúbica más 4 muy bien entonces digamos que te encuentras a alguien en la calle y ese alguien no sé a lo mejor sabe que tú ya conoces sobre sumas parciales y te dice bueno suponte que tenemos esta serie s ahí tenemos esta serie ese y tenemos la fórmula de las sumas parciales tú podrías decirme quién es la suma desde n igual a 1 hasta 6 de los genes entonces podrías intentarlo por tu cuenta de hecho te invito a que lo hagas pero por ahora vamos a hacerlo nosotros juntos verdad entonces recordemos que esto simplemente es a uno más a dos más a tres más a cuatro más a cinco más a seis estamos simplemente sumando los primeros seis términos así que si nosotros le ponemos mucho ojo a esta expresión simplemente podemos ver que es la sexta suma parcial verdad estamos sumando desde a 1 hasta a 6 verdad que corresponde como esta fórmula que tenemos aquí sin embargo tenemos que ese 6 se puede calcular mediante esta fórmula y está verdad en este caso sería 6 al cuadrado menos 3 dividido entre 6 al cubo + 4 muy bien entonces ahora simplemente nos queda calcular esto y entonces 6 al cuadrado es 36 si le quitamos 3 son 33 y dividimos entre 6 al cubo más 46 al cubo sería 6 por 6 que son 36 y 36 por 6 serían 3 por 6 son 18 verdad son 186 por 6 son 36 y la suma nos da 61 y llevamos una verdad 8 y 3 son 11 1 y 1 son 2 entonces son 216 más 4 esos son 220 y entonces aquí tenemos la expresión para la sexta suma parcial muy bien así que con este vídeo el objetivo era realmente conocer la anotación de lo que son las sumas parciales y entender realmente lo que significan
Creative Commons Attribution/Non-Commercial/Share-AlikeVideo en YouTube
A continuación: 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. 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 |
189016 | https://www.sketchy.com/mcat-lessons/common-values-of-sin-and-cos | Opens in a new window Opens an external website Opens an external website in a new window
We use cookies to provide necessary functionality and improve your experience. By remaining on this website you indicate your consent. Cookie Policy
Try for FreeLogin
GET 20% OFF SKETCHY MCAT WITH CODE REG20 | REGISTRATION DAY SALE
MCAT Curriculum
/
Physics
/
Math Fundamentals
/
Common Values of Sin and Cos
Common Values of Sin and Cos
Tags:
No items found.
Physics
It can be very help to memorize the values of sine and cosine for the most common angles (0, 30, 45, 60, and 90 degrees). For sine, as an angle increases from 0 degrees to 90 degrees, the value of sine also increases. The sine values for these angles are: sine of 0 is 0, sine of 30 is √1/2 (1/2), sine of 45 is √2/2, sine of 60 is √3/2, and sine of 90 is √4/2 (1). On the other hand, the values of cosine decrease as the angle size goes from 0 to 90 degrees. The cosine values for these angles are: cosine of 0 is √4/2 (1), cosine of 30 is √3/2, cosine of 45 is √2/2, cosine of 60 is √1/2 (1/2), and cosine of 90 is 0.
Lesson Outline
Most common angles: 0, 30, 45, 60, and 90 degrees
Starting with sine:
As angle increases from 0 to 90 degrees, value of sine increases
Sine values:
Sine of 0° = 0 (√0/2)
Sine of 30° = 1/2 (√1/2)
Sine of 45° = √2/2
Sine of 60° = √3/2
Sine of 90° = 1 (√4/2)
Moving to cosine:
As angle size goes from 0 to 90 degrees, values of cosine decrease
Cosine values:
Cosine of 0° = 1 (√4/2)
Cosine of 30° = √3/2
Cosine of 45° = √2/2 (equal to sine of 45°)
Cosine of 60° = 1/2 (√1/2)
Cosine of 90° = 0 (√0/2)
Recap: Sine values go up, cosine values go down
Don't stop here!
Get access to 29 more Physics lessons & 8 more full MCAT courses with one subscription!
Try 7 Days Free
FAQs
What are the most common trigonometric values for sin and cos that students should memorize?
Students should be familiar with sine and cosine values for standard angles such as 0°, 30°, 45°, 60°, and 90°. Common values include sin(0)=0, sin(30)=1/2, sin(45)=√2/2, sin(60)=√3/2, sin(90)=1, and for cosines: cos(0)=1, cos(30)=√3/2, cos(45)=√2/2, cos(60)=1/2, cos(90)=0.
How are the sine and cosine of an angle related to each other in basic trigonometry?
The sine and cosine functions are related through the Pythagorean identity: sin^2(x) + cos^2(x) = 1. This identity demonstrates that while the sine of an angle gives the ratio of the opposite side to the hypotenuse, the cosine gives the ratio of the adjacent side to the hypotenuse. Both sine and cosine functions are periodic, with a period of 360° or 2π radians.
What patterns can be observed when comparing sine and cosine values for different angles?
Several patterns can be seen in sine and cosine values for different angles. For every 90°, the values of sine and cosine are repeated or switched. For example, sine and cosine values for 30°, 150°, 210°, and 330° can be deduced from the values for 30° due to this pattern. Additionally, sine values correspond to the cosine values for an angle's complement, while the cosine values correspond to sine values for the angle's complement. This relationship is known as the cofunction identity.
Programs
MedicalMCATPANP
About us
CompanyBlogCareersPrivacy PolicyTerms of UseCookie Policy |
189017 | https://www.pw.live/school-prep/exams/binomial-theorem-formula | Binomial Theorem Formula: Explanation, Terms, Problems
All Courses
Competitive Exams
IIT JEE, NEET, ESE, GATE, AE/JE, Olympiad
Only IAS
UPSC, State PSC
School Preparation
Foundation (Class 6-12), Commerce, Arts, CuriousJr (1st - 8th), Science, International Boards
Govt Exam
Judiciary, SSC, Defence, Teaching, JAIIB & CAIIB, BIHAR EXAMS WALLAH, UP Exams, Railway, Nursing Exams, Banking, WB Exams
UG & PG Entrance Exams
MBA, IPMAT, IIT JAM, LAW, CUET UG, UGC NET, GMAT, Design & Architecture, Pharma, CSIR NET, CUET PG, NEET PG
FINANCE
CA, CS, Finance Courses, ACCA
Others
Online Degrees
English Proficiency Test
IELTS, TOEFL
Agriculture
Agriculture
IIT JEE NEET ESE GATE AE/JE Olympiad
VidyapeethPW SkillsPW StoreReal TestClass 1st - 8thPower Batch
Login/Register
SCHOOL
EXAMS
Binomial Theorem Formula
Binomial Theorem Formula: Explanation, Terms, Problems
The binomial theorem is a mathematical technique used for expanding expressions that have been raised to any positive integer power. This theorem serves as a versatile and valuable tool for expansion, finding applications across various fields, including algebra and probability.
Anchal Singh 16 Oct, 2023
Share
Binomial Theorem Formula – As the exponent grows, computing the expansion becomes increasingly intricate and laborious. The Binomial Theorem offers a convenient method for efficiently calculating the expansion of a binomial expression raised to a substantial power. Here, you'll gain insight into the binomial theorem's definition and statement, explore binomial expansion formulas, examine the properties of the binomial theorem, discover techniques for determining binomial coefficients, delve into the components of the binomial expansion, and explore its various applications.
Binomial Theorem Explanation
The binomial theorem is a mathematical technique used for expanding expressions that have been raised to any positive integer power. This theorem serves as a versatile and valuable tool for expansion, finding applications across various fields, including algebra and probability.
Binomial Expressions Defined
A binomial expression, in algebra, is an expression characterized by having exactly two dissimilar terms. Examples of such expressions include a + b, a 3 + b 3 , and so forth.
Binomial Expansion
The total number of terms in the expansion of (x + y) n is always (n + 1).
The sum of the exponents of x and y in each term of the expansion is always equal to n.
The binomial coefficients, denoted as nC 0 , nC 1 , nC 2 and so on (also represented as C 0 , C 1 , C 2, and so forth), play a significant role in the expansion.
Binomial coefficients that are symmetrically positioned in the expansion, both from the beginning and the end, are equal. For example, nC 0 = nC n , nC 1 = nC n-1 , nC 2 = nC n-2 , and so on.
Binomial Expansion Formula: Let n be a non-negative integer, and x, y be real numbers. The binomial expansion formula for (x + y) n is given by:
(x + y) n = n Σ r=0 nC r x n – r · y r
Where:
nCr represents the binomial coefficient, which is also expressed as Cn,r.
x and y are real numbers.
The sum is taken from r = 0 to n, covering all terms in the expansion.
Binomial Expansion Example
Expanding a Binomial Expression:
Expand (a + b)^4.
Using the Binomial Theorem, we get:
(a + b)^4 = C(4,0) a^4 b^0 + C(4,1) a^3 b^1 + C(4,2) a^2 b^2 + C(4,3) a^1 b^3 + C(4,4) a^0 b^4
Simplifying this expression gives the expanded form:
(a^4) + 4(a^3)(b) + 6(a^2)(b^2) + 4(a)(b^3) + (b^4)
Finding Specific Terms:
Determine the coefficient of the term containing x^3 in the expansion of (2x + 3)^5.
Using the Binomial Theorem, we can focus on the term with x^3:
C(5, k) (2x)^(5-k) (3)^k
To get x^3, we set (5 - k) equal to 3, so k = 2. Plug this into the expression:
C(5, 2) (2x)^(5-2) (3)^2 = C(5, 2) (2^3x^3) 9
C(5, 2) is the binomial coefficient, equal to 10:
10 8x^3 9 = 720x^3
So, the coefficient of the x^3 term is 720.
Finding the Middle Term:
Find the middle term in the expansion of (x + y)^6.
To find the middle term, we first calculate the total number of terms in the expansion using (n + 1), where n = 6:
Total terms = 6 + 1 = 7 terms
The middle term is the fourth term, as there are 3 terms on each side:
(x + y)^6 = C(6, 3) x^3 y^3
C(6, 3) is the binomial coefficient, equal to 20:
20 x^3 y^3
So, the middle term in the expansion is 20x^3y^3.
These examples illustrate how the Binomial Theorem can be used to expand expressions, find specific terms, and locate the middle term in a binomial expansion.
Also Check – Cubes Roots Formula
Binomial Expansion Formulas
The Binomial Expansion Formula allows you to expand a binomial expression raised to a positive integer power (n). Here is the general formula:
(x + y)^n = Σr=0 to n [nCr x^(n-r) y^r]
In this formula:
(x + y)^n represents the binomial expression to be expanded.
n is a non-negative integer, which is the exponent to which the binomial expression is raised.
nCr, also known as binomial coefficient or combination, is calculated as C(n, r) and represents the number of ways to choose r items from a set of n items. It can be computed as C(n, r) = n! / (r! (n - r)!), where! denotes factorial.
x^(n-r) represents x raised to the power of (n - r).
y^r represents y raised to the power of r.
The summation Σr=0 to n indicates that you need to sum up terms for all values of r from 0 to n.
Using this formula, you can expand any binomial expression (x + y)^n to find the individual terms in the expansion. These terms are often referred to as "binomial coefficients" multiplied by powers of x and y.
Here are some common binomial expansion formulas for specific values of n:
Square of a Binomial (n = 2):
(x + y)^2 = x^2 + 2xy + y^2
Cubing a Binomial (n = 3):
(x + y)^3 = x^3 + 3x^2y + 3xy^2 + y^3
Expanding a Binomial to the Fourth Power (n = 4):
(x + y)^4 = x^4 + 4x^3y + 6x^2y^2 + 4xy^3 + y^4
These are some basic examples, and you can use the general binomial expansion formula to calculate expressions for any positive integer value of n. The coefficients in these formulas are determined by binomial coefficients (nCr), which describe the combinations of terms needed to expand the binomial expression.
Some Other Useful Expansions
Sum and Difference of Binomial Expansions:
The sum of (x + y)^n and (x - y)^n is equal to 2 times the series with coefficients C0 x^n + C2 x^(n-1) y^2 + C4 x^(n-4) y^4 + ...
The difference between (x + y)^n and (x - y)^n is equal to 2 times the series with coefficients C1 x^(n-1) y + C3 x^(n-3) y^3 + C5 x^(n-5) y^5 + ...
Binomial Expansion of (1 + x)^n:
The expansion of (1 + x)^n is given by nΣr=0 [nCr x^r], which can be expressed as C0 + C1 x + C2 x^2 + ... + Cn x^n.
Sum and Difference of (1 + x)^n:
The sum of (1 + x)^n and (1 - x)^n is equal to 2 times the series with coefficients C0 + C2 x^2 + C4 x^4 + ...
The difference between (1 + x)^n and (1 - x)^n is equal to 2 times the series with coefficients C1 x + C3 x^3 + C5 x^5 + ...
Number of Terms in (x + a)^n + (x - a)^n:
If "n" is even, the number of terms in the expansion of (x + a)^n + (x - a)^n is (n + 2)/2.
If "n" is odd, the number of terms in the expansion of (x + a)^n + (x - a)^n is (n + 1)/2.
Number of Terms in (x + a)^n - (x - a)^n:
If "n" is even, the number of terms in the expansion of (x + a)^n - (x - a)^n is n/2.
If "n" is odd, the number of terms in the expansion of (x + a)^n - (x - a)^n is (n + 1)/2.
These rephrased statements provide a clearer understanding of the properties and results related to binomial expansions.
Also Check – Rational Number Formula
Terms in the Binomial Expansion
General Term: The expression that represents any term within the binomial expansion.
In the binomial expansion of (x + y)^n, each term is represented by the general term Tr+1, which can be expressed as Tr+1 = nCr x^(n-r) y^r.
General Term in (1 + x)^n:
For the binomial expansion of (1 + x)^n, the general term is simply nCr x^r.
Position of the rth Term in (x + y)^n:
In the binomial expansion of (x + y)^n, the term positioned as the rth term from the end corresponds to the (n - r + 2)th term in the expansion.
Middle Term: The term located at the center of the expansion, especially crucial when the number of terms is odd.
In cases where "n" represents an even number, the middle term corresponds to the term located at position (n/2 + 1) within the binomial expansion.
However, when "n" is an odd number, the middle terms are found at positions [(n+1)/2] and [(n+3)/2] within the expansion.
Independent Term: The term that does not have either of the variables (x or y) and often corresponds to the constant term.
The term that is independent of 'x' in the expansion of [ax^p + (b/x^q)]^n is represented by Tr+1 = nCr a^(n-r) b^r, where 'r' is an integer determined by the formula r = (np)/(p + q).
Determining a Particular Term: The process of finding a specific term in the expansion, often guided by the term's position or power.
In the expansion of (ax^p + b/x^q)^n, the coefficient of x^m is determined by the coefficient of Tr+1, where the value of r is calculated as [(np - m)/(p + q)].
In the expansion of (x + a)^n, the ratio of the coefficient of the (r+1)th term (Tr+1) to the coefficient of the rth term (Tr) is given by (n - r + 1)/r times the constant term 'a' divided by 'x'.
Numerically Greatest Term: The term with the highest absolute value or magnitude among all the terms in the expansion.
If [(n+1)|x|]/[|x|+1] = P, where P is a positive integer, then the Pth term and (P+1)th terms are the numerically greatest terms in the expansion of (1+x)^n.
However, if [(n+1)|x|]/[|x|+1] = P + F, where P is a positive integer, and 0 < F < 1, then only the (P+1)th term is numerically the greatest term in the expansion of (1+x)^n.
Ratio of Consecutive Terms/Coefficients: The relationship between consecutive terms, which can provide insights into the behavior and convergence of the expansion.
The coefficient of xr in the expansion is nCr, while the coefficient of xr + 1 is nCr + 1.
The relationship between these coefficients can be expressed as (nCr / nCr - 1) = (n - r + 1) / r.
These terms and concepts are essential when working with binomial expansions, allowing for precise calculations and a deeper understanding of the expansion's properties.
Applications of Binomial Theorem
The Binomial Theorem finds a wide array of applications in mathematics, including techniques for:
Finding Remainders Using Binomial Theorem:
For instance, to find the remainder when 7103 is divided by 25, you can utilize the Binomial Theorem to express 7103 as a binomial power of 7 and then simplify it to determine the remainder, which is 18.
Similarly, in the context of the fractional part of (2403 / 15), the Binomial Theorem aids in expressing the fraction in a simplified form, helping you find the value of K, which is 8 in this case.
Finding Digits of a Number:
The Binomial Theorem can be employed to extract the last two digits of a number, as demonstrated with (13)10, yielding the result 49.
Relation between Two Numbers:
In scenarios where you need to compare or establish a relationship between two numbers like 9950 + 10050 and 10150, the Binomial Theorem can be employed to demonstrate the inequality and provide a clear understanding of the relation.
Divisibility Tests:
To determine the divisibility of numbers, the Binomial Theorem can be applied. For example, it can be used to illustrate that 119 + 911 is divisible by 10.
Formulae:
The Binomial Theorem is accompanied by several formulae, such as the formula for the number of terms in the expansion of (x1 + x2 + … + xr)^n, which is (n + r - 1)Cr - 1. Additionally, the formula for finding the sum of the coefficients of (ax + by)^n is (a + b)^n.
Further, if you have a polynomial function f(x) = (a0 + a1x + a2x^2 + …. + amx^m)^n, the Binomial Theorem provides formulas for calculating the sum of coefficients and the sum of coefficients of even and odd powers of x.
These applications and formulae demonstrate the versatility and utility of the Binomial Theorem in solving various mathematical problems and exploring relationships between numbers and polynomials.
Also Check – Linear Equation Formula
Problems on Binomial Theorem
Question 1:
Find the positive value of λ for which the coefficient of x^2 in the expression x^2[√x + (λ/x^2)]^10 is 720.
Solution: In the expression x^2[√x + (λ/x^2)]^10, we aim to determine λ. By applying the Binomial Theorem and simplifying, we conclude that λ^2 = 16, yielding two possible solutions: λ = ±4. However, since we seek a positive value, λ = 4.
Question 2:
The sum of the real values of x for which the middle term in the binomial expansion of (x^3/3 + 3/x)^8 equals 5670 is?
Solution: To find the sum of real values of x for which the middle term equals 5670, we first identify the middle term (T5) and then solve for x. The solutions are x = ±√3. Therefore, the sum of these real values is zero.
Question 3:
Let (x + 10)^50 + (x – 10)^50 = a0 + a1x + a2x^2 + ... + a50x^50 for all x ∈R, then a2/a0 is equal to?
Solution: In the expression (x + 10)^50 + (x – 10)^50, we aim to find a2/a0. After calculation, we determine that a2/a0 = 12.25.
Question 4:
Find the coefficient of x^9 in the expansion of (1 + x)(1 + x^2)(1 + x^3) ... (1 + x^100).
Solution: To find the coefficient of x^9, we consider the ways x^9 can be formed. There are eight such ways, each contributing a coefficient of 1. Therefore, the coefficient of x^9 is 8.
Question 5:
The coefficients of three consecutive terms of (1 + x)^n+5 are in the ratio 5:10:14. Find n.
Solution: By considering the ratio of coefficients, we derive two equations involving 'n' and 'r.' Solving these equations, we find n = 6.
Question 6: Find the digit in the units place of the number 183! + 3183.
Solution: By analyzing the unit digits of 183! and 3183, we determine that the units digit of their sum is 7.
Question 7: Find the total number of terms in the expansion of (x + a)^100 + (x – a)^100.
Solution: The total number of terms in the expansion of (x + a)^100 + (x – a)^100 is 51.
Question 8: Find the coefficient of t^4 in the expansion of [(1-t^6)/(1 – t)].
Solution: To find the coefficient of t^4 in [(1-t^6)/(1 – t)], we expand the expression and extract the coefficient, which is 15.
Binomial Theorem Formula FAQs
Give the binomial theorem formula.
We use the binomial theorem to find the expansion of the algebraic terms of the form(x + y)n. The formula is (x + y)n = Σr=0n nCr xn – r · yr.
What is the general term in a binomial expansion?
The general term of a binomial expansion is Tr+1 = nCr xn-r yr.
What is the number of terms in the expansion of (x + a)n + (x-a)n ?
The number of terms in the expansion of (x + a)n + (x-a)n is (n+2)/2 if n is even or (n+1)/2 if n is odd.
List two applications of the binomial theorem.
In Mathematics, the binomial theorem is used to find the remainder and also find the digits of a number.
🔥 Trending Blogs
NCERT Solutions for Class 6 English A Pact With the Sun NCERT Solutions for Class 6 English A Pact With the Sun
NCERT Solutions for Class 9 English Chapter 5 The Snake and the Mirror NCERT Solutions for Class 9 English Chapter 5 The Snake and the Mirror
NCERT Solutions Class 9 English Chapter 4 A Truly Beautiful Mind NCERT Solutions Class 9 English Chapter 4 A Truly Beautiful Mind
NCERT Solutions Class 9 English Poem Chapter 4 The Lake Isle of Innisfree NCERT Solutions Class 9 English Poem Chapter 4 The Lake Isle of Innisfree
NCERT Solutions Class 9 English Poem Chapter 2 Wind NCERT Solutions Class 9 English Poem Chapter 2 Wind
Talk to a counsellor Have doubts? Our support team will be happy to assist you!
08448982616 0 08448982616
Check out these Related Articles
CLASS 9 EXAM NCERT Solutions Class 9 English Poem Chapter 3 Rain On The Roof Read Full Story 28 Sept 2025, 10:55 pm
CLASS 9 EXAM Important Questions for Class 9 Maths Chapter 15 Probability Read Full Story 28 Sept 2025, 10:54 pm
CLASS 7 EXAM NCERT Solutions for Class 7 Social Science History PDF Download Read Full Story 28 Sept 2025, 10:49 pm
CLASS 7 EXAM NCERT Solutions for Class 7 Science Chapter 18 Wastewater Story Read Full Story 28 Sept 2025, 10:18 pm
CLASS 7 EXAM NCERT Solutions for Class 7 Science Chapter 16 Water A Precious Resource Read Full Story 28 Sept 2025, 10:15 pm
CLASS 7 EXAM NCERT Solutions for Class 7 Science Chapter 15 Light Read Full Story 28 Sept 2025, 10:14 pm
CLASS 7 EXAM NCERT Solutions for Class 7 Science Chapter 14 Electric Current and Its Effects Read Full Story 28 Sept 2025, 10:14 pm
CLASS 7 EXAM NCERT Solutions for Class 7 Science Chapter 13 Motion and Time Read Full Story 28 Sept 2025, 10:13 pm
CLASS 7 EXAM NCERT Solutions for Class 7 Science Chapter 11 Transportation in Animals and Plants Read Full Story 28 Sept 2025, 10:12 pm
Next slide
Join 15 Million students on the app today!
Live & recorded classes available at ease
Dashboard for progress tracking
Millions of practice questions at your fingertips
Physics Wallah
We understand that every student has unique needs and abilities, that’s why our curriculum is designed to adapt to your needs and help you grow!
Let’s get social :
Company
About UsContact UsCareersUpdatesAccount Deletion
Our Centres
New DelhiPatnaKotaNoidaDhanbadVaranasiView All
Popular Exams
IIT JEENEETGATENDAUPSCSchool Prep
Connect with us
Email usTalk to a counseller
Quick Links
PW PrernaPW SIPPartner with us
Our Products
PW Learning AppPW Offline Payments (Fin-Z)
Free Learning Resources
PW Books
NEET Books
IIT JEE Books
NEET Books
PW Stationery
UPSC CSE Books
PW Merchandise
NDA/Defence Books
GATE Books
Device Courses
CUET Books
CA/Finance Books
UPPSC Books
RRB Books
BPSC Books
SSC Books
MPSC Books
MPPSC Books
Banking Books
Notes (Class 10-12)
Class 10 Math's Notes
Class 10 Chemistry Notes
Class 10 Physics Notes
Class 10 Biology Notes
Aptitude & Reasoning
Class 10 Geography
Physics Class 11 Notes
Class 11 Chemistry Notes
Maths Notes Class 11
Zoology Class 11
Class 11 Botany Notes
Physics Class 12 notes
Chemistry Class 12
Maths Notes Class 12
Zoology class 12
Class 12th Botany Notes
PW Study Materials
Arjuna JEE Study Material
JEE 7 PYQs
JEE Mind Maps
Yakeen Study Material
NCERT Punch
Arjuna Neet Study Material
NEET PYQs
CBSE Sample Paper
CBSE Question Bank
Udaan For Class 10th Study Material
UPSC Study Material
PW Stationery
PW Merchandise
PW Device Courses
Notes (Class 6-9)
Class-6 Theory & Notes
Math's Notes for class 7
Science Notes for class 7
Class 8 Math Notes
Class 8 Chemistry Notes
Class 8 Physics Notes
Class 8 Biology Notes
Class 8 SST Notes
Class 9 Math's Notes
Class 9 Physics Notes
Class 9 Chemistry Notes
Class 9 Biology Notes
Ncert Solutions
NCERT Solutions For Class 6
NCERT Solutions For class 7
NCERT Solutions for class 8
NCERT Solutions for class 9
NCERT Solutions for class 10
NCERT Solutions for class 11
NCERT Solutions for Class 12
Govt Exams
SSC Exam
Banking Exam
Teaching Exam
JAIIB CAIIB Exam
Railway Exam
Defence Exams
State PSC Exam
SSC Exam
Entrance Exams
Gate Exam
IIT JEE Exam
NEET Exam
CUET Exam
CSIR-NET Exam
MBA Exam
Nursing Exam
IIT JEE Exam
Our Other Websites
PW Store
PW Skills
PWOnlyIAS
MedEd
PW Gulf
PW IOI
CuriousJr
PW Skills Blogs
Class 6th to 12th Online Courses
Class 12 Online Course
Class 11 Online Course
Class 10 Online Course
Class 9 Online Course
Class 8 Online Course
Govt Job Exams Courses
SSC Online Coaching
Bank Exam Online Coaching
TET Online Coaching
JAIIB & CAIIB Online Coaching
Bihar Exam Online Coaching
UPSC Coaching
UPSC Online Coaching
UPSC Offline & Hybrid Coaching
State PSC Online Coaching
UPPSC Online Coaching
BPSC Online Coaching
Defence Exam Coaching
NDA Online Coaching
CDS Online Coaching
AFCAT Online Coaching
Agniveer Online Coaching
Gate Exam Coaching
Civil Engineering Courses
Mechanical Engineering Courses
CS & IT Courses
ECE Courses
EE Courses
Formulas
Science formulas
Chemistry formulas
Physics formulas
Math's Formulas
Other Exams
CA Exam
Company Secretary Exam
Know about Physics Wallah
Physics Wallah is an Indian edtech platform that provides accessible & comprehensive learning experiences to students from Class 6th to postgraduate level. We also provide extensive NCERT solutions, sample paper, NEET, JEE Mains, BITSAT previous year papers & more such resources to students. Physics Wallah also caters to over 3.5 million registered students and over 78 lakh+ Youtube subscribers with 4.8 rating on its app.
We Stand Out because
We provide students with intensive courses with India’s qualified & experienced faculties & mentors. PW strives to make the learning experience comprehensive and accessible for students of all sections of society. We believe in empowering every single student who couldn't dream of a good career in engineering and medical field earlier.
Our Key Focus Areas
Physics Wallah's main focus is to make the learning experience as economical as possible for all students. With our affordable courses like Lakshya, Udaan and Arjuna and many others, we have been able to provide a platform for lakhs of aspirants. From providing Chemistry, Maths, Physics formula to giving e-books of eminent authors like RD Sharma, RS Aggarwal and Lakhmir Singh, PW focuses on every single student's need for preparation.
What Makes Us Different
Physics Wallah strives to develop a comprehensive pedagogical structure for students, where they get a state-of-the-art learning experience with study material and resources. Apart from catering students preparing for JEE Mains and NEET, PW also provides study material for each state board like Uttar Pradesh, Bihar, and others
Privacy Policy
Terms of use
Copyright © 2025 Physicswallah Limited All rights reserved. |
189018 | https://www.chegg.com/homework-help/questions-and-answers/2-find-inverse-laplace-transform-x-s-gi-roc-re-s-0-b-x-s-845845-roc-3-re-s-q36329962 | Your solution’s ready to go!
Our expert help has broken down your problem into an easy-to-learn solution you can count on.
Question: 2. Find the inverse Laplace Transform (a) X(s)-みgi ROC: Re(s)> 0 (b) X(s) 845845 ; ROC :-3 〈 Re(s) <-2 (c) X()-C+IG2) for all the three possible choices of ROC s+1
P6 singal and system
Not the question you’re looking for?
Post any question and get expert help quickly.
Chegg Products & Services
CompanyCompany
Company
Chegg NetworkChegg Network
Chegg Network
Customer ServiceCustomer Service
Customer Service
EducatorsEducators
Educators |
189019 | http://www.geometry-math-journal.ro/pdf/Volume1-Issue1/ON%20THE%20STANDARD%20LENGTHS%20OF%20ANGLE%20BISECTORS%20AND%20THE%20ANGLE%20BISECTOR%20THEOREM.pdf | Global Journal of Advanced Research on Classical and Modern Geometries ISSN: 2284-5569, pp.15-27 ON THE STANDARD LENGTHS OF ANGLE BISECTORS AND THE ANGLE BISECTOR THEOREM G.W INDIKA SHAMEERA AMARASINGHE ABSTRACT. In this paper the author unveils several alternative proofs for the standard lengths of Angle Bisectors and Angle Bisector Theorem in any triangle, along with some new useful derivatives of them.
2010 Mathematical Subject Classification: 97G40 Keywords and phrases: Angle Bisector theorem, Parallel lines, Pythagoras Theorem, Similar triangles.
1. INTRODUCTION In this paper the author introduces alternative proofs for the standard length of An-gle Bisectors and the Angle Bisector Theorem in classical Euclidean Plane Geometry, on a concise elementary format while promoting the significance of them by acquainting some prominent generalized side length ratios within any two distinct triangles existed with some certain correlations of their corresponding angles, as new lemmas. Within this paper 8 new alternative proofs are exposed by the author on the angle bisection, 3 new proofs each for the lengths of the Angle Bisectors by various perspectives with also 5 new proofs for the Angle Bisector Theorem.
1.1. The Standard Length of the Angle Bisector Date: 1 February 2012 .
15 G.W Indika Shameera Amarasinghe The length of the angle bisector of a standard triangle such as AD in figure 1.1 is AD2 = AB · AC −BD · DC, or AD2 = bc 1 −(a2/(b + c)2) according to the standard notation of a triangle as it was initially proved by an extension of the angle bisector up to the circumcircle of the triangle. Nevertheless within this analysis the author adduces 3 new alternative methods in order to obtain the standard length of the angle bisector using some elementary Euclidean Geometry techniques without even being used trigonometry or vector Algebra at least just a little bit as follows.
Main Results 2. FIRST ALTERNATIVE METHOD: PROOF OF FIGURE 2 [ ADB = [ CAD + [ ACB = [ BAD + [ ACB [ AEX = [ ACB + [ CBE = [ ACB + [ CAD = [ ACB + [ BAD Likewise [ ADB = [ AEX and since [ BAD = [ CAD, the triangles ABD and AXE are similar.
Hence AD AE = AB AX. So, by replacing AX = AD −DX, AD2 = AD · DX + AB · AE.
Substituting AE = AC −CE, AD2 = AB · AC −AB · EC + AD · DX (2.1) The angles [ CAD and [ CBE are equal and the angle ACB is common for both the triangles ADC and BEC. Hence the triangles ADC and BEC are similar. Likewise EC DC = BC AC means EC BC = DC AC and since AD is an angle bisector, AB AC = BD DC (standard ratio). So, DC AC = BD AB.
Hence EC BC = BD AB, means that AB · EC = BC · BD (2.2) The angles [ CBE and [ BAD are equal and the angle ADB is common for both the triangles ABD and BDX. Hence, the triangles ABD and BDX are similar. Hence BD AD = DX BD . So, AD · DX = BD2 (2.3) Hence by substituting above values from (2.2) and (2.3) to (2.1) AD2 becomes AD2 = AB · AC −BC · BD + BD2 = AB · AC −BD · (BC −BD) = AB · AC −BD · DC. Likewise AD2 = AB · AC −BD · DC, and since c b = BD DC (as AD is the bisector) BD = ac b+c and DC = 16 On the Standard Lengths of Angle Bisectors and the Angle Bisector Theorem ab b+c, thus by replacing those values AD2 = bc −( a2bc (b+c)2 ), hence AD2 = bc 1 −( a2 (b+c)2 ) .
Likewise the proof is completed.
When angle [ CBE > [ ABC, means b A 2 > b B, the E point will lie on extended CA. Thus above correlation AD2 = bc 1 −( a2 (b+c)2 ) can easily be proved exactly as it has been proved earlier.
Hence it can be easily adduced the lengths of the Bisectors of angle ABC and the an-gle ACB such that h ac 1 −( b2 (a+c)2 ) i 1 2 and h ab 1 −( c2 (a+b)2 ) i 1 2 respectively, comparing with AD.
3. SECOND ALTERNATIVE METHOD: PROOF OF FIGURE 3 AEDF is a parallelogram(as DE and DF are parallel to AC and AB respectively). So, [ ADE = [ CAD and [ ADF = [ BAD, since [ BAD = [ CAD and [ ADE = [ ADF. Thus angle [ BAD = [ CAD = [ ADE = [ ADF. So, the AEDF parallelogram becomes a Rhombus and because of that AE = ED = DF = AF while AD and EF diagonals are perpendicular for each of them at O as angle [ AOE = 90◦. Moreover AO = OD = AD 2 .
BD DC = c b (As AD is the Angle Bisector). Hence BD = ac b + c (3.1) Angle [ ABC is common for both the triangles BDE and ABC. Angle [ EDB = [ ACB (DE and AC are parallel). Likewise the triangles BDE and ABC are similar. Therefore DE b = BD a , and replacing in (3.1), we get DE = bc b+c. One obtains: DE = DF = AE = AF = bc b + c.
(3.2) Angle [ BAD is common for both the triangles ADN and AOE and angle [ AOE = \ AND = 90◦. Hence the triangles ADN and AOE are similar, so that AE AD = AO AN = AD 2·AN. Thus AD2 = 2AN · AE (3.3) 17 G.W Indika Shameera Amarasinghe Angle \ AND = \ AMC = 90◦and [ BAC = [ BED (since DE and AC are parallel). Hence the triangles DEN and AMC are similar, so that EN AM = DE b , replacing by (3.2), EN = AM · c b + c (3.4) Considering the triangle BMC, a2 = CM2 + BM2 = b2 −AM2 + (c −AM)2 = b2 + c2 − 2c · AM , thus AM = b2 + c2 −a2 2c (3.5) Substituting from (3.5) to (3.4), one obtains: EN = b2 + c2 −a2 2(b + c) (3.6) AN = AE + EN, thus by replacing from (3.2) and (3.6), AN = (b + c)2 −a2 2(b + c) (3.7) Substituting from (3.2) and (3.7) to (3.3), we get: AD2 = 2bc b+c · h (b+c)2−a2 2(b+c) i .
Simplifying this furtherAD2 = bc 1 − a2 (b+c)2 . Thus the proof is successfully com-pleted.
4. THIRD ALTERNATIVE METHOD: PROOF OF FIGURE 4 BD DC = c b (as AD is the angle bisector), so that BD = ac b+c and DC = ab b + c (4.1) Considering the triangle ABX, c2 = AX2 + BX2 = AD2 −DX2 + (BD + DX)2 = AD2 + BD2 + 2BD · DX, thus 2BD · DX = c2 −AD2 −BD2 (4.2) Considering the triangle AXC, 18 On the Standard Lengths of Angle Bisectors and the Angle Bisector Theorem b2 = AX2 + CX2 = AD2 −DX2 + (DC −DX)2 = AD2 + DC2 −2DC · DX, thus 2DC · DX = AD2 + DC2 −b2 (4.3) From (4.2) and (4.3), one obtains: BD DC = c2 −AD2 −BD2 AD2 + DC2 −b2 .
(4.4) Replacing from (4.1), we get: c b = c2 −AD2 − ac b+c 2 AD2 + ab b+c 2 −b2 .
Thus simplifying this further AD2(b + c) = bc2 −b · ac b + c 2 + b2c −c · ab b + c 2 so, one obtains: AD2(b + c) = bc(b + c) −a2bc b+c . Thus AD2 = bc − a2bc (b+c)2 .
Likewise AD2 = bc 1 −( a2 (b+c)2 ) . Hence the proof is again successively completed.
Remark 1. The length of the External Angle Bisector can be assumed with the use of (4.4) as follows.
AN is a part of the external angle bisector of angle [ BAC. Let’s we imagine that the extended angle bisector AN will meet the extended BC line at D, then the length of the external angle bisector becomes AD.
Proof.
BD DC = c b (As AD is the external bisector), hence BC+DC DC = c b, so that BC DC = c−b b .
Whence CD = ab c−b.
Using (4.4) to the ABD, BC DC = c2−AC2−BC2 AC2+CD2−AD2 .
19 G.W Indika Shameera Amarasinghe Since AC = b, BC = a and CD = ab c−b, c−b b = c2−b2−a2 b2−(ab/(c−b))2−AD2 , simplifying this further from several steps, AD2 · (c −b)2 + bc · (c −b)2 = a2bc , whence the distance of the External Angle Bisector AD can be adduced such that, AD2 = bc h a c−b 2 −1 i .
5. FIRST ALTERNATIVE PROOF FOR THE ANGLE BISECTOR THEOREM Angles [ BDE and [ CDF are equals (vertically opposite angle) and [ BED = [ CFD = 90◦(BE and CF are perpendiculars), thus the triangles BDE and CDF are similar.
So that, BD DC = BE CF (5.1) Angles [ BAD and [ CAD are equals (as AD is the angle bisector) and [ AEB = [ AFC = 90◦ (BE and CF are perpendiculars), thus the triangles ABE and AFC are similar.
So that c b = BE CF (5.2) From (5.1) and (5.2) we get: BD DC = c b 20 On the Standard Lengths of Angle Bisectors and the Angle Bisector Theorem 6. SECOND ALTERNATIVE PROOF FOR THE ANGLE BISECTOR THEOREM [ BAO = [ XAO (as AD is the angle bisector) and [ AOB = [ AOX = 90◦(AO is perpendicular to BX). Therefore the triangles AOB and AOX are congruent. Hence BO = OX and AB = c = AX.
Moreover, from BO = OX and [ DOB = [ DOX = 90◦we get the congruence of the triangles BOD and XOD. So that, BD = DX. Thereafter since AB = AX , BD = DX and AD is common for both the triangles ABD and ADX, then the triangles ABD and ADX are also congruent. Hence the areas of the triangles ABD and ADX are the same which means ∆ABD ≡∆ADX.
The ratios of the respective areas of triangles are as follows: Area∆ADX Area∆ADC = AX AC = c b Since ∆ABD ≡∆ADX, Area∆ABD Area∆ADC = c b. We observe that Area∆ABD Area∆ADC = BD DC easily. Likewise BD DC = c b.
7. THIRD ALTERNATIVE PROOF FOR THE ANGLE BISECTOR THEOREM USING FIGURE 6 Using Menelaus theorem in the triangle BCX, one obtains: AX AC · DC BD · BO OX = 1 Replacing relevant values and since BO = OX we get c b · DC BD = 1.
Thus BD DC = c b.
21 G.W Indika Shameera Amarasinghe 8. FOURTH ALTERNATIVE PROOF FOR THE ANGLE BISECTOR THEOREM [ BAE = [ CAE (as AD is the Angle Bisector). Moreover, [ BAE = [ CAE = [ BCE and [ CAE = [ BAE = [ CBE. (as ABCD is a cyclic quadrilateral).
So that [ CBE = [ BCE, hence BE = EC (8.1) Since [ CBE = [ BAE and [ AEB is common for both the triangles ABE and BDE, likewise the triangles ABE and BDE are similar. So that BD c = BE AE (8.2) Since [ BCE = [ CAE and [ AEC is common for both the triangles AEC and EDC, likewise the triangles AEC and EDC are similar. So that DC b = EC AE (8.3) Since BE = EC from (8.1) then (8.2) and (8.3) coincide. Thus BD c = DC b means BD DC = c b.
9. FIFTH ALTERNATIVE PROOF FOR THE ANGLE BISECTOR THEOREM 22 On the Standard Lengths of Angle Bisectors and the Angle Bisector Theorem [ BAD = [ CAD (AD is the angle bisector) and [ APD = \ AQD = 90◦, AD is common for both the triangles APD and ADQ, likewise APD and ADQ are congruent. So that DP = DQ. Angle [ ABC is common for both the triangles BDP and BCF and [ BPD = d BFC = 90◦, likewise the triangles BDP and BCF are similar. Hence BD a = DP CF , so that BD · CF = a · DP (9.1) Angle [ ACB is common for both the triangles DCQ and BEC and [ DQC = AngleBEC = 90◦, likewise the triangles DCQ and BEC are similar. Hence DC a = DQ BE , so that BE · DC = a · DQ and since DP = DQ as it proved earlier, we get: BE · DC = a · DP (9.2) From (9.1) and (9.2), one obtains BE · DC = BD · CF, so that BD DC = BE CF (9.3) Angle [ BAC is common for both the triangles ABE and AFC and [ AFC = [ AEB = 90◦, likewise the triangles ABE and AFC are similar. Hence BE CF = c b (9.4) From (9.3) and (9.4), one obtains BD DC = c b.
Remark – When AD becomes the External Angle Bisector, it can be easily speculated and proved that BD DC = c b as before within each of those 5 new proofs presented for the Internal Angle Bisector by a slight difference of AD line in each figure.
10. SEVERAL DERIVATIVES OF THE LENGTHS OF ANGLE BISECTORS Let us imagine that ∆PQR and ∆ABC are any two distinct Euclidean triangles such that angle [ BAC = 2[ QPR = 2φ and [ ABC = [ PQR = θ. BC = a, AC = b, AB = c. Draw the AD angle bisector up to D on BC. Hereafter we divulge the following 6 significant Lemmas emerged using the Angle Bisection.
BD = ac b + c since BD DC = c b 23 G.W Indika Shameera Amarasinghe AD2 = bc 1 − a2 (b + c)2 and using the similar triangles PQR and ABD, we get PR AD = PQ c Hence, Lemma 1 can be denoted as PR PQ = AD c = h bc 1 − a2 (b+c)2 i 1 2 c = b(b + c −a)(a + b + c) c(b + c)2 1 2 Moreover PR AD = QR BD.
So that Lemma 2 can be denoted as, QR PR = BD AD = ac b + c 1 h bc 1 − a2 (b+c)2 i 1 2 = a c b(b + c −a)(a + b + c) 1 2 PQ c = QR BD, thus Lemma 3 can be denoted as, QR PQ = BD c = ac b+c c = a b + c Let us imagine again that ∆PQR and ∆XYZ are another two distinct Euclidean triangles such that [ YXZ = 2[ QPR and [ XYZ = 2[ PQR. YZ = x, XZ = y, XY = z. Draw the XD and YE angle bisectors up to D on YZ and E on XZ respectively while they intersect at I.
DY = xz y + z, DZ = xy y + z.
DI XI = DY YX = xz y+z z = x y+z , hence XI = XD · y+z x+y+z and the value of XD = h zy 1 − x2 (y+z)2 i 1 2 , thus XI = (yz((y+z)2−x2)) 1 2 y+z · y+z x+y+z = (yz((y+z)2−x2)) 1 2 x+y+z From the fact that ∆PQR and ∆XYI are similar triangles, one obtains: PQ z = PR XI and PR PQ = XI z therefore, Lemma 4 can be adduced as 24 On the Standard Lengths of Angle Bisectors and the Angle Bisector Theorem PR PQ = (yz((y + z)2 −x2)) 1 2 z(x + y + z) = s y(y + z −x) z(x + y + z) YI IE = z XE = z zy x+z = x+z y , hence YI = x+z x+y+z · YE , moreover the value of YE = xz 1 − y2 (x + z)2 1 2 , hence YI = [xz((x+z)2−y2)] 1 2 x+y+z .
PR XI = QR YI , QR PR = YI XI = xz((x + z)2 −y2) 1 2 x + y + z · x + y + z [yz((y + z)2 −x2)] 1 2 , thus Lemma 5 can be adduced as, QR PR = s x(x + z −y) y(y + z −x) QR YI = QP z , QR PQ = IY z = xz((x + z)2 −y2) 1 2 z(x + y + z) .
Therefore Lemma6 can be adduced as QR PQ = s x(x + z −y) z(x + y + z) Considering these above ratios it is very unambiguous that the ratios of one particular triangle can be adduced from the lengths of other corresponding triangle, consequently these propositions can be diffused and used towards many significant improvements of Advanced Classical Euclidean Geometry.
11. AN ELEMENTARY PROBLEM ON THE ANGLE BISECTION Suppose that KLMN is a quadrilateral at which the point P is located on its LN diagonal such that KP = PM, being both KP and PM are the angle bisectors of the opposite angles [ LKN and \ LMN respectively. Then it is proved that KL = LM and KN = NM and [ LKN = \ LMN.
25 G.W Indika Shameera Amarasinghe Let angle d LKP = α and angle [ LMP = β Proof of the correlation of KLMN Quadrilateral KP and PM are the bisectors of angle [ LKN and \ LMN respectively, whence KP2 = KL · KN −LP · PN PM2 = ML · MN −LP · PN Since KP = PM, KL · KN −LP · PN = ML · MN −LP · PN, KL · KN = ML · MN (11.1) As KP is the angle bisector, KL KN = LP PN, and as PM is the angle bisector, ML MN = LP PN, whence KL KN = ML MN (11.2) By the use (11.1) and (11.2), KL = LM and KN = MN and whence by the congruence of triangles it is proved that α = β, hence [ LKN = \ LMN.
Conclusion of Remarks- The readers are kindly encouraged to have a precise look at those interlocutory derivatives mentioned in 10 and 11 as well as particularly in , and of the references to grasp a better comprehension about the significance of the Angle Bisector Theorems on which some felicitous correlations are often emerged in Advanced Euclidean Geometry.
Acknowledgements- The author would like to render his worm gratitude to the Editor in chief for giving his invaluable comments and dedication to have successfully concluded and pub-lished this paper on the ”Global Journal of Advanced Research on Classical and Modern Geome-tries(G.J.A.R.C.M.G.)”.
REFERENCES G.W.I.S Amarasinghe, A Prominent Correlation On the Extended Angle Bisector, Journal of the World Federation of National Mathematics Competitions, 24(1), 2011, pp.33 – 36.
G.W.I.S Amarasinghe, Solutions for the problem 43.3, Mathematical Spectrum, 43(3), 2011, pp.138 – 139.
G.W.I.S. Amarasinghe, Advanced Plane Geometry Research 1, Proceedings of the 66th Annual Sessions of Sri Lanka Association for the Advancement of science(SLAAS), 66, 2010, pp.77.
26 On the Standard Lengths of Angle Bisectors and the Angle Bisector Theorem G.W.I.S. Amarasinghe, Advanced Plane Geometry Research 2, Proceedings of the 66th Annual Sessions of Sri Lanka Association for the Advancement of science(SLAAS), 66, 2010, pp.82.
G.W.I.S. Amarasinghe, The Jungle Gym, Problem 260, A Parametric Equation, Math Horizons, 18(4), 2011, pp.30.
G.W.I.S. Amarasinghe, Advanced Plane Geometry Research 3: Alternative Proofs for the Standard Theo-rems in Plane Geometry, Proceedings of the 66th Annual Sessions of Sri Lanka Association for the Advancement of Science(SLAAS), 66, 2010, pp.78.
G.W.I.S Amarasinghe, A New Theorem On Any Right Angled Cevian Triangle, Journal of the World Federation of National Mathematics Competitions, 24(2), 2011, pp.29-37.
R.H Buchholz, On Triangles with rational altitudes, angle bisectors or medians, Bulletin of the Australian Mathematical Society, 45(3), 1992, pp.525 – 526.
S.H Wu, Z.H Zhang, A Class of Inequalities Related to the Angle Bisectors and the Sides of a Triangle, Journal of Inequalities in Pure and Applied Mathematics, 7(3), Article 108, 2006.
UNIVERSITY OF KELANIYA, KELANIYA, SRI LANKA E-mail address: indshamat@gmail.com 27 |
189020 | https://www.quora.com/Among-all-the-convex-nonagons-what-is-the-maximum-number-of-points-of-intersection-of-the-diagonals-inside-the-nonagon | Something went wrong. Wait a moment and try again.
Mathematics Word Problems
Irregular Polygons
Combinatorial Math
Mathematical Problems
Geometric Mathematics
5
Among all the convex nonagons, what is the maximum number of points of intersection of the diagonals inside the nonagon?
·
To find the maximum number of points of intersection of the diagonals inside a convex nonagon (9-sided polygon), we can use a combinatorial approach.
Understanding the Diagonal Intersections
: Each intersection of the diagonals occurs between two pairs of non-adjacent vertices. For each intersection point, we need to choose 4 vertices from the 9 vertices of the nonagon, as each intersection is formed by the diagonals connecting two pairs of these vertices.
Choosing 4 Vertices
: The number of ways to choose 4 vertices from 9 is given by the binomial coefficient
(
n
k
)
, where
n
is the total nu
To find the maximum number of points of intersection of the diagonals inside a convex nonagon (9-sided polygon), we can use a combinatorial approach.
Understanding the Diagonal Intersections
: Each intersection of the diagonals occurs between two pairs of non-adjacent vertices. For each intersection point, we need to choose 4 vertices from the 9 vertices of the nonagon, as each intersection is formed by the diagonals connecting two pairs of these vertices.
Choosing 4 Vertices
: The number of ways to choose 4 vertices from 9 is given by the binomial coefficient
(
n
k
)
, where
n
is the total number of vertices and
k
is the number of vertices to choose. Thus, we need to calculate
(
9
4
)
.
(
9
4
)
=
9
!
4
!
(
9
−
4
)
!
=
9
×
8
×
7
×
6
4
×
3
×
2
×
1
=
126
3. Conclusion
: Therefore, the maximum number of points of intersection of the diagonals inside a convex nonagon is given by the number of ways to choose 4 vertices, which is 126.
Thus, the maximum number of points of intersection of the diagonals inside a convex nonagon is
126
.
Mohit Rai
Former SDE at Directi (2020–2021)
·
Updated 8y
choose any 4 vertex from all the vertices of polygon.
Diagonals made by using all the choosen vertices only make 1 intersection.Therefore points of intersection of the diagonal made inside the polygon = nC4 = n! / ( n - 4 )! 4!
here n = number of vertices.
maximum number of points of intersection of the diagonal inside the nonagon ( n => 9 ) = 9C4 = 9! / 4! 5! = 126.
choose any 4 vertex from all the vertices of polygon.
Diagonals made by using all the choosen vertices only make 1 intersection.Therefore points of intersection of the diagonal made inside the polygon = nC4 = n! / ( n - 4 )! 4!
here n = number of vertices.
maximum number of points of intersection of the diagonal inside the nonagon ( n => 9 ) = 9C4 = 9! / 4! 5! = 126.
Sponsored by Morgan & Morgan, P.A.
Are you prepared?
Most people don’t know what to do if they get into a car accident. Make sure you’re not one of them.
Mitani Chawla
10y
I find it a bit tedious to type so attached my written solution.
I hope it helps. :)
I find it a bit tedious to type so attached my written solution.
I hope it helps. :)
Related questions
What's the maximum number of points of intersection between an n-sided shape and a circle?
We draw all diagonals of a convex n-gon. Suppose no three diagonals pass through a point. Into how many parts Tn is the n-gon divided?
Trace all diagonals of an n-sided regular polygon. What is the number of distinct interior points (not on the boundary) where two or more diagonals intersect?
What is the least number of interior angles that are obtuse or right in a 11 sided convex polygon?
How many different lengths of diagonal does a regular nonagon have?
Nithin Santhosh
Field Specialist at Baker Hughes (company) (2018–present)
·
7y
Maximum possible point of intersection of diagonals of any n sided polygon is nC4. So here it's a nonagon so n=9 and ans is 9C4.
Mike Alexander
Studied A-level Mathematics
·
Author has 2.3K answers and 2.8M answer views
·
4y
Related
There is no such point in a convex n-sided polygon where more than two diagonals of the polygon pass through. How many points are there in the polygon where the diagonals intersect?
Consider the figure, a 9-vertex polygon. We will consider the number of intersections on each diagonal fanning from vertex A.
The first diagonal, AH, has one vertex to the left and 6 to the right. So there will be 1×6 diagonals intersecting AH. The next, AG, has 2 vertices to the left and 5 to the right - so will have 2×5 intersections. The next, AF, has 3 vertices to the left and 4 to the right, so 3×4 intersections. AE has 4 to the left and 3 to the right… etc.
So the total intersections for this fan from A is:
1×6 + 2×5 + 3×4 + 4×3 + 5×2 + 6×1
Now this can be seen to equal
1 + (1+2) + (1+2+3) +
Consider the figure, a 9-vertex polygon. We will consider the number of intersections on each diagonal fanning from vertex A.
The first diagonal, AH, has one vertex to the left and 6 to the right. So there will be 1×6 diagonals intersecting AH. The next, AG, has 2 vertices to the left and 5 to the right - so will have 2×5 intersections. The next, AF, has 3 vertices to the left and 4 to the right, so 3×4 intersections. AE has 4 to the left and 3 to the right… etc.
So the total intersections for this fan from A is:
1×6 + 2×5 + 3×4 + 4×3 + 5×2 + 6×1
Now this can be seen to equal
1 + (1+2) + (1+2+3) + (1+2+3+4) + (1+2+3+4+5) + (1+2+3+4+5+6)
…which is the sum of the first 6 triangular numbers: 1+3+6+10+15+21 = 56.
But not every intersection occurs on this fan. So we must multiply this number by the number of vertices, then divide by 4 (since each intersection is connected to 4 vertices).
So the answer in this case is 56 × 9/4 = 14×9 = 126.
In general, for an n-vertex polygon, each fan has n-3 diagonals, so the total intersections is given by
n/4 × f:(n-3)
where f:x denotes the sum of the first x triangle numbers.
If there's a nice formula for f:x then I'm afraid I don't know it!
EDIT:
So I've googled it, and it turns out there is a nice formula for the sum of the first x triangular numbers. It's
f:x = x(x+1)(x+2)/6
So, substituting x = n-3, our answer is:
n4×(n−3)(n−2)(n−1)6
=n(n−1)(n−2)(n−3)24
Related questions
How many different lengths of diagonal does a regular nonagon have?
What is the meaning of minimum points of intersection and maximum points of intersection?
What is the least number of interior angles that are obtuse or right in a 11 sided convex polygon?
What is the maximum number of different points where the diagonals of an irregular hexagon intersect?
What is the maximum number of diagonal in concave nonagon (9 slides) which can be drawn entirely outside of the interior of the polygon?
Roland Testa
Worked at U.S. Army (1986–1989)
·
Author has 4.6K answers and 2.5M answer views
·
4y
Related
There is no such point in a convex n-sided polygon where more than two diagonals of the polygon pass through. How many points are there in the polygon where the diagonals intersect?
First, to be clear, the statement does not apply to all “convex polygons.” This is a hexagon with a center point intersected by 3 diagonals. For all even numbers n, there will be a central point where n/2 diagonals intersect. Note there are 2n points within this polygon where exactly 2 diagonals intersect. The diagonals also intersect at their end points, but maybe we can ignore those.
A strict interpretation, that is, one counting the vertices of the polygon as intersection points of the diagonals, produces only the result n=5, with n interior points intersected by exactly 2 diagonals.
Removing
First, to be clear, the statement does not apply to all “convex polygons.” This is a hexagon with a center point intersected by 3 diagonals. For all even numbers n, there will be a central point where n/2 diagonals intersect. Note there are 2n points within this polygon where exactly 2 diagonals intersect. The diagonals also intersect at their end points, but maybe we can ignore those.
A strict interpretation, that is, one counting the vertices of the polygon as intersection points of the diagonals, produces only the result n=5, with n interior points intersected by exactly 2 diagonals.
Removing the central intersection point requires that n be an odd number. So step up to the next odd number. For n=7, there are 5n interior points of diagonal intersection.
Well then, what about n=9? Now there are 14n of these points. Feel free to count them.
Well, the easy answer is that your n must be 5, and either 5 or 10 is the answer. That satisfies the strict interpretation. Well, that or 4, where the answer is 1.
But the question is formatted such that there should me a generalized answer. So, is there an answer? That’s a 23-page paper by Bjorn Poonen and Michael Rubenstein:
I’ll have to leave that as my answer.
Sponsored by Grammarly
Is your writing working as hard as your ideas?
Grammarly’s AI brings research, clarity, and structure—so your writing gets sharper with every step.
Doug Dillon
Ph.D. Mathematics
·
Author has 12.3K answers and 11.4M answer views
·
3y
Related
In a convex 100-gon, no three diagonals pass through the same point. How many intersections of diagonals (different from vertices) are there?
In the diagram below is an intersection of two diagonals with the distracting information removed. See what determined that intersection. Well, of course, two diagonals cause the point and in a larger view, a quadrilateral determines the diagonals which determines the point. So for every choice of 4 vertices, an intersection point is had. So we count
(
100
4
)
=
3921225
.
In the diagram below is an intersection of two diagonals with the distracting information removed. See what determined that intersection. Well, of course, two diagonals cause the point and in a larger view, a quadrilateral determines the diagonals which determines the point. So for every choice of 4 vertices, an intersection point is had. So we count
(
100
4
)
=
3921225
.
Brian Peterson
Former professor at San Jose State University (1978–2015)
·
Upvoted by
Alon Amit
, Lover of math. Also, Ph.D. · Author has 670 answers and 539.4K answer views
·
2y
Related
There is no such point in a convex n-sided polygon where more than two diagonals of the polygon pass through. How many points are there in the polygon where the diagonals intersect?
This question was posted over two years ago, but landed in my feed just now. Mike Alexander answered that the number is n4 times the sum of the first n−3 triangular numbers and found that this equals n(n−1)(n−2)(n−3)24. This happens to equal the binomial coefficient (n4).
There is an easy way to see that the answer is (n4). Any crossing point is where two diagonals intersect and these have four endpoints between them (as none of their endpoints could coincide or we wouldn’t call this a crossing point). So each crossing point determines a 4-element subset of t
This question was posted over two years ago, but landed in my feed just now. Mike Alexander answered that the number is n4 times the sum of the first n−3 triangular numbers and found that this equals n(n−1)(n−2)(n−3)24. This happens to equal the binomial coefficient (n4).
There is an easy way to see that the answer is (n4). Any crossing point is where two diagonals intersect and these have four endpoints between them (as none of their endpoints could coincide or we wouldn’t call this a crossing point). So each crossing point determines a 4-element subset of the n-element set of vertices of the polygon. Conversely, given a 4-element subset of the set of vertices, among the three ways to group the 4 points into two disjoint pairs and draw the diagonal connecting each pair, for exactly one of these three ways will the two diagonals intersect, giving one of our crossings points. In this way we have mutually inverse correspondences between the set of crossing points and the set of 4-element subsets of the vertex set. There are (n4) of the latter, hence also of the former.
Sponsored by Zoho One
Run your entire small business with one software suite - Zoho One
Award-winning business software. 50M+ users across the globe trust Zoho. Start your free trial today!
Alexander Mathey
Former Chemical Engineer, retired, lives in Athens, GR
·
Author has 5.6K answers and 10.8M answer views
·
5y
Related
What is the maximum number of different points where the diagonals of an irregular hexagon intersect?
A polygon has n(n - 3)/2 diagonals.
For n = 6, this calculates to 63/2 = 9 diagonals.
So, the number of intersection points, (some of them may lie outsice the hexagon) equals the combinations of 9 items taken 2 at a time.
C(9, 2) = 9!/[2!(9 - 2)!] = 89/2 = 36
But these comprise the vertices, which are counted thrice each, since three diagonals intersect on each of them.
So, 36 = 18 intersection points coincide with the vertices.
That leaves 36 - 18 = 18 points where two diagonals intersect.
A polygon has n(n - 3)/2 diagonals.
For n = 6, this calculates to 63/2 = 9 diagonals.
So, the number of intersection points, (some of them may lie outsice the hexagon) equals the combinations of 9 items taken 2 at a time.
C(9, 2) = 9!/[2!(9 - 2)!] = 89/2 = 36
But these comprise the vertices, which are counted thrice each, since three diagonals intersect on each of them.
So, 36 = 18 intersection points coincide with the vertices.
That leaves 36 - 18 = 18 points where two diagonals intersect.
James Buddenhagen
Lives in Xico,Veracruz.Mexico (2006–present)
·
Author has 2.6K answers and 4.2M answer views
·
6y
Related
How many triangles can be formed by joining the vertices of a nonagon?
How many triangles can be formed by joining the vertices of a nonagon?
Here is what it looks like:
According to this site the total number of triangles formed is 1302.
However, if you require that each of the three vertices of each triangle is one of the vertices of the given nonagon, then there are far fewer, namely “9 take 3”, that is:
(93)=9×8×73×2×1=3×4×7=84.
How many triangles can be formed by joining the vertices of a nonagon?
Here is what it looks like:
According to this site the total number of triangles formed is 1302.
However, if you require that each of the three vertices of each triangle is one of the vertices of the given nonagon, then there are far fewer, namely “9 take 3”, that is:
(93)=9×8×73×2×1=3×4×7=84.
Sponsored by Amazon Web Services (AWS)
Want to level up your AWS game? Register for re:Invent 2025.
5 days of advanced architecture workshops, serverless deep dives, and coding labs. Las Vegas, Dec 1-5.
Dave Benson
BSc in Physics, Chemistry, and Mathematics (science grouping), University of London
·
Author has 6K answers and 2M answer views
·
4y
Related
If all the diagonals of a decagon are drawn, at how many points will they intersect?
Number of diagonals in a polygon is s(s-3)/2 where s is number of sides.
10(10–3)/2 = 35 diagonals.
Sum to n terms is Sn={n(2a+(n-d)}/2
a is first term and n is number of terms and d is arithmetic change between each.
Sn=35{(2+(35–1)}/2 so 3536/2=630 points. Answer.
Sripad Sambrani
Knows Sanskrit
·
Author has 6.8K answers and 2.9M answer views
·
2y
Related
All the diagonals are contained in the convex n-angle. It is known that no three of them intersect at any point. How many parts will the n- angle be divided into?
A line-segment which connects any pair of non-adjacent vertices of a polygon is defined as its diagonal. Leaving aside the adjacent pair of sides, (n-2) diagonals that can be drawn from a given vertex of a polygon. Hence every interior angle is divided into (n-2) non-congruent angles, an irregular polygon considered.
Trust this helps.
A line-segment which connects any pair of non-adjacent vertices of a polygon is defined as its diagonal. Leaving aside the adjacent pair of sides, (n-2) diagonals that can be drawn from a given vertex of a polygon. Hence every interior angle is divided into (n-2) non-congruent angles, an irregular polygon considered.
Trust this helps.
Amitabha Tripathi
have been teaching Discrete Mathematics for almost 40 years
·
Author has 4.7K answers and 13.9M answer views
·
1y
Related
How do I know how many incision points does a convex n-gon's diagonals have? I am particularly interested in ones with 20 edges.
Let
C
be a convex
n
-gon,
n
≥
4
. Denote the vertices of
C
by
1
,
…
,
n
, arranged clockwise. Each point
P
of intersection of two chords corresponds to a quadruple of points
P
1
,
P
2
,
P
3
,
P
4
which form the endpoints of the two intersecting chords. This correspondence sets up a bijection between the sets of quadruples of points chosen from
1
,
…
,
n
and the set of intersection points of chords. Therefore, the number of intersection points equals the number of selections of quadruples of points, and this equals
(
n
4
)
, since all permutations of
P
1
,
P
2
,
P
3
,
P
4
result in the same point of
Let C be a convex n-gon, n≥4. Denote the vertices of C by 1,…,n, arranged clockwise. Each point P of intersection of two chords corresponds to a quadruple of points P1,P2,P3,P4 which form the endpoints of the two intersecting chords. This correspondence sets up a bijection between the sets of quadruples of points chosen from 1,…,n and the set of intersection points of chords. Therefore, the number of intersection points equals the number of selections of quadruples of points, and this equals (n4), since all permutations of P1,P2,P3,P4 result in the same point of intersection.
The number of points of intersection is (n4)=124n(n−1)(n−2)(n−3).
Related questions
We draw all diagonals of a convex n-gon. Suppose no three diagonals pass through a point. Into how many parts Tn is the n-gon divided?
Trace all diagonals of an n-sided regular polygon. What is the number of distinct interior points (not on the boundary) where two or more diagonals intersect?
What's the maximum number of points of intersection between an n-sided shape and a circle?
How many diagonals are there in a 60 sided convex plane?
What are the maximum number of diagonals in a polygon?
How many different lengths of diagonal does a regular nonagon have?
What is the meaning of minimum points of intersection and maximum points of intersection?
What is the least number of interior angles that are obtuse or right in a 11 sided convex polygon?
What is the maximum number of different points where the diagonals of an irregular hexagon intersect?
What is the maximum number of diagonal in concave nonagon (9 slides) which can be drawn entirely outside of the interior of the polygon?
How many diagonals are there in a regular nonagon?
There is no such point in a convex n-sided polygon where more than two diagonals of the polygon pass through. How many points are there in the polygon where the diagonals intersect?
In a convex 100-gon, no three diagonals pass through the same point. How many intersections of diagonals (different from vertices) are there?
What is the maximum and the minimum number of points of intersection when three lines intersect?
What is the maximum possible number of intersection of a point of diagonals lying inside a hexagon?
About
·
Careers
·
Privacy
·
Terms
·
Contact
·
Languages
·
Your Ad Choices
·
Press
·
© Quora, Inc. 2025 |
189021 | https://nyjm.albany.edu/j/2002/8-2.pdf | New York Journal of Mathematics New York J. Math. 8 (2002) 9–30.
On Commuting Matrix Differential Operators Rudi Weikard Abstract. If the differential expressions P and L are polynomials (over C) of another differential expression they will obviously commute.
To have a P which does not arise in this way but satisfies [P, L] = 0 is rare. Yet the question of when it happens has received a lot of attention since Lax presented his description of the KdV hierarchy by Lax pairs (P, L). In this paper the question is answered in the case where the given expression L has matrix-valued coefficients which are rational functions bounded at infinity or simply periodic functions bounded at the end of the period strip: if Ly = zy has only meromorphic solutions then there exists a P such that [P, L] = 0 while P and L are not both polynomials of any other differential expression. The result is applied to the AKNS hierarchy where L = JD + Q is a first order expression whose coefficients J and Q are 2 × 2 matrices. It is therefore an elementary exercise to determine whether a given matrix Q with rational or simply periodic coefficients is a stationary solution of an equation in the AKNS hierarchy.
Contents 1.
Introduction 10 2.
The rational case 11 3.
The simply periodic case 16 4.
Application to the AKNS system 17 5.
The lemmas 19 Appendix A.
The theorems of Halphen and Floquet 26 Appendix B.
Higher order systems of differential equations 27 Appendix C.
Wasow’s theorem 28 References 29 Received November 8, 2000.
Mathematics Subject Classification. 34M05, 37K10.
Key words and phrases. Meromorphic solutions of differential equations, KdV-hierarchy, AKNS-hierarchy, Gelfand-Dikii-hierarchy.
Research supported in part by the US National Science Foundation under Grant No. DMS-9970299.
ISSN 1076-9803/02 9 10 Rudi Weikard 1. Introduction Consider the differential expression L = Q0 dn dxn + · · · + Qn.
When does a differential expression P exist which commutes with L? This question has drawn attention for well over one hundred years and its relationship with com-pletely integrable systems of partial differential equations has led to a heightened interest in the past quarter century. A recent survey by F. Gesztesy and myself tries to capture a part of that story and might be consulted for further references.
If L = d2/dx2 +q the problem is related to the Korteweg-de Vries (KdV) hierar-chy which, according to Lax , can be represented as the hierarchy of equations qt = [P2n+1, L] for n = 0, 1, . . . , where P2n+1 is a certain differential expression of order 2n + 1. The stationary solutions of these equations give rise to commuting differential expressions and play an important role in the solution of the Cauchy problem of the famous KdV equation (the case n = 1). Relying on a classical result of Picard , Gesztesy and myself discovered in that, when q is an elliptic func-tion, the existence of an expression P2n+1 which commutes with L is equivalent to the property that for all z ∈C all solutions of the equation Ly = zy are meromor-phic functions of the independent variable. This discovery was since generalized to cover certain 2 × 2 first order systems with elliptic coefficients (see ) and scalar n-th order equations with rational and simply periodic coefficients (see ).
According to the famous results of Burchnall and Chaundy in and a commuting pair of scalar differential expressions is associated with an algebraic curve and this fact has been one of the main avenues of attack on the problems posed by this kind of integrable systems (Its and Matveev , Krichever , , ). For this reason such differential expressions or their coefficients have been called algebro-geometric.
In this paper I will consider the case where the coefficients Q0, . . . , Qn of L are m × m matrices with rational or simply periodic entries. First let us make the following definition.1 Definition 1. A pair (P, L) of differential expressions is called a pair of nontrivially commuting differential expressions if [P, L] = 0 while there exists no differential expression A such that both P and L are in C[A].
I will give sufficient conditions for the coefficients Qj which guarantee the exis-tence of a P such that (P, L) is a nontrivially commuting pair when mn is larger than one.2 Theorem 1 covers the rational case while Theorem 2 covers the periodic case. These results are then applied to the AKNS hierarchy to obtain a character-ization of all rational and simply periodic algebro-geometric AKNS potentials (see Theorem 3).
The main ingredients in the proofs are generalizations of theorems by Halphen and Floquet , which determine the structure of the solutions of Ly = zy.
1The definition is motivated by the following observation. The expressions P and L commute if they are both polynomials of another differential expression A, i.e., if P, L ∈C[A]. Note that this does not happen in the case discussed above, i.e., when L = d2/dx2 +q and P is of odd order, unless q is constant.
2When m = n = 1 and [P, L] = 0 then P is necessarily a polynomial of L.
On Commuting Matrix Differential Operators 11 The original theorems cover the scalar case. The generalizations, which are quoted in Appendix A, are proven in and , respectively.
Algebro-geometric differential expressions with matrix coefficients have attracted a lot of attention in the past. The papers by Cherednik , Dickey , Dubrovin , van Moerbeke , Mumford , and Treibich form a (rather incomplete) list of investigations into the subject.
The organization of the paper is as follows: Sections 2 and 3 contain the state-ments and proofs of Theorems 1 and 2, respectively. Section 4 contains a short description of the AKNS hierarchy as well as Theorem 3 and its proof. The proofs of Theorems 1 and 2 rely on several lemmas which do not specifically refer to one or the other case. These lemmas are stated and proved in Section 5. Finally, for the convenience of the reader, three appendices provide the statements of the theo-rems of Halphen and Floquet, a few facts about higher order systems of differential equations, and the statement of a theorem of Wasow on the asymptotic behavior of solutions of a system of first order differential equations depending on a parameter.
Before we actually get started let us agree on a few pieces of notation. If F is a field we denote by F[x] the ring of polynomials with coefficients in F and by F(x) the associated quotient field. The ring of j ×k matrices with entries in F is denoted by Fj×k. The letter A represents the field of algebraic functions in one variable over the complex numbers. The symbol 1 denotes an identity matrix. Occasionally it is useful to indicate its dimension by a subscript as in 1k. Similarly 0 and 0j×k denote zero matrices. Polynomials are to be regarded as polynomials over C unless the contrary is explicitly stated.
2. The rational case Theorem 1. Let L be the differential expression given by Ly = Q0y(n) + Q1y(n−1) + · · · + Qny.
Suppose that the following conditions are satisfied: 1. Q0, . . . , Qn ∈C(x)m×m are bounded at infinity.
2. Q0 is constant and invertible.
3. The matrix B(λ) = λnQ0 + λn−1Q1(∞) + · · · + Qn(∞) is diagonalizable (as an element of Am×m).
4. There are linearly independent eigenvectors v1, . . . , vm ∈Am of B such that limλ→∞vj(λ), j = 1, . . . , m, exist and are linearly independent eigenvectors of Q0. In particular Q0 is diagonalizable.
If mn > 1 and if, for all z ∈C, all solutions of Ly = zy are meromorphic, then there exists a differential expression P with coefficients in C(x)m×m such that (P, L) is a pair of nontrivially commuting differential expressions.
Note that Conditions 3 and 4 are automatically satisfied if all eigenvalues of Q0 are algebraically simple.
Proof. Without loss of generality we will assume that Q0 is diagonal. Lemma 1 gives a large class of differential expressions P which commute with L. Our goal 12 Rudi Weikard is therefore to check the hypotheses of Lemma 1. After that we will address the question of finding a P which commutes nontrivially with L.
Let M = C(x). For j = 0, . . . , n let Q∞,j = Qj(∞) and let the function τ be the identity. The eigenvectors of B, which are linearly independent as elements of Am, become linearly dependent (as elements of Cm) for at most finitely many values of λ, since the determinant of the matrix whose columns are these eigenvectors is an algebraic function. Conditions 1–3 of Lemma 1 are then satisfied. Next we have to construct U such that Conditions 4 and 5 are also satisfied.
Let the characteristic polynomial of B be given by det(B(λ) −z) = ν j=1 fj(λ, z)mj where the fj ∈C[λ, z] are pairwise relatively prime. Denote the degree of fj(λ, ·) (which does not depend on λ) by kj. According to Lemma 2 we may choose λ among infinitely many values such that: 1. B(λ) is diagonalizable.
2. (f1 . . . fν)(λ, ·) has k1 + · · · + kν distinct roots.
3. if zj,k(λ) is a root of fj(λ, ·), then λ is a simple root of (f1 . . . fν)(·, zj,k(λ)).
Until further notice we will think of this value of λ as fixed and, accordingly, we will typically suppress the dependence on λ of the quantities considered.
Let zj,k be a root of fj(λ, ·), i.e., an eigenvalue of B(λ) of multiplicity mj.
The equation Ly = zy is equivalent to a first-order system ψ′ = A(z, ·)ψ where A(z, x) ∈Cmn×mn remains bounded as x tends to infinity.
By Lemma 3 the characteristic polynomial of A(z, ∞) is a constant multiple of ν j=1 fj(λ, z)mj and hence λ is an eigenvalue of A(zj,k, ∞) of algebraic multiplicity mj. But Lemma 3 implies also that the geometric multiplicity of λ is equal to mj. Theorem 2.4 of (quoted in Appendix A), which is a generalization of a theorem of Halphen, guarantees then the existence of mj linearly independent solutions ψj,k,ℓ(x) = Rj,k,ℓ(x) exp(λx), ℓ= 1, . . . , mj of ψ′ = A(zj,k, ·)ψ where the components of Rj,k,ℓare rational functions.
The common denominator q of these components is a polynomial in x whose coefficients are independent of λ and zj,k since the poles of the solutions of Ly = zy may occur only at points where one of the coefficient matrices Qj has a pole. Moreover, q may be chosen such that the entries of qQj are polynomials for all j ∈{1, . . . , n}.
The Rj,k,ℓ, ℓ= 1, . . . , mj, may have poles at infinity whose order can be deter-mined from asymptotic considerations. We denote the largest order of these poles, i.e., the largest degree of the numerators of the components of the Rj,k,ℓby s and perform the substitution ψ(x) = exp(λx) q(x) s j=0 αjxs−j.
This turns the equation ψ′ = A(z, ·)ψ into the equivalent equation s+s′ ℓ=0 xs+s′−ℓ j+k=ℓ {(s −j)qk−1 −Γk(λ, z)}αj = 0 (1) On Commuting Matrix Differential Operators 13 where s′ = deg(q) and where the Γk and qk are defined respectively by q(x)A(z, x) + (q′(x) −λq(x)) = s′ k=0 Γk(λ, z)xs′−k and q(x) = s′ k=0 qkxs′−k (quantities whose index is out of range are set equal to zero). Equation (1) repre-sents a system of (s + s′ + 1)mn linear homogeneous equations for the (s + 1)mn unknown components of the coefficients αj and is thus equivalent to the equation A(λ, z)β = 0 where A is an appropriate (s+s′+1)mn×(s+1)mn matrix and β is a vector with (s+1)mn components comprising all components of all the αj. Lemma 4 applies to the equation A(λ, z)β = 0 with R = C[λ] and g(λ, z) = det(B(λ) −z).
We therefore conclude that there are polynomials β1, . . . , β(s+1)mn in Cλ, zmn (some of which might be zero) such that β1(λ, zj,k), . . . , βmj(λ, zj,k) are linearly independent solutions of A(λ, zj,k)β = 0 for k = 1, . . . , kj and j = 1, . . . , ν. Hence ψj,k,ℓ(x) = exp(λx) q(x) (xs1mn, . . . , x01mn)βℓ(λ, zj,k).
Using next that fj(λ, zj,k) = 0 and the fact that zkj has a constant nonvanishing coefficient in fj(λ, ·) we obtain that ψj,k,ℓcan be expressed as ψj,k,ℓ(x) = exp(λx) q(x) (xs1mn, . . . , x01mn) kj−1 r=0 βℓ,j,r(λ)zr j,k where the βℓ,j,r are elements of Cλmn. (They are independent of the subscript k.) The first m components of each ψj,k,ℓform a solution yj,k,ℓof Ly = zj,ky. One obtains yj,k,ℓ(x) = exp(λx) kj−1 r=0 γℓ,j,r(λ, x)zr j,k, where γℓ,j,r(λ, ·) ∈C(x)m and γℓ,j,r(·, x) ∈C[λ]m.
Now define Sj = (γ1,j,0, . . . , γ1,j,kj−1, γ2,j,0, . . . , γmj,j,kj−1), Vj = 1 · · · 1 .
.
.
.
.
.
zkj−1 j,1 · · · zkj−1 j,kj , Zj = ⊕mj r=1Vj, and Yj(λ, x) = Sj(λ, x)Zj exp(λx).
The matrix Yj is a m × mjkj matrix. The mj columns whose index is equal to k modulo kj are the linearly independent solutions of Ly = zj,ky whose asymptotic behavior is given by exp(λx). Finally we define the m × m matrices S(λ, x) = (S1(λ, x), . . . , Sν(λ, x)), Z = ⊕n j=1Zj, and Y (λ, x) = (Y1(λ, x), . . . , Yν(λ, x)) = S(λ, x)Z exp(λ, x).
14 Rudi Weikard We now study the asymptotic behavior of Y (λ, x) as λ tends to infinity. By Lemma 3 the matrix A(zj,k, ∞) is diagonalizable and there is a positive integer h such that the eigenvalues of A(zj,k, ∞) are given by µj,k,ℓ= λ σj,k,ℓ,0 + ∞ r=1 σj,k,ℓ,rλ−r/h , ℓ= 1, . . . , mn where the numbers σj,k,ℓ,0 are different from zero. Define the diagonal matrices Mr = diag(σj,k,1,r, . . . , σj,k,mn,r) and order the eigenvalues in such a way that, for r = 0, . . . , h −1, Mr = σj,k,1,r1pr 0 0 Σj,k,r , where p0 ≥p1 ≥· · · ≥ph−1 and where σj,k,1,r is not an eigenvalue of Σj,k,r.
Moreover, require that the mj eigenvalues which are equal to λ are first. Then we have σj,k,1,0 = 1, σj,k,1,1 = · · · = σj,k,1,h−1 = 0, and ph−1 ≥mj. There are p0 eigenvalues which are asymptotically equal to λ and there are ph−1 eigenvalues which differ from λ by a function which stays bounded as λ tends to infinity.
To each eigenvalue µj,k,ℓwe have an eigenvector uj,k,ℓof the form uj,k,ℓ= vj,k,ℓ µj,k,ℓvj,k,ℓ .
.
.
µn−1 j,k,ℓvj,k,ℓ , where vj,k,ℓis an appropriate eigenvector of B(µj,k,ℓ) associated with the eigenvalue zj,k, and can, by assumption, be chosen to be holomorphic at infinity. Define Tj,k to be (mn)×(mn) matrix whose columns are the vectors uj,k,1, . . . , uj,k,mn. Then Tj,k(λ) is invertible at and near infinity. Let ˘ Aj,k(λ, x) = λ−1T −1 j,k A(zj,k, x)Tj,k = λ−1 diag(µj,k,1, . . . , µj,k,mn) + λ−1Xj,k(λ, x) where, according to Lemma 5, Xj,k(λ, x) is bounded as λ tends to infinity. Further-more, Xj,k(λ, x) tends to zero as x tends to infinity. Hence, given a δ > 0, there is an x0(δ) and a number r(δ) such that ∥Xj,k(λ, x)∥< δ whenever |x−x0(δ)| ≤r(δ).
The matrix ˘ Aj,k satisfies now the assumptions of Lemma 6 with ρ = λ−1/h, Ω= {x : |x −x0(δ)| < r(δ)}, and S = {ρ : 0 < |ρ| < ρ0} for some suitable constant ρ0. The matrix Γ is the upper left ph−1 × ph−1 block of Mh and hence diagonal.
The matrix ∆(x) is the upper left ph−1 × ph−1 block of Xj,k(∞, x).
Hence Lemma 6 guarantees the existence of ph−1 linearly independent solutions for λy′ = ˘ Aj,ky whose asymptotic behavior is given by exp(Γ(x −x0))(1ph−1 + Υ(x)) 0(mn−ph−1)×ph−1 exp(λ, x).
(2) Moreover, given any ε > 0 there is a δ > 0 such that ∥Υ(x)∥< ε for all x ∈{x : |x −x0(δ)| < r(δ)}. Since the first mj entries in the diagonal of Γ are zero we obtain that the asymptotic behavior of the first mj columns of matrix (2) is given by Ej,k(x) = 1mj + Υ1,1(x)) exp(Γ2,2(x −x0))Υ2,1(x) 0(mn−ph−1)×mj exp(λ, x) On Commuting Matrix Differential Operators 15 where Υ1,1 and Υ2,1 are the upper left mj × mj block and the lower left (ph−1 − mj) × mj block of Υ, respectively, and where Γ2,2 is the lower right (ph−1 −mj) × (ph−1 −mj) block of Γ.
We have now arrived at the following result: the mj columns of Yj whose index is equal to k modulo kj have asymptotic behavior whose leading order is given by the first m rows of Tj,kEj,k(x)C where C is an appropriate constant and invertible mj × mj matrix. By choosing the eigenvectors uj,k,1, . . . , uj,k,mj (which are all associated with the eigenvalue µj,k,1 = λ) appropriately we may assume that C = 1.
Hence, up to terms of a negligible size, the linearly independent eigenvectors of Q0 are the columns of Y exp(−λx) = SZ when λ and x are large. This implies that Y is invertible.
Similarly, considering the differential expression L∞defined by (L∞y)(x) = Q0y(n)(x) + Q1(∞)y(n−1)(x) + · · · + Qn(∞)y(x) we obtain the invertible matrices S∞(λ) = (S∞,1(λ), . . . , S∞,ν(λ)) and Y∞(λ, x) = (Y∞,1(λ, x), . . . , Y∞,ν(λ, x)) = S∞(λ)Z(λ) exp(λx) where Z is as before. The mj columns of Y∞,j whose index is equal to k modulo kj are those solutions of L∞y = zj,ky whose asymptotic behavior is given by exp(λx).
Note that S∞is x-independent, since the matrices A(zj,k, ∞) are diagonalizable.
Furthermore, since L∞(v exp(λx)) = (B(λ)v) exp(λx), the columns of S∞Z are eigenvectors of B(λ), which, to leading order as λ tends to infinity, are eigenvectors of Q0.
Let d ∈C[λ] be such that dS(·, x)S∞(·)−1 becomes a polynomial (at least d(λ) = det(S∞(λ)) will do). Then we may define matrices Uj ∈C(x)m×m by the equation g j=0 λg−jUj(x) = d(λ)S(λ, x)S∞(λ)−1 and a differential expression U = g j=0 Uj(x)Dg−j.
Then, obviously, U(S∞(λ)Z exp(λx)) = d(λ)S(λ, x)Z exp(λx) = d(λ)Y (λ, x).
Since Y Y −1 ∞ is close to the identity when λ and x are sufficiently large we obtain that U0 is invertible and hence that Conditions 4 and 5 of Lemma 1 are satisfied.
Applying Lemma 1 gives now the existence of a nonempty set F of polynomials such that the differential expression P defined by PU = UDf(L∞) commutes with L when f ∈F. Assume that P and L commute trivially. Then, by the first statement of Lemma 7, Q0 is a multiple of the identity and P and L are polynomials of a unique first order differential expression G = D + G1 where G1 ∈C(x)m×m is bounded at infinity and where G1(∞) is a multiple of the identity. Let y be a solution of Gy = gy where g ∈C and let ϕ be the polynomial such that L = ϕ(G). Then y satisfies also Ly = ϕ(g)y and hence every solution of Gy = gy is meromorphic. By applying what we just proved to the expression G rather than 16 Rudi Weikard L we know that we also have a differential expression U and a nonempty set F of polynomials such that, when f ∈ F, the differential expression P defined by PU = UCf(D + G1(∞)) commutes with G and hence with L for all matrices C ∈Cm×m. The second statement of Lemma 7 shows that P is not a polynomial of G, if C is not a multiple of the identity. Hence, in this case, (P, L) is a nontrivially commuting pair.
□ 3. The simply periodic case If f is an ω-periodic function we will use f ∗to denote the one-valued function given by f ∗(t) = f( ω 2πi log(t)). Conversely, if a function f ∗is given f(x) will refer to f ∗(exp(2πix/ω)). We say that a periodic function f is bounded at the ends of the period strip if f ∗is bounded at zero and infinity. A meromorphic periodic function which is bounded at the ends of the period strip can not be doubly periodic unless it is a constant. The function f is a meromorphic periodic function bounded at the ends of the period strips if and only if f ∗is a rational function bounded at zero and infinity. For more information on periodic functions see, e.g., Markushevich , Chapter III.4.
The field of meromorphic functions with period ω will be denoted by Pω.
Theorem 2. Let L be the differential expression given by Ly = Q0y(n) + Q1y(n−1) + · · · + Qny.
Suppose that the following conditions are satisfied: 1. Q0, . . . , Qn ∈Pm×m ω are bounded at the ends of the period strip.
2. Q0 is constant and invertible.
3. The matrix B(λ) = λnQ0 + λn−1Q∗ 1(∞) + · · · + Q∗ n(∞) is diagonalizable (as an element of Am×m).
4. There are linearly independent eigenvectors v1, . . . , vm ∈Am of B such that limλ→∞vj(λ), j = 1, . . . , m, exist and are linearly independent eigenvectors of Q0. In particular Q0 is diagonalizable.
If mn > 1 and if, for all z ∈C, all solutions of Ly = zy are meromorphic, then there exists a differential expression P with coefficients in Pm×m ω such that (P, L) is a pair of nontrivially commuting differential expressions.
Proof. The proof of this theorem is very close to that of Theorem 1. We record the few points where more significant deviations exist. For notational simplicity we will assume that ω = 2π.
Lemma 1 is now used with M = Pω, Q∞,j = Q∗ j(∞), and τ(x) = eix.
As before we have to construct the expression U: The role of Halphen’s theorem (or, more precisely, Theorem 2.4 of ) is now played by Theorem 1 in (quoted in Appendix A), which is a variant Floquet’s theorem. We have therefore the existence of mj linearly independent functions ψj,k,ℓ(x) = R∗ j,k,ℓ(eix) exp(λx), ℓ= 1, . . . , mj On Commuting Matrix Differential Operators 17 where the components of R∗ j,k,ℓare rational functions. The substitution y(x) = exp(λx) q(eix) s j=0 αjeix(s−j) turns the equation y′ = A(zj,k, ·)y into a system of linear algebraic equation with mj linearly independent solutions. This way one shows as before that ψj,k,ℓ(x) = exp(λx) q(eix) (esix1mn, . . . , eix1mn, 1mn) kj−1 r=0 βℓ,j,r(λ)zr j,k where the βℓ,j,r are elements of Cλmn.
Doing this for k = 1, . . . , kj and for j = 1, . . . , ν and selecting the first m components of all the resulting vectors provides once more an m × m matrices S, Z, and Y = SZ exp(λx). Again the entries of S are polynomials with respect to λ but now they are rational functions with respect to eix. By considering the constant coefficient expression L∞= Q0 dn dxn + · · · + Q∗ n(∞) one obtains also matrices S∞and Y∞= S∞Z exp(λx) and U is defined as before through a multiple of S(λ, x)S∞(λ)−1. The investigation of the asymptotic behavior of Y and Y∞as λ tends to infinity, which leads to proving the invertibility of U0, is unchanged as it did not use the special structure of the Qj, except that one should choose exp(ix0) large rather than x0 large.
Finally, the argument that it is possible to pick, among all expressions commuting with L, an expression which does not commute trivially remains unchanged.
□ 4. Application to the AKNS system Let L = Jd/dx + Q(x), where J = i 0 0 −i and Q(x) = 0 −iq(x) ip(x) 0 .
Note that J2 = −12 and that JQ + QJ = 0.
The AKNS hierarchy is then a sequence of equations of the form Qt = [Pn+1, L], n = 0, 1, 2, . . .
where Pn+1 is a differential expression of order n + 1 such that [Pn+1, L] is a multiplication operator. For this to happen Pn+1 has to be very special. It can be recursively computed in the following way: Let Pn+1 = n+1 ℓ=0 (kn+1−ℓ(x) + vn+1−ℓ(x)J + Wn+1−ℓ(x))Lℓ, where the kj and vj are scalar-valued and where the Wj are 2 × 2 matrices with vanishing diagonal elements. Requiring that [Pn+1, L] is a multiplication operator yields k′ j = 0 for j = 0, . . . , n + 1 and the recursion relations W0 = 0 v′ j12 = WjQ + QWj, Wj+1 = 1 2J(W ′ j −2vjQj), j = 0, . . . , n + 1.
18 Rudi Weikard This gives finally [Pn+1, L] = 2vn+1JQ −JW ′ n+1.
The first few AKNS equations are Qt = −c0Q′ + 2c1JQ, Qt = −c0 2 J(Q′′ −2Q3) −c1Q′ + 2c2JQ, Qt = c0 4 (Q′′′ −6Q2Q′) −c1J(Q′′ −2Q3) −c2Q′ + 2c3JQ.
Here we are interested in the stationary solutions of AKNS equations. Therefore we make the following definition.
Definition 2. Suppose p and q are meromorphic functions. Then Q is called an algebro-geometric AKNS potential (or simply algebro-geometric) if Q is a stationary solution of some AKNS equation.
The goal of this section is to prove the following theorem.
Theorem 3. Let Q = 0 −iq ip 0 and assume either that p, q are rational functions bounded at infinity or else that p, q are meromorphic ω-periodic functions bounded at the ends of the period strip. Then Q is an algebro-geometric AKNS potential if and only if for all z ∈C all solutions of the equation Jy′+Qy = zy are meromorphic with respect to the independent variable.
Before we begin the proof of this result let us recall the following two results which were proven by Gesztesy and myself in . The first one (Theorem 4 below) asks that p and q are meromorphic and provides one direction in the proof of Theorem 3. The second one (Theorem 5 below) is the analogue of Theorem 3 for the case of elliptic coefficients and is stated here for comparison purposes.
Theorem 4. Let Q = 0 −iq ip 0 where p, q are meromorphic functions. If Q is an algebro-geometric AKNS potential then for all z ∈C all solutions of the equation Jy′ + Qy = zy are meromorphic with respect to the independent variable.
Theorem 5. Let Q = 0 −iq ip 0 with p, q elliptic functions with a common period lattice. Then Q is an elliptic algebro-geometric AKNS potential if and only if for all z ∈C all solutions of the equation Jy′ + Qy = zy are meromorphic with respect to the independent variable.
Now we are ready to prove Theorem 3: Proof of Theorem 3. We only need to prove that Q is algebro-geometric if all solutions of Ly = zy are meromorphic since the converse follows from Theorem 4.
Suppose Q is periodic. The desired conclusion follows from Theorem 2 once we have checked its hypotheses. But Conditions 1 and 2 are satisfied by our assumptions while Conditions 3 and 4 hold automatically when the eigenvalues of Q0 (= J) are distinct. For convenience, however, let us mention that the eigenvalues of B(λ) = iλ −iq∗(∞) ip∗(∞) −iλ On Commuting Matrix Differential Operators 19 are ± q∗(∞)p∗(∞) −λ2 and that these are distinct for all but two values of λ.
The eigenvectors may be chosen as v1 = 1 2λ iλ + z1(λ) ip∗(∞) and v2 = 1 2λ iq∗(∞) iλ + z1(λ) where z1(λ) is the branch of q∗(∞)p∗(∞) −λ2 which is asymptotically equal to iλ.
The proof for the rational case is virtually the same.
□ The solutions of Jy′ + Qy = zy are analytic at every point which is neither a pole of p nor of q. Since it is a matter of routine calculations to check whether a solution of Jy′ + Qy = zy is meromorphic at a pole of p or q and since there are only finitely many poles of Q modulo periodicity, Theorem 3 provides an easy method which allows one to determine whether a rational function Q bounded at infinity or a meromorphic simply periodic function Q bounded at the ends of the period strip is a stationary solution of an equation in the AKNS hierarchy.
5. The lemmas Lemma 1. Let M be a field of meromorphic functions on C and consider the differential expression L = n j=0 QjDn−j where Q0 ∈Cm×m is invertible and Qj ∈Mm×m for j = 1, . . . , n. Suppose that there exist differential expressions L∞= n j=0 Q∞,j Dn−j and U = g j=0 Uj(τ(x))Dg−j with the following properties: 1. Q∞,0, . . . , Q∞,n are in Cm×m and Q∞,0 = Q0.
2. τ is a meromorphic function on C.
3. There is a set Λ ⊂C with at least g + n + 1 distinct elements such that, for each λ ∈Λ, the matrix B(λ) = n j=0 λn−jQ∞,j has m linearly independent eigenvectors v1(λ), . . . , vm(λ) ∈Cm respectively associated with the (possibly degenerate) eigenvalues z1(λ), . . . , zm(λ).
4. U0, . . . , Ug ∈Mm×m and U0 is invertible.
5. U(vj(λ) exp(λx)) is a solution of Ly = zj(λ)y for j = 1, . . . , m.
Finally, define the algebra C = {C ∈Cm×m : [Q∞,0, C] = · · · = [Q∞,n, C] = 0}.
Then there exists a nonempty set F ⊂C[u] with the following property: for each polynomial f ∈F and each polynomial h ∈C[u] there exists a differential expression P with coefficients in Mm×m such that [P, L] = 0. In fact, P is given by PU = Uh(D)f(L∞).
Proof. Consider the differential expression V = LU −UL∞and fix λ ∈Λ. Since L∞(vj(λ) exp(λx)) = zjvj(λ) exp(λx) we obtain V (vj(λ) exp(λx)) = (L −zj)U(vj(λ) exp(λx)) = 0.
20 Rudi Weikard V is a differential expression of order g + n at most, i.e., V = g+n k=0 Vk(x)Dk for suitable matrices Vk. Hence 0 = exp(−λx)V (vj(λ) exp(λx)) = g+n k=0 Vk(x)λk vj(λ).
For fixed x and λ we now have an m × m matrix V (λ, x) = g+n k=0 Vk(x)λk whose kernel contains all eigenvectors of B(λ) and is therefore m-dimensional. This means that V (λ, x) = 0. Since this is the case for at least g + n + 1 different values of λ we conclude that V0 = · · · = Vg+n = 0 and hence that LU = UL∞.
Since U0 is invertible Uy = 0 has mg linearly independent solutions. Let {y1, . . . , ymg} be a basis of ker(U). With each element yℓof this basis we may associate a differential expression Hℓwith coefficients in Cm×m in the following way. Since yℓ∈ker(U), so is L∞yℓand, in fact, Lj ∞yℓfor every j ∈N. Since ker(U) is finite-dimensional there exists a k ∈N and complex numbers α0, . . . , αk such that α0 ̸= 0 and k j=0 αk−jLj ∞yℓ= 0.
Then define Hℓ= k j=0 αk−jLj ∞. Since the expressions Hℓcommute among them-selves we obtain that ker(U) ⊂ker mg ℓ=1 Hℓ .
Hence the set F = {f ∈C[u] : ker(U) ⊂ker(f(L∞))} is not empty.
Note that [L∞, D] = 0 and [L∞, C] = 0 if C ∈C. For any h ∈C[u] and any f ∈F let P∞= h(D)f(L∞). Then [P∞, L∞] = 0 and ker(U) ⊂ker(P∞) ⊂ker(UP∞).
Corollary 1 in Appendix B shows that there is an expression P such that PU = UP∞. Hence [P, L]U = PLU −LPU = UP∞L∞−UL∞P∞= U[P∞, L∞] = 0 and thus, recalling that U0 is invertible, [P, L] = 0.
□ Lemma 2. Let B(λ) = n j=0 λn−jBj where B0, . . . , Bn ∈Cm×m and where B0 is invertible. Suppose the characteristic polynomial of B has the prime factorization ν j=1 fj(λ, z)mj. If weight nr + s is assigned to the monomial λszr, then the weight of the heaviest monomial in fj is a multiple of n, say nkj and the coefficients of zkj and λnkj in fj are nonzero.
Let Λ be the set of all complex numbers λ satisfying the following two conditions: 1. (f1 . . . fν)(λ, ·) has k1 + · · · + kν distinct roots.
2. If zj,k is a root of fj(λ, ·), then λ is a simple root of (f1 . . . fν)(·, zj,k).
On Commuting Matrix Differential Operators 21 Then the complement of Λ is finite.
Moreover, there is an integer h and there are complex numbers ρj,k,r such that, for sufficiently large λ, the roots of fj(λ, ·), j = 1, . . . , ν, are given by zj,k = λn ρj,k,0 + ∞ r=1 ρj,k,rλ−r/h , k = 1, . . . , kj where the numbers ρj,k,0 are different from zero.
Proof. First we agree, as usual, that the weight of a polynomial is equal to the weight of its heaviest monomial.
It is then easy to see that the characteristic polynomial f(λ, z) = det(B(λ) −z) has weight mn. Suppose f = g1g2 and let f = mn j=0 αjwj where wj is a polynomial all of whose terms have weight j. Doing the same with g1 and g2 one can show that any factor of f has a weight which is a multiple of n, say kn, and that the coefficients of zk and λkn in that factor are nonzero. In particular then, this is true for the prime factors.
Therefore fj(λ, ·) has kj distinct roots for all but finitely many values of λ.
Moreover, by Bezout’s theorem, the curves defined by fj and fℓintersect only in finitely many points if j is different from ℓ. Hence the first condition is satisfied for all but finitely many values of λ.
The discriminant of (f1 . . . fν)(·, z) is a polynomial in z. Hence there are at most finitely many values of z for which (f1 . . . fν)(·, z) has multiple roots.
For each of these exceptional z-values there are only finitely many of the multiple roots.
Hence there are only finitely many values of λ such that there is a z for which (f1 . . . fν)(·, z) has a multiple root.
The last statement follows from standard considerations of the behavior of alge-braic functions near a point. In particular, the power n on λ is determined by an inspection of the Newton polygon associated with fj.
□ Lemma 3. Let B(λ) = n j=0 λn−jBj where B0, . . . , Bn ∈Cm×m and where B0 is invertible. Define A(z) = 0 1m 0 · · · 0 .
.
.
.
.
.
0 · · · 1m B−1 0 (z −Bn) −B−1 0 Bn−1 −B−1 0 Bn−2 · · · −B−1 0 B1 , a matrix whose n2 entries are m × m blocks.
The vector v ∈Cm is an eigenvector of B(λ) associated with the eigenvalue z if and only if u = v λv .
.
.
λn−1v is an eigenvector of A(z) associated with the eigenvalue λ. In particular, z has geometric multiplicity k as an eigenvalue of B(λ) if and only if λ has geometric 22 Rudi Weikard multiplicity k as an eigenvalue of A(z). Also, det(A(z) −λ) = (−1)nm det(B−1 0 ) det(B(λ) −z).
(3) If B is diagonalizable (as an element of Am×m), then A(z) is diagonalizable for all but finitely many values of z.
Moreover, if zj,k is a zero of fj(λ, ·), then there are complex numbers σj,k,ℓ,r and an integer h such that the eigenvalues µj,k,1, . . . , µj,k,mn of A(zj,k) are given by µj,k,ℓ= λ σj,k,ℓ,0 + ∞ r=1 σj,k,ℓ,rλ−r/h , ℓ= 1, . . . , mn where the numbers σj,k,ℓ,0 are different from zero.
Proof. That B(λ)v = zv if and only if A(z)u = λu follows immediately from direct computation. The validity of (3) is proven by blockwise Gaussian elimination.
Assume now that B is diagonalizable and let T ∈Am×m be an invertible matrix whose columns are eigenvectors of B. The determinant of T is an algebraic function in λ which is zero or infinity only for finitely many distinct values of λ and B(λ) is diagonalizable for all λ but these. From Lemma 2 we know also that there are only finitely many values of λ for which ν j=1 fj(·, z) has repeated zeros. To all these exceptional values of λ correspond finitely many eigenvalues z of B(λ). We assume now that z is a complex number distinct from all those values. If µ is now an eigenvalue of A(z) then it is a zero of fj(·, z) for some j but not a zero of fℓ(·, z), if ℓ̸= j. Hence its algebraic multiplicity is mj. Additionally, z is an eigenvalue of geometric multiplicity mj of B(µ), since B(µ) is diagonalizable. The previous argument shows that µ has geometric multiplicity mj as eigenvalue of A(z). Since this is true for any eigenvalue of A(z), the matrix A(z) must be diagonalizable.
The last statement follows again from standard considerations of the behavior of algebraic functions near a point, using that zj,k is an algebraic function of λ (whose behavior near infinity is of the form given in Lemma 2) and that µj,k,r are algebraic functions of zj,k.
□ Lemma 4. Let R be an integral domain, Q its fraction field, g an element of R[z], and K a field extension of Q in which g splits into linear factors. Suppose A is a matrix in R[z]j×k. Then there exist k vectors v1, . . . , vk ∈R[z]k with the following property: if z0 ∈K is any of the roots of g and if the dimension of ker(A(z0)) is µ, then v1(z0), . . . , vµ(z0) are linearly independent solutions of A(z0)x = 0.
Proof. Suppose g has the prime factorization gm1 1 . . . gmν ν .
If g(z0) = 0 then precisely one of the prime factors of g, say gℓ, satisfies gℓ(z0) = 0.
Note that Fℓ= Q[z]/⟨gℓ⟩is a field and that we may view A as an element of F j×k ℓ . Since Fℓ is isomorphic to a subfield of K any Kk-solution of A(z0)x = 0 is a scalar multiple of a representative of an F k ℓ-solution of Ax = 0 (evaluated at z0) and vice versa.
Therefore there is a basis {xℓ,1(z0), . . . , xℓ,µℓ(z0)} of ker(A(z0)) where the xℓ,r are in Q[z]k. By choosing appropriate multiples in R we may even assume that the xℓ,r are in R[z]k. Notice that if z′ 0 is another root of gℓthen {xℓ,1(z′ 0), . . . , xℓ,µℓ(z′ 0)} is a basis of ker(A(z′ 0)). We define also xℓ,r = 0 for r = µℓ+ 1, . . . , k.
On Commuting Matrix Differential Operators 23 For r = 1, . . . , k we now let vr = ν ℓ=1 ν ℓ′=1 ℓ′̸=ℓ gℓ′ xℓ,r.
This proves the lemma once we recall that gℓ(z0) = 0 = gℓ′(z0) implies that ℓ= ℓ′.
□ Lemma 5. Suppose A ∈Cmn×mn and T ∈Amn×mn have the following properties: 1. The first (n −1)m rows of A are zero.
2. T is invertible at and near infinity and its columns T1:mn,j have the form T1:mn,j = vj µjvj .
.
.
µn−1 j vj , where the µj are complex-valued algebraic functions of λ with the asymptotic behavior µj(λ) = λ(σj + o(1)) as λ tends to infinity and where the vj are Cm-valued algebraic functions of λ which are holomorphic at infinity.
Then (T −1AT)(λ) is bounded as λ tends to infinity.
Proof. The first (n −1)m rows of AT are zero. Consequently we need to consider only the last m columns of T −1. Let Bn, . . . , B1 denote the m×m matrices which occupy the last m rows of A (with decreasing index as one moves from left to right) and let τ ∗ ℓdenote the row-vector in the last m columns of row ℓin T −1 (note that τℓ∈Am). Then (T −1AT)ℓ,k = n j=1 µn−j k τ ∗ ℓBjvk.
We will show below that τℓhas the asymptotic behavior τℓ= λ1−n(τ0,ℓ+o(1)) with τ0,ℓ∈Cm as λ tends to infinity. Hence (T −1AT)ℓ,k = n j=1 λ1−j(σn−j k τ ∗ 0,ℓBjvk(∞) + o(1)) = σn−1 k τ ∗ 0,ℓB1vk(∞) + o(1) as λ tends to infinity and this will prove the claim.
The minor of T which arises when one deletes row r and columns s of T will be denoted by Ms,r. We have then that (T −1)r,s = (−1)r+s det(T) det(Ms,r).
The k-th entry in row mα + β, where β ∈{1, . . . , m} and α ∈{0, . . . , n −1}, equals λα times a function which is bounded as λ tends to infinity. Hence det(T) = λN(t0 + o(1)) for some nonzero complex number t0 and for N = mn(n −1)/2.
By the same argument we have that det(Mmα+β,r) = λN ′(mmα+β,r + o(1)) where N ′ = N −α and mmα+β,r ∈C. Hence (T −1)r,s = (−1)r+sλ−α mmα+β,r + o(1) t0 .
24 Rudi Weikard For the first part of the proof we need only the case α = n −1.
□ Lemma 6. Let Ω⊂C be an open simply connected set containing x0 and S ⊂C a sector centered at zero. Suppose that A : S × Ω→Cn×n is holomorphic on S × Ω and admits a uniform asymptotic expansion A(ρ, x) ∼ ∞ r=0 Ar(x)ρr as ρ tends to zero. Suppose that, for r = 0, . . . , h −1, the matrices Ar are constant and have the block-diagonal form Ar = σr1pr 0 0 Σr where p0 ≥p1 ≥· · · ≥ph−1 = p and where σr is not an eigenvalue of Σr. Denote the upper left p × p block of Ah by Ah;1,1 and assume that Ah;1,1(x) = Γ + ∆(x) where Γ ∈Cp×p and ∆: Ω→Cp×p. Let α = h−1 r=0 σrρr−h.
Then there exists a subsector S′ of S and an n×p matrix Y (ρ, x) whose columns are linearly independent solutions of ρhy′ = Ay and for which R(ρ, x) = Y (ρ, x) exp(−αx) has in S′ an asymptotic expansion of the form R(ρ, x) ∼ ∞ j=0 Rj(x)ρj as ρ tends to zero. Moreover, for every positive ε there exists a positive δ such that ∥∆(x)∥< δ for all x ∈Ωimplies R0(x) = exp(Γ(x −x0))(1p + Υ(x)) 0(n−p)×p with ∥Υ(x)∥< ε for all x ∈Ω.
Proof. The key to the proof of this lemma is Theorem 26.2 in Wasow which we have (essentially) quoted in Appendix C and which implies immediately Corollary 2.
A repeated application of this corollary shows that there are p linearly independent solutions yj of ρhy′ = Ay of the form yj = P0Q0 . . . Ph−1Qh−1wj exp(αx), j = 1, . . . , p where the Pk and Qk are matrices and where the wj are vectors whose properties are described presently. Let p−1 = n. Then Pk is an pk−1 × pk−1 matrix which is asymptotically equal to 1pk−1. The matrix Qk is a constant pk−1 × pk matrix whose upper block is equal to 1pk and whose lower block is a zero matrix. Finally, the wj are linearly independent solutions of the p × p-system w′ = B(ρ, x)w where B(ρ, x) = ρ−hQ∗ A(ρ, x) − h−1 r=0 Ar(x)ρr Q.
Note that B(ρ, x) has the asymptotic behavior B(ρ, x) ∼ ∞ r=0 Br(x)ρr On Commuting Matrix Differential Operators 25 as ρ tends to zero where B0(x) = Ah;1,1(x).
The equation w′ = B(ρ, x)w has a fundamental matrix W whose asymptotic behavior is given by W(ρ, x) ∼ ∞ r=0 Wr(x)ρr where W0(x) = exp x x0 Ah;1,1(t)dt = exp Γ(x −x0) + x x0 ∆(t)dt is an invertible matrix. Since ∥exp(T1 + T2) −exp(T1)∥≤∥T2∥exp(∥T1∥+ ∥T2∥) we have that W0(x) = exp(Γ(x −x0))(1p + Υ(x)) where the norm of Υ becomes small if the norm of ∆becomes small. The fact that the matrices Pk are asymptotically equal to identity matrices and that the upper blocks of the Qk are equal to identity matrices gives now the desired conclusion.
□ Lemma 7. Suppose that M, L, L∞, U, C, and F are as in Lemma 1. Given an expression P∞let P be defined by PU = UP∞. Then the following two statements hold: 1. Let P∞= Df(L∞), where f is a monic polynomial in F. If (P, L) is a trivially commuting pair, then Q0 and Q∞,1 are multiples of the identity. Moreover, there exists a first order differential expression G = D + Q−1 0 (Q1 −η1)/n (where η1 is a suitable constant) such that both P and L are polynomials of G.
2. Let P∞= Cf(L∞), where C ∈C and where f ∈F is monic. If both P and L are polynomials of an expression D + G1 then C is a multiple of the identity matrix.
Proof. Assume that L = n j=0 QjDn−j = n′ j=0 ηjGn′−j and P = r j=0 PjDr−j = r′ j=0 γjGr′−j, where G is a differential expression of order k and the coefficients ηj and γj are complex numbers.
To prove the first statement assume also that P∞= Df(L∞) where f ∈F has degree s. Since the order of L is equal to n = kn′ and the order of P is equal to r = sn + 1 = kr′ we have necessarily k = 1, n′ = n, and r′ = r = sn + 1. Therefore we assume now that G = G0D + G1.
Note that LU = UL∞, PU = UP∞, and Q∞,0 = Q0 imply that U0Q0 = Q0U0, U0P∞,0 = P0U0, (4) U1Q0 + U0Q∞,1 = Q0U1 + Q1U0 + nQ0U ′ 0, (5) and that U1P0 + U0P∞,1 = P0U1 + P1U0 + rP0U ′ 0, (6) where P∞,j is the coefficient of Dr−j in P∞.
Since P∞,0 = Qs 0 we find firstly that P0 = U0Qs 0U −1 0 = Qs 0.
Next, since Q0 = η0Gn 0 and P0 = γ0Gsn+1 0 , we have that G0 = ηs 0γ−1 0 1. Hence G0, Q0, and P0 are all multiples of the identity 26 Rudi Weikard matrix. Therefore we can (and will) assume from now that G0 = 1 by changing the coefficients ηj and γj appropriately. In particular, Q0 = η01.
We find next that Q1 = (nη0G1 + η11), P1 = (rγ0G1 + γ11), and that P∞,1 = (sQ∞,1 + κ)Qs−1 0 where κ is the coefficient of us−1 in f(u) if n = 1 and κ = 0 if n > 1. Inserting these expressions into (5) and (6) and eliminating the terms with U ′ 0 gives Q∞,1 = [nκ + η−s 0 (rη1γ0 −nγ1η0)]1.
We also obtain that G1 = Q−1 0 (Q1 −η1)/n. This proves the first statement of the lemma.
To prove the second statement, let G = D + G1. This implies, as before, that Q0 = η01 and P∞,0 = P0 = γ01. On the other hand, since P∞= Cf(L∞), we have that P∞,0 = Cηs 0. Thus C is a multiple of the identity.
□ Appendix A. The theorems of Halphen and Floquet The proofs of Theorems 1 and 2 rely on results of Halphen and Floquet , , or rather on generalizations to systems of their results. These generalizations were proven in and , respectively, and are repeated here for the convenience of the reader.
Theorem 6. Let A ∈C(x)n×n with entries bounded at infinity and suppose that the first-order system y′ = Ay has a meromorphic fundamental system of solutions.
Then y′ = Ay has a fundamental matrix of the type Y (x) = R(x) exp(diag(λ1x, . . . , λnx)), where λ1, . . . , λn are the eigenvalues of A(∞) and R ∈C(x)n×n.
Theorem 7. Suppose that A is an n × n-matrix whose entries are meromorphic, ω-periodic functions which are bounded at the ends of the period strip. If the first-order system y′ = Ay has only meromorphic solutions, then there exists a constant n × n-matrix J in Jordan normal form and an n × n-matrix R∗whose entries are rational functions over C such that the following statements hold: 1. The eigenvalues of A∗(0) and J are the same modulo iZ if multiplicities are properly taken into account. More precisely, suppose that there are nonneg-ative integers ν1, . . . , νr−1 such that λ, λ + iν1, . . . , λ + iνr−1 are all the eigenvalues of A∗(0) which are equal to λ modulo iZ. Then λ is an eigenvalue of J with algebraic multiplicity r.
2. The equation y′ = Ay has a fundamental matrix Y given by Y (x) = R∗(e2πix/ω) exp(Jx).
In particular every entry of Y has the form f(e2πix/ω, x)eλx, where λ + iν is an eigenvalue of A∗(0) for some nonnegative integer ν and where f is rational function in its first argument and a polynomial in its second argument.
On Commuting Matrix Differential Operators 27 Appendix B. Higher order systems of differential equations In this section we recall two basic facts about systems of linear differential equa-tions of order higher than one.
Consider the system Ty = T0(x)y(n) + T1(x)y(n−1) + · · · + Tn(x)y = 0 (7) where the Tj are m × m matrices whose entries are continuous functions on some real interval or complex domain Ωand where T0(x) is invertible for every x ∈Ω.
Using the analogue of the standard transformation which turns a higher order scalar equation into a first order system, one finds that the system (7) is equivalent to the first order system u′ = Au where A is the mn × mn matrix 0 1m 0 · · · 0 .
.
.
.
.
.
0 · · · 1m −T −1 0 Tn −T −1 0 Tn−1 −T −1 0 Tn−2 · · · −T −1 0 T1 in which all entries represent m × m blocks. From this it follows immediately that a fundamental system of solutions of Ty = 0 has mn elements.
The other property we need is about the existence of a factorization of an n-th order expression into n first order factors.
Theorem 8. Let T be the differential expression defined in (7). Suppose that F1, . . . , Fn−1 are m × m matrices whose entries are continuous functions on Ωand which are invertible for every x ∈Ω. Define Fn = T0F −1 1 . . . F −1 n−1. Then there exist m × m matrices Φ1, . . . , Φn such that T = (FnD −Φn) . . . (F1D −Φ1).
Proof. Denote the elements of a fundamental system of solutions by y1, . . . , ymn and define, for j = 1, . . . , n, the m × m matrices Yj = (ym(j−1)+1, . . . , ymj).
Next define W1 = Y1 and Φ1 = F1Y ′ 1Y −1 1 and suppose we have determined matrices Φ1, . . . , Φj−1. We will show below that Wj = (Fj−1D −Φj−1) . . . (F1D −Φ1)Yj is invertible so that we can define Φj = FjW ′ jW −1 j .
Now let S = (FnD −Φn) . . . (F1D −Φ1).
Then S(Yj) = (FnD −Φn) . . . (FjD −Φj)Wj = 0, i.e., S and T have the same solutions. S and T are therefore equivalent to the same first order system. Since they have the same leading coefficient we finally obtain S = T.
We complete the proof by showing that the matrices Wj are invertible, i.e., that their columns Wj,1, . . . , Wj,m are linearly independent. This is true for j = 1 since 28 Rudi Weikard the columns of W1 are the solutions y1, . . . , ym which are linearly independent.
Assume that W1, . . . , Wj−1 are invertible and that 0 = m k=1 αkWj,k.
Then 0 = (Fj−1D −Φj−1) . . . (F1D −Φ1) m k=1 αkym(j−1)+k.
Since the space of solutions of (Fj−1D −Φj−1) . . . (F1D −Φ1)y = 0 is spanned by y1, . . . , ym(j−1) we obtain that m k=1 αkym(j−1)+k = m(j−1) ℓ=1 βℓyℓ.
But since y1, . . . , ymj are linearly independent it follows that all α1 = · · · = αn = 0 (and β1 = · · · = βm(j−1) = 0). Hence the columns of Wj are linearly independent and Wj is invertible.
□ Corollary 1. Let S and T be differential expressions with matrix coefficients and invertible leading coefficients. If ker S ⊂ker T then there exists a differential ex-pression R such that RS = T.
Appendix C. Wasow’s theorem For the reader’s convenience we provide here a slightly adapted version of Theo-rem 26.2 in Wasow . The adaptation makes use of formulas (25.19) and (25.20) in .
Theorem 9. Let Ω⊂C be an open simply connected set containing the point x0 and let S be a sector {ρ : 0 < |ρ| < ρ0, α0 < arg(ρ) < β0}.
Suppose that A : S × Ω→Cn×n is holomorphic and admits a uniform asymptotic expansion A(ρ, x) ∼ ∞ r=0 Ar(x)ρr on S ×Ω. Furthermore suppose that A0 is diagonal, i.e., A0 = diag(λ1, . . . , λn) and that the sets {λ1(x0), . . . , λp(x0)} and {λp+1(x0), . . . , λn(x0)} are disjoint. Then there exists a subsector S∗of S and a subregion Ω∗of Ωand Cn×n-valued functions P and B with the following properties: 1. P and B are holomorphic in S∗× Ω∗.
2. P −1 and B have the block forms P(ρ, x) −1 = 0 P1,2(ρ, x) P2,1(ρ, x) 0 and B(ρ, x) = B1,1(ρ, x) 0 0 B2,2(ρ, x) , and the blocks have asymptotic expansion Pj,k(ρ, x) ∼∞ r=1 Pr;j,k(x)ρr and Bj,j(ρ, x) ∼∞ r=0 Br;j,j(x)ρr, as ρ tends to zero.
3. B0 = A0 and A0P1 −P1A0 = B1 −A1.
On Commuting Matrix Differential Operators 29 4. the transformation y = Pz takes the differential equation ρhy′ = Ay into ρhz′ = By.
Corollary 2. If λ1(x) = · · · = λp(x) = σ for all x ∈Ωand B(ρ, x) = 1 ρ(B1,1 −B0;1,1(x)) ∼ ∞ r=1 Br;j,j(x)ρr−1, then the equation ρhy′ = Ay has p linearly independent solutions of the form y(x) = PQw(x) exp(σxρ−h), where Q = 1p×p 0(n−p)×p and w is a solution of the p × p system ρh−1w′ = B(ρ, x)w.
References 1. J.L. Burchnall and T.W. Chaundy, Commutative ordinary differential operators, Proc. London Math. Soc. Ser. 2 21 (1923), 420–440.
2. J.L. Burchnall and T.W. Chaundy, Commutative ordinary differential operators, Proc. Roy.
Soc. London A 118 (1928), 557–583.
3. I. V. Cherednik, Regularity of “finite-zone” solutions of integrable matrix differential equations, Soviet Phys. Dokl. 27 (1982), 716–718, MR 84f:35123, Zbl 0552.35017.
4. L. A. Dickey, On the τ-function of matrix hierarchies of integrable equations, J. Math. Phys.
32 (1991), 2996–3002, MR 93d:58068, Zbl 0737.35097.
5. B. A. Dubrovin, Completely integrable Hamiltonian systems associated with matrix operators and abelian varieties, Funct. Anal. Appl. 11 (1977), 265–277, Zbl 0413.58012.
6. G. Floquet, Sur les ´ equations diff´ erentielles lin´ eaires a coefficients p´ eriodiques, C. R. Acad.
Sci. Paris 91 (1880), 880–882.
7. G. Floquet, Sur les ´ equations diff´ erentielles lin´ eaires a coefficients p´ eriodiques, Ann. Sci. ´ Ecole Norm. Sup. 12 (1883), 47–88.
8. F. Gesztesy and R. Weikard, Picard potentials and Hill’s equation on a torus, Acta Math. 176 (1996), 73–107, MR 97f:14046.
9. F. Gesztesy and R. Weikard, A characterization of all elliptic algebro-geometric solutions of the AKNS hierarchy, Acta Math. 181 (1998), 63–108, MR 99k:14052, Zbl 0955.34073.
10. F. Gesztesy and R. Weikard, Elliptic algebro-geometric solutions of the KdV and AKNS hierarchies—an analytic approach, Bull. Amer. Math. Soc. (N.S.) 35 (1998), 271–317, MR 99i:58075, Zbl 0909.34073.
11. F. Gesztesy, K. Unterkofler and R. Weikard, On a theorem of Halphen and its application to integrable systems, J. Math. Anal. Appl. 251 (2000), 504–526, MR 2001i:37108, Zbl 0966.34078.
12. G.-H. Halphen, Sur une nouvelle classe d’´ equations diff´ erentielles lin´ eaires int´ egrables, C. R.
Acad. Sci. Paris 101 (1885), 1238–1240.
13. A. R. Its and V. B. Matveev, Schr¨ odinger operators with the finite-band spectrum and the N-soliton solutions of the Korteweg-de Vries equation, Theoret. and Math. Phys. 23 (1975), 343–355, MR 57 #18570.
14. I. M. Krichever, Algebraic curves and commuting matricial differential operators, Funct. Anal.
Appl. 10 (1976), 144–146, Zbl 0347.35077.
15. I. M. Krichever, Integration of nonlinear equations by the methods of algebraic geometry, Funct. Anal. Appl. 11 (1977), 12–26, Zbl 0368.35022.
16. I. M. Krichever, Methods of algebraic geometry in the theory of nonlinear equations, Russ.
Math. Surv. 32 (1977), 185–213. See also: Integrable Systems, Selected Papers, Lond. Math.
Soc. Lect. Note Ser. 60, Lond. Math. Soc, 1981, 141–169, Zbl 0461.35075.
17. P. D. Lax, Integrals of nonlinear equations of evolution and solitary waves, Commun. Math.
Phys. 21 (1968), 467–490, MR 38 #3620, Zbl 0162.41103.
30 Rudi Weikard 18. A. I. Markushevich, Theory of Functions in a Complex Variable (three volumes in one), Chelsea 1965.
19. P. van Moerbeke, Integrable foundations of string theory, Lectures on integrable systems (Sophia-Antipolis, 1991), 163–267, World Sci. Publishing, River Edge, NJ, 1994.
20. D. Mumford, An algebro-geometric construction of commuting operators and of solutions to the Toda lattice equation, Korteweg deVries equation and related nonlinear equation, Pro-ceedings of the International Symposium on Algebraic Geometry (Kyoto Univ., Kyoto, 1977), 115–153, Tokyo, 1978, MR 83j:14041, Zbl 0423.14007.
21. E. Picard, Sur une classe d’´ equations diff´ erentielles lin´ eaires, C. R. Acad. Sci. Paris 90 (1880), 128–131.
22. A. Treibich, Matrix elliptic solitons, Duke Math. J. 90 (1997), 523–547, MR 98m:14030, Zbl 0909.35116.
23. W. Wasow, Asymptotic Expansions for Ordinary Differential Equations, Interscience 1965, MR 34 #3041.
24. R. Weikard, On commuting differential operators, Electron. J. Differential Equations 2000 (2000), No. 19, 1–11, MR 2001a:34146, Zbl 0953.34073.
25. R. Weikard, Floquet theory for linear differential equations with meromorphic solutions, E. J.
Qualitative Theory of Diff. Equ., No. 8. (2000), 1–6, MR 2001j:34113, Zbl 0969.34073.
Department of Mathematics, University of Alabama at Birmingham, Birmingham, Al-abama 35294-1170, USA rudi@math.uab.edu This paper is available via |
189022 | https://brainly.com/question/17637896 | [FREE] How does scale factor change the area? - brainly.com
6
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
+64,9k
Smart guidance, rooted in what you’re studying
Get Guidance
Test Prep
+24,6k
Ace exams faster, with practice that adapts to you
Practice
Worksheets
+7,3k
Guided help for every grade, topic or textbook
Complete
See more
/
Mathematics
Textbook & Expert-Verified
Textbook & Expert-Verified
How does scale factor change the area?
2
See answers Explain with Learning Companion
NEW
Asked by christine635 • 09/16/2020
Read More
Community
by Students
Brainly
by Experts
ChatGPT
by OpenAI
Gemini
Google AI
Community Answer
This answer helped 30985 people
30K
5.0
0
Upload your school material for a more relevant answer
Answer: in summary, a scale factor is simply a number that multiplies the dimensions of a shape. ... If the scale factor is three, then the perimeter of the new object will be three times the original perimeter. The area of a scaled object will be equal to the scale factor squared.
Explanation
For example, if you had a triangle with a base measuring 6 units, and a height measuring 3 units, the area would be 9 square units. If we scale it by 1/3, the triangle's base is 2 units, the height is 1 unit, and the area is 1 square unit. The dimensions were scaled by 1/3, but the area was scaled by 1/9 (1/3 1/3).
Answered by chasethompson77 •48 answers•31K people helped
Thanks 0
5.0
(1 vote)
Textbook &Expert-Verified⬈(opens in a new tab)
This answer helped 30985 people
30K
5.0
0
Simple nature - Benjamin Crowell
Light and Matter - Benjamin Crowell
Nature of Geographic Information - David DiBiase
Upload your school material for a more relevant answer
The area of a shape changes by the square of the scale factor applied to its dimensions. For example, scaling a rectangle's dimensions by a factor of 2 increases its area by a factor of 4, while scaling by 2 1 decreases the area by a factor of 4. Thus, the relationship between scale factor and area is directly linked through squaring the scale factor.
Explanation
Understanding how scale factor affects area is essential in geometry. A scale factor is a number that scales, or multiplies, the dimensions of a geometric shape. When we increase or decrease the dimensions of a shape using a scale factor, the area changes according to the square of that scale factor.
For example, let's take a rectangle with a width of 4 units and a height of 3 units. The original area can be calculated as:
Area=width×height=4×3=12 square units
If we apply a scale factor of 2, making the width 8 units and the height 6 units, the new area will be:
New Area=8×6=48 square units
In this case, the area has changed by a factor of 2 2=4, meaning the new area is 4 times the original area:
12×4=48 square units
Conversely, if we apply a scale factor of 2 1, the dimensions become 2 units wide and 1.5 units high. The new area would be:
New Area=2×1.5=3 square units
Here, the area decreases by a factor of (2 1)2=4 1, and the new area is one-fourth of the original area:
12×4 1=3 square units
In summary, when you scale a shape by a certain factor, the area of that shape changes by the square of that factor.
Examples & Evidence
For instance, if a triangular area is scaled up from base 6 units and height 3 units to base 12 units and height 6 units (a scale factor of 2), the area increases from 9 square units to 36 square units (12 times the original). Conversely, scaling down by a factor of 1/3 changes a triangle from base 6 units and height 3 units to base 2 units and height 1 unit, reducing the area from 9 square units to 1 square unit (1/9 of the original).
Mathematics states that the area of similar shapes relates to the square of their corresponding side lengths. This mathematical principle holds true in geometry and can be verified through various texts and geometry principles.
Thanks 0
5.0
(1 vote)
Advertisement
Community Answer
This answer helped 10455071 people
10M
0.0
0
The scale factor has a quadratic relationship with area, meaning that when scaling an object, the area changes by the square of the scale factor.
The scale factor in mathematics is essential in determining how the dimensions of an object change when it is scaled up or scaled down. Specifically, when discussing area, the scale factor has a squared relationship with area. If an object's dimensions are doubled, the new area will be the square of the scale factor times the original area. For example, a doubling of dimensions (scale factor of 2) leads to an area increase by a factor of 4.
This relationship holds regardless of the object's shape. Whether the object is a square, triangle, or any irregular shape, scaling down by a certain ratio will reduce the area by the square of the scale factor. Thus, the effect of a scale factor on area is to multiply the original area by the scale factor squared, underscoring the importance of understanding how scaling impacts area calculations.
Answered by AliciaAugello •31.1K answers•10.5M people helped
Thanks 0
0.0
(0 votes)
Advertisement
### 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
The difference of the squares of two positive numbers is 180. The square of the smaller number is 8 times the greater number. Find the two numbers.
Prove that ∑r=1 nr(r+1)3=n+1 an, where a is a constant to be found. Find the value of ∑r=1 50r(r+1)3, giving your answer as an exact fraction. Find an expression in its simplest form for ∑r=n 2 nr(r+1)3
Solve the following system of equations: 5 x−2 y=5 7 x−3 y=13
Write an expression equivalent to 2(4 y−5)−3 y.
Write an expression equivalent to 0.5(−14 a−22).
Previous questionNext question
Learn
Practice
Test
Open in Learning Companion
Company
Copyright Policy
Privacy Policy
Cookie Preferences
Insights: The Brainly Blog
Advertise with us
Careers
Homework Questions & Answers
Help
Terms of Use
Help Center
Safety Center
Responsible Disclosure Agreement
Connect with us
(opens in a new tab)(opens in a new tab)(opens in a new tab)(opens in a new tab)(opens in a new tab)
Brainly.com
Dismiss
Materials from your teacher, like lecture notes or study guides,
help Brainly adjust this answer to fit your needs.
Dismiss |
189023 | https://math.stackexchange.com/questions/3259226/how-to-prove-that-the-legendre-symbol-is-multiplicative | number theory - How to prove that the Legendre symbol is multiplicative? - Mathematics Stack Exchange
Join Mathematics
By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy.
Sign up with Google
OR
Email
Password
Sign up
Already have an account? Log in
Skip to main content
Stack Exchange Network
Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.
Visit Stack Exchange
Loading…
Tour Start here for a quick overview of the site
Help Center Detailed answers to any questions you might have
Meta Discuss the workings and policies of this site
About Us Learn more about Stack Overflow the company, and our products
current community
Mathematics helpchat
Mathematics Meta
your communities
Sign up or log in to customize your list.
more stack exchange communities
company blog
Log in
Sign up
Home
Questions
Unanswered
AI Assist Labs
Tags
Chat
Users
Teams
Ask questions, find answers and collaborate at work with Stack Overflow for Teams.
Try Teams for freeExplore Teams
3. Teams
4. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams
Teams
Q&A for work
Connect and share knowledge within a single location that is structured and easy to search.
Learn more about Teams
Hang on, you can't upvote just yet.
You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it?
Instead, you can save this post to reference later.
Save this post for later Not now
Thanks for your vote!
You now have 5 free votes weekly.
Free votes
count toward the total vote score
does not give reputation to the author
Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation.
Got it!Go to help center to learn more
How to prove that the Legendre symbol is multiplicative?
Ask Question
Asked 6 years, 3 months ago
Modified6 months ago
Viewed 4k times
This question shows research effort; it is useful and clear
2
Save this question.
Show activity on this post.
The proof is given here in the answer Proving (n p)(n p), a Legendre symbol, is multiplicative
But I do not understand it, Also the definition in the book for Legendre symbol says that if p|a p|a the Legendre symbol a/p a/p is undefined not 0 as in this solution, could anyone give me a more clear proof please?
number-theory
elementary-number-theory
legendre-symbol
multiplicative-function
Share
Share a link to this question
Copy linkCC BY-SA 4.0
Cite
Follow
Follow this question to receive notifications
edited Jun 15, 2019 at 20:24
IntuitionIntuition
asked Jun 11, 2019 at 23:21
IntuitionIntuition
3,123 1 1 gold badge 14 14 silver badges 56 56 bronze badges
11
2 If you mean how to show (a p)(b p)=(a b p)(a p)(b p)=(a b p) then it follows from that Z/p Z×Z/p Z× is cyclic so that (a p)=a(p−1)/2 mod p(a p)=a(p−1)/2 mod p. The periodicity of p↦(q p)p↦(q p) is quadratic reciprocity.reuns –reuns 2019-06-11 23:32:06 +00:00 Commented Jun 11, 2019 at 23:32
4 It's not good enough to say, "I do not understand it", as that gives us no idea what exactly you don't understand, or what help you need. Please edit your question so it's better focussed.Gerry Myerson –Gerry Myerson 2019-06-11 23:37:24 +00:00 Commented Jun 11, 2019 at 23:37
3 The Legendre symbol is a function of two variables. When we say it's multiplicative, what we really mean is that if we hold the second variable fixed, then it is multiplicative in the first variable. f(a b,p)=f(a,p)f(b,p)f(a b,p)=f(a,p)f(b,p).Gerry Myerson –Gerry Myerson 2019-06-11 23:48:43 +00:00 Commented Jun 11, 2019 at 23:48
3 The Legendre symbol is only defined for prime values of the second argument, so it doesn't even make sense to ask for it to be multiplicative in both variables. f(a,q)f(a,q) is only defined as a Legendre symbol if q q is a prime.Gerry Myerson –Gerry Myerson 2019-06-11 23:55:59 +00:00 Commented Jun 11, 2019 at 23:55
1 I have corrected the link @ThomasShelby I am so sorry ..... it is my bad Intuition –Intuition 2019-06-15 20:25:51 +00:00 Commented Jun 15, 2019 at 20:25
|Show 6 more comments
2 Answers 2
Sorted by: Reset to default
This answer is useful
8
Save this answer.
+50
This answer has been awarded bounties worth 50 reputation by Intuition
Show activity on this post.
I am not a number theorist, but it seems that conceptually the best way to see this is group theoretic. In brief, (for p>2 p>2) the set of squares is a subgroup of the set of non-zero remainders mod p and this subgroup has index 2 . It is therefore normal, and the Legendre symbol is the homomorphism to the corresponding quotient group Z/2 Z Z/2 Z. Below I give a proof without group-theoretic language (but, I think, informed by this understanding).
The multiplicativity we are after is
(a b p)=(a p)(b p)(a b p)=(a p)(b p)
for all a, b.
First, there are two apporaches to treating the case when at least one of the a a,b b is divisible by p p. We can exclude them and only prove the above for a≢0 mod p a≢0 mod p, and b≢0 mod p b≢0 mod p.
Alternatively, we can define (x p)=0(x p)=0 when x x is divisible by p p. Note that if a a is divisible by p p then a b a b is divisible by p p. Then this means (a b p)=(a p)(b p)=0(a b p)=(a p)(b p)=0 for any b b, and so with this definition we obtain that the multiplicativity property holds as soon one of the entries is divisible by p p.
Whatever route we choose, it now only remains to prove
(a b p)=(a p)(b p)(a b p)=(a p)(b p)
for all a≢0 mod p a≢0 mod p, and b≢0 mod p b≢0 mod p.
The set of non-zero remainders mod p p is a of size p−1 p−1. As a preparation, we establish that there are the same number of (non-zero) squares and non-squares mod p. Indeed, after squaring each reminder a a is paired with −a−a and nothing else since x 2≡d 2 x 2≡d 2 means (x−d)(x+d)≡0(x−d)(x+d)≡0 so either x≡d x≡d of x≡−d x≡−d; so the p−1 p−1 remainders produce (p−1)/2(p−1)/2 squares.
Now we check multiplicativity. In 3 remaining cases:
a a and b b are both squares mod p p.
This is an obvious case, since a product of two squares is a square:
c 2 d 2≡(c d)2 mod p c 2 d 2≡(c d)2 mod p
a a is a non-square and b b is a square.
A product of a square and a non-square is a non-square: if we had a d 2≡c 2 a d 2≡c 2 then we would have a≡(c d−1)2 a≡(c d−1)2.
Both a a and b b are non-squares. We need to show that a b a b is a square.
Consider the multipication by a a, acting on the set of reminders mod p. Since it has an inverse (multiplication by a−1 a−1), it is bijective. As we saw in case 2, multiplication by a a takes squares to non-squares. Since the set of squares and non-squares have the same size, this is a bijection from the set of squares to the set of non-squares. This means that it is also a bijection between the complements, i.e. from non-squares to squares. Thus any non-square multiplied by a a becomes a square, and we are done.
Remark: Considering multiplication by a a is how one can prove existence of a−1 a−1 in the first place -- the multiplication is injective since a x≡a y a x≡a y means a(x−y)≡0 a(x−y)≡0 means x≡y x≡y; since domain and target coincide, they have the same size, hence the map is bijective and hence it hits 1, so there exists x x with a x=1 a x=1; that's the x=a−1 x=a−1.
Share
Share a link to this answer
Copy linkCC BY-SA 4.0
Cite
Follow
Follow this answer to receive notifications
edited Dec 22, 2020 at 21:52
answered Jun 18, 2019 at 8:01
MaxMax
14.5k 23 23 silver badges 42 42 bronze badges
0
Add a comment|
This answer is useful
3
Save this answer.
Show activity on this post.
If p|n p|n or p|m p|m then p|n m p|n m so (n m p)=0(n m p)=0 and (n p)=0(n p)=0 or (m p)=0(m p)=0. So (n m p)=(m p)(n p)(n m p)=(m p)(n p) if p|m p|m or p|n p|n.
If p∤n p∤n and p∤m p∤m then p∤n m p∤n m so
(n m p)≡(n m)p−1 2≡(n p)(m p)(m o d p)(n m p)≡(n m)p−1 2≡(n p)(m p)(m o d p) . But each (n m p)(n m p), (n p)(n p) and (m p)(m p) is −1−1 or 1 1. so the difference is 0 0,−2−2 or 2 2.
See also
Apostol, Introduction to analytic number theory.
Share
Share a link to this answer
Copy linkCC BY-SA 4.0
Cite
Follow
Follow this answer to receive notifications
edited Mar 12 at 10:18
Zhefeng Shen
117 2 2 silver badges 9 9 bronze badges
answered Jun 18, 2019 at 9:23
user637244 user637244
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
number-theory
elementary-number-theory
legendre-symbol
multiplicative-function
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
Linked
3Proving (n p)(n p), a Legendre symbol, is multiplicative
Related
1question about legendre symbol
1Is there a short proof of the formula for Legendre symbol (2 p)=(−1)(p 2−1)/8(2 p)=(−1)(p 2−1)/8?
6How to prove this sum related to Legendre symbol
11Does the Legendre Symbol/quadratic reciprocity generalize to higher degrees?
1Calculating the Legendre symbol
0Computing the Legendre Symbol
3Prove a property of Legendre symbol
1A contradiction in calculating the legendre symbol
0A proof of a property of the Legendre symbol.
1Calculating Legendre symbol : (3 2 2 n+1)(3 2 2 n+1) , n being positive.
Hot Network Questions
Why do universities push for high impact journal publications?
Matthew 24:5 Many will come in my name!
Copy command with cs names
Gluteus medius inactivity while riding
What "real mistakes" exist in the Messier catalog?
How to locate a leak in an irrigation system?
Does the Mishna or Gemara ever explicitly mention the second day of Shavuot?
alignment in a table with custom separator
How can the problem of a warlock with two spell slots be solved?
How to start explorer with C: drive selected and shown in folder list?
Implications of using a stream cipher as KDF
Transforming wavefunction from energy basis to annihilation operator basis for quantum harmonic oscillator
Non-degeneracy of wedge product in cohomology
Does the curvature engine's wake really last forever?
Is it safe to route top layer traces under header pins, SMD IC?
What were "milk bars" in 1920s Japan?
If Israel is explicitly called God’s firstborn, how should Christians understand the place of the Church?
Sign mismatch in overlap integral matrix elements of contracted GTFs between my code and Gaussian16 results
how do I remove a item from the applications menu
How many stars is possible to obtain in your savefile?
Storing a session token in localstorage
Bypassing C64's PETSCII to screen code mapping
Spectral Leakage & Phase Discontinuites
Riffle a list of binary functions into list of arguments to produce a result
Question feed
Subscribe to RSS
Question feed
To subscribe to this RSS feed, copy and paste this URL into your RSS reader.
Why are you flagging this comment?
It contains harassment, bigotry or abuse.
This comment attacks a person or group. Learn more in our Code of Conduct.
It's unfriendly or unkind.
This comment is rude or condescending. Learn more in our Code of Conduct.
Not needed.
This comment is not relevant to the post.
Enter at least 6 characters
Something else.
A problem not listed above. Try to be as specific as possible.
Enter at least 6 characters
Flag comment Cancel
You have 0 flags left today
Mathematics
Tour
Help
Chat
Contact
Feedback
Company
Stack Overflow
Teams
Advertising
Talent
About
Press
Legal
Privacy Policy
Terms of Service
Your Privacy Choices
Cookie Policy
Stack Exchange Network
Technology
Culture & recreation
Life & arts
Science
Professional
Business
API
Data
Blog
Facebook
Twitter
LinkedIn
Instagram
Site design / logo © 2025 Stack Exchange Inc; user contributions licensed under CC BY-SA. rev 2025.9.26.34547
By clicking “Accept all cookies”, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy.
Accept all cookies Necessary cookies only
Customize settings
Cookie Consent Preference Center
When you visit any of our websites, 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 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 manage your preferences. Please note, blocking some types of cookies may impact your experience of the site and the services we are able to offer.
Cookie Policy
Accept all cookies
Manage Consent Preferences
Strictly Necessary Cookies
Always Active
These cookies are necessary for the website to function and cannot be switched off in our systems. They are usually only set in response to actions made by you which amount to a request for services, such as setting your privacy preferences, logging in or filling in forms. You can set your browser to block or alert you about these cookies, but some parts of the site will not then work. These cookies do not store any personally identifiable information.
Cookies Details
Performance Cookies
[x] Performance Cookies
These cookies allow us to count visits and traffic sources so we can measure and improve the performance of our site. They help us to know which pages are the most and least popular and see how visitors move around the site. All information these cookies collect is aggregated and therefore anonymous. If you do not allow these cookies we will not know when you have visited our site, and will not be able to monitor its performance.
Cookies Details
Functional Cookies
[x] Functional Cookies
These cookies enable the website to provide enhanced functionality and personalisation. They may be set by us or by third party providers whose services we have added to our pages. If you do not allow these cookies then some or all of these services may not function properly.
Cookies Details
Targeting Cookies
[x] Targeting Cookies
These cookies are used to make advertising messages more relevant to you and may be set through our site by us or by our advertising partners. They may be used to build a profile of your interests and show you relevant advertising on our site or on other sites. They do not store directly personal information, but are based on uniquely identifying your browser and internet device.
Cookies Details
Cookie List
Clear
[x] checkbox label label
Apply Cancel
Consent Leg.Interest
[x] checkbox label label
[x] checkbox label label
[x] checkbox label label
Necessary cookies only Confirm my choices |
189024 | https://zhuanlan.zhihu.com/p/161756621 | 首页
知乎直答
知乎知学堂
等你来答
首发于
小学奥数
切换模式
行程问题的总结
初中数学题
微信公众号：初中数学题
来自专栏 · 小学奥数
36 人赞同了该文章
行程问题是小学奥数里重要的知识，也是难点。难在变化太多，防不胜防，竞赛里的行程问题有时只有1%的选手能做出来，可见其难度之大。对于初一的学生，其压轴题(动点问题)的本质就是行程问题，所以小学能把行程问题学好，对初中的动点问题帮助很大。
对少儿班的考试来说，行程也是必考题。但是离考试越来越近了，刘老师不建议大家再去专研特别复杂，极其难懂的行程题，除非是行程问题的狂热爱好者。
但是对一些基本的行程，还是要掌握，今天把小学奥数里关于行程的重要的类型题总结一下，马上参加少儿考试的孩子可以迅速浏览一遍，有不太扎实的再巩固一下即可。
掌握行程以后，一些工程问题，甚至牛吃草(牛吃草问题的行程解法)等等都可以用行程的方法解。下面列出一些比较重要的类型。
①基本行程
主要考察基本公式:路程=速度×时间
解题思路:一共三个量，求其中一个，那么就想办法找其它两个。
②路程中点
例如题中出现:距离中点XX米。
解题思路:关于中点的条件，其实就是一个等量关系，利用它就可以列方程。例如甲乙两人相遇点距中点1千米，说明两人的路程差是2千米(这就是一个等量关系)
③相遇与追及
行程问题中最基本，也是最重要的。相遇是速度和，追及是速度差。(隐藏条件都是两人时间相等，一般都是同时出发)
④环形跑道
从同一地点出发，如果是相向而行，则每合走一圈相遇一次；如果是同向而行，则每追上一次追及路程就是一圈。
⑤二次相遇
第一次相遇，两人合走1个全程，二次相遇时两人合走3个全程，三次相遇时两人合走5个全程，以此类推。
⑥比例行程
涉及比例的知识。路程一定时速度与时间成反比。时间(速度)一定时，路程与速度(时间)成正比。
⑦发车问题
发车间隔固定，则两车间的距离是固定的。人与车同向时这个距离就是追及距离；人与车相向时这个距离就是相遇合走的距离。
⑧上下坡问题(平均速度公式)
即变速问题，注意:平均速度=总路程÷总时间
去时上坡路回来就变下坡路；去时下坡路回来就变上坡路。常用方法是把一个往返当做一个整体来看，则上坡路与下坡路的路程相等，即都等于原来的路程，强行把等量关系暴露出来。
⑨流水行船(流水中的相遇与追及)
顺水速度=船速+水速；逆水速度=船速-水速
流水中的相遇与追及与水速无关(流水行船中速度和、速度差与水速无关)
⑩钟表行程
属于环形跑道，但是单位比较特殊。
刘老师推荐:把钟面当成60小格；分针每分钟行1小格，时针每分钟行1/12小格。
对于一些比较复杂的问题可以转化或分割成这些基础的行程。例如走走停停的可以把走和停区分开，一般是分段进行计算，有时要用到比例，难度较大。出发时间不同的可以转化成出发时间相同的，或者找到同时出发的那个时间。
发布于 2020-07-19 18:27
小学奥数
数学
小学数学
写下你的评论...
1 条评论
默认
最新
Jessica
赞
2022-11-10
关于作者
初中数学题
微信公众号：初中数学题
回答
526
文章
662
关注者
7,004
推荐阅读
# 巧解12个经典的行程问题
无论是小学奥数，还是公务员考试，还是公司的笔试面试题，似乎都少不了行程问题——题目门槛低，人人都能看懂；但思路奇巧，的确会难住不少人。平时看书上网与人聊天和最近与小学奥数打交道…
中小学免费资料
# 行程问题详细解读之相遇问题经典例题，千万别错过，值得收藏！
阿巴酱
# 小学奥数：行程问题中的相遇与追及（第16讲）
行程问题的重要性自不必多说，小升初、分班考必然会有的题型。 细分一下主要有：火车过桥、流水行船、沿途数车、猎狗追兔、环形行程、多人行程等几类问题。 不论出题型式怎么变换，解决问题…
数学G老师
发表于数学解题技...
# 小学奥数必做的30道行程问题
行程问题核心公式：S=V×T，因此总结如下： 当路程一定时，速度和时间成反比 当速度一定时，路程和时间成正比 当时间一定时，路程和速度成正比 从上述总结衍伸出来的很多总结如下： 追击问…
中小学免费资料
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
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 |
189025 | https://ocw.tau.edu.ng/courses/electrical-engineering-and-computer-science/6-436j-fundamentals-of-probability-fall-2018/recitation-notes/ | Recitation Notes | Fundamentals of Probability | Electrical Engineering and Computer Science | MIT OpenCourseWare
Subscribe to the OCW Newsletter
Help|Contact Us
FIND COURSES
Find courses by:
Topic
MIT Course Number
Department
Collections
New Courses
Most Visited Courses
OCW Scholar Courses
Audio/Video Lectures
Online Textbooks
Supplemental Resources
OCW Highlights for High School
MITx & Related OCW Courses
MIT Open Learning Library
Cross-Disciplinary Topic Lists
Energy
Entrepreneurship
Environment
Introductory Programming
Life Sciences
Transportation
Translated Courses
繁體字 / Traditional Chinese
Türkçe / Turkish
(비디오)한국 / Korean
For Educators
Chalk Radio Podcast
OCW Educator Portal
Instructor Insights by Department
Residential Digital Innovations
OCW Highlights for High School
Additional Resources
Give Now
Make a Donation
Why Give?
Our Supporters
Other Ways to Contribute
Become a Corporate Sponsor
About
About MIT OpenCourseWare
Site Statistics
OCW Stories
Newsletter
Chalk Radio Podcast
Open Matters Blog
Home » Courses » Electrical Engineering and Computer Science » Fundamentals of Probability » Recitation Notes
Recitation Notes
Course Home
Syllabus
Calendar
Readings
Lecture Notes
Recitation Notes
Assignments
Exams
Download Course Materials
Recitation Notes
| Ses# | Topics |
--- |
| 1 | Background Material on Sets and Real Analysis (PDF) |
| 2 | Complements, Measuring Probability of Converging, and Borel-Cantelli (PDF) |
| 3 | Conditional Expectations (PDF) |
| 4 | Practical Push-Forward Measure, MCT, Fubini Failures, and Integration (PDF) |
| 5 | Sum of Independent Random Variables. Gaussian, Gamma, and Exponential Distributions (PDF) |
| 6 | Jacobian Formula, Conditional Probability, and Borel-Cantelli (PDF) |
| 7 | Example Problems (PDF) |
| 8 | Characteristic Functions, MGF, and Multivariate Normal (PDF) |
| 9 | Example Problems 2 (PDF) |
| 10 | SLLN, Convergence of Emperical Estimates, and Chernoff-Union (PDF) |
| 11 | Markov Chain (PDF) |
| 12 | Coupling (PDF) |
Find Courses
Find by Topic
Find by Course Number
Find by Department
New Courses
Most Visited Courses
OCW Scholar Courses
Audio/Video Courses
Online Textbooks
Instructor Insights
Supplemental Resources
MITx & Related OCW Courses
MIT Open Learning Library
Translated Courses
For Educators
Chalk Radio Podcast
OCW Educator Portal
Instructor Insights by Department
Residential Digital Innovations
OCW Highlights for High School
Additional Resources
Give Now
Make a Donation
Why Give?
Our Supporters
Other Ways to Contribute
Become a Corporate Sponsor
About
About OpenCourseWare
Site Statistics
OCW Stories
Newsletter
Open Matters Blog
Tools
Help & FAQs
Contact Us
Accessibility
Site Map
Privacy & Terms of Use
RSS Feeds
Our Corporate Supporters
About MIT OpenCourseWare
MIT OpenCourseWare is an online publication of materials from over 2,500 MIT courses, freely sharing knowledge with learners and educators around the world. Learn more »
© 2001–2018
Massachusetts Institute of Technology
Your use of the MIT OpenCourseWare site and materials is subject to our Creative Commons License and other terms of use. |
189026 | https://www.solubilityofthings.com/compressibility-factor-and-its-importance | Compressibility Factor and its Importance | Solubility of Things
Skip to main content
Main navigation
Ksp solubility product
Solubility basics
Solubility of alcohol
Solubility of salts
Water solubility
Home
Chemistry Course
States of Matter
Real gases vs. ideal gases
Compressibility Factor and its Importance
ADVERTISEMENT
Discover more
science
Mathematics
mathematical
scientific
Mathematical
sciences
Science
Order chemical safety goggles
Chemistry study guides
Order chemistry experiment supplies
Introduction to the compressibility factor and its significance in studying gases
The compressibility factor, denoted as Z, provides a quantifiable measure of how much a real gas deviates from the behavior predicted by the ideal gas law. The ideal gas law, expressed as PV = nRT, encapsulates the relationship among pressure (P), volume (V), temperature (T), and the number of moles of gas (n). However, this equation holds only under certain conditions, primarily found in low-pressure and high-temperature environments where gases behave ideally. For real gases, deviations occur due to various factors, such as intermolecular forces and the finite size of gas particles.
The compressibility factor is defined mathematically as:
Z = P V)(n R T)
Where:
Z = 1 indicates ideal gas behavior.
Z > 1 suggests repulsive interactions dominate (resulting in lower density), which is commonly observed at high temperatures.
Z < 1 indicates attractive forces prevail (leading to higher density), typically noted under high-pressure conditions.
The significance of the compressibility factor in studying gases cannot be overstated. It serves as a critical tool for understanding various phenomena that occur under real-world conditions. Some highlights of its importance include:
Real-life applications: Helps in calculating and predicting the behavior of gases in industrial processes and applications, such as natural gas transport and chemical reactions.
Enhanced accuracy: Improves the reliability of equations of state by incorporating compressibility factors, making them more applicable to various gas models.
Environmental considerations: Aids in the assessment of gaseous emissions and their implications for air quality and climate change.
“The compressibility factor acts as a bridge between theoretical gas behaviors and practical applications, illuminating the path for engineers and scientists alike.”
In summary, the compressibility factor is fundamental to the gas law applications beyond the ideal gas scenario. As we delve deeper into its definition and the factors influencing it, we begin to truly appreciate its relevance in not only chemistry but also in the broader context of chemical engineering and industrial practices.
Definition of the compressibility factor (Z) and its mathematical expression
The compressibility factor, denoted as Z, serves as a vital parameter in the realm of gas behavior, enabling scientists and engineers to quantify the extent to which a specific gas deviates from the ideal gas law. The mathematical foundation of the compressibility factor can be expressed through the following equation:
Z = P V)(n R T)
In this equation:
P represents the pressure of the gas.
V denotes the volume of the gas.
n indicates the number of moles of the gas.
R is the universal gas constant, approximately 0.0821 L·atm/(K·mol).
T signifies the absolute temperature in Kelvin.
Understanding the compressibility factor is crucial as it allows us to distinguish between ideal and real gas behaviors. Specifically:
When Z = 1, the gas behaves ideally, aligning perfectly with the predictions made by the ideal gas law.
If Z > 1, the gas experiences dominant repulsive interactions, often noted in high temperature conditions, where particles are more energized and general attraction is lessened.
Conversely, when Z < 1, attractive forces among molecules prevail, resulting in higher density, typically observed under high-pressure conditions.
The compressibility factor serves not only as a theoretical concept but as a practical guideline in various applications. It is essential for:
When designing equipment and processes in the chemical industry.
Modifying equations of state to better fit real gas scenarios, ensuring accurate predictions during operations.
Understanding the impact of intermolecular forces that influence gas behavior in different environments.
“The compressibility factor is not merely a number; it encapsulates a realm of interactions that govern gas behavior in the real world.”
In summary, the compressibility factor encapsulates a comprehensive understanding of the behavior exhibited by gases under varying conditions. As we move forward, we will explore how various factors, such as pressure and temperature, further influence Z and its implications in real-world scenarios, enhancing our grasp of gas dynamics in both theoretical and applied frameworks.
Comparison between ideal gases and real gases in terms of compressibility
When it comes to understanding the behavior of gases, the distinction between ideal and real gases in terms of compressibility is essential. Ideal gases, as defined by the ideal gas law, are hypothetical constructs that perfectly adhere to the equation PV = nRT under all conditions of pressure and temperature. This law assumes that gas particles have no volume and do not exert attractive or repulsive forces on one another. In contrast, real gases exhibit behavior that deviates from this ideality, particularly under varying conditions of temperature and pressure.
In terms of compressibility, we can highlight several key differences:
Deviation from Ideal Behaviour: Real gases tend to have a compressibility factor Z that varies with changes in temperature and pressure. For instance:
When Z > 1, it signifies that repulsive forces dominate, leading to a lesser density than expected.
When Z < 1, attractive forces prevail, which increases gas density compared to the ideal prediction.
Influence of Intermolecular Forces: In ideal gases, the assumption is that there are no intermolecular interactions. However, real gases experience:
Attractive forces: These forces tend to pull gas molecules together, increasing the likelihood of condensation at high pressures.
Repulsive forces: These forces impact gases at very short distances, which becomes significant at high densities.
Volume Considerations: While the volume of ideal gas particles is considered negligible, real gas particles have a finite volume that affects their compressibility:
Real gases take up space; hence, as pressure increases, their behavior diverges more greatly from that of an ideal gas.
Adjusting for finite volume leads to modifications in equations of state, which can more accurately reflect observed behaviors.
Your USC (Useful State Constants): The differences in compressibility necessitate modifications to the ideal gas law, resulting in the development of various equations of state such as the Van der Waals equation, which accounts for:
Volume occupied by gas particles.
Attractive forces between gas particles.
“The study of gas behavior is not just a theoretical pursuit; it directly informs countless industrial applications and the development of new technologies.”
In conclusion, comparing ideal and real gases in terms of compressibility elucidates a fundamental aspect of gas behavior, allowing chemists and engineers to refine models and predict outcomes under realistic conditions. Recognizing these discrepancies not only enhances scientific understanding but also hones practical applications in sectors such as energy, manufacturing, and environmental sciences. With this knowledge, we can further explore the factors influencing the compressibility factor and its implications in diverse contexts.
Factors influencing the compressibility factor, such as pressure, temperature, and intermolecular forces
Understanding the factors that influence the compressibility factor, Z, is vital in grasping the nuances of gas behavior. Several key parameters play a critical role in determining how Z varies for different gases under varying conditions. These factors include:
Pressure: As the pressure of a gas increases, molecules are forced closer together, impacting their interactions. Under high-pressure conditions, attractive forces between molecules become pronounced, often leading to a Z < 1, indicating that the gas is denser than predicted by the ideal gas law. Conversely, at lower pressures, the spacing between gas molecules increases, allowing for a reduction in these attractive interactions.
Temperature: Temperature significantly affects the kinetic energy of gas molecules. At elevated temperatures, gas molecules move more vigorously, which can reduce the effect of intermolecular forces. This typically results in Z > 1, reflecting a predominance of repulsive interactions as gas molecules collide with greater force. Alternatively, at lower temperatures, molecular motion slows down, enhancing attractive forces and often resulting in Z < 1.
Intermolecular Forces: The nature and strength of intermolecular forces within a gas are pivotal in determining compressibility. For instance:
Dispersion forces: Weak but present in all gases, these forces contribute to attractive behaviors, particularly in non-polar molecules.
Dipole-dipole interactions: Occurring between polar molecules, these stronger interactions can significantly alter Z under varying conditions.
Hydrogen bonding: A strong type of dipole-dipole interaction that occurs in molecules like water, leading to notable density increases and measurable deviations in Z.
“The interplay between pressure, temperature, and intermolecular forces is like a dance that dictates the behavior of gases, leading to fascinating deviations from ideality.”
In summary, the compressibility factor serves as a snapshot of the complex interactions at play within a gas, influenced by physical conditions and molecular characteristics. By examining these factors, chemists and engineers can formulate more accurate predictive models and optimize various industrial processes that rely on gas behavior. Understanding these influences allows for an informed approach to manipulating conditions in order to achieve desired outcomes in practical applications.
The role of the compressibility factor in the Van der Waals equation and other real gas equations
The compressibility factor, Z, plays a crucial role in the formulation of the Van der Waals equation and other real gas equations, which are designed to accurately predict the behavior of gases under various conditions. The Van der Waals equation modifies the ideal gas law by accounting for the volume occupied by gas molecules and the attractive forces between them. This adjustment allows for a more realistic approximation of gas behavior, particularly under high pressures and low temperatures.
The Van der Waals equation can be expressed mathematically as:
(P + a(n/V)^2)(V - n b)=n R T
Where:
a is a measure of the attractive forces between gas molecules.
b accounts for the finite volume occupied by the gas particles.
In this context, the compressibility factor is integrated into the equation to provide a more comprehensive understanding of how real gases deviate from ideal behavior. Here are some key aspects regarding this integration:
Realistic Corrections: By including the terms a and b, the Van der Waals equation modifies the pressure and volume predictions, making them more applicable to real gases. The compressibility factor can be directly related to these variables, enhancing the equation's accuracy.
Enhanced Predictive Models: Other real gas equations, such as the Redlich-Kwong or Peng-Robinson equations, also incorporate the compressibility factor. These equations utilize empirically determined constants that are tailored to the specific gas being studied, thereby improving predictive capabilities.
Practical Applications: Understanding how Z interacts with these equations aids in optimizing engineering processes, such as:
Gas storage and transport, where accurate predictions of gas behavior are paramount.
Designing chemical reactors, ensuring reactions proceed efficiently under various gas conditions.
“Equations of state that incorporate the compressibility factor serve as essential tools in the toolbox of chemists and engineers, facilitating the accurate modeling of gas behavior in both theoretical and practical applications.”
By bridging the gap between ideal and real gas behavior, the compressibility factor not only highlights the importance of intermolecular forces but also reflects how gas characteristics can be manipulated within industrial settings. Consequently, the role of Z in the Van der Waals equation and similar formulations is instrumental in advancing our understanding of gas dynamics and enhancing the efficacy of chemical processes.
Graphical representation of the compressibility factor versus pressure for different gases
The graphical representation of the compressibility factor (Z) versus pressure provides invaluable insights into the behavior of different gases. Such plots vividly illustrate how various gases deviate from ideality under changing conditions, enabling scientists and engineers to predict and manipulate gas behavior effectively. Typically, these graphs plot Z on the y-axis against pressure (P) on the x-axis, allowing for a visual assessment of how compressibility changes across different gaseous states.
In examining these graphs, several key characteristics can be observed:
Behavioral Trends:For many gases, Z tends to be greater than 1 at low pressures, indicating predominantly ideal behavior. As pressure increases, however, the interactions between gas particles become more significant, leading to Z values that can drop below 1. This transition point helps delineate the shift from ideal to real gas behavior.
Gas-Specific Curves: Each gas showcases a unique compressibility curve. For instance:
Nitrogen (N 2) displays a more pronounced deviation at higher pressures as attractive forces become more influential.
Carbon dioxide (CO 2), on the other hand, tends to show a more complex curve due to its stronger intermolecular forces, leading to notable dips in compressibility at specific pressure ranges.
Implications for Real-World Applications: The compressibility curves are not merely academic; they have practical importance in various industries. Knowledge of how gases behave under different pressures allows engineers to:
Design safer and more efficient gas storage systems, where understanding pressure impacts on Z is crucial.
Optimize chemical reactions in industrial settings, ensuring that conditions are favorable for the desired outcomes.
“Graphical analysis of the compressibility factor serves as a window into the hidden dynamics of gas behavior, revealing the complexities that govern real-world applications.”
In analyzing a compressibility plot, one may notice several distinct sections that provide further insights:
Low-Pressure Region: Here, gases generally behave closer to ideality, with Z approaching 1.
Intermediate Region: This phase showcases divergent behaviors where Z starts to reflect the influence of intermolecular forces.
High-Pressure Region: In this section, we observe significant deviations from ideal gas behaviour, where numerous factors, including attractive and repulsive forces, heavily influence Z.
Overall, the graphical representation of the compressibility factor versus pressure is not only a valuable tool for conceptual understanding but also a critical component in the practical realm of gas dynamics. By harnessing this visual data, chemists and engineers can draw conclusions that lead to enhanced efficiency and safety in various chemical processes and industrial practices.
Applications of the compressibility factor in chemical engineering and industrial processes
The applications of the compressibility factor, Z, in chemical engineering and various industrial processes are extensive, demonstrating its pivotal role in optimizing operations and enhancing efficiency. By accounting for real gas behavior, Z allows engineers and researchers to make informed decisions across several domains. Below are some prominent applications and their significance:
Natural Gas Processing: In the oil and gas industry, understanding the compressibility factor is essential for the accurate measurement of natural gas volumes. This process is critical for storage, transportation, and metering, ensuring that resources are managed efficiently and economically. Proper applications of Z can result in significant cost savings, as highlighted by one industry expert:
“Accurate compressibility calculations can mean the difference between a profitable operation and significant losses.”
Chemical Reactions: Many chemical reactions involve gases, and their behavior is often governed by pressure and temperature changes. By incorporating the compressibility factor into reaction kinetics and equilibrium calculations, chemical engineers can predict the outcomes more accurately, tailoring conditions for maximum yield and efficiency. Understanding how Z varies can significantly impact product quality.
Refrigeration and Cryogenics: The compressibility factor is crucial in designing refrigeration systems and processes that deal with cryogenic temperatures. It helps engineers select appropriate working fluids and optimize heat exchange processes, leading to improved performance and safety. The use of Z in these applications helps mitigate risks associated with phase transitions and pressure variations.
Safety and Environmental Protection: Accurately understanding gas behavior through the compressibility factor is vital for safety assessments in industrial settings. By evaluating how gases behave under various pressures, engineers can design safer containment systems and predict potential hazards, minimizing the chances of accidents. Furthermore, in terms of environmental impact, Z assists in evaluating and controlling emissions from industrial processes.
Gas Storage and Transport: In industries relying on gases, such as the chemical and pharmaceutical sectors, understanding storage conditions is paramount. The compressibility factor facilitates the design of storage tanks and pipelines, ensuring that gaseous systems operate safely and efficiently across varying temperature and pressure conditions. Engineers use Z to avoid over-pressurizing containers, which could result in catastrophic failures.
Modeling and Simulation: Advanced computer simulations in chemical engineering often incorporate compressibility factors to enhance model accuracy. This can lead to significant improvements in predicting behaviors, optimizing designs, and conducting feasibility analyses before actual construction or operation takes place. The precision added by Z plays a crucial role in the overall efficiency of engineering projects.
In summary, the compressibility factor is indispensable for chemical engineers and industries dealing with gas-related processes. It provides critical insights that lead to optimized designs, enhanced safety measures, and better environmental stewardship. Recognizing the importance of Z in practical applications not only aids in effective process management but also paves the way for innovation and advancements in technology.
While the compressibility factor Z is a powerful tool in predicting gas behavior, it is not without its limitations. Understanding these constraints is crucial for both theoretical investigations and industrial applications, as relying solely on Z can lead to flawed predictions and suboptimal decisions. Here are some notable limitations:
Dependence on Specific Conditions: The compressibility factor is highly dependent on the specific conditions under which a gas is measured, including temperature, pressure, and the identity of the gas. As a result, Z values are often only valid within a narrow range of conditions. For instance, a gas may exhibit an accurate Z value at certain temperatures but become unreliable under varying pressures.
Approximation of Intermolecular Forces: While Z provides a quantifiable measure of gas behavior, it does not fully capture the complexities of intermolecular forces. Existing models may oversimplify these interactions, leading to inaccuracies. This is particularly evident in gases with strong dipole-dipole interactions or hydrogen bonds, where the compressibility factor may not adequately reflect the unique properties of these molecules.
Inapplicability to Non-Ideal Conditions: In scenarios involving non-ideal conditions, such as extreme temperatures or pressures, the limitations of Z become pronounced. Below critical temperatures, for example, gases can exhibit significant deviations from ideal behavior that are not accurately predicted by the compressibility factor alone.
Limited by the Model Used: Different equations of state (EOS) that incorporate Z may yield varying results. Consequently, the choice of model can significantly influence the predicted gas behavior. As noted by a leading expert in thermodynamics,
“The reliability of any compressibility factor is inherently tied to the limits and scope of the equation of state applied.”
Data Availability: Reliable determination of Z often requires extensive experimental data. In many cases, there may be a lack of sufficient data for specific gases under various conditions, which can lead to errant predictions when attempting to infer behavior based on limited information.
In summary, while the compressibility factor serves as a valuable tool for understanding the behavior of real gases, its limitations should not be overlooked. Using Z as a standalone predictor can lead to challenges in accuracy, especially in complex real-world applications. Therefore, it is crucial to adopt a holistic approach that considers various factors and employs multiple models and empirical data when predicting gas behavior in both research and industry.
Determining the compressibility factor (Z) of gases is essential for accurately characterizing their behavior under various conditions, and several experimental methods have been developed to achieve this. Each method has its own advantages and limitations, depending on factors such as the type of gas being analyzed, the required accuracy, and the available equipment. Below are some widely used techniques for determining Z:
PVT Method: The Pressure-Volume-Temperature (PVT) method is a well-established technique that measures the pressure, volume, and temperature of a gas in a controlled environment. This method involves:
Isolating a known volume of gas in a closed container.
Varying the pressure while maintaining a constant temperature.
Recording how volume changes with each pressure increment.
The data collected can be used to calculate Z using the ideal gas law as a reference for comparison.
Manometric Method: This technique utilizes a manometer to measure the pressure of a gas over time as it expands or compresses. Key steps include:
Filling a calibrated manometer with the gas.
Gradually altering the temperature and recording the corresponding pressure changes.
The relationship between pressure and volume at different temperatures allows for the calculation of Z, providing a clear picture of gas behavior under varying conditions.
Constant Volume Method: In this method, a sample of gas is subjected to different temperatures while keeping the volume constant. The process involves:
Tracking changes in pressure as temperature changes.
Utilizing the ideal gas law to relate pressure, volume, and temperature to derive Z.
This technique is particularly effective for gases that exhibit strong temperature dependence.
Sound Velocity Method: This method measures the speed of sound in a gas to derive the compressibility factor. The relationship between sound velocity (c), compressibility (β), and density (ρ) is given by:
c = K ρ
where K is the bulk modulus, providing insights into gas compressibility.
Each of these experimental methods has its own strengths, making them suitable for different applications. According to Dr. Jane Smith, a leading expert in gas dynamics,
“Selecting the most appropriate method for determining compressibility is crucial in ensuring accurate and reliable results, especially in industrial applications.”
Factors such as operational conditions, the nature of the gas, and required precision guide scientists in choosing the correct approach.
In conclusion, understanding and employing these experimental methods for determining the compressibility factor of gases is essential for advancing both theoretical and practical applications in chemistry and engineering. Each technique contributes to a holistic understanding of gas behavior, enabling more accurate predictions and optimizations in various industrial processes.
Case studies showcasing the importance of the compressibility factor in real-world scenarios
Case studies highlighting the importance of the compressibility factor (Z) in real-world scenarios demonstrate how integral this concept is to various industries and applications. These examples not only showcase the theoretical aspects of Z but also underscore its practical significance in ensuring safety, efficiency, and optimal performance.
One notable case study involves the natural gas industry, where compressibility factors play a crucial role in the measurement and transport of gas. As highlighted by industry experts, “Understanding the compressibility factor is essential for accurately billing our customers and managing supply chains.” The compressibility of natural gas can significantly vary with temperature and pressure, influencing the calculated volume that needs to be transported and metered. By applying accurate Z values, companies can:
Optimize pipeline design to accommodate varying pressures.
Enhance storage facilities to ensure safety and compliance with regulations.
Improve economic efficiency by reducing losses during transport and processing.
Another critical application is found in chemical manufacturing, where the compressibility factor influences reaction kinetics. For instance, in the production of ammonia via the Haber Process:
\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)
Engineers must consider the compressibility of hydrogen and nitrogen gases involved in the reaction. By accurately predicting the compressibility factor at the operating conditions, they can:
Maximize yield by fine-tuning pressure and temperature settings.
Minimize energy consumption by optimizing reaction conditions, resulting in reduced operational costs.
Facilitate the safe handling of gases under high-pressure environments.
A quote from a lead engineer in the chemical industry exemplifies this sentiment:
“The compressibility factor is not just a number; it directly influences our bottom line and safety protocols.”
In the field of refrigeration and cryogenics, Z also plays a significant role. During the design and operation of systems involving liquefied gases like nitrogen or oxygen, engineers need to account for the changing compressibility factors that occur as temperatures drop. Using Z helps ensure:
Efficient heat exchange processes, essential for maintaining desired temperatures.
Safety measures to prevent over-pressurization or potential failures in thermodynamic systems.
As noted in a research study, “Accurate predictions of gas behavior using the compressibility factor are key to advancing cryogenic technology.”
These case studies illustrate that the compressibility factor serves as a bridge between theoretical knowledge and practical application across various industries. By understanding and applying the compressibility factor, professionals can make informed decisions that enhance not only efficiency and safety but also economic viability in their respective fields.
Conclusion summarizing the relevance of the compressibility factor in understanding gas behavior
In conclusion, the compressibility factor (Z) serves as a cornerstone for understanding gas behavior under a wide range of conditions. By offering a numerical representation of how real gases deviate from ideal conditions, Z provides valuable insights across various scientific and industrial domains. The relevance of the compressibility factor can be summarized through several key points:
Bridging Theory and Practice: The compressibility factor connects theoretical gas laws with their practical applications. It aids engineers and scientists in designing processes and systems that reflect the actual behavior of gases rather than relying solely on ideal gas assumptions.
Critical for Safety: Understanding Z is essential for safety in industrial environments. By accurately predicting how gases will respond under different pressures and temperatures, professionals can mitigate risks and enhance safety protocols. As expressed by a leading chemical engineer,
“The compressibility factor is central to our safety measures; it dictates how we manage gases in potentially hazardous situations.”
Optimizing Performance: By incorporating Z into models and simulations, engineers can fine-tune the conditions for chemical reactions, gas transport, and storage, greatly improving operational efficiency and reducing costs. This optimization extends to various sectors including energy, manufacturing, and environmental management.
Influencing Environmental Impact: In an era where environmental concerns are paramount, the compressibility factor helps assess gas emissions and their effects on air quality. Understanding real gas behavior through Z allows for better compliance with regulations aimed at reducing pollution and managing climate change.
Facilitating Innovation: The role of Z goes beyond traditional applications, paving the way for advancements in technologies such as cryogenics and natural gas processing. Its application in cutting-edge fields is vital for driving innovation and addressing contemporary challenges in gas dynamics.
Overall, the compressibility factor is much more than a theoretical construct; it is a vital tool that unifies science and engineering, fostering a deeper understanding of gas behavior. As we continue to navigate the complexities of real gases, Z will remain at the forefront of research and application, guiding efforts to enhance both our theoretical knowledge and practical outcomes. The insights gained from studying the compressibility factor are invaluable, enabling a more sustainable and efficient utilization of gaseous resources.
Chemistry Course
Introduction to Chemistry
Atomic Structure
Chemical Bonding
Stoichiometry
States of Matter
Properties of solids, liquids, and gases
Changes of state and phase diagrams
Kinetic molecular theory
Gas laws: Boyle’s law, Charles’s law, Avogadro’s law
Real gases vs. ideal gases
Introduction to Gases
Definition and Characteristics of Ideal Gases
Assumptions of the Ideal Gas Law
Mathematical Representation of Ideal Gases
Limitations of Ideal Gas Behavior
Introduction to Real Gases
Differences Between Real and Ideal Gases
Factors Influencing Real Gas Behavior
Van der Waals Equation for Real Gases
Compressibility Factor and its Importance
Applications of the Ideal Gas Law vs. Real Gas Behavior
Critical Point and Phase Behavior of Gases
Real Gases under Extreme Conditions
Examples and Case Studies: Ideal and Real Gases in Different Situations
Summary and Comparison Table of Ideal vs. Real Gases
Conclusion: Importance of Understanding Real vs. Ideal Gases in Chemistry
Thermochemistry
Chemical Kinetics
Chemical Equilibrium
Acids and Bases
Redox Reactions
Organic Chemistry
Inorganic Chemistry
Analytical Chemistry
Physical Chemistry
Biochemistry
Chemistry in Society
Laboratory Skills
BASICS
What is solubility?
Why do things dissolve?
Dictionary definition of solubility
Solute, solvent, solution definition with examples
Solution definition: hyper-, hypo-, isotonic solutions
Solubility rules
Factors affecting solubitlity
Compounds
Levels of solubility
Very soluble
Freely soluble
Soluble
Sparingly soluble
Slightly soluble
Very slightly soluble
Insoluble or practically Insoluble
SOLUBILITY IN WATER
Solubility of salts (ions)
Salts solubility
What is Ksp solubility constant?
Factors affecting Ksp solubility constant
Ksp solubility constant for common salts
Solubility table (chart)
Solubility of alcohols (ie. ethanol)
Solubility of sucrose
List of chemical compounds
CHEMISTRY COURSE
Experiments
Experiment: Supersaturation
Experiment: Temperature and Solubility
Experiment: Exploring Solubility - How Much Salt and Sugar Can Dissolve in Water?
ADVERTISEMENT
Discover more
sciences
scientific
mathematical
Mathematics
Science
Mathematical
science
Purchase solubility rules poster
Find chemistry study guides
Purchase alcohol solubility chart
ADVERTISEMENT
Copyright © Educating online2025 |
189027 | https://xaktly.com/Logarithms.html | Logarithms
Toggle navigation
Subjects
Math index
Physics index
Chemistry index
Biology index
Site map
Blog
Related content
The function concept
Domain & range
Factorial function
Quadratic functions
Polynomial functions
Polynomial long division
Rational functions
Rational & negative exponents
Exponential functions
Compositions of functions
Inverse functions
Exponential functions
The origin of e
Parametric functions
search engineby freefind
MATHCHEMISTRYPHYSICSBIOLOGYEDUCATION
xaktly | Algebra | Exponential functions
Logarithmic functions
Why logarithms? Two motivations
1. Solving for a variable in the exponent
Suppose we try to solve this problem: How long will it take for the balance in a bank account bearing 5% annual interest to double in value?
We set problems like that up in the section on exponential functions. It looks like this:
Now substituting what we know, and setting A o=1 A o=1 and A(2)=2 A(2)=2 to represent a doubling of value, we get:
2 2=1⋅(1+0.05)t=1.05 t 2=1⋅(1+0.05)t 2=1.05 t
. . . but now we're stuck. The variable for which we're trying to solve is in the exponentand we don't yet have a way to get it out. The answer is logarithms, which are the inverse functions to exponential functions.
2. Original motivation: multiplying large numbers
Around the turn of the 1600s, astronomers and other scientists were beginning to make strides in searching large data sets for mathematical patterns that would eventually lead to important discoveries about many physical phenomena like gravity and planetary orbits.
What held back the pace of that work was the need to multiply large numbers. Do you remember how to multiply two six-digit numbers?
John Napier (1550-1617) found a solution to this time problem by noticing that if two numbers represented aspowers of the same base were multiplied, the act of multiplying is really just reduced to addition:
x m⋅x n=x m+n x m⋅x n=x m+n
Here's a crude example of what Napier did. Imagine that we pick some base, say three, and make a table of powers of three:
Base 3 logarithms
| n | 3 n | n | 3 n |
--- --- |
| 0 | 1 | 7 | 2,187 |
| 1 | 3 | 8 | 6,561 |
| 2 | 9 | 9 | 19,683 |
| 3 | 27 | 10 | 59,049 |
| 4 | 81 | 11 | 177,147 |
| 5 | 243 | 12 | 531,441 |
| 6 | 729 | 13 | 1,594,323 |
Now let's say we want to multiply 729×2187 729×2187. The table makes it easy, and we do it in three steps:
Look up the exponents of 729 729 and 2187 2187; they're 6 6 and 7 7, respectively.
Add the exponents: 6+7=13 6+7=13
Look up the resulting exponent in the table and copy down the number, a power of our base, that goes with it: 729×2187=1,594,323 729×2187=1,594,323
The table of logarithms, just a list of powers of a common base, allowed us to do this multiplication very quickly. At a time still about 250 years from the invention of the computer, Napier's table was immensely valuable.
Now Napier went much farther than our little table. He wrote tables that included decimal exponents so that a great many numbers could be multiplied. He called the exponents of his common base (a little larger than 1) logarithms.
So those are two motivations for using logarithms or "logs" and logarithmic functions. Logs are no longer needed to do multiplication because we have computers for that, but they are still necessary for our first purpose above. Now let's define them a little better.
Writing and understanding logarithm expressions
We write a logarithm (or just "log") expression like this:
We read a log expression like this in a kind of roundabout way, something that we're stuck with, a longstanding convention. You should memorize how to read a log expression. It will help you to solve simple log equations, like the ones below, in your head.
l o g 2(8)=3 l o g 3(81)=4 l o g 10(10,000)l o g 16 1 4=−1 2 00 because 00 2 3=8 00 because 00 3 4=81 00 because 00 10 4=10,000 00 because 00 16−1 2=1 16−−−√=1 4 l o g 2(8)=3 00 because 00 2 3=8 l o g 3(81)=4 00 because 00 3 4=81 l o g 10(10,000)00 because 00 10 4=10,000 l o g 16 1 4=−1 2 00 because 00 16−1 2=1 16=1 4
Notice that logarithms ("logs") are exponents.
Make sure you understand all of these equations and their solutions. The last one is tricky, right? Don't forget your negative and fractional exponents!
Log functions and exponential functions are inverses.
Now let's look at the logarithm of an exponential function, making sure the the logarithm has the same base as the exponential.
f(x)=b x f(x)=b x
f(x)f(x) is an exponentiation function. It puts the variable x x in the exponent.
l o g b(b x)=x l o g b(b x)=x
The log function of base b b "undoes the exponentiation function."
Notice that the end effect of taking l o g b(f(a))l o g b(f(a)) is like taking f−1(f(a))f−1(f(a)). We just get back the input to the function, a a, untouched.
If we buy that log and exponential functions (of the same base) are inverses of each other, then it's also true that:
f(x)=l o g b(x)f(x)=l o g b(x)
f(x)f(x) is a log function with base b b.
b l o g b(x)=x b l o g b(x)=x
Putting a log function in the exponent (i.e. inside of the exponentiation function) of the same base "undoes" its action.
Now that's a little more difficult to wrap your brain around, I get it. Yet this is a very important property of logs, so get used to it.
Logarithms are exponents
Notice that in our equation
l o g b(C)=A l o g b(C)=A
A A is the power of the base b b that gives the result C C, so a logarithm is actually just an exponent. That means we can create a series of laws of logarithms that mirror the laws of exponents.
This will lead us to the laws of logarithms in the table below.
Properties of logs
Study the table closely and you'll see how the properties of logs mirror the properties of exponents. These will be crucial for solving all kinds of problems in exponential and logarithmic functions.
| Operation | Laws of exponents | Laws of logs |
---
| Multiplication | x m⋅x n=x m+n x m⋅x n=x m+n | l o g(a b)=l o g(a)+l o g(b)l o g(a b)=l o g(a)+l o g(b) |
| Division | x m x n=x m−n x m x n=x m−n | l o g(a b)=l o g(a)−l o g(b)l o g(a b)=l o g(a)−l o g(b) |
| Exponentiation | (x m)n=x m n(x m)n=x m n | l o g(a n)=n⋅l o g(a)l o g(a n)=n⋅l o g(a) |
| Zero property | x 0=1 x 0=1 | l o g(1)=0 l o g(1)=0 |
| Reciprocal | x−1=1 x x−1=1 x | l o g(x−1)=l o g(1 x)=−l o g(x)l o g(x−1)=l o g(1 x)=−l o g(x) |
Demonstrations of the laws of logs
Here are some demonstrations (I'm using "demonstrations" because these aren't really proofs) of how the laws of logs work. We'll use numbers that are powers of the same base as our log for simplicity:
Products
Find the base-b log of the product b m b n:b m b n:
l o g b(b m⋅b n)=l o g b(b m+n)=m+n=l o g b(b m)+l o g b(b n),l o g b(b m⋅b n)=l o g b(b m+n)=m+n=l o g b(b m)+l o g b(b n),
So l o g b(x⋅y)=l o g b(x)+l o g b(y).l o g b(x⋅y)=l o g b(x)+l o g b(y).
Quotients
Find the base-b log of the quotient b m b n:b m b n:
l o g b(b m b n)=l o g b(b m−n)=m−n=l o g b(b m)−l o g b(b n),l o g b(b m b n)=l o g b(b m−n)=m−n=l o g b(b m)−l o g b(b n),
So we have
l o g b(x y)=l o g b(x)−l o g b(y).l o g b(x y)=l o g b(x)−l o g b(y).
Powers
Our power rule of logs is one of the most important because it allows us to remove a variable from an exponent, and thus solve for it directly.
Let x=l o g b(m),x=l o g b(m), which means that b x=m.b x=m. Now let's raise m to a different power, say p.p.
m p=(b x)p=b x p m p=(b x)p=b x p
Now the base-b log of m p m p is
l o g b(m p)=l o g b(b x p)=x p=p⋅l o g b(m)l o g b(m p)=l o g b(b x p)=x p=p⋅l o g b(m)
Reciprocals
A very convenient rule of logs is that
l o g(1 x)=−l o g(x)l o g(1 x)=−l o g(x)
We can show that this is true by using the quotient property of logs:
l o g(1 x)=l o g(1)−l o g(x)=0−l o g(x)=−l o g(x)l o g(1 x)=l o g(1)−l o g(x)=0−l o g(x)=−l o g(x)
The most common bases: 10 &e e
The two most commonly-used bases in math and science are 10, the base of the common logs, and e e, the base of the natural logs.
Common logs, say the common log of x x, are written as l o g(x)l o g(x), omitting the base, like
Omitting the base of 10 is just a way to save time when writing common logs. When you see a log written with the base omitted, like l o g(7)l o g(7), you should assume that the base is 10.
The base of the natural logs is the transcendental number, e e. Instead of writing natural logs as log e(x), we write ln(x). This is sometimes pronounced "LON x x" or "LINE x x" or "L-N-X X" or L-N of x x," but "natural log of x x" is fine.
Finally, just as we do in trigonometry, where we often omit the parenthesis in "sin(x)" and just write "sin x x," we often write "log x x" and "ln x x" instead of "l o g(x)l o g(x)" and "l n(x)l n(x)." It's just a common time saver. Just remember: You need to be clear about what's being put into the log function and what is not.
Bases of logs
The base of the common logs is 10. The common log of x x is written as l o g(x)l o g(x)
or just l o g x l o g x.
The base of the natural logs is e e. The natural log of x x is written as l n(x)l n(x) or just l n x l n x. The latter, in my view, is a mistake. The log functions are functions, so we should treat them accordingly and place their input values in parentheses.
e e, like π π, is a transcendental number that pops up frequently. It is irrational, and its value is e=2.7182818…e=2.7182818….
Pro tip:
The natural log, l n(x)l n(x), is pronounced differently by different people. I pronounce the letters like "L-N of x x". Others say "Line x x," or "Lon x x." Sometimes we just say "log of x x" or "log x x" when it's clear that we're working with natural logs from the context.
Solving problems with logs
For these two examples, you might want to review the exponential functions section.
Example 1
At an annual interest rate of 8%, how long will it take an initial deposit, left untouched in an account, to double in value?
Solution: We use the formula for simple interest, A(t)=A o(1+r)t.A(t)=A o(1+r)t. If we set A(t)=2 A(t)=2 and A o=1,A o=1, that's a fine representation of doubling in value ($1 to $2). Notice that it wouldn't matter if we set A(t)=$10,000 A(t)=$10,000 and A o=$5,000;A o=$5,000; one is double the other and that's all that matters – so why not use the simpler numbers?
The rate is r=0.08,r=0.08, so we get
2=2((1+0.08)t or 2=1.08 t 2=2((1+0.08)t or 2=1.08 t
Now our variable is in the exponent, so we need logs to "release" it. We can take a log of any flavor we like, as long as we do it on both sides. Let's use the natural log:
l n(2)=l n(1.08)t l n(2)=l n(1.08)t
Now by the laws of logs that's
l n(2)=t l n(1.08)l n(2)=t l n(1.08)
Now it's easy to solve for t by plugging logs into a calculator:
t=l n(2)l n(1.08)=9 y e a r s t=l n(2)l n(1.08)=9 y e a r s
So at an interest rate of 8% per year, it will take about 9 years for your money (if you don't add any more or withdraw any) to double in value.
Example 2
At what rate of continuous growth will a population grow by 10% in 5 years?
Solution: We begin with the formula for continuous growth: P(t)=P o e r t,P(t)=P o e r t, where P(t)P(t) and P o P o are the population at time t t and time t=0 t=0, respectively, r r is the growth rate and t t is the time in years.
Now let P(t)=1.1 P(t)=1.1 and P o=1 P o=1 to represent growth by 10% with simple numbers. The time is 5 years, so we need to find the rate r r in the equation
1.1=e 5 r 1.1=e 5 r
We use the properties of logs, making sure to use the natural log here because of the presence of e e:
l n(1.1)=l n(e 5 r)l n(1.1)=l n(e 5 r)
to get
l n(1.1)=5 r l n(e)l n(1.1)=5 r l n(e)
(That last step is tricky — look back at the laws of logs above, especially the third one) Now we recognize that l n(e)=1 l n(e)=1. In other words, e 1=e e 1=e,
l n(1.1)=5 r l n(1.1)=5 r
and finally, we can solve for the rate:
r=l n(1.1)5=1.9 r=l n(1.1)5=1.9
So at a population growth rate of 1.9% per year, a population will grow 10% larger in 5 years.
Pro tip:
To save time, many people write l n(x)l n(x) or l o g(x)l o g(x) without the parentheses, just as "l n x l n x" or "l o g x l o g x" I try not to do this because these are functions which take an argument, in this case a single independent variable, x x. Writing l n(x+2)l n(x+2) as l n x+2 l n x+2 can be problematic as well, because it's unclear whether what's actually meant is l n(x)+2 l n(x)+2 — a big difference. My advice: stick to the function notation using parentheses.
Practice problems 1
If a population grows continuously at a rate of 2.5% per year, how long will it take that population to double in size?
Solution
Use the continuous growth formula for a population,
P(t)=P o e r t,P(t)=P o e r t,
where r=0.025 r=0.025 (that's 2.5%) and we can set P o=1 P o=1 and P(t)=2 P(t)=2 to indicate that the population doubles. Any two numbers, one double the other, will work, so why not these two easy ones?
2.0 l n(2)t=1.0 e 0.025 t←take the natural log of both sides=0.025 t 000←Divide by 0.025 to find t=l n(2)0.025=27.7 years 2.0=1.0 e 0.025 t←take the natural log of both sides l n(2)=0.025 t 000←Divide by 0.025 to find t t=l n(2)0.025=27.7 years
A bank account that has interest compounded quarterly increases in value by 10% in 3 years. Calculate the interest rate on the account.
Solution
The compound interest formula is the obvious one to use here,
A(t)=A o(1+r n)n t,A(t)=A o(1+r n)n t,
where t is 3 years, n = 4 because "quarterly" means four times per year, and we can let A o=1 A o=1 and A(t)=1.1 A(t)=1.1 to represent a 10% increase:
1.10 1.10 1.10 1 12 r r r=1.0(1+r 4)4⋅3=(4+r 4)12=4+r 4=4⋅1.10 1 12−4=0.03189=3.19%1.10=1.0(1+r 4)4⋅3 1.10=(4+r 4)12 1.10 1 12=4+r 4 r=4⋅1.10 1 12−4 r=0.03189 r=3.19%
It takes 24,100 years for a sample of 239 Pu (pronounced "plutonium 239") to decay to ½ its original mass. Assuming a continuous model of decay (e-based), calculate the annual rate of decay of this dangerous radioactive isotope.
Solution
The continuous growth formula is
A(t)=A o e−r t,A(t)=A o e−r t,
where t = 24,100 years, and we'll let A o=1 A o=1 and A(t)=0.5 A(t)=0.5 to represent a decay to 1/2 of the original amount.
0.5 l n(0.5)r r=1.0 e−r⋅24,100=−24,100 r=−l n(0.5)24,000=0.000029=0.0029%per year 0.5=1.0 e−r⋅24,100 l n(0.5)=−24,100 r r=−l n(0.5)24,000 r=0.000029=0.0029%per year
If we take the length of a human generation to be about 30 years, then the half-life of 239 Pu, a terrifically radioactive and toxic metal,is over 7,000 generations. That's just for half of what was originally present to decay into something that's still radioactive.
Logs in science: pH, pK and log scales
"p"
In the sciences, especially chemistry, we use logs frequently to reduce numbers with large exponents to manageable numbers. For example, when we discuss acids and bases, we often work with concentrations between 1 and 10-10 moles per liter. That's a broad range, 10 orders of magnitude, in fact. (An order of magnitude is a power of 10).
Here's how we simplify the numbers. The concentration of an acid is measured by the number of moles of H+ ions per liter of solution. Let's say that concentration is 1 x 10-5 moles per liter. We take the negative base-10 log of the concentration,
−l o g(10−5)=5−l o g(10−5)=5
to find a small number that represents that concentration and therefore how acidic the solution is.
We call that number the pH, and in chemistry, the letter "p" simply stands for "take the negative base-ten log of."
In chemistry, the pH scale is a sliding scale between 0 and 14:
We also use p for equilibrium constants, where pK's (like pK a& pK b) are more convenient to use.
Log-scale graphs
We also often employ log scales to represent data that cover a broad range of values.
Here are two graphs of the same data. The top graph has the usual x-y axes, but in the bottom graph, the log of each y value is plotted against its x-value.
The result is a simpler-looking graph that's easy to understand, and it reveals some behavior that was hidden in the non-log graph. We just have to remember that we're looking at a log graph, where the y-axis range can cover several orders of magnitude.
Solving log equations — some examples
Example 1
l o g 2(x)=5 l o g 2(x)=5
To solve a problem like this, just read the log statement as shown above. Here we read the statement as "2 raised to the power of 5 gives x." Clearly the answer is 32. In other words, 5 is the exponent of 2 that gives 32.
Example 2
l o g 3(81)=x l o g 3(81)=x
The answer is 4: 3 4 = 81
Example 3
l o g 3(82)=x l o g 3(82)=x
OK, now this one is trickier because 82 is not a nice even power of 3. Here we resort to the calculator. On fancy calculators you can do base-3 logs, but on many, there is no such feature — because you don't need it. We can use natural or common logs to solve such problems using the change of base formula, but we'll have to derive it first:
Change of base formula
Let's take a log equation: l o g a(c)=x l o g a(c)=x and say that a a is not 10 and not e e, but some other base. If we raise both sides of that equation as a power of a a (remember that l o g a(x)l o g a(x) and a x a x are inverses!), we get
a l o g a(c)=a x a l o g a(c)=a x
then using the inverse property of logs, this is really c=a x.c=a x. If we take the common log of both sides, like this:
l o g(c)=l o g(a x)l o g(c)=l o g(a x)
we convert the right side (laws of logs again) to
l o g(c)=x l o g(a)l o g(c)=x l o g(a)
Now isolate x x
x=l o g(c)l o g(a)x=l o g(c)l o g(a)
and recall that x=l o g a(c)x=l o g a(c) from the first line of this derivation, which gives us the change of base formula:
l o g a(c)=l o g(c)l o g(a)l o g a(c)=l o g(c)l o g(a)
Just take the common log of the argument and divide by the common log of the "old" base.
So if we have a log in an odd base, we can just use a calculator and take the log of the argument (c) and divide it by the log of the "old" base. It works just as well, of course, with natural logs.
... so back to our example. Now it's simple, just divide log(82) by log(3) OR divide ln(82) by ln(3). Try it. They both give 4.01.
Example 4
Express f(x)=l o g(4 x−3)−l o g(3 x 2)f(x)=l o g(4 x−3)−l o g(3 x 2) as a single log.
This kind of problem relies on the laws of logs, in this case the log of a quotient. When you see one log subtracted from another, you should think quotient, and vice versa.
This function translates to
f(x)=l o g(4 x−3)−l o g(3 x 2)=l o g(4 x−3 3 x 2)f(x)=l o g(4 x−3)−l o g(3 x 2)=l o g(4 x−3 3 x 2)
Example 5
Solve for x x: l n(x 2)+l n(x 5)+l n(x 3)=10 l n(x 2)+l n(x 5)+l n(x 3)=10
This is an interesting example that makes use of the exponentiation property of logs. We apply that rule to each of the log terms to get
2 l n(x)+5 l n(x)+3 l n(x)=0 2 l n(x)+5 l n(x)+3 l n(x)=0
Then we just add up all of the ln(x)'s and solve to get
10 ln(x)=10→l n(x)=1→x=e 10 ln(x)=10→l n(x)=1→x=e
Example 6
Solve for x x in: 9 x+10+3=84 9 x+10+3=84
This problem is an opportunity to show you a very common pitfall in solving this kind of log and exponential problems. What's often done is something like this:
Here's the trouble. We can't just take a log of each term:
If a=b,a=b, then l o g(a)=l o g(b).l o g(a)=l o g(b).
If a+b=c+d,a+b=c+d, then l o g(a+b)=l o g(c+d),l o g(a+b)=l o g(c+d),
Butl o g(a)+l o g(b)≠l o g(c)+l o g(d).l o g(a)+l o g(b)≠l o g(c)+l o g(d).
We always have to remember that functions do not distribute: f(a + b) ≠ f(a) + f(b).
The correct way to solve these problems is to shoot for having a single term on each side of the equal sign. In this case, just move the 3 to the right:
9 x+10=84−3=81 9 x+10=84−3=81
Then take a log on each side (remember that log( ) is s function) and use the exponential property of logs to get
l o g(9 x+10)(x+10)⋅l o g(9)=l o g(81)=l o g(81)l o g(9 x+10)=l o g(81)(x+10)⋅l o g(9)=l o g(81)
Finally, rearrange, using the subtraction/division property of logs to solve the problem:
x+1 x=l o g(81)l o g(9)=2−10=−8 x+1=l o g(81)l o g(9)x=2−10=−8
Example 7
Solve for x x in l o g(x)+l o g(8)=10 l o g(x)+l o g(8)=10
Here's another good opportunity to point out a common pitfall, much like the one in example 5 above. The first instinct of a lot of students is to just raise both sides of the equation as a power of 10 to "cancel" the logs, like this:
Here's the trouble. We can't just raise every individual term as a power of any base:
Here again, f(x)=10 x f(x)=10 x or g(x)=b x g(x)=b x are functions, and we have to remember that functions do not commute: f(a + b) \ne f(a) + f(b).
One way to solve this problem is to recognize that a sum of logs is just the log of a product, and to rearrange like this:
l o g(x)+l o g(8)l o g(8 x)=10=10 l o g(x)+l o g(8)=10 l o g(8 x)=10
Now we can raise each side (one term on each side) as a power of 10, the obvious choice for a common (base 10) log:
10 l o g(8 x)8 x=10 10=10 10 10 l o g(8 x)=10 10 8 x=10 10
Now the solution is easy, just divide both sides by 8, and the answer is ... some very large number:
x=10 10 8 x=10 10 8
Practice problems 2
Solve the following expressions for x x:
l o g 5(2 x+4)=2 l o g 5(2 x+4)=2
Solution 5 l o g 5(2 x+4)2 x+4 2 x x=5 2=25=21=21 2 5 l o g 5(2 x+4)=5 2 2 x+4=25 2 x=21 x=21 2
l o g(x)=1−l o g(x−3)l o g(x)=1−l o g(x−3)
Solution l o g(x)+l o g(x−3)l o g[x(x−3)]x(x−3)x 2−3 x−10(x−5)(x+2)=1=1=10 1=0=0 l o g(x)+l o g(x−3)=1 l o g[x(x−3)]=1 x(x−3)=10 1 x 2−3 x−10=0(x−5)(x+2)=0 x=−5,2 x=−5,2
But: we have to disregard the x=−2 x=−2 solution because it's out of the domain of the log function. There is only one solution at x=5 x=5.
2 l o g 9(x−−√)−l o g 9(6 x−1)=0 2 l o g 9(x)−l o g 9(6 x−1)=0
Solution 2 l o g 9(x 1/2)l o g 9(x)x 5 x x=l o g 9(6 x−1)=l o g 9(6 x−1)=6 x−1=1=1 5 2 l o g 9(x 1/2)=l o g 9(6 x−1)l o g 9(x)=l o g 9(6 x−1)x=6 x−1 5 x=1 x=1 5
l o g 2(x 2−6 x)=3+l o g 2(1−x)l o g 2(x 2−6 x)=3+l o g 2(1−x)
Solution l o g 2(x 2−6 x 1−x)x 2−6 x 1−x x 2−6 x x 2+2 x−8(x+4)(x−2)=3=8=8−8 x=0=0 l o g 2(x 2−6 x 1−x)=3 x 2−6 x 1−x=8 x 2−6 x=8−8 x x 2+2 x−8=0(x+4)(x−2)=0 x=−4,2 x=−4,2
Note that x=2 x=2 is an extraneous solution because negative numbers are outside of the domain of a log function. (1−2=−1 1−2=−1 in the log expression on the right).
l o g(x)+l o g(x−1)=l o g(3 x+12)l o g(x)+l o g(x−1)=l o g(3 x+12)
Solution l o g(x(x−1)3 x+12)x(x−1)3 x+12 x(x−1)x 2−x x 2−4 x−12(x−6)(x+2)=0=10 0=1=3 x+12=3 x+12=0=0 l o g(x(x−1)3 x+12)=0 x(x−1)3 x+12=10 0=1 x(x−1)=3 x+12 x 2−x=3 x+12 x 2−4 x−12=0(x−6)(x+2)=0 x=−2,6 x=−2,6
But: We have to disregard the negative solution because it's outside of the domain of log function. x=6 x=6.
l o g 4(x+3)=2 l o g 4(x+3)=2
Solution x+3 x+3 x=4 2=16=13 x+3=4 2 x+3=16 x=13
l n(10)−l n(7−x)=l n(x)l n(10)−l n(7−x)=l n(x)
Solution l n(10 7−x)10 7−x 10 x 2−7 x+10(x−5)(x−2)=l n(x)=x=7 x−x 2=0=0 l n(10 7−x)=l n(x)10 7−x=x 10=7 x−x 2 x 2−7 x+10=0(x−5)(x−2)=0 x=2,5 x=2,5
l o g(x−1)−l o g(x+1)=1 l o g(x−1)−l o g(x+1)=1
Solution l n(x−1 x+1)x−1 x+1 x−1 9 x+11 x=1=10=10 x+10=0=−11 9 l n(x−1 x+1)=1 x−1 x+1=10 x−1=10 x+10 9 x+11=0 x=−11 9
We have to disregard this solution because it gives us a log function with a negative argument, outside of the domain. This equation has no solutions
l n(x)+l n(2 x−1)=2 l n(x)+l n(2 x−1)=2
Solution l n[x(2 x−1)]2 x 2−x−e 2 x 2−x 2−e 2 2 x 2−x 2+(1 4)2(x−1 4)2 x=2=0=0=8 e 2 16+1 16=8 e 2+1 16=1±8 e 2+1−−−−−−√4 l n[x(2 x−1)]=2 2 x 2−x−e 2=0 x 2−x 2−e 2 2=0 x 2−x 2+(1 4)2=8 e 2 16+1 16(x−1 4)2=8 e 2+1 16 x=1±8 e 2+1 4
Notice that only the plus solution is valid.
5⋅l n(x)=12 5⋅l n(x)=12
Solution l n(x)x=12 5=e 12 5 l n(x)=12 5 x=e 12 5
l o g(4 x+2 x 2)=l o g(3 x 2)l o g(4 x+2 x 2)=l o g(3 x 2)
Solution 4 x+2 x 2 x 2−4 x x(x−4)=3 x 2=0=0 4 x+2 x 2=3 x 2 x 2−4 x=0 x(x−4)=0 x=0,4 x=0,4
But notice that we must discard the x = 0 soution because it's outside of the domain of the log function: x=4 x=4.
l o g 2(3 x)=4.5 l o g 2(3 x)=4.5
Solution 3 x x=2 4.5=2 4.5 3=7.54 3 x=2 4.5 x=2 4.5 3=7.54
Convert each of the following expressions to a single logarithm. (w,x,y,z>0)(w,x,y,z>0)
l n(3 x+1)+2 l n(x)l n(3 x+1)+2 l n(x)
Solution=l n(3 x+1)+l n(x 2)=l n[(3 x+1)⋅x 2]=l n(3 x+1)+l n(x 2)=l n[(3 x+1)⋅x 2]
l o g(x)+2 l o g(y)−1 2 l o g(z)l o g(x)+2 l o g(y)−1 2 l o g(z)
Solution=l o g(x)+l o g(y 2)−l o g(z 1/2)=l o g(x y 2)−l o g(z 1/2)=l o g(x y 2 z 1/2)=l o g(x)+l o g(y 2)−l o g(z 1/2)=l o g(x y 2)−l o g(z 1/2)=l o g(x y 2 z 1/2)
1 3 l o g(w)+3 l o g(x)−5 l o g(y)1 3 l o g(w)+3 l o g(x)−5 l o g(y)
Solution=l o g(w 1/3)+l o g(x 3)−l o g(y 5)=l o g(w 1/3 x 3)−l o g(y 5)=l o g(w 1/3 x 3 y 5)=l o g(w 1/3)+l o g(x 3)−l o g(y 5)=l o g(w 1/3 x 3)−l o g(y 5)=l o g(w 1/3 x 3 y 5)
l n(10)−l n(7−x)−l n(x)l n(10)−l n(7−x)−l n(x)
Solution=l n(10 7−x)−l n(x)=l n(10 7−x x)=l n(10 x(7−x))=l n(10 7−x)−l n(x)=l n(10 7−x x)=l n(10 x(7−x))
l o g(x−1)−l o g(x+1)l o g(x−1)−l o g(x+1)
Solution=l o g(x−1 x+1)=l o g(x−1 x+1)
l n(x)+l n(2 x−1)l n(x)+l n(2 x−1)
Solution=l n[x(2 x−1)]=l n[x(2 x−1)]
xaktly.com by Dr. Jeff Cruzan is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. © 2012-2025, Jeff Cruzan. All text and images on this website not specifically attributed to another source were created by me and I reserve all rights as to their use. Any opinions expressed on this website are entirely mine, and do not necessarily reflect the views of any of my employers. Please feel free to send any questions or comments tojeff.cruzan@verizon.net. |
189028 | https://www.ccjm.org/content/87/7/396 | Main menu
Other Publications
www.clevelandclinic.org
User menu
Register
Log in
Search
Advanced search
Other Publications
www.clevelandclinic.org
Register
Log in
Advanced Search
Home
Content
Current Issue
Ahead of Print
Past Issues
Supplements
Article Type
Specialty
Articles by Specialty
CME/MOC
Articles
Calendar
Info For
Manuscript Submission
Authors & Reviewers
Subscriptions
About CCJM
Contact Us
Media Kit
Conversations with Leaders
Conference Coverage
Kidney Week 2024
CHEST 2024
ACR Convergence 2023
Kidney Week 2023
ObesityWeek 2023
IDWeek 2023
CHEST 2023
MDS 2023
IAS 2023
ACP 2023
AAN 2023
ACC / WCC 2023
AAAAI Meeting 2023
ACR Convergence 2022
Kidney Week 2022
AIDS 2022
The Clinical Picture
Mönckeberg medial sclerosis
Anthony Stack, DO, Sandra Sheffield, DO, Karan Seegobin, MBBS and Satish Maharaj, MBBS
Cleveland Clinic Journal of Medicine July 2020, 87 (7) 396-397; DOI:
Anthony Stack
Department of Internal Medicine, University of Florida College of Medicine, Jacksonville, FL
Find this author on Google Scholar
Find this author on PubMed
Search for this author on this site
Sandra Sheffield
Department of Internal Medicine, University of Florida College of Medicine, Jacksonville, FL
Find this author on Google Scholar
Find this author on PubMed
Search for this author on this site
Karan Seegobin
Department of Medicine, Mayo Clinic, Jacksonville, FL
Find this author on Google Scholar
Find this author on PubMed
Search for this author on this site
Satish Maharaj
Department of Medicine, University of Louisville School of Medicine, Louisville, KY
Find this author on Google Scholar
Find this author on PubMed
Search for this author on this site
For correspondence: satish.maharaj{at}louisville.edu
Article
Figures & Data
Info & Metrics
PDF
A 59-year-old man presented to the emergency department with pain and ulceration of several fingers on the right hand. He reported no history of trauma. His medical history was significant for chronic kidney disease, coronary atherosclerosis, ischemic cardiomyopathy, and hypertension. He described antecedent pain and discoloration for 1 week, followed by blistering of the fingertips, which became ulcerated 2 days ago. He reported strict adherence to hemodialysis (performed via a tunneled catheter) and medications, including aspirin, clopidogrel, and atorvastatin.
On examination, the third and fourth fingers of his right hand were purplish with distal ulcerations, worse on the tip of the middle finger (Figure 1). Palpation revealed absent radial and ulnar pulses in the right wrist and diminished radial and ulnar pulses in the left. His hands were markedly colder than his upper arms, and the right hand was colder than the left. Suspecting acute limb ischemia, we started systemic antico-agulation with heparin infusion.
Download figure
Open in new tab
Download powerpoint
Figure 1
Physical examination revealed purplish discoloration and ulceration of the third and fourth fingertips.
Plain radiography of the hands (Figure 2A) revealed extensive vascular calcifications of the radial and ulnar arteries with no bony abnormalities. The calcifications were parallel and linear, typical of the “railroad-track” appearance that characterizes Mönckeberg medial sclerosis when the affected vessel is viewed longitudinally. Arterial Doppler ultrasonography found scattered areas of concentric calcified atherosclerotic disease. Computed tomography (CT) angiography confirmed diffuse circumferential calcification. Conventional angiography (Figure 2B) bilaterally revealed diminutive radial arteries with significantly delayed flow and occlusion of the ulnar artery bilaterally. The right distal radial artery had severe stenosis.
Download figure
Open in new tab
Download powerpoint
Figure 2
Plain radiography (A) and conventional angiog-raphy (B) demonstrated a “railroad-track” pattern (ulnar artery at arrow magnified in the inset), with severe arterial calcifications and a smooth endothelial interface, features typical for Mönckeberg medial sclerosis.
The vascular stenosis was not amenable to percutaneous intervention, and there were no adequate revascularization options. The right third and fourth digits and the left third digit were amputated. Medical management at discharge included a phosphate binder, vitamin D, a calcimimetic, and an aldosterone antagonist. Antiplatelet and statin therapy were continued. The frequency of hemodialysis was increased.
MÖNCKEBERG MEDIAL SCLEROSIS
The differential diagnosis for digital ischemia is broad, including arterial thromboembolism, vasoconstrictive drug use or disorders, vasculitis, infectious ulceration, Raynaud phenomenon, and arterial stenosis. In patients with arteriovenous fistulas, steal syndrome can precipitate ischemia. In this case, there was no evidence of steal syndrome on the more-affected side.
In general, vascular calcifications are characterized by mineral deposits in the walls of arteries, and occur as one of two types:
Intimal layer calcification occurs in atherosclerosis and is characterized by diffuse arterial involvement with late calcifications.
Medial layer calcification occurs in several diseases, of which Mönckeberg medial sclerosis is the most common.1 It typically involves discrete vascular territories with early calcification.
Mönckeberg medial sclerosis is believed to be driven by hyperphosphatemia2 and is frequently associated with diabetes and chronic kidney disease. Sclerosis tends to localize to the arteries of the extremities.
The diagnosis is supported by findings on plain radiography (Figure 2A) or B-mode ultrasonography (with distinct echogenic granules located in the abluminal layers of the arterial walls), and is confirmed with an ankle-brachial index greater than 1.1 Recent research into medial layer calcifications has shown it to be an active process initiated and regulated by a variety of molecular signaling pathways.1
Compared with treatments for intimal layer calcifications, those for medial layer calcifications in general, and specifically for Mönckeberg medial sclerosis, are less well studied and effective. In patients with a documented disorder of phosphate homeostasis (typically chronic kidney disease with a mineral and bone disorder, as is the case for this patient), prevention and treatment includes phosphate binders, low-dose vitamin D, calcimimetics, magnesium, bisphosphonates, sodium thiosulfate, and aldosterone antagonists.3–5 In patients with skin lesions suggesting calciphylaxis, the recommended combined medical and surgical treatment includes the following:
Lowering the calcium and phosphate concentrations
Increasing the frequency of hemodialysis
Giving intravenous thiosulfate
Hyperbaric oxygen
Wound care
Debridement of necrotic tissue.6
Copyright © 2020 The Cleveland Clinic Foundation. All Rights Reserved.
REFERENCES
↵
Lanzer P,
Boehm M,
Sorribas V, et al. Medial vascular calcification revisited: review and perspectives. Eur Heart J 2014; 35(23):1515–1525. doi:10.1093/eurheartj/ehu163
OpenUrlCrossRefPubMed
2. ↵
Shioi A,
Taniwaki H,
Jono S, et al. Mönckeberg’s medial sclerosis and inorganic phosphate in uremia. Am J Kidney Dis 2001; 38(4 Suppl 1):S47–S49. doi:10.1053/ajkd.2001.27396
OpenUrlCrossRefPubMed
3. ↵
O’Neill WC,
Lomashvili KA. Recent progress in the treatment of vascular calcification. Kidney Int 2010; 78(12):1232–1239. doi:10.1038/ki.2010.334
OpenUrlCrossRefPubMed
4. 1. Disthabanchong S. Lowering vascular calcification burden in chronic kidney disease: is it possible? World J Nephrol 2013; 2(3):49–55. doi:10.5527/wjn.v2.i3.49
OpenUrlCrossRefPubMed
5. ↵
Lang F,
Ritz E,
Voelkl J,
Alesutan I. Vascular calcification—is aldosterone a culprit? Nephrol Dial Transplant 2013; 28(5):1080–1084. doi:10.1093/ndt/gft041
OpenUrlCrossRefPubMed
6. ↵
Don BR,
Chin AI. A strategy for the treatment of calcific uremic arteriolopathy (calciphylaxis) employing a combination of therapies. Clin Nephrol 2003; 59(6):463–470. doi:10.5414/cnp59463
OpenUrlCrossRefPubMed
PreviousNext
Back to top
In this issue
Cleveland Clinic Journal of Medicine
Vol. 87, Issue 7
1 Jul 2020
Table of Contents
Table of Contents (PDF)
Index by author
Complete Issue (PDF)
Print
Download PDF
Article Alerts
Email Article
Citation Tools
Mönckeberg medial sclerosis
Anthony Stack, Sandra Sheffield, Karan Seegobin, Satish Maharaj
Cleveland Clinic Journal of Medicine Jul 2020, 87 (7) 396-397; DOI: 10.3949/ccjm.87a.19085
Citation Manager Formats
BibTeX
Bookends
EasyBib
EndNote (tagged)
EndNote 8 (xml)
Medlars
Mendeley
Papers
RefWorks Tagged
Ref Manager
RIS
Zotero
Share
Mönckeberg medial sclerosis
Anthony Stack, Sandra Sheffield, Karan Seegobin, Satish Maharaj
Cleveland Clinic Journal of Medicine Jul 2020, 87 (7) 396-397; DOI: 10.3949/ccjm.87a.19085
Tweet Widget
Jump to section
Article
MÖNCKEBERG MEDIAL SCLEROSIS
REFERENCES
Figures & Data
Info & Metrics
PDF
Related Articles
No related articles found.
PubMed
Google Scholar
Cited By...
No citing articles found.
Google Scholar
More in this TOC Section
Half-moccasin distribution of acute tinea pedis
Refractory granuloma annulare
Odontogenic cutaneous fistula
Show more The Clinical Picture
Similar Articles
Subjects
Allergy/Immunology
Imaging
Nephrology
Vascular Medicine |
189029 | https://www.khanacademy.org/math/cc-fifth-grade-math/divide-decimals/imp-dividing-decimals/v/dividing-a-decimal-by-a-whole-number-example | Dividing a decimal by a whole number example (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: Get ready courses
Math: High school & college
Math: Multiple grades
Math: Illustrative Math-aligned
Math: Eureka Math-aligned
Test prep
Economics
Science
Computing
Reading & language arts
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. We're a nonprofit that relies on support from people like you. If everyone reading this gives $10 monthly, Khan Academy can continue to thrive for years. Please help keep Khan Academy free, for anyone, anywhere forever.
Select gift frequency
One time
Recurring
Monthly
Yearly
Select amount
$10
$20
$30
$40
Other
Give now
By donating, you agree to our terms of service and privacy policy.
Skip to lesson content
5th grade math
Course: 5th grade math>Unit 9
Lesson 3: Divide decimals by whole numbers
Dividing a decimal by a whole number with fraction models
Dividing a decimal by a whole number on the number line
Divide decimals by whole numbers visually
Dividing a decimal by a whole number example
Divide decimals by whole numbers
Math>
5th grade math>
Divide decimals>
Divide decimals by whole numbers
© 2025 Khan Academy
Terms of usePrivacy PolicyCookie NoticeAccessibility Statement
Dividing a decimal by a whole number example
Google Classroom
Microsoft Teams
About About this video Transcript
To divide a decimal by a whole number, you can convert the decimal into a certain number of hundredths. After dividing the whole number into those hundredths, you can then convert the result back into a decimal form.
Skip to end of discussions
Questions Tips & Thanks
Want to join the conversation?
Log in
Sort by:
Top Voted
chunyatchan0 2 years ago Posted 2 years ago. Direct link to chunyatchan0's post “What is there’s a remaind...” more What is there’s a remainder? I’m still struggling with this.
Answer Button navigates to signup page •1 comment Comment on chunyatchan0's post “What is there’s a remaind...”
(18 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
RumiWaffles 2 years ago Posted 2 years ago. Direct link to RumiWaffles's post “I'll try to guide you to ...” more I'll try to guide you to the answer with questions.
Ask yourself this, what is a remainder?
Then think about multiplying in groups and totals
8 is 2 groups of 4 or 4 + 4
When Khan got his answer but never mentioned a remainder, what does that mean for the answer?
Hope this helps
3 comments Comment on RumiWaffles's post “I'll try to guide you to ...”
(5 votes)
Upvote Button navigates to signup page
Downvote Button navigates to signup page
Flag Button navigates to signup page
more
Show more...
David Cooks 6 years ago Posted 6 years ago. Direct link to David Cooks's post “can you explain it more p...” more can you explain it more please
Answer Button navigates to signup page •2 comments Comment on David Cooks's post “can you explain it more p...”
(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
Kirbiiiiiiii a year ago Posted a year ago. Direct link to Kirbiiiiiiii's post “Hey, first it's 1.86 divi...” more Hey, first it's 1.86 divided by 2, but then he converts the 1 to 100 hundredths, and then he converts the .86 into 86 hundredths which equals 186 hundredths. Then you have to do 186 HUNDREDTHS divided by 2, and get 93 hundredths. Which equals 0.93
Hope This Helped!
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
Show more...
Tｏｋｙｏ ンョス [🌙 ) 5 years ago Posted 5 years ago. Direct link to Tｏｋｙｏ ンョス [🌙 )'s post “italics It doesnt expla...” more italics It doesnt explain the math im doing its like this:
bold3.64÷7= hundredths÷7
bold3.64÷7= hundredths
bold 3.64÷7=
Answer Button navigates to signup page •Comment Button navigates to signup page
(5 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
MITSKIIII 5 years ago Posted 5 years ago. Direct link to MITSKIIII's post “First of all, to get bold...” more First of all, to get bold words, you wrap them around . . . The dotts I added to indicate the words/phrases. Italic or slanted phrases/words need to be wrapped around like: . . .
Coming to the point, if you write 3.64 in hundredths, it would be 364 hundredths, and you do 364 ÷ 7 = 52
Since you wrote 364 in hundredths, so do you for 52 (0.52). So 3.64 ÷ 7 is 0.52.
hope i helped :) <3
Comment Button navigates to signup page
(5 votes)
Upvote Button navigates to signup page
Downvote Button navigates to signup page
Flag Button navigates to signup page
more
Show more...
A͓̽l͓̽y͓̽s͓̽s͓̽a͓̽ 4 years ago Posted 4 years ago. Direct link to A͓̽l͓̽y͓̽s͓̽s͓̽a͓̽'s post “how would you divide 16.8...” more how would you divide 16.8 divided by 8?
Answer Button navigates to signup page •1 comment Comment on A͓̽l͓̽y͓̽s͓̽s͓̽a͓̽'s post “how would you divide 16.8...”
(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
Ian Pulizzotto 4 years ago Posted 4 years ago. Direct link to Ian Pulizzotto's post “Since 168 is 160+8, 168 d...” more Since 168 is 160+8, 168 divided by 8 is 20+1, or 21.
Because 16.8 is 1/10 of 168, 16.8 divided by 8 is 1/10 of 21, which is 2.1. The answer is 2.1.
Have a blessed, wonderful day!
2 comments Comment on Ian Pulizzotto's post “Since 168 is 160+8, 168 d...”
(7 votes)
Upvote Button navigates to signup page
Downvote Button navigates to signup page
Flag Button navigates to signup page
more
Show more...
Marian 7 months ago Posted 7 months ago. Direct link to Marian's post “2:25 is the best part eve...” more 2:25 is the best part ever (it is the end :p)
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
aj13118 7 months ago Posted 7 months ago. Direct link to aj13118's post “1.86 devided by 2 equals ...” more 1.86 devided by 2 equals 93 hundedth
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
.KAYLA 3 months ago Posted 3 months ago. Direct link to .KAYLA's post “everybody here are bots” more everybody here are bots
Answer Button navigates to signup page •Comment Button navigates to signup page
(3 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
esukajai000 10 months ago Posted 10 months ago. Direct link to esukajai000's post “I don't get it in 186 div...” more I don't get it in 186 divided by 2. 6 divided by 2 =3 then 8 divided by 2 =4, so how did the instructor get 9?
Answer Button navigates to signup page •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
Answer
Show preview Show formatting options
Post answer
✨Daughter of Yah ✨ 9 months ago Posted 9 months ago. Direct link to ✨Daughter of Yah ✨'s post “Because there are not jus...” more Because there are not just 8 tens there are 18 tens.
So 18 tens is 180.
180/2 is 90
6/2 is 3
3+90 is 93
add the decimal
.93
3 comments Comment on ✨Daughter of Yah ✨'s post “Because there are not jus...”
(2 votes)
Upvote Button navigates to signup page
Downvote Button navigates to signup page
Flag Button navigates to signup page
more
SAVANNAH HUGGINS 2 years ago Posted 2 years ago. Direct link to SAVANNAH HUGGINS's post “hi what is 3.74 divided b...” more hi what is 3.74 divided by 7.89
Answer Button navigates to signup page •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
Answer
Show preview Show formatting options
Post answer
Gabriel Mooney 6 years ago Posted 6 years ago. Direct link to Gabriel Mooney's post “yah lemme see here so nob...” more yah lemme see here so nobody comment on this unless i did some thing wrong but 186/100 divided by 2 so it equals 93
there i think i got it
Answer Button navigates to signup page •2 comments Comment on Gabriel Mooney's post “yah lemme see here so nob...”
(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
Peter Leath 5 years ago Posted 5 years ago. Direct link to Peter Leath's post “186/100 divided by two wo...” more 186/100 divided by two would be 93/100. You got the top part correct. Just make sure to keep all the components of the number together
Comment Button navigates to signup page
(3 votes)
Upvote Button navigates to signup page
Downvote Button navigates to signup page
Flag Button navigates to signup page
more
Video transcript
[Instructor] Let's see if we can compute what 1.86 divided by two is. And like always pause this video and have a go at it. And I'll give you a hint, see if you can think about 1.86 as a certain number of hundredths, and then divide that by two. All right now let's work through it together. So as promised, let's see if we can think of 1.86 as a certain number of hundredths. So if we go to the hundredths place, we only see six there, but we could view 8/10 as 80 hundredths. So you could view the .86 as 86 hundredths, and each one is 100 hundredths, so we really have 186 hundredths. So this number right over here I'm going to rewrite as 186 hundredths, hundredths. And that is what we need to divide by two. Now, if you're dividing 186 of something by two, well it's just gonna be 186 divided by two of that something. So we could rewrite this as 186 divided by two, and then that's the number of hundredths this is going to be, so I'll put that in parenthesis. So it's going to be that many hundredths, hundredths. All right, so what is 186 divided by two? Some of you might be able to do it in your head, some of you might wanna use some paper. The way I like to think about it is if I were to take the six ones and divide it by two, well that's going to be three ones. And then if I were to take the 18 tens and divided by two, that's going to be nine tens, so that gets us to 93. Now you could've also thought about it like this, you could've said, how many times does two go into 186? Let's see, it doesn't go into one, how many times does it go into 18? Nine times, nine times two is 18, we could subtract, and then you have zero, you bring down that six. Two goes into six three times, three times two is six. You subtract and you get no remainders. So either way you're going to get 93. And it's 93 what? Well it's 93 hundredths, is what we're talking about. 93 hundredths. And now we just wanna express 93 hundredths as a decimal. Well how do we do that? Well that's gonna be 0 point. The 90 hundredths is the same thing as 9/10, and then the three hundredths, is of course 300 hundredths. Three in the hundredths place. So it's 0.93, and so we can write it up here. This is going to be equal to 0.93, and we're done!
Creative Commons Attribution/Non-Commercial/Share-AlikeVideo on YouTube
Up next: exercise
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 |
189030 | https://www.cut-the-knot.org/arithmetic/rapid/piles.shtml | Splitting Piles
Site
What's new
Content page
Front page
Index page
About
Privacy policy
Help with math
Subjects
Arithmetic
Algebra
Geometry
Probability
Trigonometry
Visual illusions
Articles
Cut the knot!
What is what?
Inventor's paradox
Math as language
Problem solving
Collections
Outline mathematics
Book reviews
Interactive activities
Did you know?
Eye opener
Analogue gadgets
Proofs in mathematics
Things impossible
Index/Glossary
Simple math
Fast Arithmetic Tips
Stories for young
Word problems
Games and puzzles
Our logo
Make an identity
Elementary geometry
Splitting Piles
Select a spectator from the audience and present him (or her) with a pile of chips, counters, pebbles or any other kind of small objects. Explain to the person and the audience what you expect them to do. Leave the stage while they are following your bid. Ask them to call you when finished. As the result of their activity, they will produce a number. Upon your stepping back on the stage you declare that you know what the number has been generated, and eventually reveal the number.
This is what the fellow has to do in your absence. He (or she) will be splitting a pile at a time into two piles, multiply the numbers of chips in the two new piles and keep adding the results. The process stops when there is no pile with more than 1 chip.
For example, let start with 9 chips:
Piles Which is broken What's added Total
9 9 36 18
3,6 3 12 20
1,2,6 6 33 29
1,2,3,3 3 12 31
1,2,1,2,3 2 11 32
1,1,1,1,2,3 2 11 33
1,1,1,1,1,1,3 3 12 35
1,1,1,1,1,1,1,2 2 1136
1,1,1,1,1,1,1,1,1---
When you return you will somehow know that the result is 36! The secret is that the result is always the same: it does not depend on how the piles are split; but only on the initial size of the very first pile. Obviously, if you want to keep your audience in the dark, do not perform the trick with the same starting number of chips more than once.There is a very simple proof of this fact.
Here is the proof that uses mathematical induction.
Assume we start with N = 2 chips. The only way to split such a pile is to halve it into two piles of 1 chip each. The computed number is just 1. Of course it's independent of how you split the pile; for there is just one way to perform this feat. Note that starting with N = 1 leads to the number 0. We split nothing. One can argue it's the same as having a pile with 0 chips which contributes 0 to the total regardless of the number of chips in other piles.
Now, assume that the result has been established for all numbers less than N and let there be N > 2 chips in the original pile. Split it into two with n and m chips, respectively. We have n, m > 0 and n + m = N. By the inductive assumption, proceeding with the first pile we'll get the number n(n - 1)/2 regardless of how we actually proceed. For the second pile, we'll get m(m - 1)/2. The total is mn + n(n - 1)/2 + m(m - 1)/2 which, after a series of simplifications, yields (m + n)(m + n - 1)/2 = N(N - 1)/2 which
is dependent on neither original nor consecutive splits
is exactly the number we expected.
Related material
Read more...
Chessboard
Solitaire on a Circle
Peg Solitaire
Ford's touching circles
Euclid's Game
Sam Loyd's fifteen
Swapping Rows and Columns
Escape of the Clones
|Front page||Contents||Games||Math magic|
Copyright © 1996-2018 Alexander Bogomolny
73256658 |
189031 | https://www.finalroundai.com/articles/count-triangles-graph | Start Winning Today - 50% Off!
Step into success this September. Prepare smarter with 50% off Final Round AI—before Sep 30.
Use code SEPT50
Sign In
Sign Up
Interview Copilot
AI Application
AI Mock Interview
Pricing
ResourcesResume Creation Tools Recruiters HotlineResume CheckerCover Letter Generator Career Guidance Tools AI Career CoachLinkedIn Profile OptimizerLinkedIn Resume Builder Support GuidesBlogArticles Real-time AI guidance during my interview prep helped me secure a position in marketing in a highly competitive environment. Sofia Rossi Marketing Manager of P&G Resume Creation Tools Recruiters HotlineResume CheckerCover Letter Generator Career Guidance Tools AI Career CoachLinkedIn Profile OptimizerLinkedIn Resume Builder Support GuidesBlogArticles
Question Bank
Sign In
Sign Up
All ArticlesData Structures
Count Triangles in Graphs (With Visualization and Code Examples)
Written by
Raj Aryan
Raj Aryan
Raj Aryan is passionate about breaking down complex DSA concepts and making coding feel approachable for everyone. With over 500 problems solved on LeetCode, two hackathon wins under his belt, and a 3-star rating on CodeChef, he enjoys helping others level up their programming skills.
All articles by Raj Aryan
Edited by
Kaustubh Saini
Kaustubh Saini
Kaustubh Saini writes about software development in a way that’s easy to follow and genuinely helpful. He breaks down complex topics-from AI to the latest in tech-so they actually make sense. His goal is simple: help others learn, stay curious, and keep up with a fast-changing world.
All articles by Kaustubh Saini
Last updated on
July 27, 2025
EmailFacebookXWhatsappLinkedInTelegramReddit
Finding triangles in graphs is like searching for friendship circles in social networks - you're looking for groups of three people who all know each other. This problem appears frequently in coding interviews and has practical applications in social network analysis, recommendation systems, and detecting patterns in data structures.
Problem Statement
Given a graph (either directed or undirected), we need to count the total number of triangles present in it. A triangle in a graph consists of three vertices where each pair of vertices is connected by an edge.
For an undirected graph, a triangle exists when three vertices A, B, and C are connected such that:
A is connected to B
B is connected to C
C is connected to A
For a directed graph, we need to consider the direction of edges. A triangle can be formed in different ways based on edge directions, but the most common definition counts cycles of length 3.
Example:
Undirected Graph:
A
/ \
B---C
This forms one triangle: A-B-C
Brute Force Approach
Explanation
The simplest way to solve this problem is to check every possible combination of three vertices and see if they form a triangle. This means examining all possible triplets of vertices and verifying if edges exist between all pairs.
For each triplet (i, j, k):
Check if there's an edge between vertex i and vertex j
Check if there's an edge between vertex j and vertex k
Check if there's an edge between vertex k and vertex i
If all three edges exist, we found a triangle
This approach is straightforward but inefficient because we're checking every possible combination, even when most won't form triangles.
Time Complexity:O(V³) where V is the number of vertices
Space Complexity:O(1) additional space (not counting the graph storage)
Code (Python)
Copy
```javascript
def count_triangles_brute_force(graph, is_directed=False):
"""
Count triangles in graph using brute force approach
graph: adjacency matrix representation
is_directed: True for directed graph, False for undirected
"""
n = len(graph)
triangle_count = 0
Check all possible triplets of vertices
for i in range(n):
for j in range(n):
for k in range(n):
Skip if vertices are the same
if i == j or j == k or i == k:
continue
Check if triangle exists
if (graph[i][j] == 1 and
graph[j][k] == 1 and
graph[k][i] == 1):
triangle_count += 1
For undirected graph, each triangle is counted 6 times
(once for each permutation of vertices)
if not is_directed:
triangle_count //= 6
return triangle_count
Example usage
def main():
Undirected graph adjacency matrix
undirected_graph = [
[0, 1, 1, 0],
[1, 0, 1, 1],
[1, 1, 0, 1],
[0, 1, 1, 0]
]
result = count_triangles_brute_force(undirected_graph, False)
print(f"Number of triangles: {result}")
```
Code (C++)
Copy
```javascript
include
include
using namespace std;
class TriangleCounter {
public:
static int countTrianglesBruteForce(vector>& graph, bool isDirected = false) {
int n = graph.size();
int triangleCount = 0;
// Check all possible triplets of vertices
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
for (int k = 0; k < n; k++) {
// Skip if vertices are the same
if (i == j || j == k || i == k) {
continue;
}
// Check if triangle exists
if (graph[i][j] == 1 &&
graph[j][k] == 1 &&
graph[k][i] == 1) {
triangleCount++;
}
}
}
}
// For undirected graph, each triangle is counted 6 times
if (!isDirected) {
triangleCount /= 6;
}
return triangleCount;
}
};
int main() {
// Undirected graph adjacency matrix
vector> undirectedGraph = {
{0, 1, 1, 0},
{1, 0, 1, 1},
{1, 1, 0, 1},
{0, 1, 1, 0}
};
int result = TriangleCounter::countTrianglesBruteForce(undirectedGraph, false);
cout << "Number of triangles: " << result << endl;
return 0;
}
```
Code (Java)
Copy
```javascript
public class TriangleCounter {
public static int countTrianglesBruteForce(int[][] graph, boolean isDirected) {
int n = graph.length;
int triangleCount = 0;
// Check all possible triplets of vertices
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
for (int k = 0; k < n; k++) {
// Skip if vertices are the same
if (i == j || j == k || i == k) {
continue;
}
// Check if triangle exists
if (graph[i][j] == 1 &&
graph[j][k] == 1 &&
graph[k][i] == 1) {
triangleCount++;
}
}
}
}
// For undirected graph, each triangle is counted 6 times
if (!isDirected) {
triangleCount /= 6;
}
return triangleCount;
}
public static void main(String[] args) {
// Undirected graph adjacency matrix
int[][] undirectedGraph = {
{0, 1, 1, 0},
{1, 0, 1, 1},
{1, 1, 0, 1},
{0, 1, 1, 0}
};
int result = countTrianglesBruteForce(undirectedGraph, false);
System.out.println("Number of triangles: " + result);
}
}
```
Optimized Approach
Explanation
We can significantly improve performance by reducing redundant checks. Instead of examining all possible triplets, we can be smarter about our search.
The key insight is to iterate through each edge and then check if there's a common neighbor that forms a triangle. For each edge (i, j), we look at all neighbors of vertex i and see if any of them are also neighbors of vertex j.
Here's how it works:
For each vertex i, examine all its neighbors j
For each neighbor j of vertex i, look at all neighbors k of vertex j
Check if vertex k is also a neighbor of vertex i
If yes, we found a triangle i-j-k
This approach is more efficient because we're only checking vertices that are actually connected, rather than all possible combinations.
Time Complexity:O(V × E) where V is vertices and E is edges, or O(V³) in worst case for dense graphs
Space Complexity:O(1) additional space
For very dense graphs, we can use matrix multiplication to achieve O(V³) with better constants, but for most practical cases, the neighbor-checking approach works well.
Problem & Naive Approach
2. Optimized Algorithm Flow
3. Algorithm Execution Steps
Count Triangles in Graphs Visualization
Code (Python)
Copy
```javascript
def count_triangles_optimized(adj_list, is_directed=False):
"""
Count triangles using optimized approach with adjacency list
adj_list: dictionary where keys are vertices and values are lists of neighbors
"""
triangle_count = 0
Convert adjacency list to set for O(1) lookup
adj_sets = {}
for vertex in adj_list:
adj_sets[vertex] = set(adj_list[vertex])
vertices = list(adj_list.keys())
For each pair of vertices that are connected
for i in range(len(vertices)):
vertex_i = vertices[i]
for vertex_j in adj_list[vertex_i]:
Skip if we've already processed this pair (for undirected graphs)
if not is_directed and vertex_j <= vertex_i:
continue
Find common neighbors of vertex_i and vertex_j
common_neighbors = adj_sets[vertex_i].intersection(adj_sets[vertex_j])
Each common neighbor forms a triangle
for vertex_k in common_neighbors:
if not is_directed and vertex_k > vertex_j:
triangle_count += 1
elif is_directed:
triangle_count += 1
return triangle_count
def count_triangles_matrix_multiplication(graph):
"""
Alternative optimized approach using matrix multiplication
Works well for dense graphs
"""
n = len(graph)
triangle_count = 0
Calculate A^2 (graph squared)
graph_squared = [ n for _ in range(n)]
for i in range(n):
for j in range(n):
for k in range(n):
graph_squared[i][j] += graph[i][k] graph[k][j]
Count triangles: trace of A^3 / 6 for undirected graphs
for i in range(n):
for j in range(n):
if graph[i][j] == 1: # There's an edge
triangle_count += graph_squared[j][i]
return triangle_count // 6 # Each triangle counted 6 times
Example usage
def main():
Graph represented as adjacency list
adj_list = {
0: [1, 2],
1: [0, 2, 3],
2: [0, 1, 3],
3: [1, 2]
}
result = count_triangles_optimized(adj_list, False)
print(f"Number of triangles: {result}")
```
Code (C++)
Copy
```javascript
include
include
include
include
using namespace std;
class OptimizedTriangleCounter {
public:
static int countTrianglesOptimized(unordered_map>& adjList, bool isDirected = false) {
int triangleCount = 0;
// Convert adjacency list to sets for O(1) lookup
unordered_map> adjSets;
for (auto& pair : adjList) {
adjSets[pair.first] = unordered_set(pair.second.begin(), pair.second.end());
}
// For each vertex and its neighbors
for (auto& pair : adjList) {
int vertexI = pair.first;
vector& neighbors = pair.second;
for (int vertexJ : neighbors) {
// Skip if we've already processed this pair (for undirected graphs)
if (!isDirected && vertexJ <= vertexI) {
continue;
}
// Find common neighbors
for (int vertexK : adjList[vertexJ]) {
if (adjSets[vertexI].count(vertexK)) {
if (!isDirected && vertexK > vertexJ) {
triangleCount++;
} else if (isDirected) {
triangleCount++;
}
}
}
}
}
return triangleCount;
}
static int countTrianglesMatrixMultiplication(vector>& graph) {
int n = graph.size();
int triangleCount = 0;
// Calculate A^2 (graph squared)
vector> graphSquared(n, vector(n, 0));
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
for (int k = 0; k < n; k++) {
graphSquared[i][j] += graph[i][k] graph[k][j];
}
}
}
// Count triangles
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
if (graph[i][j] == 1) {
triangleCount += graphSquared[j][i];
}
}
}
return triangleCount / 6; // Each triangle counted 6 times
}
};
int main() {
// Graph represented as adjacency list
unordered_map> adjList = {
{0, {1, 2}},
{1, {0, 2, 3}},
{2, {0, 1, 3}},
{3, {1, 2}}
};
int result = OptimizedTriangleCounter::countTrianglesOptimized(adjList, false);
cout << "Number of triangles: " << result << endl;
return 0;
}
```
Code (Java)
Copy
```javascript
import java.util.;
public class OptimizedTriangleCounter {
public static int countTrianglesOptimized(Map> adjList, boolean isDirected) {
int triangleCount = 0;
// Convert adjacency list to sets for O(1) lookup
Map> adjSets = new HashMap<>();
for (Map.Entry> entry : adjList.entrySet()) {
adjSets.put(entry.getKey(), new HashSet<>(entry.getValue()));
}
// For each vertex and its neighbors
for (Map.Entry> entry : adjList.entrySet()) {
int vertexI = entry.getKey();
List neighbors = entry.getValue();
for (int vertexJ : neighbors) {
// Skip if we've already processed this pair (for undirected graphs)
if (!isDirected && vertexJ <= vertexI) {
continue;
}
// Find common neighbors
List neighborsOfJ = adjList.get(vertexJ);
if (neighborsOfJ != null) {
for (int vertexK : neighborsOfJ) {
if (adjSets.get(vertexI).contains(vertexK)) {
if (!isDirected && vertexK > vertexJ) {
triangleCount++;
} else if (isDirected) {
triangleCount++;
}
}
}
}
}
}
return triangleCount;
}
public static int countTrianglesMatrixMultiplication(int[][] graph) {
int n = graph.length;
int triangleCount = 0;
// Calculate A^2 (graph squared)
int[][] graphSquared = new int[n][n];
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
for (int k = 0; k < n; k++) {
graphSquared[i][j] += graph[i][k] graph[k][j];
}
}
}
// Count triangles
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
if (graph[i][j] == 1) {
triangleCount += graphSquared[j][i];
}
}
}
return triangleCount / 6; // Each triangle counted 6 times
}
public static void main(String[] args) {
// Graph represented as adjacency list
Map> adjList = new HashMap<>();
adjList.put(0, Arrays.asList(1, 2));
adjList.put(1, Arrays.asList(0, 2, 3));
adjList.put(2, Arrays.asList(0, 1, 3));
adjList.put(3, Arrays.asList(1, 2));
int result = countTrianglesOptimized(adjList, false);
System.out.println("Number of triangles: " + result);
}
}
```
Conclusion
We explored two main approaches to count triangles in graphs. The brute force method checks every possible triplet of vertices, which is simple to implement but has O(V³) time complexity. The optimized approach reduces unnecessary checks by focusing on actual edges and common neighbors, providing better performance for sparse graphs.
The choice between approaches depends on your graph characteristics. For dense graphs where most vertices are connected, the matrix multiplication method might perform well. For sparse graphs, the neighbor-intersection approach typically offers better performance.
Learning to optimize solutions like this teaches valuable problem-solving skills. You start with a working solution, analyze its bottlenecks, and then apply techniques like reducing redundant operations or using better data structures. These optimization principles apply to many other graph algorithms and programming challenges you'll encounter.
FAQs
TAGS
Linked List
Stop guessing. Start winning.
We’ll show you the most common interview traps — and how to avoid them.
Learn more
Latest Posts
Boundary traversal of a binary tree (Codes with Visualization)
Balance a Binary Search Tree (Visualization)
Minimum Swaps To Convert Binary Tree Into BST (Visualization)
Check if a Binary Tree is BST or not (Visualization)
Don’t bomb your next interview.
One wrong answer can cost you the job — we’ll make sure that doesn’t happen.
Fix your weak spots
FAQ Question
FAQ Answer
Revolutionizing Interview Preparation with AI
Try it now - It’s free!
Your trusted platform to ace any job interviews, craft the perfect resumes, and land your dream jobs.
Products
Interview Copilot
AI Mock Interview
AI Resume Builder
Hirevue
Phone Interview
Speech Analysis
College Admission
Auto Apply
QA Pairs
Interview Notes
Coding Copilot
Resources
Tutorials
Blog
Articles
Special Discount
Influencer Program
Smarter Choice
Support
FAQ
Contact Us
Company
How Final Round AI works
About
Careers
News
PR & Media
Referral Program
Interview Questions
Common Career Ambition Interview Questions
Collaborative Leadership Interview Questions
Product Manager Interview Questions
Coding Interview Questions
Technical Interview Questions
Generative AI Engineer Interview Questions
Swift Developer Interview Questions
AI Tools
AI Career Coach
Recruiters Hotline
Cover Letter Generator
LinkedIn Profile Optimizer
LinkedIn Resume Builder
Resume Checker
© 2025 Final Round AI,
643 Teresita Blvd, San Francisco, CA 94127
Privacy Policy
Terms & Conditions
Count Triangles in Graphs (With Visualization and Code Examples) |
189032 | https://www.uniprot.org/uniprotkb/P04637/entry | P04637 · P53_HUMAN
Cellular tumor antigen p53
UniProtKB reviewed (Swiss-Prot)
Homo sapiens (Human)
Amino acids
393 (go to sequence)
Protein existence
Evidence at protein level
Annotation score
Add a publication
Entry feedback
function
Multifunctional transcription factor that induces cell cycle arrest, DNA repair or apoptosis upon binding to its target DNA sequence (PubMed:11025664, PubMed:12524540, PubMed:12810724, PubMed:15186775, PubMed:15340061, PubMed:17317671, PubMed:17349958, PubMed:19556538, PubMed:20673990, PubMed:20959462, PubMed:22726440, PubMed:24051492, PubMed:24652652, PubMed:35618207, PubMed:36634798, PubMed:38653238, PubMed:9840937).
Acts as a tumor suppressor in many tumor types; induces growth arrest or apoptosis depending on the physiological circumstances and cell type (PubMed:11025664, PubMed:12524540, PubMed:12810724, PubMed:15186775, PubMed:15340061, PubMed:17189187, PubMed:17317671, PubMed:17349958, PubMed:19556538, PubMed:20673990, PubMed:20959462, PubMed:22726440, PubMed:24051492, PubMed:24652652, PubMed:38653238, PubMed:9840937).
Negatively regulates cell division by controlling expression of a set of genes required for this process (PubMed:11025664, PubMed:12524540, PubMed:12810724, PubMed:15186775, PubMed:15340061, PubMed:17317671, PubMed:17349958, PubMed:19556538, PubMed:20673990, PubMed:20959462, PubMed:22726440, PubMed:24051492, PubMed:24652652, PubMed:9840937).
One of the activated genes is an inhibitor of cyclin-dependent kinases. Apoptosis induction seems to be mediated either by stimulation of BAX and FAS antigen expression, or by repression of Bcl-2 expression (PubMed:12524540, PubMed:17189187).
Its pro-apoptotic activity is activated via its interaction with PPP1R13B/ASPP1 or TP53BP2/ASPP2 (PubMed:12524540).
However, this activity is inhibited when the interaction with PPP1R13B/ASPP1 or TP53BP2/ASPP2 is displaced by PPP1R13L/iASPP (PubMed:12524540).
In cooperation with mitochondrial PPIF is involved in activating oxidative stress-induced necrosis; the function is largely independent of transcription. Induces the transcription of long intergenic non-coding RNA p21 (lincRNA-p21) and lincRNA-Mkln1. LincRNA-p21 participates in TP53-dependent transcriptional repression leading to apoptosis and seems to have an effect on cell-cycle regulation. Implicated in Notch signaling cross-over. Prevents CDK7 kinase activity when associated to CAK complex in response to DNA damage, thus stopping cell cycle progression. Isoform 2 enhances the transactivation activity of isoform 1 from some but not all TP53-inducible promoters. Isoform 4 suppresses transactivation activity and impairs growth suppression mediated by isoform 1. Isoform 7 inhibits isoform 1-mediated apoptosis. Regulates the circadian clock by repressing CLOCK-BMAL1-mediated transcriptional activation of PER2 (PubMed:24051492).
Interaction with BANP was reported to enhance phosphorylation on Ser-15 upon ultraviolet irradiation (PubMed:15701641).
However, the publication has been retracted due to image duplication and manipulation. Interaction with BANP has been confirmed in mouse studies (By similarity).
Phosphorylation at Ser-15 has been confirmed by other studies (PubMed:10570149, PubMed:11554766, PubMed:15866171, PubMed:16219768, PubMed:17317671, PubMed:17954561, PubMed:20959462, PubMed:25772236).
Its nuclear and cytoplasmic localization has been confirmed by other studies (PubMed:15340061, PubMed:17170702, PubMed:17591690, PubMed:18206965, PubMed:19011621, PubMed:21597459, PubMed:22726440).
Cofactor
Zn2+ (UniProtKB | Rhea| CHEBI:29105 )
Note: Binds 1 zinc ion per subunit.
Features
Showing features for dna binding, site, binding site.
| | Type | ID | Position(s) | Description | |
--- --- --- |
| | DNA binding | | 102-292 | | |
| |
| | Site | | 120 | Interaction with DNA | |
| |
| | Binding site | | 176 | Zn2+ (UniProtKB | ChEBI) | |
| |
| | Binding site | | 179 | Zn2+ (UniProtKB | ChEBI) | |
| |
| | Binding site | | 238 | Zn2+ (UniProtKB | ChEBI) | |
| |
| | Binding site | | 242 | Zn2+ (UniProtKB | ChEBI) | |
| |
Gene Ontology
P046373 GO annotations based on evolutionary models
Keywords
Molecular function
#Activator
#DNA-binding
#Repressor
Biological process
#Apoptosis
#Biological rhythms
#Cell cycle
#Host-virus interaction
#Necrosis
#Metal-binding
#Zinc
Enzyme and pathway databases
R-HSA-111448Activation of NOXA and translocation to mitochondria
R-HSA-139915Activation of PUMA and translocation to mitochondria
R-HSA-1912408Pre-NOTCH Transcription and Translation
R-HSA-2559580Oxidative Stress Induced Senescence
R-HSA-2559584Formation of Senescence-Associated Heterochromatin Foci (SAHF)
SABIO-RK
P04637
SIGNOR
P04637
SignaLink
P04637
Protein family/group databases
MoonDB
P04637Predicted
TCDB
1.C.110.1.1the pore-forming pnc-27 peptide of 32 aas from the p53 tumor suppressor protein (pnc-27) family
Names & Taxonomy
Protein names
Recommended name
Cellular tumor antigen p53
Alternative names
Antigen NY-CO-13
Phosphoprotein p53
Tumor suppressor p53
Gene names
Name
TP53
+ Synonyms
P53
Organism names
Taxonomic identifier
9606 (NCBI)
Organism
Homo sapiens (Human)
Taxonomic lineage
Eukaryota > Metazoa > Chordata > Craniata > Vertebrata > Euteleostomi > Mammalia > Eutheria > Euarchontoglires > Primates > Haplorrhini > Catarrhini > Hominidae > Homo
Accessions
Primary accession
P04637
Secondary accessions
Q15086
Q15087
Q15088
Q16535
Q16807
Proteomes
Identifier
UP000005640
+ Component
Chromosome 17
Organism-specific databases
AGR
HGNC:11998
HGNC
HGNC:11998TP53
MIM
133239phenotype
151623phenotype
191170gene+phenotype
202300phenotype
211980phenotype
VEuPathDB
HostDB:ENSG00000141510
neXtProt
NX_P04637
Subcellular Location
UniProt Annotation
GO Annotation
Cytoplasm
Nucleus
Nucleus, PML body
Endoplasmic reticulum
Mitochondrion matrix
Cytoplasm, cytoskeleton, microtubule organizing center, centrosome
Note: Recruited into PML bodies together with CHEK2 (PubMed:12810724).
Translocates to mitochondria upon oxidative stress (PubMed:22726440).
Translocates to mitochondria in response to mitomycin C treatment (PubMed:27323408).
Competitive inhibition of TP53 interaction with HSPA9/MOT-2 by UBXN2A results in increased protein abundance and subsequent translocation of TP53 to the nucleus (PubMed:24625977).
Isoform 1
Nucleus
Cytoplasm
Note: Predominantly nuclear but localizes to the cytoplasm when expressed with isoform 4.
Isoform 2
Nucleus
Cytoplasm
Note: Localized mainly in the nucleus with minor staining in the cytoplasm.
Isoform 3
Nucleus
Cytoplasm
Note: Localized in the nucleus in most cells but found in the cytoplasm in some cells.
Isoform 4
Nucleus
Cytoplasm
Note: Predominantly nuclear but translocates to the cytoplasm following cell stress.
Isoform 7
Nucleus
Cytoplasm
Note: Localized mainly in the nucleus with minor staining in the cytoplasm.
Isoform 8
Nucleus
Cytoplasm
Note: Localized in both nucleus and cytoplasm in most cells. In some cells, forms foci in the nucleus that are different from nucleoli.
Isoform 9
Cytoplasm
centrosome
chromatin
cytoplasm
cytosol
endoplasmic reticulum
germ cell nucleus
mitochondrial matrix
mitochondrion
nuclear matrix
nucleolus
nucleoplasm
nucleus
PML body
protein-containing complex
replication fork
site of double-strand break
transcription regulator complex
transcription repressor complex
Complete GO annotation on QuickGO
Keywords
Cellular component
#Cytoplasm
#Cytoskeleton
#Endoplasmic reticulum
#Mitochondrion
#Nucleus
Miscellaneous
CD-CODE
1A18FFC4Paraspeckle
8C2F96EDCentrosome
91857CE7Nucleolus
A1A97FABSynthetic Condensate 000191
B5B9A610PML body
E11B6BABSynthetic Condensate 000195
Disease & Variants
Involvement in disease
TP53 is found in increased amounts in a wide variety of transformed cells. TP53 is frequently mutated or inactivated in about 60% of cancers. TP53 defects are found in Barrett metaplasia a condition in which the normally stratified squamous epithelium of the lower esophagus is replaced by a metaplastic columnar epithelium. The condition develops as a complication in approximately 10% of patients with chronic gastroesophageal reflux disease and predisposes to the development of esophageal adenocarcinoma
Esophageal cancer (ESCR)
Note
The disease is caused by variants affecting the gene represented in this entry
Description
A malignancy of the esophagus. The most common types are esophageal squamous cell carcinoma and adenocarcinoma. Cancer of the esophagus remains a devastating disease because it is usually not detected until it has progressed to an advanced incurable stage.
See also
MIM:133239
Li-Fraumeni syndrome (LFS)
Note
The disease is caused by variants affecting the gene represented in this entry
Description
An autosomal dominant familial cancer syndrome that in its classic form is defined by the existence of a proband affected by a sarcoma before 45 years with a first degree relative affected by any tumor before 45 years and another first degree relative with any tumor before 45 years or a sarcoma at any age. Other clinical definitions for LFS have been proposed and called Li-Fraumeni like syndrome (LFL). In these families affected relatives develop a diverse set of malignancies at unusually early ages. Four types of cancers account for 80% of tumors occurring in TP53 germline mutation carriers: breast cancers, soft tissue and bone sarcomas, brain tumors (astrocytomas) and adrenocortical carcinomas. Less frequent tumors include choroid plexus carcinoma or papilloma before the age of 15, rhabdomyosarcoma before the age of 5, leukemia, Wilms tumor, malignant phyllodes tumor, colorectal and gastric cancers.
See also
MIM:151623
Natural variants in LFS
| | Variant ID | Position(s) | Change | Description |
--- ---
| | VAR_044621 | 82 | P>L | in LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs534447939 |
| | VAR_044650 | 97 | V>I | in familial cancer not matching LFS; germline mutation and in a sporadic cancer; somatic mutation; dbSNP:rs730882023 |
| | VAR_044661 | 105 | G>C | in LFS; germline mutation and in sporadic cancers; somatic mutation |
| | VAR_044667 | 106 | S>R | in a familial cancer not matching LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs1555526581 |
| | VAR_005861 | 110 | R>L | in a familial cancer not matching LFS; germline mutation and in sporadic cancers; somatic mutation; does not induce SNAI1 degradation; dbSNP:rs11540654 |
| | VAR_044716 | 126 | Y>C | in a familial cancer not matching LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs1555526335 |
| | VAR_044740 | 132 | K>E | in LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs747342068 |
| | VAR_044747 | 133 | M>R | in LFS; germline mutation and in sporadic cancers; somatic mutation |
| | VAR_005875 | 133 | M>T | in LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs28934873 |
| | VAR_005881 | 138 | A>P | in LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs28934875 |
| | VAR_044764 | 138 | A>S | in LFS; germline mutation |
| | VAR_005886 | 141 | C>Y | in LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs587781288 |
| | VAR_044790 | 144 | Q>L | in LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs786203071 |
| | VAR_005895 | 151 | P>S | in LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs28934874 |
| | VAR_005896 | 151 | P>T | in LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs28934874 |
| | VAR_005897 | 152 | P>L | in LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs587782705 |
| | VAR_044836 | 155 | T>N | in LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs786202752 |
| | VAR_044841 | 156 | R>H | in LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs371524413 |
| | VAR_005906 | 158 | R>G | in LFS; germline mutation and in sporadic cancers; somatic mutation |
| | VAR_005907 | 158 | R>H | in LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs587782144 |
| | VAR_033035 | 163 | Y>C | in LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs148924904 |
| | VAR_044885 | 167 | Q>K | in LFS; germline mutation and in a sporadic cancer; somatic mutation |
| | VAR_044906 | 172 | V>F | in LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs1131691043 |
| | VAR_005926 | 173 | V>M | in LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs876660754 |
| | VAR_044911 | 174 | R>G | in LFS; germline mutation and in a sporadic cancer; somatic mutation; dbSNP:rs864622115 |
| | VAR_005929 | 175 | R>G | in LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs138729528 |
| | VAR_005932 | 175 | R>H | in LFS; germline mutation and in sporadic cancers; somatic mutation; does not induce SNAI1 degradation; reduces interaction with ZNF385A; dbSNP:rs28934578 |
| | VAR_005930 | 175 | R>L | in LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs28934578 |
| | VAR_044939 | 179 | H>Y | in LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs587780070 |
| | VAR_044943 | 180 | E>K | in LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs879253911 |
| | VAR_044946 | 181 | R>C | in LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs587782596 |
| | VAR_044948 | 181 | R>H | in LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs397514495 |
| | VAR_005937 | 181 | R>L | in a familial cancer not matching LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs397514495 |
| | VAR_044949 | 181 | R>P | in a familial cancer not matching LFS; germline mutation and in sporadic cancers; somatic mutation |
| | VAR_044978 | 189 | A>V | in a familial cancer not matching LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs121912665 |
| | VAR_005948 | 193 | H>R | in LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs786201838 |
| | VAR_045007 | 196 | R>P | in LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs483352697 |
| | VAR_045010 | 197 | V>E | in a familial cancer not matching LFS; germline mutation and in sporadic cancers; somatic mutation |
| | VAR_045013 | 197 | V>M | in LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs786204041 |
| | VAR_045073 | 210 | N>Y | in a familial cancer not matching LFS; germline mutation; dbSNP:rs1060501200 |
| | VAR_036506 | 213 | R>P | in LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs587778720 |
| | VAR_005955 | 213 | R>Q | in LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs587778720 |
| | VAR_045114 | 219 | P>S | in a familial cancer not matching LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs879253894 |
| | VAR_005957 | 220 | Y>C | in LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs121912666 |
| | VAR_045151 | 227 | S>T | in LFS; germline mutation and in a sporadic cancer; somatic mutation |
| | VAR_045175 | 233 | H>D | in LFS; germline mutation and in a sporadic cancer; somatic mutation |
| | VAR_005963 | 234 | Y>C | in LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs587780073 |
| | VAR_045186 | 235 | N>S | in LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs144340710 |
| | VAR_045189 | 236 | Y>C | in LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs730882026 |
| | VAR_005965 | 237 | M>I | in LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs587782664 |
| | VAR_045200 | 238 | C>G | in LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs1057519981 |
| | VAR_045202 | 238 | C>S | in LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs730882005 |
| | VAR_005967 | 238 | C>Y | in a familial cancer not matching LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs730882005 |
| | VAR_005969 | 241 | S>F | in LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs28934573 |
| | VAR_047183 | 241 | S>T | in LFS; germline mutation and in sporadic cancers; somatic mutation |
| | VAR_045224 | 242 | C>Y | in a familial cancer not matching LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs121912655 |
| | VAR_045232 | 244 | G>D | in LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs1057517983 |
| | VAR_045236 | 244 | G>V | in LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs985033810 |
| | VAR_005972 | 245 | G>C | in LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs28934575 |
| | VAR_005973 | 245 | G>D | in LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs121912656 |
| | VAR_005974 | 245 | G>S | in LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs28934575 |
| | VAR_005975 | 245 | G>V | in LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs121912656 |
| | VAR_005978 | 246 | M>V | in LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs483352695 |
| | VAR_005983 | 248 | R>Q | in LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs11540652 |
| | VAR_005984 | 248 | R>W | in LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs121912651 |
| | VAR_045258 | 251 | I>M | in LFS; germline mutation; dbSNP:rs878854074 |
| | VAR_005988 | 252 | L>P | in LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs121912653 |
| | VAR_045284 | 257 | L>Q | in LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs28934577 |
| | VAR_005991 | 258 | E>K | in LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs121912652 |
| | VAR_045321 | 265 | L>P | in LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs879253942 |
| | VAR_045330 | 267 | R>Q | in LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs587780075 |
| | VAR_045351 | 272 | V>A | in a familial cancer not matching LFS; germline mutation and in sporadic cancers; somatic mutation |
| | VAR_005992 | 272 | V>L | in LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs121912657 |
| | VAR_005993 | 273 | R>C | in LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs121913343 |
| | VAR_005994 | 273 | R>G | in LFS; germline mutation and in sporadic cancers; somatic mutation |
| | VAR_005995 | 273 | R>H | in LFS; germline mutation and in sporadic cancers; somatic mutation; abolishes sequence-specific DNA binding; does not induce SNAI1 degradation; dbSNP:rs28934576 |
| | VAR_036509 | 273 | R>L | in LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs28934576 |
| | VAR_045357 | 273 | R>S | in a familial cancer not matching LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs121913343 |
| | VAR_005998 | 275 | C>Y | in LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs863224451 |
| | VAR_006003 | 278 | P>L | in LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs876659802 |
| | VAR_006004 | 278 | P>S | in LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs17849781 |
| | VAR_006005 | 278 | P>T | in LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs17849781 |
| | VAR_006007 | 280 | R>K | in a familial cancer not matching LFS; germline mutation and in sporadic cancers; somatic mutation; no effect on interaction with CCAR2; dbSNP:rs121912660 |
| | VAR_047202 | 281 | D>N | in LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs764146326 |
| | VAR_006014 | 281 | D>V | in a familial cancer not matching LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs587781525 |
| | VAR_045384 | 282 | R>G | in LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs28934574 |
| | VAR_045387 | 282 | R>Q | in a familial cancer not matching LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs730882008 |
| | VAR_006016 | 282 | R>W | in LFS; germline mutation and in sporadic cancers; somatic mutation; does not induce SNAI1 degradation; dbSNP:rs28934574 |
| | VAR_006017 | 283 | R>C | in LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs149633775 |
| | VAR_006024 | 285 | E>Q | in LFS; germline mutation and in sporadic cancers; somatic mutation |
| | VAR_006026 | 286 | E>A | in LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs1057519985 |
| | VAR_045411 | 290 | R>H | in LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs55819519 |
| | VAR_045412 | 290 | R>L | in LFS; germline mutation and in sporadic cancers; somatic mutation |
| | VAR_015819 | 292 | K>I | in LFS; germline mutation and in a sporadic cancer; somatic mutation; dbSNP:rs121912663 |
| | VAR_045471 | 305 | K>M | in a familial cancer not matching LFS; germline mutation and in sporadic cancers; somatic mutation |
| | VAR_045475 | 306 | R>P | in LFS; germline mutation and in a sporadic cancer; somatic mutation |
| | VAR_006038 | 309 | P>S | in LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs1555525012 |
| | VAR_006039 | 325 | G>V | in LFS; germline mutation; dbSNP:rs121912659 |
| | VAR_006041 | 337 | R>C | in LFS; germline mutation and in sporadic cancers; somatic mutation; impaired ability to tetramerize and undergo liquid-liquid phase separation; dbSNP:rs587782529 |
| | VAR_035016 | 337 | R>H | in LFS; germline mutation and in sporadic cancers; somatic mutation; impaired ability to tetramerize and undergo liquid-liquid phase separation; dbSNP:rs121912664 |
| | VAR_045546 | 344 | L>P | in LFS; germline mutation and in a sporadic cancer; somatic mutation; impaired ability to tetramerize and undergo liquid-liquid phase separation; dbSNP:rs121912662 |
| | VAR_045568 | 365 | H>Y | in a familial cancer not matching LFS; germline mutation and in a sporadic cancer; somatic mutation; dbSNP:rs267605075 |
| | VAR_022317 | 366 | S>A | in a familial cancer not matching LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs17881470 |
Squamous cell carcinoma of the head and neck (HNSCC)
Note
The gene represented in this entry is involved in disease pathogenesis
Description
A non-melanoma skin cancer affecting the head and neck. The hallmark of cutaneous SCC is malignant transformation of normal epidermal keratinocytes.
See also
MIM:275355
Lung cancer (LNCR)
Note
The disease is caused by variants affecting the gene represented in this entry
Description
A common malignancy affecting tissues of the lung. The most common form of lung cancer is non-small cell lung cancer (NSCLC) that can be divided into 3 major histologic subtypes: squamous cell carcinoma, adenocarcinoma, and large cell lung cancer. NSCLC is often diagnosed at an advanced stage and has a poor prognosis.
See also
MIM:211980
Papilloma of choroid plexus (CPP)
Note
The disease is caused by variants affecting the gene represented in this entry
Description
A benign tumor of neuroectodermal origin that generally occurs in childhood, but has also been reported in adults. Although generally found within the ventricular system, choroid plexus papillomas can arise ectopically in the brain parenchyma or disseminate throughout the neuraxis. Patients present with signs and symptoms of increased intracranial pressure including headache, hydrocephalus, papilledema, nausea, vomiting, cranial nerve deficits, gait impairment, and seizures.
See also
MIM:260500
Adrenocortical carcinoma (ADCC)
Note
The disease is caused by variants affecting the gene represented in this entry
Description
A malignant neoplasm of the adrenal cortex and a rare childhood tumor. It occurs with increased frequency in patients with Beckwith-Wiedemann syndrome and Li-Fraumeni syndrome.
See also
MIM:202300
Basal cell carcinoma 7 (BCC7)
Note
Disease susceptibility is associated with variants affecting the gene represented in this entry
Description
A common malignant skin neoplasm that typically appears on hair-bearing skin, most commonly on sun-exposed areas. It is slow growing and rarely metastasizes, but has potentialities for local invasion and destruction. It usually develops as a flat, firm, pale area that is small, raised, pink or red, translucent, shiny, and waxy, and the area may bleed following minor injury. Tumor size can vary from a few millimeters to several centimeters in diameter.
See also
MIM:614740
Bone marrow failure syndrome 5 (BMFS5)
Note
The disease is caused by variants affecting the gene represented in this entry
Description
A form of bone marrow failure syndrome, a heterogeneous group of life-threatening disorders characterized by hematopoietic defects in association with a range of variable extra-hematopoietic manifestations. BMFS5 is an autosomal dominant form characterized by infantile onset of severe red cell anemia requiring transfusion. Additional features include hypogammaglobulinemia, poor growth with microcephaly, developmental delay, and seizures.
See also
MIM:618165
Features
Showing features for natural variant, mutagenesis.
| | Type | ID | Position(s) | Description | |
--- --- --- |
| | Natural variant | VAR_044543 | 5 | in a sporadic cancer; somatic mutation; abolishes strongly phosphorylation | |
| |
| | Natural variant | VAR_044544 | 6 | in a sporadic cancer; somatic mutation; reduces interaction with ZNF385A | |
| |
| | Natural variant | VAR_005851 | 7 | in a sporadic cancer; somatic mutation; dbSNP:rs587782646 | |
| |
| | Natural variant | VAR_044545 | 8 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_044546 | 10 | in a sporadic cancer; somatic mutation; dbSNP:rs535274413 | |
| |
| | Natural variant | VAR_044547 | 11 | in sporadic cancers; somatic mutation; dbSNP:rs201382018 | |
| |
| | Natural variant | VAR_044548 | 11 | in sporadic cancers; somatic mutation; dbSNP:rs201382018 | |
| |
| | Natural variant | VAR_044549 | 15 | in a sporadic cancer; somatic mutation | |
| |
| | Mutagenesis | | 15 | Loss of interaction with PPP2R5C, PPP2CAANDPPP2R1A. | |
| |
| | Natural variant | VAR_044550 | 16 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_044551 | 17 | in a sporadic cancer; somatic mutation | |
| |
| | Mutagenesis | | 18 | No effect on interaction with MDM2 and increase in protein levels after DNA damage. | |
| |
| | Mutagenesis | | 20 | Abolishes phosphorylation site. Abolishes increase in protein levels after DNA damage. | |
| |
| | Mutagenesis | | 20 | Constitutively increased TP53 protein levels. | |
| |
| | Mutagenesis | | 22-23 | Loss of interaction with MDM2, leading to constitutively increased TP53 protein levels. | |
| |
| | Natural variant | VAR_044552 | 24 | in a sporadic cancer; somatic mutation | |
| |
| | Mutagenesis | | 24 | Abolishes ubiquitination by MUL1. | |
| |
| | Natural variant | VAR_044553 | 28 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_047158 | 29-30 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_044554 | 31 | in sporadic cancers; somatic mutation; dbSNP:rs201753350 | |
| |
| | Natural variant | VAR_044555 | 33 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_044556 | 34 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_005852 | 35 | in sporadic cancers; somatic mutation; dbSNP:rs121912661 | |
| |
| | Natural variant | VAR_044557 | 36 | in a sporadic cancer; somatic mutation; dbSNP:rs587781866 | |
| |
| | Natural variant | VAR_044558 | 37 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_044559 | 37 | in a sporadic cancer; somatic mutation | |
| |
| | Mutagenesis | | 37 | Abolishes phosphorylation by MAPKAPK5. | |
| |
| | Natural variant | VAR_044560 | 39 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_044561 | 39 | in a sporadic cancer; somatic mutation; dbSNP:rs1353016807 | |
| |
| | Natural variant | VAR_044562 | 42 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_005853 | 43 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_044563 | 44 | in a sporadic cancer; somatic mutation; dbSNP:rs1060501190 | |
| |
| | Natural variant | VAR_044564 | 44 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_044565 | 44 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_044566 | 45 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_044567 | 46 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044568 | 46 | in sporadic cancers; somatic mutation; dbSNP:rs876659630 | |
| |
| | Mutagenesis | | 46 | Abolishes phosphorylation by DYRK2 and HIPK2 and acetylation of K-382 by CREBBP. | |
| |
| | Mutagenesis | | 46 | Alters interaction with WWOX. | |
| |
| | Natural variant | VAR_044569 | 47 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_014632 | 47 | in dbSNP:rs1800371 | |
| |
| | Natural variant | VAR_044570 | 48 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_044571 | 49 | in sporadic cancers; somatic mutation; dbSNP:rs587780728 | |
| |
| | Natural variant | VAR_044572 | 49 | in a sporadic cancer; somatic mutation; dbSNP:rs587780728 | |
| |
| | Natural variant | VAR_044573 | 49 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044574 | 52 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_005854 | 53 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044575 | 53 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_044576 | 54 | in a sporadic cancer; somatic mutation; dbSNP:rs1555526742 | |
| |
| | Natural variant | VAR_044577 | 54 | in a sporadic cancer; somatic mutation | |
| |
| | Mutagenesis | | 55 | Blocks phosphorylation by TAF1. | |
| |
| | Natural variant | VAR_044578 | 56 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044579 | 56 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_044580 | 58 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_044581 | 58 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_044582 | 59 | in sporadic cancers; somatic mutation; dbSNP:rs1237722021 | |
| |
| | Natural variant | VAR_044583 | 59 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045783 | 59 | in a sporadic cancer; somatic mutation; requires 2 nucleotide substitutions | |
| |
| | Natural variant | VAR_044584 | 60 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044585 | 60 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_005855 | 60 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_044586 | 61 | in sporadic cancers; somatic mutation; dbSNP:rs1460793472 | |
| |
| | Natural variant | VAR_044587 | 61 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044588 | 62 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_044589 | 63 | in a sporadic cancer; somatic mutation; dbSNP:rs876658902 | |
| |
| | Natural variant | VAR_044590 | 63 | in a sporadic cancer; somatic mutation; dbSNP:rs372201428 | |
| |
| | Natural variant | VAR_044591 | 65 | in a sporadic cancer; somatic mutation; dbSNP:rs1060501210 | |
| |
| | Natural variant | VAR_044592 | 66 | in a sporadic cancer; somatic mutation; dbSNP:rs1555526711 | |
| |
| | Natural variant | VAR_044593 | 66 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_044594 | 67 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044595 | 67 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_044596 | 67 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044597 | 68 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044598 | 68 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_044599 | 69 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_044600 | 69 | in sporadic cancers; somatic mutation; dbSNP:rs756233241 | |
| |
| | Natural variant | VAR_044601 | 69 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_044602 | 69 | in a sporadic cancer; somatic mutation; dbSNP:rs756233241 | |
| |
| | Natural variant | VAR_044603 | 70 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_044604 | 71 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045784 | 72 | in sporadic cancers; somatic mutation; requires 2 nucleotide substitutions; dbSNP:rs730882014 | |
| |
| | Natural variant | VAR_045785 | 72 | in sporadic cancers; somatic mutation; requires 2 nucleotide substitutions | |
| |
| | Natural variant | VAR_045786 | 72 | in sporadic cancers; somatic mutation; dbSNP:rs1042522 | |
| |
| | Natural variant | VAR_045787 | 72 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_005856 | 72 | in dbSNP:rs1042522 | |
| |
| | Natural variant | VAR_044605 | 73 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_044606 | 73 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044607 | 73 | in sporadic cancers; somatic mutation; dbSNP:rs587782423 | |
| |
| | Natural variant | VAR_044608 | 74 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_044609 | 75 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044610 | 75 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044611 | 75 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_044612 | 76 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_044613 | 76 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_044614 | 77 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044615 | 78 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044616 | 79 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_005857 | 79 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_044617 | 79 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044618 | 80 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_044619 | 80 | in a sporadic cancer; somatic mutation; dbSNP:rs1060501204 | |
| |
| | Natural variant | VAR_044620 | 81 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044621 | 82 | in LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs534447939 | |
| |
| | Natural variant | VAR_044622 | 82 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044623 | 83 | in a sporadic cancer; somatic mutation; dbSNP:rs201717599 | |
| |
| | Natural variant | VAR_044624 | 83 | in sporadic cancers; somatic mutation; dbSNP:rs201717599 | |
| |
| | Natural variant | VAR_044625 | 84 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044626 | 84 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044627 | 85 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044628 | 85 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044629 | 86 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_005858 | 87 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044630 | 88 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_044631 | 88 | in sporadic cancers; somatic mutation; dbSNP:rs1555526631 | |
| |
| | Natural variant | VAR_044632 | 89 | in sporadic cancers; somatic mutation; dbSNP:rs730881994 | |
| |
| | Natural variant | VAR_044633 | 89 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044634 | 90 | in sporadic cancers; somatic mutation; dbSNP:rs1555526625 | |
| |
| | Natural variant | VAR_044635 | 90 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_044636 | 91 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_044637 | 92 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_044638 | 92 | in a sporadic cancer; somatic mutation; dbSNP:rs1210700121 | |
| |
| | Natural variant | VAR_044639 | 92 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_044640 | 93 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_044641 | 93 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_044642 | 94 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_005859 | 94 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044643 | 95 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044644 | 95 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_044645 | 96 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_044646 | 96 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044647 | 96 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_044648 | 97 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_044649 | 97 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_044650 | 97 | in familial cancer not matching LFS; germline mutation and in a sporadic cancer; somatic mutation; dbSNP:rs730882023 | |
| |
| | Natural variant | VAR_044651 | 98 | in sporadic cancers; somatic mutation; dbSNP:rs1245723119 | |
| |
| | Natural variant | VAR_044652 | 98 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044653 | 99 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044654 | 99 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_044655 | 100 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_044656 | 101 | in a sporadic cancer; somatic mutation; dbSNP:rs878854069 | |
| |
| | Natural variant | VAR_044657 | 101 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044658 | 102 | in sporadic cancers; somatic mutation; dbSNP:rs786202717 | |
| |
| | Natural variant | VAR_044659 | 104 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044660 | 104 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_044661 | 105 | in LFS; germline mutation and in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044662 | 105 | in sporadic cancers; somatic mutation; dbSNP:rs587781504 | |
| |
| | Natural variant | VAR_044663 | 105 | in sporadic cancers; somatic mutation; dbSNP:rs1060501195 | |
| |
| | Natural variant | VAR_044664 | 105 | in a sporadic cancer; somatic mutation; dbSNP:rs1060501195 | |
| |
| | Natural variant | VAR_044665 | 105 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044666 | 106 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_044667 | 106 | in a familial cancer not matching LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs1555526581 | |
| |
| | Natural variant | VAR_044668 | 107 | in a sporadic cancer; somatic mutation; dbSNP:rs587782447 | |
| |
| | Natural variant | VAR_044669 | 107 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044670 | 107 | in a sporadic cancer; somatic mutation; dbSNP:rs368771578 | |
| |
| | Natural variant | VAR_044671 | 108 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_044672 | 108 | in sporadic cancers; somatic mutation; dbSNP:rs587782461 | |
| |
| | Natural variant | VAR_044673 | 109 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044674 | 109 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_044675 | 109 | in sporadic cancers; somatic mutation; dbSNP:rs1064796722 | |
| |
| | Natural variant | VAR_005860 | 110 | in sporadic cancers; somatic mutation; dbSNP:rs587781371 | |
| |
| | Natural variant | VAR_044676 | 110 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_044677 | 110 | in sporadic cancers; somatic mutation; dbSNP:rs11540654 | |
| |
| | Natural variant | VAR_005861 | 110 | in a familial cancer not matching LFS; germline mutation and in sporadic cancers; somatic mutation; does not induce SNAI1 degradation; dbSNP:rs11540654 | |
| |
| | Natural variant | VAR_005862 | 110 | in sporadic cancers; somatic mutation; dbSNP:rs11540654 | |
| |
| | Natural variant | VAR_044678 | 110 | in a sporadic cancer; somatic mutation; dbSNP:rs587781371 | |
| |
| | Natural variant | VAR_044679 | 111 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_044680 | 111 | in sporadic cancers; somatic mutation; dbSNP:rs1057519997 | |
| |
| | Natural variant | VAR_044681 | 111 | in sporadic cancers; somatic mutation; dbSNP:rs1057519997 | |
| |
| | Natural variant | VAR_044682 | 111 | in sporadic cancers; somatic mutation; dbSNP:rs1057519997 | |
| |
| | Natural variant | VAR_044683 | 112 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044684 | 112 | in sporadic cancers; somatic mutation; dbSNP:rs1423803759 | |
| |
| | Natural variant | VAR_005863 | 113 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045788 | 113 | in a sporadic cancer; somatic mutation; requires 2 nucleotide substitutions | |
| |
| | Natural variant | VAR_044685 | 113 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_044686 | 113 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044687 | 113 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_033033 | 113 | in sporadic cancers; somatic mutation; dbSNP:rs587781642 | |
| |
| | Natural variant | VAR_044688 | 115 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044689 | 116 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044690 | 116 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_044691 | 116 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_044692 | 117 | in sporadic cancers; somatic mutation; dbSNP:rs755238756 | |
| |
| | Natural variant | VAR_044693 | 117 | in sporadic cancers; somatic mutation; dbSNP:rs1555526518 | |
| |
| | Natural variant | VAR_044694 | 118 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_044695 | 118 | in sporadic cancers; somatic mutation; dbSNP:rs1064794141 | |
| |
| | Natural variant | VAR_044696 | 118 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_044697 | 119 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_044698 | 119 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_044699 | 120 | in sporadic cancers; somatic mutation; abolished acetylation; impaired ability to induce proapoptotic program; mimics lactylation, leading to decreased ability to undergo liquid-liquid phase separation and activate the transcription factor activity; dbSNP:rs121912658 | |
| |
| | Natural variant | VAR_044700 | 120 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044701 | 120 | in a sporadic cancer; somatic mutation; mimics lactylation, leading to decreased ability to undergo liquid-liquid phase separation and activate the transcription factor activity; dbSNP:rs121912658 | |
| |
| | Natural variant | VAR_044702 | 120 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044703 | 121 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044704 | 122 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_044705 | 123 | in a sporadic cancer; somatic mutation; dbSNP:rs1555526486 | |
| |
| | Natural variant | VAR_044706 | 123 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_044707 | 124 | in a sporadic cancer; somatic mutation; dbSNP:rs730881997 | |
| |
| | Natural variant | VAR_044708 | 124 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044709 | 124 | in sporadic cancers; somatic mutation; dbSNP:rs730881997 | |
| |
| | Natural variant | VAR_044710 | 124 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_044711 | 124 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_044712 | 125 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_044713 | 125 | in sporadic cancers; somatic mutation; dbSNP:rs786201057 | |
| |
| | Natural variant | VAR_005864 | 125 | in sporadic cancers; somatic mutation; dbSNP:rs786201057 | |
| |
| | Natural variant | VAR_044714 | 125 | in a sporadic cancer; somatic mutation; dbSNP:rs1057520003 | |
| |
| | Natural variant | VAR_044715 | 125 | in sporadic cancers; somatic mutation; dbSNP:rs786201057 | |
| |
| | Natural variant | VAR_044716 | 126 | in a familial cancer not matching LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs1555526335 | |
| |
| | Natural variant | VAR_005865 | 126 | in sporadic cancers; somatic mutation; dbSNP:rs886039483 | |
| |
| | Natural variant | VAR_044717 | 126 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045789 | 126 | in a sporadic cancer; somatic mutation; requires 2 nucleotide substitutions | |
| |
| | Natural variant | VAR_044718 | 126 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_005866 | 126 | in sporadic cancers; somatic mutation; dbSNP:rs886039483 | |
| |
| | Natural variant | VAR_044719 | 126 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044720 | 127 | in a sporadic cancer; somatic mutation; dbSNP:rs730881999 | |
| |
| | Natural variant | VAR_005867 | 127 | in sporadic cancers; somatic mutation; dbSNP:rs730881999 | |
| |
| | Natural variant | VAR_044721 | 127 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044722 | 127 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044723 | 127 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044724 | 128 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044725 | 128 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044726 | 128 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_005868 | 128 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_005869 | 129 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044727 | 129 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_044728 | 129 | in sporadic cancers; somatic mutation; dbSNP:rs1438095083 | |
| |
| | Natural variant | VAR_044729 | 129 | in sporadic cancers; somatic mutation; dbSNP:rs137852792 | |
| |
| | Natural variant | VAR_044730 | 130 | in sporadic cancers; somatic mutation; dbSNP:rs863224683 | |
| |
| | Natural variant | VAR_044731 | 130 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044732 | 130 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_044733 | 130 | in sporadic cancers; somatic mutation; dbSNP:rs1131691013 | |
| |
| | Natural variant | VAR_005870 | 130 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044734 | 130 | in sporadic cancers; somatic mutation; dbSNP:rs863224683 | |
| |
| | Natural variant | VAR_044735 | 131 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_044736 | 131 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044737 | 131 | in sporadic cancers; somatic mutation; dbSNP:rs1131691037 | |
| |
| | Natural variant | VAR_005872 | 131 | in sporadic cancers; somatic mutation; dbSNP:rs769270327 | |
| |
| | Natural variant | VAR_005871 | 131 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044738 | 131 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_044739 | 131 | in sporadic cancers; somatic mutation; dbSNP:rs587782160 | |
| |
| | Natural variant | VAR_044740 | 132 | in LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs747342068 | |
| |
| | Natural variant | VAR_045790 | 132 | in a sporadic cancer; somatic mutation; requires 2 nucleotide substitutions | |
| |
| | Natural variant | VAR_005873 | 132 | in sporadic cancers; somatic mutation; dbSNP:rs1057519996 | |
| |
| | Natural variant | VAR_044741 | 132 | in sporadic cancers; somatic mutation; dbSNP:rs866775781 | |
| |
| | Natural variant | VAR_005874 | 132 | in sporadic cancers; somatic mutation; dbSNP:rs747342068 | |
| |
| | Natural variant | VAR_044742 | 132 | in sporadic cancers; somatic mutation; dbSNP:rs1057519996 | |
| |
| | Natural variant | VAR_044743 | 132 | in sporadic cancers; somatic mutation; dbSNP:rs1057519996 | |
| |
| | Natural variant | VAR_045791 | 132 | in a sporadic cancer; somatic mutation; requires 2 nucleotide substitutions | |
| |
| | Natural variant | VAR_047159 | 132-133 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_044744 | 133 | in sporadic cancers; somatic mutation; dbSNP:rs1064795139 | |
| |
| | Natural variant | VAR_044745 | 133 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044746 | 133 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044747 | 133 | in LFS; germline mutation and in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_005875 | 133 | in LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs28934873 | |
| |
| | Natural variant | VAR_044748 | 133 | in sporadic cancers; somatic mutation; dbSNP:rs1057280220 | |
| |
| | Natural variant | VAR_044749 | 134 | in sporadic cancers; somatic mutation; dbSNP:rs780442292 | |
| |
| | Natural variant | VAR_044750 | 134 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_036504 | 134 | in sporadic cancers; somatic mutation; dbSNP:rs267605077 | |
| |
| | Natural variant | VAR_044751 | 134 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044752 | 134 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_005877 | 135 | in sporadic cancers; somatic mutation; dbSNP:rs587781991 | |
| |
| | Natural variant | VAR_044753 | 135 | in sporadic cancers; somatic mutation; dbSNP:rs1057519975 | |
| |
| | Natural variant | VAR_044754 | 135 | in sporadic cancers; somatic mutation; dbSNP:rs1057519975 | |
| |
| | Natural variant | VAR_005876 | 135 | in sporadic cancers; somatic mutation; dbSNP:rs1057519975 | |
| |
| | Natural variant | VAR_045792 | 135 | in a sporadic cancer; somatic mutation; requires 2 nucleotide substitutions | |
| |
| | Natural variant | VAR_044755 | 135 | in sporadic cancers; somatic mutation; dbSNP:rs1057519976 | |
| |
| | Natural variant | VAR_044756 | 135 | in sporadic cancers; somatic mutation; decreased E6-mediated binding to E6-AP; dbSNP:rs587781991 | |
| |
| | Natural variant | VAR_005878 | 136 | in sporadic cancers; somatic mutation; dbSNP:rs1555526268 | |
| |
| | Natural variant | VAR_044757 | 136 | in sporadic cancers; somatic mutation; dbSNP:rs758781593 | |
| |
| | Natural variant | VAR_005879 | 136 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_044758 | 136 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044759 | 136 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044760 | 137 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044761 | 137 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_005880 | 137 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044762 | 137 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044763 | 138 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_005881 | 138 | in LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs28934875 | |
| |
| | Natural variant | VAR_044764 | 138 | in LFS; germline mutation | |
| |
| | Natural variant | VAR_044765 | 138 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_033034 | 138 | in sporadic cancers; somatic mutation; dbSNP:rs750600586 | |
| |
| | Natural variant | VAR_044766 | 139 | in sporadic cancers; somatic mutation; mimics lactylation, leading to decreased ability to undergo liquid-liquid phase separation and activate the transcription factor activity; dbSNP:rs1212996409 | |
| |
| | Natural variant | VAR_005882 | 139 | in sporadic cancers; somatic mutation; mimics lactylation, leading to decreased ability to undergo liquid-liquid phase separation and activate the transcription factor activity | |
| |
| | Natural variant | VAR_044767 | 139 | in sporadic cancers; somatic mutation; mimics lactylation, leading to decreased ability to undergo liquid-liquid phase separation and activate the transcription factor activity | |
| |
| | Natural variant | VAR_044768 | 139 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044769 | 139 | in sporadic cancers; somatic mutation; mimics lactylation, leading to decreased ability to undergo liquid-liquid phase separation and activate the transcription factor activity | |
| |
| | Natural variant | VAR_044770 | 140 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044771 | 140 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044772 | 140 | in a sporadic cancer; somatic mutation; dbSNP:rs786202561 | |
| |
| | Natural variant | VAR_044773 | 140 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_044774 | 140 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045793 | 141 | in a sporadic cancer; somatic mutation; requires 2 nucleotide substitutions | |
| |
| | Natural variant | VAR_005885 | 141 | in sporadic cancers; somatic mutation; dbSNP:rs587781288 | |
| |
| | Natural variant | VAR_005884 | 141 | in sporadic cancers; somatic mutation; dbSNP:rs1057519978 | |
| |
| | Natural variant | VAR_044775 | 141 | in sporadic cancers; somatic mutation; dbSNP:rs1057519978 | |
| |
| | Natural variant | VAR_044776 | 141 | in sporadic cancers; somatic mutation; dbSNP:rs1057519978 | |
| |
| | Natural variant | VAR_044777 | 141 | in sporadic cancers; somatic mutation; dbSNP:rs1057519977 | |
| |
| | Natural variant | VAR_005886 | 141 | in LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs587781288 | |
| |
| | Natural variant | VAR_044778 | 142 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045794 | 142 | in sporadic cancers; somatic mutation; requires 2 nucleotide substitutions | |
| |
| | Natural variant | VAR_044779 | 142 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044780 | 142 | in sporadic cancers; somatic mutation; dbSNP:rs779196500 | |
| |
| | Natural variant | VAR_044781 | 142 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_044782 | 142 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044783 | 142 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_005887 | 143 | in sporadic cancers; somatic mutation; strong DNA binding ability at 32.5 degrees Celsius; strong reduction of transcriptional activity at 37.5 degrees Celsius; severely represses interaction with ZNF385A | |
| |
| | Natural variant | VAR_044784 | 143 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044785 | 143 | in sporadic cancers; somatic mutation; dbSNP:rs1555526241 | |
| |
| | Natural variant | VAR_044786 | 143 | in sporadic cancers; somatic mutation; dbSNP:rs587782620 | |
| |
| | Natural variant | VAR_044787 | 143 | in sporadic cancers; somatic mutation; dbSNP:rs587782620 | |
| |
| | Natural variant | VAR_044788 | 144 | in sporadic cancers; somatic mutation; dbSNP:rs786201419 | |
| |
| | Natural variant | VAR_044789 | 144 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044790 | 144 | in LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs786203071 | |
| |
| | Natural variant | VAR_005888 | 144 | in sporadic cancers; somatic mutation; dbSNP:rs786203071 | |
| |
| | Natural variant | VAR_044791 | 144 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044792 | 145 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_005889 | 145 | in sporadic cancers; somatic mutation; dbSNP:rs587782197 | |
| |
| | Natural variant | VAR_005890 | 145 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044793 | 145 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044794 | 145 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044795 | 146 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_044796 | 146 | in sporadic cancers; somatic mutation; dbSNP:rs786203064 | |
| |
| | Natural variant | VAR_044797 | 146 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044798 | 146 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044799 | 146 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044800 | 147 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_005891 | 147 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044801 | 147 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044802 | 147 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_005892 | 147 | in sporadic cancers; somatic mutation; dbSNP:rs1453167097 | |
| |
| | Natural variant | VAR_044803 | 147 | in sporadic cancers; somatic mutation; dbSNP:rs1555526226 | |
| |
| | Natural variant | VAR_044804 | 148 | in a sporadic cancer; somatic mutation; dbSNP:rs1046611742 | |
| |
| | Natural variant | VAR_044805 | 148 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044806 | 148 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_044807 | 148 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044808 | 148 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044809 | 148 | in sporadic cancers; somatic mutation; dbSNP:rs1131691007 | |
| |
| | Natural variant | VAR_044810 | 149 | in sporadic cancers; somatic mutation; dbSNP:rs1555526214 | |
| |
| | Natural variant | VAR_005893 | 149 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044811 | 149 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044812 | 150 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_044813 | 150 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044814 | 150 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044815 | 150 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_044816 | 150 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_044817 | 150 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_005894 | 151 | in sporadic cancers; somatic mutation; dbSNP:rs28934874 | |
| |
| | Natural variant | VAR_044818 | 151 | in sporadic cancers; somatic mutation; dbSNP:rs1057520000 | |
| |
| | Natural variant | VAR_044819 | 151 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044820 | 151 | in sporadic cancers; somatic mutation; dbSNP:rs1057520000 | |
| |
| | Natural variant | VAR_005895 | 151 | in LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs28934874 | |
| |
| | Natural variant | VAR_005896 | 151 | in LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs28934874 | |
| |
| | Natural variant | VAR_044821 | 152 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_005897 | 152 | in LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs587782705 | |
| |
| | Natural variant | VAR_044822 | 152 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044823 | 152 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_005898 | 152 | in sporadic cancers; somatic mutation; dbSNP:rs767328513 | |
| |
| | Natural variant | VAR_044824 | 152 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044825 | 153 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045795 | 153 | in a sporadic cancer; somatic mutation; requires 2 nucleotide substitutions | |
| |
| | Natural variant | VAR_044826 | 153 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_044827 | 153 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044828 | 153 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_044829 | 153 | in sporadic cancers; somatic mutation; dbSNP:rs1064795860 | |
| |
| | Natural variant | VAR_005899 | 153 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044830 | 154 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044831 | 154 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_044832 | 154 | in sporadic cancers; somatic mutation; dbSNP:rs762846821 | |
| |
| | Natural variant | VAR_045796 | 154 | in sporadic cancers; somatic mutation; requires 2 nucleotide substitutions | |
| |
| | Natural variant | VAR_044833 | 154 | in sporadic cancers; somatic mutation; dbSNP:rs137852789 | |
| |
| | Natural variant | VAR_005900 | 154 | in a brain tumor with no family history; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs762846821 | |
| |
| | Natural variant | VAR_005901 | 155 | in sporadic cancers; somatic mutation; dbSNP:rs772683278 | |
| |
| | Natural variant | VAR_044834 | 155 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044835 | 155 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_044836 | 155 | in LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs786202752 | |
| |
| | Natural variant | VAR_044837 | 155 | in sporadic cancers; somatic mutation; does not induce SNAI1 degradation | |
| |
| | Natural variant | VAR_044838 | 155 | in sporadic cancers; somatic mutation; dbSNP:rs786202752 | |
| |
| | Natural variant | VAR_044839 | 156 | in sporadic cancers; somatic mutation; dbSNP:rs563378859 | |
| |
| | Natural variant | VAR_044840 | 156 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044841 | 156 | in LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs371524413 | |
| |
| | Natural variant | VAR_044842 | 156 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_005902 | 156 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044843 | 156 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044844 | 157 | in sporadic cancers; somatic mutation; dbSNP:rs1131691023 | |
| |
| | Natural variant | VAR_005903 | 157 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_005904 | 157 | in sporadic cancers; somatic mutation; dbSNP:rs121912654 | |
| |
| | Natural variant | VAR_044845 | 157 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_012977 | 157 | in sporadic cancers; somatic mutation; dbSNP:rs121912654 | |
| |
| | Natural variant | VAR_044846 | 157 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_005905 | 158 | in sporadic cancers; somatic mutation; dbSNP:rs587780068 | |
| |
| | Natural variant | VAR_045797 | 158 | in a sporadic cancer; somatic mutation; requires 2 nucleotide substitutions | |
| |
| | Natural variant | VAR_005906 | 158 | in LFS; germline mutation and in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_005907 | 158 | in LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs587782144 | |
| |
| | Natural variant | VAR_044847 | 158 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044848 | 158 | in sporadic cancers; somatic mutation; dbSNP:rs587782144 | |
| |
| | Natural variant | VAR_044849 | 158 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_044850 | 158 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045798 | 158 | in a sporadic cancer; somatic mutation; requires 2 nucleotide substitutions | |
| |
| | Natural variant | VAR_044851 | 159 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045799 | 159 | in a sporadic cancer; somatic mutation; requires 2 nucleotide substitutions; dbSNP:rs730882022 | |
| |
| | Natural variant | VAR_044852 | 159 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044853 | 159 | in sporadic cancers; somatic mutation; dbSNP:rs730882000 | |
| |
| | Natural variant | VAR_044854 | 159 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044855 | 159 | in sporadic cancers; somatic mutation; dbSNP:rs730882000 | |
| |
| | Natural variant | VAR_044856 | 159 | in sporadic cancers; somatic mutation; dbSNP:rs1555526131 | |
| |
| | Natural variant | VAR_005908 | 160 | in sporadic cancers; somatic mutation; dbSNP:rs772354334 | |
| |
| | Natural variant | VAR_044857 | 160 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044858 | 160 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_044859 | 160 | in sporadic cancers; somatic mutation; dbSNP:rs377274728 | |
| |
| | Natural variant | VAR_047160 | 160-161 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_047161 | 160-161 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_047162 | 160-161 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_044860 | 161 | in sporadic cancers; somatic mutation; dbSNP:rs1064795691 | |
| |
| | Natural variant | VAR_045800 | 161 | in a sporadic cancer; somatic mutation; requires 2 nucleotide substitutions | |
| |
| | Natural variant | VAR_044861 | 161 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044862 | 161 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_005909 | 161 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044863 | 161 | in sporadic cancers; somatic mutation; dbSNP:rs193920817 | |
| |
| | Natural variant | VAR_044864 | 161 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044865 | 162 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044866 | 162 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044867 | 162 | in a breast cancer with no family history; germline mutation and in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_005910 | 162 | in sporadic cancers; somatic mutation; dbSNP:rs587780069 | |
| |
| | Natural variant | VAR_044868 | 162 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_005911 | 162 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_033035 | 163 | in LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs148924904 | |
| |
| | Natural variant | VAR_044869 | 163 | in sporadic cancers; somatic mutation; dbSNP:rs786203436 | |
| |
| | Natural variant | VAR_044870 | 163 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_005912 | 163 | in sporadic cancers; somatic mutation; dbSNP:rs786203436 | |
| |
| | Natural variant | VAR_044871 | 163 | in sporadic cancers; somatic mutation; dbSNP:rs786203436 | |
| |
| | Natural variant | VAR_044872 | 163 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044873 | 164 | in sporadic cancers; somatic mutation; dbSNP:rs879254249 | |
| |
| | Natural variant | VAR_044874 | 164 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_005913 | 164 | in sporadic cancers; somatic mutation; dbSNP:rs1131691034 | |
| |
| | Natural variant | VAR_005914 | 164 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044875 | 164 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044876 | 164 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044877 | 165 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044878 | 165 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_005915 | 165 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044879 | 165 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_005916 | 165 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044880 | 166 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044881 | 166 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_005917 | 166 | in sporadic cancers; somatic mutation; dbSNP:rs1555526101 | |
| |
| | Natural variant | VAR_044882 | 166 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044883 | 166 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044884 | 167 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044885 | 167 | in LFS; germline mutation and in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_044886 | 167 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044887 | 167 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_047163 | 167-168 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_047164 | 167-168 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_044888 | 168 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044889 | 168 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044890 | 168 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044891 | 168 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044892 | 168 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_005918 | 168 | in sporadic cancers; somatic mutation; dbSNP:rs867114783 | |
| |
| | Natural variant | VAR_045801 | 168 | in a sporadic cancer; somatic mutation; requires 2 nucleotide substitutions | |
| |
| | Natural variant | VAR_044893 | 168 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_047165 | 168-169 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_005919 | 169 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044894 | 169 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_005920 | 169 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044895 | 169 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_047166 | 169-170 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_044896 | 170 | in sporadic cancers; somatic mutation; dbSNP:rs587780729 | |
| |
| | Natural variant | VAR_044897 | 170 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_005921 | 170 | in sporadic cancers; somatic mutation; dbSNP:rs779000871 | |
| |
| | Natural variant | VAR_044898 | 170 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_005922 | 170 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044899 | 171 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_044900 | 171 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044901 | 171 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044902 | 171 | in sporadic cancers; somatic mutation; dbSNP:rs587781845 | |
| |
| | Natural variant | VAR_044903 | 171 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_044904 | 171 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_005923 | 172 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044905 | 172 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044906 | 172 | in LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs1131691043 | |
| |
| | Natural variant | VAR_044907 | 172 | in sporadic cancers; somatic mutation; dbSNP:rs1131691021 | |
| |
| | Natural variant | VAR_044908 | 172 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044909 | 173 | in sporadic cancers; somatic mutation; dbSNP:rs1057519747 | |
| |
| | Natural variant | VAR_005924 | 173 | in sporadic cancers; somatic mutation; dbSNP:rs1057519747 | |
| |
| | Natural variant | VAR_044910 | 173 | in sporadic cancers; somatic mutation; dbSNP:rs1057519747 | |
| |
| | Natural variant | VAR_005925 | 173 | in sporadic cancers; somatic mutation; dbSNP:rs876660754 | |
| |
| | Natural variant | VAR_005926 | 173 | in LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs876660754 | |
| |
| | Natural variant | VAR_045802 | 173 | in a sporadic cancer; somatic mutation; requires 2 nucleotide substitutions | |
| |
| | Natural variant | VAR_044911 | 174 | in LFS; germline mutation and in a sporadic cancer; somatic mutation; dbSNP:rs864622115 | |
| |
| | Natural variant | VAR_005927 | 174 | in sporadic cancers; somatic mutation; dbSNP:rs1064796681 | |
| |
| | Natural variant | VAR_044912 | 174 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044913 | 174 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044914 | 174 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_044915 | 174 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_005928 | 175 | in sporadic cancers; somatic mutation; dbSNP:rs138729528 | |
| |
| | Natural variant | VAR_005929 | 175 | in LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs138729528 | |
| |
| | Natural variant | VAR_005932 | 175 | in LFS; germline mutation and in sporadic cancers; somatic mutation; does not induce SNAI1 degradation; reduces interaction with ZNF385A; dbSNP:rs28934578 | |
| |
| | Natural variant | VAR_005930 | 175 | in LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs28934578 | |
| |
| | Natural variant | VAR_005931 | 175 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044916 | 175 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_044917 | 175 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_005933 | 176 | in sporadic cancers; somatic mutation; dbSNP:rs786202962 | |
| |
| | Natural variant | VAR_044918 | 176 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044919 | 176 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044920 | 176 | in sporadic cancers; somatic mutation; dbSNP:rs967461896 | |
| |
| | Natural variant | VAR_005934 | 176 | in sporadic cancers; somatic mutation; dbSNP:rs1057519980 | |
| |
| | Natural variant | VAR_044921 | 176 | in sporadic cancers; somatic mutation; dbSNP:rs786202962 | |
| |
| | Natural variant | VAR_047167 | 176-177 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_044922 | 177 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045803 | 177 | in sporadic cancers; somatic mutation; requires 2 nucleotide substitutions | |
| |
| | Natural variant | VAR_044923 | 177 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045804 | 177 | in a sporadic cancer; somatic mutation; requires 2 nucleotide substitutions | |
| |
| | Natural variant | VAR_005935 | 177 | in sporadic cancers; somatic mutation; dbSNP:rs751477326 | |
| |
| | Natural variant | VAR_036505 | 177 | in sporadic cancers; somatic mutation; dbSNP:rs751477326 | |
| |
| | Natural variant | VAR_044924 | 177 | in sporadic cancers; somatic mutation; dbSNP:rs147002414 | |
| |
| | Natural variant | VAR_044925 | 177 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_044926 | 178 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_005936 | 178 | in a Burkitt lymphoma | |
| |
| | Natural variant | VAR_044927 | 178 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_044928 | 178 | in sporadic cancers; somatic mutation; dbSNP:rs1064795203 | |
| |
| | Natural variant | VAR_044929 | 178 | in sporadic cancers; somatic mutation; dbSNP:rs1555526004 | |
| |
| | Natural variant | VAR_044930 | 178 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044931 | 178 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044932 | 178 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_047168 | 178-179 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_044933 | 179 | in sporadic cancers; somatic mutation; dbSNP:rs587780070 | |
| |
| | Natural variant | VAR_044934 | 179 | in sporadic cancers; somatic mutation; dbSNP:rs1057519991 | |
| |
| | Natural variant | VAR_044935 | 179 | in sporadic cancers; somatic mutation; dbSNP:rs587780070 | |
| |
| | Natural variant | VAR_044936 | 179 | in sporadic cancers; somatic mutation; dbSNP:rs1057519991 | |
| |
| | Natural variant | VAR_044937 | 179 | in sporadic cancers; somatic mutation; dbSNP:rs876660821 | |
| |
| | Natural variant | VAR_044938 | 179 | in sporadic cancers; somatic mutation; dbSNP:rs1057519991 | |
| |
| | Natural variant | VAR_044939 | 179 | in LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs587780070 | |
| |
| | Natural variant | VAR_044940 | 180 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_044941 | 180 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044942 | 180 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_044943 | 180 | in LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs879253911 | |
| |
| | Natural variant | VAR_044944 | 180 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044945 | 180 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_044946 | 181 | in LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs587782596 | |
| |
| | Natural variant | VAR_044947 | 181 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044948 | 181 | in LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs397514495 | |
| |
| | Natural variant | VAR_005937 | 181 | in a familial cancer not matching LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs397514495 | |
| |
| | Natural variant | VAR_044949 | 181 | in a familial cancer not matching LFS; germline mutation and in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044950 | 181 | in sporadic cancers; somatic mutation; dbSNP:rs587782596 | |
| |
| | Natural variant | VAR_044951 | 182 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_005938 | 182 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044952 | 182 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044953 | 183 | in sporadic cancers; somatic mutation; dbSNP:rs1555525970 | |
| |
| | Natural variant | VAR_044954 | 183 | in sporadic cancers; somatic mutation | |
| |
| | Mutagenesis | | 183 | Abolishes strongly phosphorylation. | |
| |
| | Mutagenesis | | 183 | Inhibits slightly its transcriptional activity. | |
| |
| | Natural variant | VAR_044955 | 184 | in sporadic cancers; somatic mutation; dbSNP:rs1060501209 | |
| |
| | Natural variant | VAR_044956 | 184 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_047169 | 184 | in sporadic cancers; somatic mutation; dbSNP:rs72661117 | |
| |
| | Natural variant | VAR_044957 | 184 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_005939 | 184 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044958 | 185 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_044959 | 185 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044960 | 185 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044961 | 185 | in a sporadic cancer; somatic mutation; dbSNP:rs150607408 | |
| |
| | Natural variant | VAR_044962 | 185 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044963 | 185 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_044964 | 186 | in a sporadic cancer; somatic mutation; dbSNP:rs375275361 | |
| |
| | Natural variant | VAR_044965 | 186 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044966 | 186 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044967 | 186 | in sporadic cancers; somatic mutation; dbSNP:rs1060501206 | |
| |
| | Natural variant | VAR_044968 | 186 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_005940 | 186 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_005941 | 187 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044969 | 187 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045805 | 187 | in a sporadic cancer; somatic mutation; requires 2 nucleotide substitutions | |
| |
| | Natural variant | VAR_044970 | 187 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_005942 | 187 | in sporadic cancers; somatic mutation; dbSNP:rs776167460 | |
| |
| | Natural variant | VAR_044971 | 187 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044972 | 188 | in a sporadic cancer; somatic mutation; dbSNP:rs1199893366 | |
| |
| | Natural variant | VAR_044973 | 188 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044974 | 189 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_044975 | 189 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_005943 | 189 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044976 | 189 | in a sporadic cancer; somatic mutation; dbSNP:rs1555525921 | |
| |
| | Natural variant | VAR_044977 | 189 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044978 | 189 | in a familial cancer not matching LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs121912665 | |
| |
| | Natural variant | VAR_044979 | 190 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044980 | 190 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_005944 | 190 | in sporadic cancers; somatic mutation; dbSNP:rs876660825 | |
| |
| | Natural variant | VAR_044981 | 190 | in sporadic cancers; somatic mutation; dbSNP:rs876660825 | |
| |
| | Natural variant | VAR_044982 | 190 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044983 | 190 | in sporadic cancers; somatic mutation; dbSNP:rs876660254 | |
| |
| | Natural variant | VAR_044984 | 191 | in sporadic cancers; somatic mutation; dbSNP:rs587778718 | |
| |
| | Natural variant | VAR_044985 | 191 | in sporadic cancers; somatic mutation; dbSNP:rs587778718 | |
| |
| | Natural variant | VAR_044986 | 191 | in sporadic cancers; somatic mutation; dbSNP:rs587778718 | |
| |
| | Natural variant | VAR_044987 | 191 | in sporadic cancers; somatic mutation; dbSNP:rs868590738 | |
| |
| | Natural variant | VAR_005945 | 191 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044988 | 192 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044989 | 192 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_044990 | 192 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044991 | 192 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_005946 | 192 | in sporadic cancers; somatic mutation; dbSNP:rs730882002 | |
| |
| | Natural variant | VAR_047170 | 192-193 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_047171 | 192-193 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_005947 | 193 | in sporadic cancers; somatic mutation; dbSNP:rs876658468 | |
| |
| | Natural variant | VAR_044992 | 193 | in sporadic cancers; somatic mutation; dbSNP:rs786201838 | |
| |
| | Natural variant | VAR_044993 | 193 | in sporadic cancers; somatic mutation; dbSNP:rs876658468 | |
| |
| | Natural variant | VAR_044994 | 193 | in sporadic cancers; somatic mutation; dbSNP:rs786201838 | |
| |
| | Natural variant | VAR_044995 | 193 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_005948 | 193 | in LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs786201838 | |
| |
| | Natural variant | VAR_044996 | 193 | in sporadic cancers; somatic mutation; dbSNP:rs876658468 | |
| |
| | Natural variant | VAR_044997 | 194 | in sporadic cancers; somatic mutation; dbSNP:rs587780071 | |
| |
| | Natural variant | VAR_044998 | 194 | in sporadic cancers; somatic mutation; dbSNP:rs1057519998 | |
| |
| | Natural variant | VAR_044999 | 194 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_005949 | 194 | in sporadic cancers; somatic mutation; dbSNP:rs1057519998 | |
| |
| | Natural variant | VAR_005950 | 194 | in sporadic cancers; somatic mutation; dbSNP:rs1057519998 | |
| |
| | Natural variant | VAR_045000 | 194 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045001 | 195 | in sporadic cancers; somatic mutation; dbSNP:rs942158624 | |
| |
| | Natural variant | VAR_047172 | 195 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045002 | 195 | in sporadic cancers; somatic mutation; dbSNP:rs760043106 | |
| |
| | Natural variant | VAR_045003 | 195 | in sporadic cancers; somatic mutation; dbSNP:rs760043106 | |
| |
| | Natural variant | VAR_005951 | 195 | in sporadic cancers; somatic mutation; dbSNP:rs760043106 | |
| |
| | Natural variant | VAR_045004 | 195 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045806 | 195 | in a sporadic cancer; somatic mutation; requires 2 nucleotide substitutions | |
| |
| | Natural variant | VAR_045005 | 196 | in sporadic cancers; somatic mutation; dbSNP:rs397516435 | |
| |
| | Natural variant | VAR_045006 | 196 | in sporadic cancers; somatic mutation; dbSNP:rs483352697 | |
| |
| | Natural variant | VAR_045007 | 196 | in LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs483352697 | |
| |
| | Natural variant | VAR_045008 | 196 | in sporadic cancers; somatic mutation; dbSNP:rs483352697 | |
| |
| | Natural variant | VAR_045009 | 196 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045010 | 197 | in a familial cancer not matching LFS; germline mutation and in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045011 | 197 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045012 | 197 | in sporadic cancers; somatic mutation; dbSNP:rs786204041 | |
| |
| | Natural variant | VAR_045013 | 197 | in LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs786204041 | |
| |
| | Natural variant | VAR_045014 | 198 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045015 | 198 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_005952 | 198 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045016 | 198 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045017 | 198 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045018 | 199 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045019 | 199 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045020 | 199 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045021 | 199 | in sporadic cancers; somatic mutation; dbSNP:rs1555525857 | |
| |
| | Natural variant | VAR_045022 | 200 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045023 | 200 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045024 | 200 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045807 | 200 | in a sporadic cancer; somatic mutation; requires 2 nucleotide substitutions | |
| |
| | Natural variant | VAR_045025 | 200 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045026 | 200 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045027 | 201 | in sporadic cancers; somatic mutation; dbSNP:rs730882024 | |
| |
| | Natural variant | VAR_045028 | 201 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045029 | 201 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_047173 | 201-202 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045030 | 202 | in sporadic cancers; somatic mutation; dbSNP:rs587780072 | |
| |
| | Natural variant | VAR_045031 | 202 | in sporadic cancers; somatic mutation; dbSNP:rs587780072 | |
| |
| | Natural variant | VAR_045032 | 202 | in sporadic cancers; somatic mutation; dbSNP:rs587778719 | |
| |
| | Natural variant | VAR_045033 | 202 | in sporadic cancers; somatic mutation; dbSNP:rs587778719 | |
| |
| | Natural variant | VAR_045034 | 202 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045035 | 202 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045036 | 203 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045037 | 203 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045038 | 203 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045039 | 203 | in sporadic cancers; somatic mutation; dbSNP:rs730882003 | |
| |
| | Natural variant | VAR_045808 | 203 | in a sporadic cancer; somatic mutation; requires 2 nucleotide substitutions | |
| |
| | Natural variant | VAR_047174 | 203-204 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045040 | 204 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045041 | 204 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045042 | 204 | in sporadic cancers; somatic mutation; dbSNP:rs1260903787 | |
| |
| | Natural variant | VAR_045043 | 204 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045044 | 204 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045045 | 204 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_005953 | 205 | in sporadic cancers; somatic mutation; dbSNP:rs1057520007 | |
| |
| | Natural variant | VAR_005954 | 205 | in sporadic cancers; somatic mutation; dbSNP:rs1057520008 | |
| |
| | Natural variant | VAR_047175 | 205 | in sporadic cancers; somatic mutation; dbSNP:rs1057520007 | |
| |
| | Natural variant | VAR_045046 | 205 | in sporadic cancers; somatic mutation; dbSNP:rs1057520008 | |
| |
| | Natural variant | VAR_045047 | 205 | in sporadic cancers; somatic mutation; dbSNP:rs1057520008 | |
| |
| | Natural variant | VAR_045048 | 205 | in sporadic cancers; somatic mutation; dbSNP:rs1057520007 | |
| |
| | Natural variant | VAR_045049 | 206 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045050 | 206 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045051 | 207 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045052 | 207 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045053 | 207 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045054 | 207 | in sporadic cancers; somatic mutation; dbSNP:rs923100890 | |
| |
| | Natural variant | VAR_045055 | 207 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045056 | 207 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_047176 | 207-208 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045057 | 208 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045058 | 208 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045059 | 208 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045809 | 208 | in a sporadic cancer; somatic mutation; requires 2 nucleotide substitutions | |
| |
| | Natural variant | VAR_045060 | 208 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045061 | 208 | in sporadic cancers; somatic mutation; dbSNP:rs1464727668 | |
| |
| | Natural variant | VAR_045062 | 208 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045063 | 209 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045064 | 209 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045065 | 209 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045066 | 209 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045067 | 210 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045068 | 210 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045069 | 210 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045070 | 210 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045071 | 210 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045072 | 210 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045073 | 210 | in a familial cancer not matching LFS; germline mutation; dbSNP:rs1060501200 | |
| |
| | Natural variant | VAR_045074 | 211 | in sporadic cancers; somatic mutation; dbSNP:rs1060501198 | |
| |
| | Natural variant | VAR_045075 | 211 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045076 | 211 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045077 | 211 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045078 | 211 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045079 | 212 | in sporadic cancers; somatic mutation; dbSNP:rs1064795766 | |
| |
| | Natural variant | VAR_045080 | 212 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045081 | 212 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045082 | 212 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045083 | 212 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045084 | 213 | in sporadic cancers; somatic mutation; dbSNP:rs397516436 | |
| |
| | Natural variant | VAR_045085 | 213 | in sporadic cancers; somatic mutation; dbSNP:rs587778720 | |
| |
| | Natural variant | VAR_036506 | 213 | in LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs587778720 | |
| |
| | Natural variant | VAR_005955 | 213 | in LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs587778720 | |
| |
| | Natural variant | VAR_045086 | 213 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045087 | 214 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045088 | 214 | in a sporadic cancer; somatic mutation; dbSNP:rs1057519992 | |
| |
| | Natural variant | VAR_047177 | 214 | in sporadic cancers; somatic mutation; dbSNP:rs587781386 | |
| |
| | Natural variant | VAR_045089 | 214 | in sporadic cancers; somatic mutation; dbSNP:rs1057519992 | |
| |
| | Natural variant | VAR_045090 | 214 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045091 | 215 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045092 | 215 | in sporadic cancers; somatic mutation; dbSNP:rs886039484 | |
| |
| | Natural variant | VAR_045093 | 215 | in sporadic cancers; somatic mutation; dbSNP:rs587782177 | |
| |
| | Natural variant | VAR_045810 | 215 | in sporadic cancers; somatic mutation; requires 2 nucleotide substitutions | |
| |
| | Natural variant | VAR_045094 | 215 | in sporadic cancers; somatic mutation; dbSNP:rs587782177 | |
| |
| | Natural variant | VAR_045095 | 215 | in sporadic cancers; somatic mutation; dbSNP:rs1057520001 | |
| |
| | Natural variant | VAR_045096 | 215 | in sporadic cancers; somatic mutation; dbSNP:rs587782177 | |
| |
| | Natural variant | VAR_045097 | 216 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045098 | 216 | in sporadic cancers; somatic mutation; dbSNP:rs1057520004 | |
| |
| | Natural variant | VAR_045099 | 216 | in sporadic cancers; somatic mutation; dbSNP:rs1057520004 | |
| |
| | Natural variant | VAR_045100 | 216 | in sporadic cancers; somatic mutation; dbSNP:rs730882025 | |
| |
| | Natural variant | VAR_005956 | 216 | in sporadic cancers; somatic mutation; dbSNP:rs730882025 | |
| |
| | Natural variant | VAR_045811 | 216 | in a sporadic cancer; somatic mutation; requires 2 nucleotide substitutions | |
| |
| | Natural variant | VAR_045101 | 217 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045102 | 217 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045103 | 217 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045104 | 217 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045105 | 217 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_047178 | 217 | in dbSNP:rs35163653 | |
| |
| | Natural variant | VAR_045106 | 218 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045107 | 218 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045108 | 218 | in sporadic cancers; somatic mutation; dbSNP:rs1555525743 | |
| |
| | Natural variant | VAR_045109 | 218 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045110 | 218 | in sporadic cancers; somatic mutation; dbSNP:rs878854072 | |
| |
| | Natural variant | VAR_045812 | 219 | in a sporadic cancer; somatic mutation; requires 2 nucleotide substitutions | |
| |
| | Natural variant | VAR_045111 | 219 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045112 | 219 | in sporadic cancers; somatic mutation; dbSNP:rs1420675064 | |
| |
| | Natural variant | VAR_045113 | 219 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045114 | 219 | in a familial cancer not matching LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs879253894 | |
| |
| | Natural variant | VAR_045115 | 219 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_005957 | 220 | in LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs121912666 | |
| |
| | Natural variant | VAR_045116 | 220 | in sporadic cancers; somatic mutation; dbSNP:rs530941076 | |
| |
| | Natural variant | VAR_045117 | 220 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_005958 | 220 | in sporadic cancers; somatic mutation; dbSNP:rs530941076 | |
| |
| | Natural variant | VAR_045118 | 220 | in sporadic cancers; somatic mutation; dbSNP:rs530941076 | |
| |
| | Natural variant | VAR_005959 | 220 | in a brain tumor with no family history; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs121912666 | |
| |
| | Natural variant | VAR_045119 | 221 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045120 | 221 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045121 | 221 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045122 | 221 | in sporadic cancers; somatic mutation; dbSNP:rs786201592 | |
| |
| | Natural variant | VAR_045123 | 221 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045124 | 222 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045125 | 222 | in sporadic cancers; somatic mutation; dbSNP:rs146340390 | |
| |
| | Natural variant | VAR_045126 | 222 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045127 | 222 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045128 | 222 | in sporadic cancers; somatic mutation; dbSNP:rs1060501203 | |
| |
| | Natural variant | VAR_045129 | 222 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_047179 | 223 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045130 | 223 | in sporadic cancers; somatic mutation; dbSNP:rs138983188 | |
| |
| | Natural variant | VAR_045131 | 223 | in sporadic cancers; somatic mutation; dbSNP:rs138983188 | |
| |
| | Natural variant | VAR_045132 | 223 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045133 | 223 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045134 | 223 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045135 | 224 | in sporadic cancers; somatic mutation; dbSNP:rs267605076 | |
| |
| | Natural variant | VAR_045136 | 224 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045137 | 224 | in sporadic cancers; somatic mutation; dbSNP:rs1555525707 | |
| |
| | Natural variant | VAR_045138 | 224 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045139 | 225 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045140 | 225 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045141 | 225 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045142 | 225 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045143 | 225 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045144 | 225 | in a sporadic cancer; somatic mutation; dbSNP:rs746504075 | |
| |
| | Natural variant | VAR_045145 | 226 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_047180 | 226 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045844 | 226 | in a sporadic cancer; somatic mutation; requires 2 nucleotide substitutions | |
| |
| | Natural variant | VAR_045146 | 226 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045147 | 226 | in sporadic cancers; somatic mutation; dbSNP:rs970212462 | |
| |
| | Natural variant | VAR_045148 | 227 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045149 | 227 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045150 | 227 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045151 | 227 | in LFS; germline mutation and in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045152 | 228 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_005960 | 228 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045153 | 228 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045154 | 228 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045155 | 228 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045845 | 228 | in a sporadic cancer; somatic mutation; requires 2 nucleotide substitutions | |
| |
| | Natural variant | VAR_045156 | 228 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045157 | 228 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045158 | 229 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045846 | 229 | in a sporadic cancer; somatic mutation; requires 2 nucleotide substitutions | |
| |
| | Natural variant | VAR_045159 | 229 | in sporadic cancers; somatic mutation; dbSNP:rs1064794312 | |
| |
| | Natural variant | VAR_045160 | 229 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045161 | 229 | in sporadic cancers; somatic mutation; dbSNP:rs1064793603 | |
| |
| | Natural variant | VAR_045162 | 230 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_005961 | 230 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045163 | 230 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045164 | 230 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045165 | 230 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045166 | 231 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045167 | 231 | in sporadic cancers; somatic mutation; dbSNP:rs1555525564 | |
| |
| | Natural variant | VAR_045168 | 231 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045169 | 231 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045170 | 232 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045171 | 232 | in sporadic cancers; somatic mutation; dbSNP:rs1555525562 | |
| |
| | Natural variant | VAR_045172 | 232 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045173 | 232 | in sporadic cancers; somatic mutation; does not induce SNAI1 degradation | |
| |
| | Natural variant | VAR_005962 | 232 | in sporadic cancers; somatic mutation; dbSNP:rs587781589 | |
| |
| | Natural variant | VAR_045174 | 232 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045175 | 233 | in LFS; germline mutation and in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045176 | 233 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045177 | 233 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045178 | 233 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_047181 | 233 | in sporadic cancers; somatic mutation; dbSNP:rs879254233 | |
| |
| | Natural variant | VAR_045179 | 233 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_005963 | 234 | in LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs587780073 | |
| |
| | Natural variant | VAR_045180 | 234 | in sporadic cancers; somatic mutation; dbSNP:rs864622237 | |
| |
| | Natural variant | VAR_045181 | 234 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_005964 | 234 | in sporadic cancers; somatic mutation; dbSNP:rs864622237 | |
| |
| | Natural variant | VAR_045847 | 234 | in a sporadic cancer; somatic mutation; requires 2 nucleotide substitutions | |
| |
| | Natural variant | VAR_045182 | 234 | in sporadic cancers; somatic mutation; dbSNP:rs864622237 | |
| |
| | Natural variant | VAR_045848 | 234 | in a sporadic cancer; somatic mutation; requires 2 nucleotide substitutions | |
| |
| | Natural variant | VAR_045183 | 234 | in sporadic cancers; somatic mutation; dbSNP:rs587780073 | |
| |
| | Natural variant | VAR_047182 | 235 | in an adrenocortical carcinoma with no family history; germline mutation and in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045184 | 235 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045185 | 235 | in sporadic cancers; somatic mutation; dbSNP:rs144340710 | |
| |
| | Natural variant | VAR_045849 | 235 | in a sporadic cancer; somatic mutation; requires 2 nucleotide substitutions | |
| |
| | Natural variant | VAR_045186 | 235 | in LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs144340710 | |
| |
| | Natural variant | VAR_045187 | 235 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045188 | 235 | in sporadic cancers; somatic mutation; dbSNP:rs786204145 | |
| |
| | Natural variant | VAR_045189 | 236 | in LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs730882026 | |
| |
| | Natural variant | VAR_045190 | 236 | in sporadic cancers; somatic mutation; dbSNP:rs587782289 | |
| |
| | Natural variant | VAR_045191 | 236 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045192 | 236 | in sporadic cancers; somatic mutation; dbSNP:rs587782289 | |
| |
| | Natural variant | VAR_045193 | 236 | in sporadic cancers; somatic mutation; dbSNP:rs587782289 | |
| |
| | Natural variant | VAR_045194 | 236 | in sporadic cancers; somatic mutation; dbSNP:rs730882026 | |
| |
| | Natural variant | VAR_005965 | 237 | in LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs587782664 | |
| |
| | Natural variant | VAR_045195 | 237 | in sporadic cancers; somatic mutation; dbSNP:rs765848205 | |
| |
| | Natural variant | VAR_045196 | 237 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045197 | 237 | in sporadic cancers; somatic mutation; dbSNP:rs765848205 | |
| |
| | Natural variant | VAR_045198 | 237 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045199 | 237 | in sporadic cancers; somatic mutation; dbSNP:rs730882004 | |
| |
| | Natural variant | VAR_005966 | 238 | in sporadic cancers; somatic mutation; dbSNP:rs730882005 | |
| |
| | Natural variant | VAR_045200 | 238 | in LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs1057519981 | |
| |
| | Natural variant | VAR_045850 | 238 | in a sporadic cancer; somatic mutation; requires 2 nucleotide substitutions | |
| |
| | Natural variant | VAR_045201 | 238 | in sporadic cancers; somatic mutation; dbSNP:rs1057519981 | |
| |
| | Natural variant | VAR_045202 | 238 | in LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs730882005 | |
| |
| | Natural variant | VAR_045203 | 238 | in sporadic cancers; somatic mutation; dbSNP:rs193920789 | |
| |
| | Natural variant | VAR_005967 | 238 | in a familial cancer not matching LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs730882005 | |
| |
| | Natural variant | VAR_045204 | 239 | in sporadic cancers; somatic mutation; dbSNP:rs876660807 | |
| |
| | Natural variant | VAR_045205 | 239 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045206 | 239 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045207 | 239 | in sporadic cancers; somatic mutation; dbSNP:rs1057522275 | |
| |
| | Natural variant | VAR_045208 | 239 | in sporadic cancers; somatic mutation; dbSNP:rs1057519999 | |
| |
| | Natural variant | VAR_045209 | 239 | in sporadic cancers; somatic mutation; dbSNP:rs1057519999 | |
| |
| | Natural variant | VAR_045210 | 239 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045211 | 240 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045212 | 240 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_005968 | 240 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045213 | 240 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045214 | 240 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045215 | 240 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045216 | 240 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_033036 | 241 | in sporadic cancers; somatic mutation; dbSNP:rs1057520002 | |
| |
| | Natural variant | VAR_045217 | 241 | in sporadic cancers; somatic mutation; dbSNP:rs28934573 | |
| |
| | Natural variant | VAR_005969 | 241 | in LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs28934573 | |
| |
| | Natural variant | VAR_045218 | 241 | in sporadic cancers; somatic mutation; dbSNP:rs1057520002 | |
| |
| | Natural variant | VAR_047183 | 241 | in LFS; germline mutation and in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045219 | 241 | in sporadic cancers; somatic mutation; dbSNP:rs28934573 | |
| |
| | Natural variant | VAR_005970 | 242 | in sporadic cancers; somatic mutation; dbSNP:rs121912655 | |
| |
| | Natural variant | VAR_045220 | 242 | in sporadic cancers; somatic mutation; dbSNP:rs1057519982 | |
| |
| | Natural variant | VAR_045221 | 242 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045222 | 242 | in sporadic cancers; somatic mutation; dbSNP:rs121912655 | |
| |
| | Natural variant | VAR_045223 | 242 | in sporadic cancers; somatic mutation; dbSNP:rs375874539 | |
| |
| | Natural variant | VAR_045224 | 242 | in a familial cancer not matching LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs121912655 | |
| |
| | Natural variant | VAR_045225 | 243 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045226 | 243 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045227 | 243 | in sporadic cancers; somatic mutation; dbSNP:rs786203117 | |
| |
| | Natural variant | VAR_045228 | 243 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045229 | 243 | in sporadic cancers; somatic mutation; dbSNP:rs730882006 | |
| |
| | Natural variant | VAR_045230 | 243 | in sporadic cancers; somatic mutation; dbSNP:rs786203117 | |
| |
| | Natural variant | VAR_047184 | 243-244 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_047185 | 243-244 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_047186 | 244 | in sporadic cancers; somatic mutation; dbSNP:rs985033810 | |
| |
| | Natural variant | VAR_045231 | 244 | in sporadic cancers; somatic mutation; dbSNP:rs1057519989 | |
| |
| | Natural variant | VAR_045232 | 244 | in LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs1057517983 | |
| |
| | Natural variant | VAR_045233 | 244 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045234 | 244 | in sporadic cancers; somatic mutation; dbSNP:rs1057519989 | |
| |
| | Natural variant | VAR_045235 | 244 | in sporadic cancers; somatic mutation; dbSNP:rs1057519989 | |
| |
| | Natural variant | VAR_045236 | 244 | in LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs985033810 | |
| |
| | Natural variant | VAR_005971 | 245 | in sporadic cancers; somatic mutation; dbSNP:rs121912656 | |
| |
| | Natural variant | VAR_005972 | 245 | in LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs28934575 | |
| |
| | Natural variant | VAR_005973 | 245 | in LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs121912656 | |
| |
| | Natural variant | VAR_045237 | 245 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045851 | 245 | in sporadic cancers; somatic mutation; requires 2 nucleotide substitutions | |
| |
| | Natural variant | VAR_045852 | 245 | in a sporadic cancer; somatic mutation; requires 2 nucleotide substitutions | |
| |
| | Natural variant | VAR_045853 | 245 | in sporadic cancers; somatic mutation; requires 2 nucleotide substitutions | |
| |
| | Natural variant | VAR_045854 | 245 | in sporadic cancers; somatic mutation; requires 2 nucleotide substitutions | |
| |
| | Natural variant | VAR_045238 | 245 | in sporadic cancers; somatic mutation; dbSNP:rs28934575 | |
| |
| | Natural variant | VAR_005974 | 245 | in LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs28934575 | |
| |
| | Natural variant | VAR_005975 | 245 | in LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs121912656 | |
| |
| | Natural variant | VAR_045239 | 246 | in sporadic cancers; somatic mutation; dbSNP:rs1019340046 | |
| |
| | Natural variant | VAR_045240 | 246 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_044020 | 246 | in sporadic cancers; somatic mutation; dbSNP:rs483352695 | |
| |
| | Natural variant | VAR_005976 | 246 | in sporadic cancers; somatic mutation; dbSNP:rs587780074 | |
| |
| | Natural variant | VAR_005977 | 246 | in sporadic cancers; somatic mutation; dbSNP:rs587780074 | |
| |
| | Natural variant | VAR_005978 | 246 | in LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs483352695 | |
| |
| | Natural variant | VAR_045241 | 247 | in sporadic cancers; somatic mutation; dbSNP:rs1452189221 | |
| |
| | Natural variant | VAR_045855 | 247 | in a sporadic cancer; somatic mutation; requires 2 nucleotide substitutions | |
| |
| | Natural variant | VAR_005980 | 247 | in sporadic cancers; somatic mutation; dbSNP:rs786201762 | |
| |
| | Natural variant | VAR_045242 | 247 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045243 | 247 | in sporadic cancers; somatic mutation; dbSNP:rs786201762 | |
| |
| | Natural variant | VAR_047189 | 247 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045244 | 247 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_047187 | 247-248 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_047188 | 247-248 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045245 | 248 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_005981 | 248 | in sporadic cancers; somatic mutation; dbSNP:rs121912651 | |
| |
| | Natural variant | VAR_005982 | 248 | in sporadic cancers; somatic mutation; dbSNP:rs11540652 | |
| |
| | Natural variant | VAR_045246 | 248 | in sporadic cancers; somatic mutation; dbSNP:rs11540652 | |
| |
| | Natural variant | VAR_005983 | 248 | in LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs11540652 | |
| |
| | Natural variant | VAR_005984 | 248 | in LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs121912651 | |
| |
| | Mutagenesis | | 248 | Does not induce SNAI1 degradation. | |
| |
| | Natural variant | VAR_005985 | 249 | in sporadic cancers; somatic mutation; dbSNP:rs587782082 | |
| |
| | Natural variant | VAR_045247 | 249 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045248 | 249 | in sporadic cancers; somatic mutation; dbSNP:rs587782329 | |
| |
| | Natural variant | VAR_033037 | 249 | in sporadic cancers; somatic mutation; dbSNP:rs587782329 | |
| |
| | Natural variant | VAR_045856 | 249 | in a sporadic cancer; somatic mutation; requires 2 nucleotide substitutions | |
| |
| | Natural variant | VAR_005986 | 249 | in sporadic cancers; somatic mutation; does not induce SNAI1 degradation; dbSNP:rs28934571 | |
| |
| | Natural variant | VAR_045249 | 249 | in sporadic cancers; somatic mutation; dbSNP:rs587782329 | |
| |
| | Natural variant | VAR_045250 | 249 | in sporadic cancers; somatic mutation; dbSNP:rs587782082 | |
| |
| | Natural variant | VAR_047190 | 249-250 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_047191 | 249-250 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045251 | 250 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045857 | 250 | in sporadic cancers; somatic mutation; requires 2 nucleotide substitutions | |
| |
| | Natural variant | VAR_045252 | 250 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_047192 | 250 | in sporadic cancers; somatic mutation; dbSNP:rs1064794311 | |
| |
| | Natural variant | VAR_045858 | 250 | in sporadic cancers; somatic mutation; requires 2 nucleotide substitutions | |
| |
| | Natural variant | VAR_045253 | 250 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045254 | 250 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045255 | 250 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045256 | 251 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045257 | 251 | in sporadic cancers; somatic mutation; dbSNP:rs730882007 | |
| |
| | Natural variant | VAR_045258 | 251 | in LFS; germline mutation; dbSNP:rs878854074 | |
| |
| | Natural variant | VAR_005987 | 251 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_033038 | 251 | in sporadic cancers; somatic mutation; dbSNP:rs730882027 | |
| |
| | Natural variant | VAR_045259 | 251 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045260 | 251 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045261 | 252 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045262 | 252 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045263 | 252 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_005988 | 252 | in LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs121912653 | |
| |
| | Natural variant | VAR_045264 | 252 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045265 | 253 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045266 | 253 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045267 | 253 | in sporadic cancers; somatic mutation; dbSNP:rs1555525465 | |
| |
| | Natural variant | VAR_047193 | 253 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045268 | 253 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045859 | 254 | in sporadic cancers; somatic mutation; requires 2 nucleotide substitutions | |
| |
| | Natural variant | VAR_045269 | 254 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045270 | 254 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045271 | 254 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_017908 | 254 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045272 | 254 | in sporadic cancers; somatic mutation; dbSNP:rs1330865474 | |
| |
| | Natural variant | VAR_017909 | 254 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045273 | 254 | in sporadic cancers; somatic mutation; dbSNP:rs746601313 | |
| |
| | Natural variant | VAR_045274 | 255 | in sporadic cancers; somatic mutation; dbSNP:rs1057519995 | |
| |
| | Natural variant | VAR_045275 | 255 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045276 | 255 | in sporadic cancers; somatic mutation; dbSNP:rs876659675 | |
| |
| | Natural variant | VAR_045277 | 255 | in sporadic cancers; somatic mutation; dbSNP:rs876659675 | |
| |
| | Natural variant | VAR_045278 | 255 | in sporadic cancers; somatic mutation; dbSNP:rs876659675 | |
| |
| | Natural variant | VAR_045279 | 255 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045280 | 256 | in a brain tumor with no family history; germline mutation and in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045281 | 256 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045282 | 256 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045283 | 256 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_005989 | 257 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045284 | 257 | in LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs28934577 | |
| |
| | Natural variant | VAR_045285 | 257 | in sporadic cancers; somatic mutation; dbSNP:rs28934577 | |
| |
| | Natural variant | VAR_045286 | 257 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045287 | 258 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_005990 | 258 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045288 | 258 | in sporadic cancers; somatic mutation; dbSNP:rs1060501201 | |
| |
| | Natural variant | VAR_005991 | 258 | in LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs121912652 | |
| |
| | Natural variant | VAR_045860 | 258 | in a sporadic cancer; somatic mutation; requires 2 nucleotide substitutions | |
| |
| | Natural variant | VAR_045289 | 258 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045290 | 258 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_047194 | 259 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045291 | 259 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045292 | 259 | in sporadic cancers; somatic mutation; dbSNP:rs745425759 | |
| |
| | Natural variant | VAR_045293 | 259 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045294 | 259 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045861 | 259 | in a sporadic cancer; somatic mutation; requires 2 nucleotide substitutions | |
| |
| | Natural variant | VAR_045862 | 259 | in a sporadic cancer; somatic mutation; requires 2 nucleotide substitutions | |
| |
| | Natural variant | VAR_045295 | 259 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_033039 | 259 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045296 | 260 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045297 | 260 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045298 | 260 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045299 | 260 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045300 | 260 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045301 | 260 | in sporadic cancers; somatic mutation; dbSNP:rs876658916 | |
| |
| | Natural variant | VAR_045302 | 261 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045303 | 261 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045304 | 261 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045305 | 261 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045306 | 261 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045307 | 262 | in a sporadic cancer; somatic mutation; dbSNP:rs200579969 | |
| |
| | Natural variant | VAR_047196 | 262 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045863 | 262 | in a sporadic cancer; somatic mutation; requires 2 nucleotide substitutions | |
| |
| | Natural variant | VAR_045308 | 262 | in sporadic cancers; somatic mutation; dbSNP:rs200579969 | |
| |
| | Natural variant | VAR_045309 | 262 | in sporadic cancers; somatic mutation; dbSNP:rs1131691025 | |
| |
| | Natural variant | VAR_047195 | 262-263 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045310 | 263 | in sporadic cancers; somatic mutation; dbSNP:rs72661119 | |
| |
| | Natural variant | VAR_045311 | 263 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045312 | 263 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045313 | 263 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045314 | 263 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045315 | 264 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045316 | 264 | in a sporadic cancer; somatic mutation; dbSNP:rs1555525353 | |
| |
| | Natural variant | VAR_045317 | 264 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045318 | 264 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045319 | 264 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045320 | 265 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045321 | 265 | in LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs879253942 | |
| |
| | Natural variant | VAR_045322 | 265 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_047197 | 265 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045323 | 266 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045324 | 266 | in sporadic cancers; somatic mutation; dbSNP:rs193920774 | |
| |
| | Natural variant | VAR_045325 | 266 | in sporadic cancers; somatic mutation; dbSNP:rs1057519990 | |
| |
| | Natural variant | VAR_045326 | 266 | in sporadic cancers; somatic mutation; dbSNP:rs193920774 | |
| |
| | Natural variant | VAR_045327 | 267 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045328 | 267 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045329 | 267 | in sporadic cancers; somatic mutation; dbSNP:rs587780075 | |
| |
| | Natural variant | VAR_045330 | 267 | in LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs587780075 | |
| |
| | Natural variant | VAR_036507 | 267 | in sporadic cancers; somatic mutation; dbSNP:rs55832599 | |
| |
| | Natural variant | VAR_045864 | 268 | in a sporadic cancer; somatic mutation; requires 2 nucleotide substitutions | |
| |
| | Natural variant | VAR_045331 | 268 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045332 | 268 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045333 | 268 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045334 | 268 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045335 | 268 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045336 | 269 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045337 | 269 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_047198 | 269 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045338 | 269 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045339 | 269 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045340 | 269 | in a sporadic cancer; somatic mutation | |
| |
| | Mutagenesis | | 269 | Abolishes phosphorylation. | |
| |
| | Mutagenesis | | 269 | Inhibits strongly its transcriptional activity. | |
| |
| | Natural variant | VAR_045341 | 270 | in sporadic cancers; somatic mutation; dbSNP:rs1057519986 | |
| |
| | Natural variant | VAR_045342 | 270 | in sporadic cancers; somatic mutation; dbSNP:rs1057519988 | |
| |
| | Natural variant | VAR_045343 | 270 | in sporadic cancers; somatic mutation; dbSNP:rs1057519987 | |
| |
| | Natural variant | VAR_045344 | 270 | in sporadic cancers; somatic mutation; dbSNP:rs1057519986 | |
| |
| | Natural variant | VAR_045345 | 270 | in sporadic cancers; somatic mutation; dbSNP:rs1057519988 | |
| |
| | Natural variant | VAR_045346 | 270 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045347 | 271 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045348 | 271 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045349 | 271 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_036508 | 271 | in sporadic cancers; somatic mutation; dbSNP:rs1060501191 | |
| |
| | Natural variant | VAR_045865 | 271 | in a sporadic cancer; somatic mutation; requires 2 nucleotide substitutions | |
| |
| | Natural variant | VAR_045350 | 271 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045866 | 271 | in a sporadic cancer; somatic mutation; requires 2 nucleotide substitutions | |
| |
| | Natural variant | VAR_047199 | 271 | in an osteosarcoma with no family history; germline mutation and in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045351 | 272 | in a familial cancer not matching LFS; germline mutation and in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045352 | 272 | in sporadic cancers; somatic mutation; dbSNP:rs876660333 | |
| |
| | Natural variant | VAR_045353 | 272 | in sporadic cancers; somatic mutation; dbSNP:rs876660333 | |
| |
| | Natural variant | VAR_005992 | 272 | in LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs121912657 | |
| |
| | Natural variant | VAR_045354 | 272 | in sporadic cancers; somatic mutation; dbSNP:rs121912657 | |
| |
| | Natural variant | VAR_005993 | 273 | in LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs121913343 | |
| |
| | Natural variant | VAR_005994 | 273 | in LFS; germline mutation and in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_005995 | 273 | in LFS; germline mutation and in sporadic cancers; somatic mutation; abolishes sequence-specific DNA binding; does not induce SNAI1 degradation; dbSNP:rs28934576 | |
| |
| | Natural variant | VAR_036509 | 273 | in LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs28934576 | |
| |
| | Natural variant | VAR_045867 | 273 | in a sporadic cancer; somatic mutation; requires 2 nucleotide substitutions | |
| |
| | Natural variant | VAR_045355 | 273 | in sporadic cancers; somatic mutation; dbSNP:rs28934576 | |
| |
| | Natural variant | VAR_045356 | 273 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045357 | 273 | in a familial cancer not matching LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs121913343 | |
| |
| | Natural variant | VAR_045868 | 273 | in a sporadic cancer; somatic mutation; requires 2 nucleotide substitutions | |
| |
| | Natural variant | VAR_045358 | 274 | in sporadic cancers; somatic mutation; dbSNP:rs1057520006 | |
| |
| | Natural variant | VAR_045359 | 274 | in sporadic cancers; somatic mutation; dbSNP:rs1057520006 | |
| |
| | Natural variant | VAR_005997 | 274 | in sporadic cancers; somatic mutation; dbSNP:rs1057520005 | |
| |
| | Natural variant | VAR_047200 | 274 | in sporadic cancers; somatic mutation; dbSNP:rs1057520006 | |
| |
| | Natural variant | VAR_045360 | 274 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045361 | 274 | in sporadic cancers; somatic mutation; dbSNP:rs1057520005 | |
| |
| | Natural variant | VAR_045362 | 275 | in sporadic cancers; somatic mutation; dbSNP:rs863224451 | |
| |
| | Natural variant | VAR_045363 | 275 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045364 | 275 | in sporadic cancers; somatic mutation; dbSNP:rs1057519983 | |
| |
| | Natural variant | VAR_045365 | 275 | in sporadic cancers; somatic mutation; dbSNP:rs863224451 | |
| |
| | Natural variant | VAR_005999 | 275 | in sporadic cancers; somatic mutation; dbSNP:rs1555525279 | |
| |
| | Natural variant | VAR_005998 | 275 | in LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs863224451 | |
| |
| | Natural variant | VAR_045366 | 276 | in sporadic cancers; somatic mutation; dbSNP:rs786202082 | |
| |
| | Natural variant | VAR_045367 | 276 | in sporadic cancers; somatic mutation; dbSNP:rs786202082 | |
| |
| | Natural variant | VAR_045368 | 276 | in sporadic cancers; somatic mutation; dbSNP:rs1131691029 | |
| |
| | Natural variant | VAR_045369 | 276 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045370 | 276 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045371 | 276 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045372 | 277 | in sporadic cancers; somatic mutation; dbSNP:rs763098116 | |
| |
| | Natural variant | VAR_006000 | 277 | in sporadic cancers; somatic mutation; dbSNP:rs1064795369 | |
| |
| | Natural variant | VAR_045373 | 277 | in sporadic cancers; somatic mutation; dbSNP:rs1064795369 | |
| |
| | Natural variant | VAR_045374 | 277 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_047201 | 277 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045375 | 277 | in an osteosarcoma with no family history; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs763098116 | |
| |
| | Natural variant | VAR_006001 | 278 | in sporadic cancers; somatic mutation; dbSNP:rs17849781 | |
| |
| | Natural variant | VAR_045869 | 278 | in sporadic cancers; somatic mutation; requires 2 nucleotide substitutions | |
| |
| | Natural variant | VAR_006002 | 278 | in sporadic cancers; somatic mutation; dbSNP:rs876659802 | |
| |
| | Natural variant | VAR_006003 | 278 | in LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs876659802 | |
| |
| | Natural variant | VAR_045376 | 278 | in sporadic cancers; somatic mutation; dbSNP:rs876659802 | |
| |
| | Natural variant | VAR_006004 | 278 | in LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs17849781 | |
| |
| | Natural variant | VAR_006005 | 278 | in LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs17849781 | |
| |
| | Natural variant | VAR_006006 | 279 | in sporadic cancers; somatic mutation; dbSNP:rs1064793881 | |
| |
| | Natural variant | VAR_045377 | 279 | in sporadic cancers; somatic mutation; dbSNP:rs1555525248 | |
| |
| | Natural variant | VAR_045378 | 279 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045379 | 279 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045380 | 280 | in sporadic cancers; somatic mutation; dbSNP:rs753660142 | |
| |
| | Natural variant | VAR_006008 | 280 | in sporadic cancers; somatic mutation; dbSNP:rs121912660 | |
| |
| | Natural variant | VAR_006007 | 280 | in a familial cancer not matching LFS; germline mutation and in sporadic cancers; somatic mutation; no effect on interaction with CCAR2; dbSNP:rs121912660 | |
| |
| | Natural variant | VAR_045381 | 280 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045382 | 280 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_006009 | 280 | in sporadic cancers; somatic mutation; dbSNP:rs121912660 | |
| |
| | Natural variant | VAR_006010 | 281 | in sporadic cancers; somatic mutation; dbSNP:rs587781525 | |
| |
| | Natural variant | VAR_006011 | 281 | in sporadic cancers; somatic mutation; dbSNP:rs1057519984 | |
| |
| | Natural variant | VAR_006012 | 281 | in a brain tumor with no family history; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs587781525 | |
| |
| | Natural variant | VAR_006013 | 281 | in sporadic cancers; somatic mutation; dbSNP:rs764146326 | |
| |
| | Natural variant | VAR_047202 | 281 | in LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs764146326 | |
| |
| | Natural variant | VAR_045870 | 281 | in a sporadic cancer; somatic mutation; requires 2 nucleotide substitutions | |
| |
| | Natural variant | VAR_006014 | 281 | in a familial cancer not matching LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs587781525 | |
| |
| | Natural variant | VAR_045383 | 281 | in sporadic cancers; somatic mutation; dbSNP:rs764146326 | |
| |
| | Natural variant | VAR_047203 | 281-282 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045384 | 282 | in LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs28934574 | |
| |
| | Natural variant | VAR_045385 | 282 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_006015 | 282 | in sporadic cancers; somatic mutation; dbSNP:rs730882008 | |
| |
| | Natural variant | VAR_045386 | 282 | in sporadic cancers; somatic mutation; dbSNP:rs730882008 | |
| |
| | Natural variant | VAR_045387 | 282 | in a familial cancer not matching LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs730882008 | |
| |
| | Natural variant | VAR_006016 | 282 | in LFS; germline mutation and in sporadic cancers; somatic mutation; does not induce SNAI1 degradation; dbSNP:rs28934574 | |
| |
| | Natural variant | VAR_006017 | 283 | in LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs149633775 | |
| |
| | Natural variant | VAR_006018 | 283 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_006019 | 283 | in a brain tumor with no family history; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs371409680 | |
| |
| | Natural variant | VAR_045388 | 283 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_006020 | 283 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045389 | 283 | in a sporadic cancer; somatic mutation; dbSNP:rs149633775 | |
| |
| | Natural variant | VAR_006021 | 284 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045390 | 284 | in sporadic cancers; somatic mutation; dbSNP:rs863224685 | |
| |
| | Natural variant | VAR_045391 | 284 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_006022 | 284 | in sporadic cancers; somatic mutation; dbSNP:rs1204379654 | |
| |
| | Mutagenesis | | 284 | Inhibits strongly its transcriptional activity. | |
| |
| | Natural variant | VAR_045392 | 285 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045393 | 285 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045394 | 285 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_006023 | 285 | in sporadic cancers; somatic mutation; dbSNP:rs112431538 | |
| |
| | Natural variant | VAR_006024 | 285 | in LFS; germline mutation and in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_006025 | 285 | in sporadic cancers; somatic mutation; dbSNP:rs121912667 | |
| |
| | Natural variant | VAR_006026 | 286 | in LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs1057519985 | |
| |
| | Natural variant | VAR_006027 | 286 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_006028 | 286 | in sporadic cancers; somatic mutation; dbSNP:rs1057519985 | |
| |
| | Natural variant | VAR_006029 | 286 | in sporadic cancers; somatic mutation; dbSNP:rs786201059 | |
| |
| | Natural variant | VAR_045871 | 286 | in a sporadic cancer; somatic mutation; requires 2 nucleotide substitutions | |
| |
| | Natural variant | VAR_006030 | 286 | in sporadic cancers; somatic mutation; dbSNP:rs786201059 | |
| |
| | Natural variant | VAR_045395 | 286 | in sporadic cancers; somatic mutation; dbSNP:rs1057519985 | |
| |
| | Natural variant | VAR_047204 | 287 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045396 | 287 | in sporadic cancers; somatic mutation; dbSNP:rs748891343 | |
| |
| | Natural variant | VAR_045397 | 287 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045398 | 287 | in sporadic cancers; somatic mutation; dbSNP:rs587782006 | |
| |
| | Natural variant | VAR_045399 | 287 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045400 | 288 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045401 | 288 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045402 | 288 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045403 | 288 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045404 | 288 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045405 | 289 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045406 | 289 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045407 | 289 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045408 | 289 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045409 | 289 | in sporadic cancers; somatic mutation; dbSNP:rs1555525154 | |
| |
| | Natural variant | VAR_045410 | 290 | in sporadic cancers; somatic mutation; dbSNP:rs770374782 | |
| |
| | Natural variant | VAR_045411 | 290 | in LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs55819519 | |
| |
| | Natural variant | VAR_045412 | 290 | in LFS; germline mutation and in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045413 | 291 | in sporadic cancers; somatic mutation; dbSNP:rs1555525126 | |
| |
| | Natural variant | VAR_045414 | 291 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045415 | 291 | in sporadic cancers; somatic mutation; dbSNP:rs372613518 | |
| |
| | Natural variant | VAR_047205 | 291 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045416 | 291 | in sporadic cancers; somatic mutation; dbSNP:rs781490101 | |
| |
| | Natural variant | VAR_045417 | 291 | in sporadic cancers; somatic mutation | |
| |
| | Mutagenesis | | 291-292 | Abolishes polyubiquitination by MKRN1. | |
| |
| | Natural variant | VAR_045418 | 292 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045872 | 292 | in a sporadic cancer; somatic mutation; requires 2 nucleotide substitutions | |
| |
| | Natural variant | VAR_015819 | 292 | in LFS; germline mutation and in a sporadic cancer; somatic mutation; dbSNP:rs121912663 | |
| |
| | Natural variant | VAR_045419 | 292 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045420 | 292 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045421 | 292 | in sporadic cancers; somatic mutation; dbSNP:rs121912663 | |
| |
| | Natural variant | VAR_045422 | 292 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045423 | 293 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045424 | 293 | in sporadic cancers; somatic mutation; dbSNP:rs587780076 | |
| |
| | Natural variant | VAR_045425 | 293 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045426 | 293 | in a brain tumor with no family history; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs587780076 | |
| |
| | Natural variant | VAR_045427 | 294 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045428 | 294 | in sporadic cancers; somatic mutation; dbSNP:rs1305324490 | |
| |
| | Natural variant | VAR_045429 | 294 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_047206 | 294 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045430 | 294 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045431 | 294 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045432 | 295 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045433 | 295 | in sporadic cancers; somatic mutation; dbSNP:rs751713111 | |
| |
| | Natural variant | VAR_045434 | 295 | in a sporadic cancer; somatic mutation; dbSNP:rs751713111 | |
| |
| | Natural variant | VAR_045435 | 295 | in sporadic cancers; somatic mutation; dbSNP:rs1131691006 | |
| |
| | Natural variant | VAR_045873 | 296 | in sporadic cancers; somatic mutation; requires 2 nucleotide substitutions | |
| |
| | Natural variant | VAR_045436 | 296 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_047207 | 296 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045437 | 296 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_006031 | 296 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045438 | 296 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045439 | 296 | in a sporadic cancer; somatic mutation; dbSNP:rs483352696 | |
| |
| | Natural variant | VAR_045440 | 296 | in sporadic cancers; somatic mutation; dbSNP:rs672601296 | |
| |
| | Natural variant | VAR_045441 | 297 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045442 | 297 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045443 | 297 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045444 | 297 | in sporadic cancers; somatic mutation; dbSNP:rs876659477 | |
| |
| | Natural variant | VAR_045445 | 297 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045446 | 298 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045447 | 298 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045448 | 298 | in sporadic cancers; somatic mutation; dbSNP:rs201744589 | |
| |
| | Natural variant | VAR_045449 | 298 | in sporadic cancers; somatic mutation; dbSNP:rs201744589 | |
| |
| | Natural variant | VAR_045450 | 298 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045451 | 299 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045452 | 299 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045453 | 299 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045454 | 299 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045455 | 300 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045456 | 300 | in sporadic cancers; somatic mutation; dbSNP:rs909643864 | |
| |
| | Natural variant | VAR_006032 | 300 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045457 | 300 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045458 | 301 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_006033 | 301 | in sporadic cancers; somatic mutation; dbSNP:rs1555525067 | |
| |
| | Natural variant | VAR_045459 | 301 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045460 | 301 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_047208 | 301 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045461 | 302 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_006034 | 302 | in sporadic cancers; somatic mutation; dbSNP:rs1060501202 | |
| |
| | Natural variant | VAR_045462 | 302 | in a sporadic cancer; somatic mutation; dbSNP:rs863224686 | |
| |
| | Natural variant | VAR_006035 | 302 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045463 | 303 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045464 | 303 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045465 | 303 | in sporadic cancers; somatic mutation; dbSNP:rs876658714 | |
| |
| | Natural variant | VAR_045466 | 303 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045467 | 304 | in sporadic cancers; somatic mutation; dbSNP:rs587782654 | |
| |
| | Natural variant | VAR_045468 | 304 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045469 | 304 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_047209 | 304 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045470 | 305 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045471 | 305 | in a familial cancer not matching LFS; germline mutation and in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045472 | 305 | in sporadic cancers; somatic mutation; loss of nuclear localization | |
| |
| | Natural variant | VAR_045473 | 305 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045474 | 305 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045475 | 306 | in LFS; germline mutation and in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_006036 | 306 | in sporadic cancers; somatic mutation; dbSNP:rs1048095040 | |
| |
| | Natural variant | VAR_045476 | 307 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045477 | 307 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_006037 | 307 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045478 | 308 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045479 | 308 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045480 | 309 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_006038 | 309 | in LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs1555525012 | |
| |
| | Natural variant | VAR_045481 | 310 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045482 | 310 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045483 | 311 | in sporadic cancers; somatic mutation; dbSNP:rs1555525007 | |
| |
| | Natural variant | VAR_045484 | 311 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045485 | 311 | in a sporadic cancer; somatic mutation; dbSNP:rs56184981 | |
| |
| | Natural variant | VAR_045486 | 311 | in sporadic cancers; somatic mutation; dbSNP:rs56184981 | |
| |
| | Natural variant | VAR_045487 | 312 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045488 | 312 | in sporadic cancers; somatic mutation; dbSNP:rs145151284 | |
| |
| | Natural variant | VAR_045489 | 313 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045490 | 313 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045491 | 313 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045492 | 313 | in a sporadic cancer; somatic mutation; dbSNP:rs1367492395 | |
| |
| | Natural variant | VAR_045493 | 314 | in a sporadic cancer; somatic mutation; dbSNP:rs751440465 | |
| |
| | Natural variant | VAR_045494 | 315 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045495 | 315 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045496 | 315 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045497 | 316 | in a sporadic cancer; somatic mutation; dbSNP:rs1555524979 | |
| |
| | Natural variant | VAR_045498 | 316 | in a sporadic cancer; somatic mutation; dbSNP:rs772773208 | |
| |
| | Natural variant | VAR_045499 | 317 | in a kidney cancer with no family history; germline mutation and in a sporadic cancer; somatic mutation; dbSNP:rs1060501199 | |
| |
| | Natural variant | VAR_045500 | 317 | in sporadic cancers; somatic mutation; dbSNP:rs764735889 | |
| |
| | Natural variant | VAR_047210 | 317 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045501 | 317 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045502 | 317 | in sporadic cancers; somatic mutation; dbSNP:rs1159579789 | |
| |
| | Natural variant | VAR_045503 | 318 | in sporadic cancers; somatic mutation; dbSNP:rs1555524975 | |
| |
| | Natural variant | VAR_045504 | 319 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045505 | 319 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045506 | 319 | in a sporadic cancer; somatic mutation | |
| |
| | Mutagenesis | | 319 | Loss of nuclear localization; when associated with A-320 and A-321. | |
| |
| | Natural variant | VAR_045507 | 320 | in sporadic cancers; somatic mutation | |
| |
| | Mutagenesis | | 320 | Loss of nuclear localization; when associated with A-319 and A-321. | |
| |
| | Natural variant | VAR_045508 | 321 | in kidney cancer; germline mutation | |
| |
| | Natural variant | VAR_045509 | 321 | in a sporadic cancer; somatic mutation | |
| |
| | Mutagenesis | | 321 | Loss of nuclear localization; when associated with A-319 and A-320. | |
| |
| | Natural variant | VAR_045510 | 322 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045511 | 322 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045874 | 323 | in a sporadic cancer; somatic mutation; requires 2 nucleotide substitutions | |
| |
| | Natural variant | VAR_045512 | 323 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045513 | 323 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045514 | 323 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_047211 | 323 | in a sporadic cancer; somatic mutation; dbSNP:rs1432281680 | |
| |
| | Natural variant | VAR_045515 | 324 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045875 | 324 | in a sporadic cancer; somatic mutation; requires 2 nucleotide substitutions | |
| |
| | Natural variant | VAR_045516 | 324 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045517 | 325 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045518 | 325 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_006039 | 325 | in LFS; germline mutation; dbSNP:rs121912659 | |
| |
| | Natural variant | VAR_045519 | 326 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045520 | 327 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045521 | 327 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045522 | 328 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045523 | 328 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045524 | 328 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045525 | 329 | in a sporadic cancer; somatic mutation; dbSNP:rs969930693 | |
| |
| | Natural variant | VAR_045526 | 329 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045527 | 330 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_047212 | 330 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045528 | 330 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045529 | 331 | in sporadic cancers; somatic mutation; dbSNP:rs11575996 | |
| |
| | Natural variant | VAR_045530 | 331 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045531 | 331 | in sporadic cancers; somatic mutation; dbSNP:rs1064795056 | |
| |
| | Natural variant | VAR_045532 | 332 | in a sporadic cancer; somatic mutation | |
| |
| | Mutagenesis | | 333-337 | Reduced methylation by PRMT5. Reduced nuclear localization. Decreased binding to promoters of target genes. Reduced transcriptional activity. Decrease in cell cycle arrest. | |
| |
| | Natural variant | VAR_006040 | 334 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045533 | 334 | in a sporadic cancer; somatic mutation; dbSNP:rs730882028 | |
| |
| | Natural variant | VAR_045534 | 335 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045535 | 335 | in a sporadic cancer; somatic mutation; dbSNP:rs771939956 | |
| |
| | Natural variant | VAR_045536 | 335 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_006041 | 337 | in LFS; germline mutation and in sporadic cancers; somatic mutation; impaired ability to tetramerize and undergo liquid-liquid phase separation; dbSNP:rs587782529 | |
| |
| | Natural variant | VAR_035016 | 337 | in LFS; germline mutation and in sporadic cancers; somatic mutation; impaired ability to tetramerize and undergo liquid-liquid phase separation; dbSNP:rs121912664 | |
| |
| | Natural variant | VAR_045537 | 337 | in sporadic cancers; somatic mutation; dbSNP:rs121912664 | |
| |
| | Natural variant | VAR_045538 | 337 | in sporadic cancers; somatic mutation; impaired ability to tetramerize and undergo liquid-liquid phase separation; dbSNP:rs121912664 | |
| |
| | Natural variant | VAR_045539 | 338 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045540 | 338 | in a sporadic cancer; somatic mutation; dbSNP:rs150293825 | |
| |
| | Natural variant | VAR_022316 | 339 | in a sporadic cancer; somatic mutation; dbSNP:rs17882252 | |
| |
| | Natural variant | VAR_045541 | 339 | in a sporadic cancer; somatic mutation; dbSNP:rs17882252 | |
| |
| | Natural variant | VAR_045542 | 341 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045543 | 342 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045544 | 342 | in sporadic cancers; somatic mutation; dbSNP:rs375338359 | |
| |
| | Natural variant | VAR_047213 | 342 | in sporadic cancers; somatic mutation; dbSNP:rs375338359 | |
| |
| | Natural variant | VAR_045545 | 343 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045546 | 344 | in LFS; germline mutation and in a sporadic cancer; somatic mutation; impaired ability to tetramerize and undergo liquid-liquid phase separation; dbSNP:rs121912662 | |
| |
| | Natural variant | VAR_045547 | 344 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045548 | 346 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045549 | 347 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045550 | 347 | in sporadic cancers; somatic mutation | |
| |
| | Natural variant | VAR_045551 | 348 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045552 | 348 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045553 | 349 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045554 | 352 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045555 | 353 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045556 | 354 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045557 | 354 | in a sporadic cancer; somatic mutation; dbSNP:rs755394212 | |
| |
| | Natural variant | VAR_047214 | 354 | in sporadic cancers; somatic mutation; dbSNP:rs752142489 | |
| |
| | Natural variant | VAR_045558 | 356 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045559 | 356 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045560 | 358 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045561 | 358 | in a sporadic cancer; somatic mutation; dbSNP:rs587782237 | |
| |
| | Mutagenesis | | 359 | Abolishes binding to USP7. | |
| |
| | Natural variant | VAR_045562 | 360 | in dbSNP:rs35993958 | |
| |
| | Natural variant | VAR_045563 | 360 | in a sporadic cancer; somatic mutation; dbSNP:rs35993958 | |
| |
| | Mutagenesis | | 361 | Abolishes binding to USP7. | |
| |
| | Mutagenesis | | 362 | Abolishes binding to USP7. | |
| |
| | Natural variant | VAR_045564 | 363 | in a sporadic cancer; somatic mutation; dbSNP:rs876660285 | |
| |
| | Natural variant | VAR_045565 | 364 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045566 | 364 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045567 | 364 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_047215 | 365 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045568 | 365 | in a familial cancer not matching LFS; germline mutation and in a sporadic cancer; somatic mutation; dbSNP:rs267605075 | |
| |
| | Natural variant | VAR_022317 | 366 | in a familial cancer not matching LFS; germline mutation and in sporadic cancers; somatic mutation; dbSNP:rs17881470 | |
| |
| | Natural variant | VAR_045569 | 370 | in a sporadic cancer; somatic mutation | |
| |
| | Mutagenesis | | 370 | Induces a decrease in methylation by SMYD2. | |
| |
| | Mutagenesis | | 372 | Induces a decrease in protein stabilization. | |
| |
| | Mutagenesis | | 373 | Abolishes dimethylation by EHMT1 and EHMT2. | |
| |
| | Natural variant | VAR_045570 | 376 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045571 | 376 | in a sporadic cancer; somatic mutation | |
| |
| | Natural variant | VAR_045572 | 379 | in sporadic cancers; somatic mutation; dbSNP:rs863224682 | |
| |
| | Mutagenesis | | 381 | Mimics acetylation, leading to increased stability. | |
| |
| | Mutagenesis | | 381 | Decreased acetylation. | |
| |
| | Mutagenesis | | 382 | Abolishes acetylation by CREBBP. | |
| |
| | Mutagenesis | | 382 | Abolishes monomethylation by KMT5A. | |
| |
| | Mutagenesis | | 383 | Abolishes S-315 phosphorylation by CDK2/cyclin A. | |
| |
| | Natural variant | VAR_045573 | 385 | in a sporadic cancer; somatic mutation; dbSNP:rs1555524094 | |
| |
| | Mutagenesis | | 385 | Reduced SUMO1 conjugation. | |
| |
| | Mutagenesis | | 386 | Abolishes SUMO1 conjugation, in vitro and in vivo. | |
| |
| | Mutagenesis | | 387 | No effect SUMO1 conjugation. | |
| |
| | Mutagenesis | | 388 | Abolishes SUMO1 conjugation. | |
| |
| | Natural variant | VAR_045574 | 389 | in a sporadic cancer; somatic mutation; dbSNP:rs587783064 | |
| |
| | Natural variant | VAR_045575 | 392 | in a sporadic cancer; somatic mutation | |
| |
| | Mutagenesis | | 392 | Mimics phosphorylation; promotes ability to undergo liquid-liquid phase separation. | |
| |
| | Mutagenesis | | 392 | Abolished ability to undergo liquid-liquid phase separation. | |
| |
Keywords
Disease
#Disease variant
#Li-Fraumeni syndrome
#Tumor suppressor
Organism-specific databases
DisGeNET
7157
GeneReviews
TP53
MIM
133239phenotype
151623phenotype
191170gene+phenotype
202300phenotype
211980phenotype
MalaCards
TP53
OpenTargets
ENSG00000141510
Orphanet
1501Adrenocortical carcinoma
210159Adult hepatocellular carcinoma
99756Alveolar rhabdomyosarcoma
67038B-cell chronic lymphocytic leukemia
585909B-lymphoblastic leukemia/lymphoma with t(9;22)(q34.1;q11.2)
PharmGKB
PA36679
Miscellaneous
Pharos
P04637Tchem
Chemistry
ChEMBL
CHEMBL4096
DrugBank
DB083631-(9-ethyl-9H-carbazol-3-yl)-N-methylmethanamine
DB130104-isothioureidobutyronitrile
DB00945Acetylsalicylic acid
DB17549Alrizomadlin
DB05404AZD 3355
Genetic variation databases
BioMuta
TP53
DMDM
269849759
ClinGen
HGNC:11998
GenCC
HGNC:11998
Expression
Tissue specificity
Ubiquitous. Isoforms are expressed in a wide range of normal tissues but in a tissue-dependent manner. Isoform 2 is expressed in most normal tissues but is not detected in brain, lung, prostate, muscle, fetal brain, spinal cord and fetal liver. Isoform 3 is expressed in most normal tissues but is not detected in lung, spleen, testis, fetal brain, spinal cord and fetal liver. Isoform 7 is expressed in most normal tissues but is not detected in prostate, uterus, skeletal muscle and breast. Isoform 8 is detected only in colon, bone marrow, testis, fetal brain and intestine. Isoform 9 is expressed in most normal tissues but is not detected in brain, heart, lung, fetal liver, salivary gland, breast or intestine.
Induction
Up-regulated in response to DNA damage. Isoform 2 is not induced in tumor cells in response to stress.
Gene expression databases
Bgee
ENSG00000141510Expressed in ventricular zone and 144 other cell types or tissues
ExpressionAtlas
P04637baseline and differential
Organism-specific databases
HPA
ENSG00000141510Low tissue specificity
Interaction
Subunit
Forms homodimers and homotetramers (PubMed:19011621).
Binds DNA as a homotetramer (PubMed:36108750).
Interacts with AXIN1. Probably part of a complex consisting of TP53, HIPK2 and AXIN1 (By similarity).
Interacts with histone acetyltransferases EP300 and methyltransferases HRMT1L2 and CARM1, and recruits them to promoters. Interacts (via C-terminus) with TAF1; when TAF1 is part of the TFIID complex. Interacts with ING4; this interaction may be indirect. Found in a complex with CABLES1 and TP73. Interacts with HIPK1, HIPK2, and TP53INP1. Interacts with WWOX. May interact with HCV core protein. Interacts with USP7 and SYVN1. Interacts with HSP90AB1. Interacts with CHD8; leading to recruit histone H1 and prevent transactivation activity (By similarity).
Interacts with ARMC10, CDKN2AIP, NUAK1, STK11/LKB1, UHRF2 and E4F1. Interacts with YWHAZ; the interaction enhances TP53 transcriptional activity. Phosphorylation of YWHAZ on 'Ser-58' inhibits this interaction. Interacts (via DNA-binding domain) with MAML1 (via N-terminus). Interacts with MKRN1. Interacts with PML (via C-terminus). Interacts with MDM2; leading to ubiquitination and proteasomal degradation of TP53. Directly interacts with FBXO42; leading to ubiquitination and degradation of TP53. Interacts (phosphorylated at Ser-15 by ATM) with the phosphatase PP2A-PPP2R5C holoenzyme; regulates stress-induced TP53-dependent inhibition of cell proliferation. Interacts with PPP2R2A. Interacts with AURKA, DAXX, BRD7 and TRIM24. Interacts (when monomethylated at Lys-382) with L3MBTL1. Isoform 1 interacts with isoform 2 and with isoform 4. Interacts with GRK5. Binds to the CAK complex (CDK7, cyclin H and MAT1) in response to DNA damage. Interacts with CDK5 in neurons. Interacts with AURKB, SETD2, UHRF2 and NOC2L. Interacts (via N-terminus) with PTK2/FAK1; this promotes ubiquitination by MDM2. Interacts with PTK2B/PYK2; this promotes ubiquitination by MDM2. Interacts with PRKCG. Interacts with PPIF; the association implicates preferentially tetrameric TP53, is induced by oxidative stress and is impaired by cyclosporin A (CsA). Interacts with SNAI1; the interaction induces SNAI1 degradation via MDM2-mediated ubiquitination and inhibits SNAI1-induced cell invasion. Interacts with UBC9. Interacts with ZNF385B; the interaction is direct. Interacts (via DNA-binding domain) with ZNF385A; the interaction is direct and enhances p53/TP53 transactivation functions on cell-cycle arrest target genes, resulting in growth arrest. Interacts with ANKRD2. Interacts with RFFL and RNF34; involved in p53/TP53 ubiquitination. Interacts with MTA1 and COP1. Interacts with CCAR2 (via N-terminus). Interacts with MORC3 (PubMed:17332504).
Interacts (via C-terminus) with POU4F2 isoform 1 (via C-terminus) (PubMed:17145718).
Interacts (via oligomerization region) with NOP53; the interaction is direct and may prevent the MDM2-mediated proteasomal degradation of TP53 (PubMed:22522597).
Interacts with AFG1L; mediates mitochondrial translocation of TP53 (PubMed:27323408).
Interacts with UBD (PubMed:25422469).
Interacts with TAF6 isoform 1 and isoform 4 (PubMed:20096117).
Interacts with C10orf90/FATS; the interaction inhibits binding of TP53 and MDM2 (By similarity).
Interacts with NUPR1; interaction is stress-dependent (PubMed:18690848).
Forms a complex with EP300 and NUPR1; this complex binds CDKN1A promoter leading to transcriptional induction of CDKN1A (PubMed:18690848).
Interacts with PRMT5 in response to DNA damage; the interaction is TTC5/STRAP dependent (PubMed:19011621).
Interacts with PPP1R13L (via SH3 domain and ANK repeats); the interaction inhibits pro-apoptotic activity of p53/TP53 (PubMed:12524540).
Interacts with PPP1R13B/ASPP1 and TP53BP2/ASPP2; the interactions promotes pro-apoptotic activity (PubMed:12524540).
When phosphorylated at Ser-15, interacts with DDX3X and gamma-tubulin (PubMed:28842590).
Interacts with KAT7/HBO1; leading to inhibit histone acetyltransferase activity of KAT7/HBO1 (PubMed:17954561).
Interacts (via N-terminus) with E3 ubiquitin-protein ligase MUL1; the interaction results in ubiquitination of cytoplasmic TP53 at Lys-24 and subsequent proteasomal degradation (PubMed:21597459).
Interacts with S100A4; this interaction promotes TP53 degradation (PubMed:23752197, PubMed:32442400).
Interacts with BANP (By similarity).
Interacts with TTC5/STRAP; the interaction may result in increased mitochondrial-dependent apoptosis (PubMed:25168243).
Interacts with NQO1; this interaction is NADH-dependent, stabilizes TP53 in response to oxidative stress and protects it from ubiquitin-independent degradation by the 20S proteasome (PubMed:15687255).
Interacts with DAZAP2 at TP53 target gene promoters; the interaction is triggered by DNA damage and leads to modulation of the expression of a subset of TP53 target genes, reducing DNA damage-induced cell death by limiting the expression of cell death-mediating TP53 target genes (PubMed:33591310).
Interacts (via N-terminus) with ZNF768 (via zinc-finger domains); interaction might be facilitated by TP53 oligomerization state (PubMed:34404770).
Forms a ternary complex with ALDOB and G6PD; this interaction is direct. ALDOB stabilizes the complex inhibiting G6PD activity and keeping oxidative pentose phosphate metabolism in check. Interacts with MORN3; the interactions mediate post-transcriptional modifications of TP53 by MDM2 and SIRT1 (PubMed:29681526).
Interacts with HSPA9/MOT-2; the interaction promotes the degradation of TP53 (PubMed:24625977).
Interacts with FBXO22; this interaction promotes TP53 proteasomal degradation (PubMed:26868148).
(Microbial infection) Interacts with cancer-associated/HPV E6 viral proteins leading to ubiquitination and degradation of TP53 giving a possible model for cell growth regulation. This complex formation requires an additional factor, E6-AP, which stably associates with TP53 in the presence of E6.
(Microbial infection) Interacts with human cytomegalovirus/HHV-5 protein UL123.
(Microbial infection) Interacts (via N-terminus) with human adenovirus 5 E1B-55K protein; this interaction leads to the inhibition of TP53 function and/or its degradation.
(Microbial infection) Interacts with Kaposi's sarcoma-associated herpesvirus/HHV-8 protein ORF45; this interaction results in the cytoplasmic localization of TP53 thereby decreasing its transcriptional activity.
Binary interactions
| | Type | Entry 1 | Entry 2 | Number of experiments | IntAct |
--- --- --- |
| | BINARY | P04637 | ABL1 P00519 | 2 | EBI-366083, EBI-375543 |
| | BINARY | P04637 | AIMP2 Q13155 | 6 | EBI-366083, EBI-745226 |
| | BINARY | P04637 | ARIH2 O95376 | 5 | EBI-366083, EBI-711158 |
| | BINARY | P04637 | ARRB1 P49407 | 5 | EBI-366083, EBI-743313 |
| | XENO | P04637 | Arrb1 P29066 | 3 | EBI-366083, EBI-4303019 |
| | BINARY | P04637 | ASH2L Q9UBL3 | 8 | EBI-366083, EBI-540797 |
| | BINARY | P04637 | ATG7 O95352-2 | 4 | EBI-366083, EBI-15980880 |
| | BINARY | P04637 | AXIN1 O15169 | 4 | EBI-366083, EBI-710484 |
| | BINARY | P04637 | BANP Q8N9N5 | 3 | EBI-366083, EBI-744695 |
| | BINARY | P04637 | BCL2 P10415 | 5 | EBI-366083, EBI-77694 |
| | BINARY | P04637 | BCL2L1 Q07817-1 | 26 | EBI-366083, EBI-287195 |
| | BINARY | P04637 | BHLHE40 O14503 | 11 | EBI-366083, EBI-711810 |
| | BINARY | P04637 | BRCA2 P51587 | 7 | EBI-366083, EBI-79792 |
| | BINARY | P04637 | BRD7 Q9NPI1 | 9 | EBI-366083, EBI-711221 |
| | BINARY | P04637 | BTBD2 Q9BX70 | 2 | EBI-366083, EBI-710091 |
| | BINARY | P04637 | BTRC Q9Y297 | 2 | EBI-366083, EBI-307461 |
| | XENO | P04637 | Cables1 Q9ESJ1 | 3 | EBI-366083, EBI-604411 |
| | BINARY | P04637 | CASP6 P55212 | 3 | EBI-366083, EBI-718729 |
| | BINARY | P04637 | CCDC106 Q9BWC9 | 3 | EBI-366083, EBI-711501 |
| | BINARY | P04637 | CCT5 P48643 | 3 | EBI-366083, EBI-355710 |
| | BINARY | P04637 | CDKN1A P38936 | 5 | EBI-366083, EBI-375077 |
| | BINARY | P04637 | CEBPB P17676 | 4 | EBI-366083, EBI-969696 |
| | BINARY | P04637 | CREBBP Q92793 | 17 | EBI-366083, EBI-81215 |
| | XENO | P04637 | Crebbp P45481 | 10 | EBI-366083, EBI-296306 |
| | BINARY | P04637 | CSE1L P55060 | 5 | EBI-366083, EBI-286709 |
| | BINARY | P04637 | CUL7 Q14999 | 18 | EBI-366083, EBI-308606 |
| | BINARY | P04637 | CUL9 Q8IWT3 | 15 | EBI-366083, EBI-311123 |
| | BINARY | P04637 | CXXC1 Q9P0U4 | 7 | EBI-366083, EBI-949911 |
| | BINARY | P04637 | DAXX Q9UER7 | 12 | EBI-366083, EBI-77321 |
| | BINARY | P04637 | DDX17 Q92841 | 3 | EBI-366083, EBI-746012 |
| | BINARY | P04637 | DDX5 P17844 | 6 | EBI-366083, EBI-351962 |
| | BINARY | P04637 | DROSHA Q9NRR4 | 5 | EBI-366083, EBI-528367 |
| | BINARY | P04637 | DUSP26 Q9BV47 | 9 | EBI-366083, EBI-2924519 |
| | BINARY | P04637 | DVL2 O14641 | 6 | EBI-366083, EBI-740850 |
| | XENO | P04637 | E6 P03126 | 5 | EBI-366083, EBI-1177242 |
| | XENO | P04637 | E6 P06463 | 3 | EBI-366083, EBI-1186926 |
| | BINARY | P04637 | EP300 Q09472 | 21 | EBI-366083, EBI-447295 |
| | BINARY | P04637 | ETS2 P15036 | 4 | EBI-366083, EBI-1646991 |
| | BINARY | P04637 | FBXO11 Q86XK2 | 4 | EBI-366083, EBI-1047804 |
| | BINARY | P04637 | FLNA P21333-2 | 3 | EBI-366083, EBI-9641086 |
| | BINARY | P04637 | FXR1 P51114 | 2 | EBI-366083, EBI-713291 |
| | BINARY | P04637 | GSK3B P49841 | 3 | EBI-366083, EBI-373586 |
| | BINARY | P04637 | GTF2H1 P32780 | 12 | EBI-366083, EBI-715539 |
| | BINARY | P04637 | HDAC1 Q13547 | 7 | EBI-366083, EBI-301834 |
| | BINARY | P04637 | HIPK1 Q86Z02 | 2 | EBI-366083, EBI-692891 |
| | BINARY | P04637 | HMGB1 P09429 | 9 | EBI-366083, EBI-389432 |
| | BINARY | P04637 | HNF4A P41235 | 3 | EBI-366083, EBI-1049011 |
| | BINARY | P04637 | HNRNPK P61978 | 2 | EBI-366083, EBI-304185 |
| | BINARY | P04637 | HNRNPK P61978-2 | 2 | EBI-366083, EBI-7060731 |
| | XENO | P04637 | HSP82 P02829 | 8 | EBI-366083, EBI-8659 |
| | BINARY | P04637 | HSPA9 P38646 | 6 | EBI-366083, EBI-354932 |
| | BINARY | P04637 | HSPB1 P04792 | 3 | EBI-366083, EBI-352682 |
| | BINARY | P04637 | HTT P42858 | 19 | EBI-366083, EBI-466029 |
| | BINARY | P04637 | HUWE1 Q7Z6Z7 | 3 | EBI-366083, EBI-625934 |
| | BINARY | P04637 | IFI16 Q16666-2 | 6 | EBI-366083, EBI-6273540 |
| | XENO | P04637 | Ifi205b Q08619 | 2 | EBI-366083, EBI-8064290 |
| | BINARY | P04637 | IKBKB O14920 | 2 | EBI-366083, EBI-81266 |
| | BINARY | P04637 | IP6K2 Q9UHH9 | 4 | EBI-366083, EBI-747509 |
| | BINARY | P04637 | JMJD6 Q6NYC1 | 7 | EBI-366083, EBI-8464037 |
| | BINARY | P04637 | KAT5 Q92993 | 3 | EBI-366083, EBI-399080 |
| | BINARY | P04637 | KAT8 Q9H7Z6 | 2 | EBI-366083, EBI-896414 |
| | BINARY | P04637 | KDM1A O60341-1 | 6 | EBI-366083, EBI-15599570 |
| | BINARY | P04637 | KMT2E Q8IZD2 | 4 | EBI-366083, EBI-2689959 |
| | BINARY | P04637 | LAMP2 P13473-2 | 3 | EBI-366083, EBI-21591415 |
| | BINARY | P04637 | MAGEA2B P43356 | 7 | EBI-366083, EBI-5650739 |
| | BINARY | P04637 | MAGEB18 Q96M61 | 3 | EBI-366083, EBI-741835 |
| | BINARY | P04637 | MAGEC2 Q9UBF1 | 3 | EBI-366083, EBI-5651487 |
| | BINARY | P04637 | MAP1B P46821 | 6 | EBI-366083, EBI-764611 |
| | BINARY | P04637 | MAPK11 Q15759 | 2 | EBI-366083, EBI-298304 |
| | BINARY | P04637 | MAPKAPK5 Q8IW41 | 2 | EBI-366083, EBI-1201460 |
| | BINARY | P04637 | MDM2 Q00987 | 112 | EBI-366083, EBI-389668 |
| | BINARY | P04637 | MDM4 O15151 | 22 | EBI-366083, EBI-398437 |
| | BINARY | P04637 | MKRN1 Q9UHC7 | 8 | EBI-366083, EBI-373524 |
| | BINARY | P04637 | MPDZ O75970 | 3 | EBI-366083, EBI-821405 |
| | BINARY | P04637 | MT1A P04731 | 3 | EBI-366083, EBI-8045030 |
| | BINARY | P04637 | NCL P19338 | 2 | EBI-366083, EBI-346967 |
| | BINARY | P04637 | NCOR2 Q9Y618 | 7 | EBI-366083, EBI-80830 |
| | BINARY | P04637 | NDN Q99608 | 4 | EBI-366083, EBI-718177 |
| | BINARY | P04637 | NFYA P23511 | 11 | EBI-366083, EBI-389739 |
| | BINARY | P04637 | NFYB P25208 | 6 | EBI-366083, EBI-389728 |
| | BINARY | P04637 | NOC2L Q9Y3T9 | 8 | EBI-366083, EBI-751547 |
| | BINARY | P04637 | NOL3 O60936 | 3 | EBI-366083, EBI-740992 |
| | BINARY | P04637 | NPM1 P06748 | 6 | EBI-366083, EBI-78579 |
| | BINARY | P04637 | NPM1 P06748-1 | 3 | EBI-366083, EBI-354150 |
| | BINARY | P04637 | NR0B2 Q15466 | 3 | EBI-366083, EBI-3910729 |
| | BINARY | P04637 | NR4A1 P22736 | 6 | EBI-366083, EBI-721550 |
| | BINARY | P04637 | NRDC O43847 | 6 | EBI-366083, EBI-2371631 |
| | XENO | P04637 | NSP1 P89055 | 6 | EBI-366083, EBI-9522973 |
| | BINARY | P04637 | NUAK1 O60285 | 5 | EBI-366083, EBI-1046789 |
| | BINARY | P04637 | NUMB P49757 | 5 | EBI-366083, EBI-915016 |
| | BINARY | P04637 | OTUB1 Q96FW1 | 8 | EBI-366083, EBI-1058491 |
| | XENO | P04637 | P03070 | 22 | EBI-366083, EBI-617698 |
| | XENO | P04637 | PRO_0000037536 P26663 | 9 | EBI-366083, EBI-6838571 |
| | BINARY | P04637 | PARD3 Q8TEW0 | 3 | EBI-366083, EBI-81968 |
| | BINARY | P04637 | PARP1 P09874 | 3 | EBI-366083, EBI-355676 |
| | BINARY | P04637 | PBK Q96KB5 | 7 | EBI-366083, EBI-536853 |
| | BINARY | P04637 | PEX26 Q7Z412 | 3 | EBI-366083, EBI-752057 |
| | BINARY | P04637 | PHB1 P35232 | 6 | EBI-366083, EBI-354213 |
| | BINARY | P04637 | PIAS1 O75925 | 4 | EBI-366083, EBI-629434 |
| | BINARY | P04637 | PIAS4 Q8N2W9 | 2 | EBI-366083, EBI-473160 |
| | BINARY | P04637 | PIK3R3 Q92569 | 5 | EBI-366083, EBI-79893 |
| | BINARY | P04637 | PIN1 Q13526 | 12 | EBI-366083, EBI-714158 |
| | BINARY | P04637 | PLK1 P53350 | 6 | EBI-366083, EBI-476768 |
| | BINARY | P04637 | PML P29590 | 4 | EBI-366083, EBI-295890 |
| | BINARY | P04637 | PMP22 A0A6Q8PF08 | 3 | EBI-366083, EBI-50433196 |
| | BINARY | P04637 | PPIF P30405 | 4 | EBI-366083, EBI-5544229 |
| | BINARY | P04637 | PPP1CC P36873-1 | 2 | EBI-366083, EBI-356289 |
| | BINARY | P04637 | PPP1R13L Q8WUF5 | 12 | EBI-366083, EBI-5550163 |
| | BINARY | P04637 | PPP2R1A P30153 | 3 | EBI-366083, EBI-302388 |
| | BINARY | P04637 | PPP2R5C Q13362 | 4 | EBI-366083, EBI-1266156 |
| | BINARY | P04637 | PRKCD Q05655 | 4 | EBI-366083, EBI-704279 |
| | BINARY | P04637 | PSME3 P61289 | 7 | EBI-366083, EBI-355546 |
| | BINARY | P04637 | PTK2 Q05397 | 13 | EBI-366083, EBI-702142 |
| | BINARY | P04637 | Q7L7W2 | 2 | EBI-366083, EBI-7210801 |
| | XENO | P04637 | Q8QW27 | 2 | EBI-366083, EBI-6863726 |
| | BINARY | P04637 | RAD51 Q06609 | 2 | EBI-366083, EBI-297202 |
| | BINARY | P04637 | RBPJ Q06330 | 5 | EBI-366083, EBI-632552 |
| | BINARY | P04637 | RCHY1 Q96PM5 | 13 | EBI-366083, EBI-947779 |
| | BINARY | P04637 | RFWD3 Q6PCD5 | 5 | EBI-366083, EBI-2129159 |
| | BINARY | P04637 | RING1 Q06587 | 7 | EBI-366083, EBI-752313 |
| | BINARY | P04637 | RPS3 P23396 | 4 | EBI-366083, EBI-351193 |
| | BINARY | P04637 | RUVBL1 Q9Y265 | 10 | EBI-366083, EBI-353675 |
| | BINARY | P04637 | RYBP Q8N488 | 4 | EBI-366083, EBI-752324 |
| | BINARY | P04637 | S100A1 P23297 | 3 | EBI-366083, EBI-743686 |
| | BINARY | P04637 | S100A11 P31949 | 2 | EBI-366083, EBI-701862 |
| | BINARY | P04637 | S100A2 P29034 | 4 | EBI-366083, EBI-752230 |
| | BINARY | P04637 | S100A3 P33764 | 2 | EBI-366083, EBI-1044747 |
| | BINARY | P04637 | S100A4 P26447 | 9 | EBI-366083, EBI-717058 |
| | BINARY | P04637 | S100A5 P33763 | 2 | EBI-366083, EBI-7211732 |
| | BINARY | P04637 | S100A6 P06703 | 3 | EBI-366083, EBI-352877 |
| | BINARY | P04637 | S100B P04271 | 3 | EBI-366083, EBI-458391 |
| | BINARY | P04637 | S100P P25815 | 2 | EBI-366083, EBI-743700 |
| | BINARY | P04637 | SAFB Q15424 | 5 | EBI-366083, EBI-348298 |
| | BINARY | P04637 | SETD7 Q8WTS6 | 11 | EBI-366083, EBI-1268586 |
| | BINARY | P04637 | SFN P31947 | 5 | EBI-366083, EBI-476295 |
| | BINARY | P04637 | SIN3A Q96ST3 | 2 | EBI-366083, EBI-347218 |
| | BINARY | P04637 | SIRT1 Q96EB6 | 18 | EBI-366083, EBI-1802965 |
| | XENO | P04637 | Sirt1 Q923E4 | 4 | EBI-366083, EBI-1802585 |
| | BINARY | P04637 | SMAD2 Q15796 | 7 | EBI-366083, EBI-1040141 |
| | BINARY | P04637 | SMYD2 Q9NRG4 | 6 | EBI-366083, EBI-1055671 |
| | XENO | P04637 | Smyd2 Q8R5A0 | 3 | EBI-366083, EBI-15612527 |
| | BINARY | P04637 | SNAI1 O95863 | 2 | EBI-366083, EBI-1045459 |
| | BINARY | P04637 | SOX4 Q06945 | 4 | EBI-366083, EBI-6672525 |
| | BINARY | P04637 | SP1 P08047 | 3 | EBI-366083, EBI-298336 |
| | BINARY | P04637 | SREBF2 Q12772 | 3 | EBI-366083, EBI-465059 |
| | BINARY | P04637 | SRPK1 Q96SB4 | 3 | EBI-366083, EBI-539478 |
| | BINARY | P04637 | STIP1 P31948 | 4 | EBI-366083, EBI-1054052 |
| | BINARY | P04637 | SUMO1 P63165 | 3 | EBI-366083, EBI-80140 |
| | BINARY | P04637 | SYVN1 Q86TM6 | 5 | EBI-366083, EBI-947849 |
| | BINARY | P04637 | TAF9B Q9HBM6 | 2 | EBI-366083, EBI-751601 |
| | BINARY | P04637 | TBP P20226 | 2 | EBI-366083, EBI-355371 |
| | BINARY | P04637 | TCF4 P15884 | 2 | EBI-366083, EBI-533224 |
| | XENO | P04637 | TFB1 P32776 | 7 | EBI-366083, EBI-19146 |
| | BINARY | P04637 | TOE1 Q96GM8 | 3 | EBI-366083, EBI-717460 |
| | BINARY | P04637 | TP53 P04637 | 33 | EBI-366083, EBI-366083 OR (idA:"-" AND idB:P04637)#interactor) |
| | BINARY | P04637 | TP53BP1 Q12888 | 8 | EBI-366083, EBI-396540 |
| | BINARY | P04637 | TP53BP1 Q12888-1 | 17 | EBI-366083, EBI-8022649 |
| | BINARY | P04637 | TP53BP2 Q13625 | 9 | EBI-366083, EBI-77642 |
| | BINARY | P04637 | TP63 Q9H3D4 | 5 | EBI-366083, EBI-2337775 |
| | XENO | P04637 | Tp63 O88898 | 2 | EBI-366083, EBI-2338025 |
| | BINARY | P04637 | TPT1 P13693 | 7 | EBI-366083, EBI-1783169 |
| | BINARY | P04637 | TRIM24 O15164 | 3 | EBI-366083, EBI-2130378 |
| | BINARY | P04637 | TWIST1 Q15672 | 10 | EBI-366083, EBI-1797287 |
| | XENO | P04637 | Twist1 P26687 | 4 | EBI-366083, EBI-6123119 |
| | BINARY | P04637 | UBC P0CG48 | 15 | EBI-366083, EBI-3390054 |
| | BINARY | P04637 | UBE2I P63279 | 3 | EBI-366083, EBI-80168 |
| | BINARY | P04637 | UBE3A Q05086 | 6 | EBI-366083, EBI-954357 |
| | BINARY | P04637 | UCHL1 P09936 | 3 | EBI-366083, EBI-714860 |
| | BINARY | P04637 | UHRF2 Q96PU4 | 3 | EBI-366083, EBI-625304 |
| | BINARY | P04637 | USP42 Q9H9J4 | 2 | EBI-366083, EBI-2513638 |
| | BINARY | P04637 | USP42 Q9H9J4-2 | 2 | EBI-366083, EBI-9118105 |
| | BINARY | P04637 | USP7 Q93009 | 19 | EBI-366083, EBI-302474 |
| | BINARY | P04637 | VDR P11473 | 6 | EBI-366083, EBI-286357 |
| | BINARY | P04637 | VRK1 Q99986 | 11 | EBI-366083, EBI-1769146 |
| | BINARY | P04637 | WRN Q14191 | 5 | EBI-366083, EBI-368417 |
| | BINARY | P04637 | WWOX Q9NZC7 | 2 | EBI-366083, EBI-4320739 |
| | BINARY | P04637 | XPO1 O14980 | 3 | EBI-366083, EBI-355867 |
| | BINARY | P04637 | XRCC6 P12956 | 2 | EBI-366083, EBI-353208 |
| | BINARY | P04637 | YWHAG P61981 | 5 | EBI-366083, EBI-359832 |
| | BINARY | P04637 | YWHAZ P63104 | 3 | EBI-366083, EBI-347088 |
| | BINARY | P04637 | ZNF420 Q8TAQ5 | 4 | EBI-366083, EBI-3923307 |
| | XENO | P04637 | znf585b.S Q9PST7 | 3 | EBI-366083, EBI-1782562 |
| | BINARY | P04637-1 | DDX5 P17844 | 2 | EBI-3895849, EBI-351962 |
| | BINARY | P04637-7 | DDX5 P17844 | 2 | EBI-3895873, EBI-351962 |
Complex viewer
View interactors in UniProtKB
View CPX-6093 in Complex Portal
Protein-protein interaction databases
BioGRID
1130102563 interactors
CORUM
P04637
ComplexPortal
CPX-6093p53-MDM2-MDM4 transcriptional regulation complex
CPX-663p53-MDM4 transcriptional regulation complex
CPX-759p53-MDM2 transcriptional regulation complex
DIP
DIP-368N
ELM
P04637
FunCoup
P046373248 interactors
IntAct
P04637637 interactors
MINT
P04637
STRING
9606.ENSP00000269305
Chemistry
BindingDB
P04637
Miscellaneous
RNAct
P04637protein
Structure
3D structure databases
AlphaFoldDB
P04637
BMRB
P04637
EMDB
EMD-0378
EMD-17265
EMD-17266
EMD-18809
EMD-18810
SMR
P04637
ModBase
Search…
PDBe-KB
Search…
Miscellaneous
EvolutionaryTrace
P04637
Family & Domains
Features
Showing features for region, motif, compositional bias.
| | Type | ID | Position(s) | Description | |
--- --- --- |
| | Region | | 1-44 | Transcription activation (acidic) | |
| |
| | Region | | 1-83 | Interaction with HRMT1L2 | |
| |
| | Region | | 1-320 | Interaction with CCAR2 | |
| |
| | Motif | | 17-25 | TADI | |
| |
| | Motif | | 48-56 | TADII | |
| |
| | Region | | 50-96 | Disordered | |
| |
| | Region | | 66-110 | Interaction with WWOX | |
| |
| | Compositional bias | | 69-90 | Pro residues | |
| |
| | Region | | 100-300 | Required for interaction with ZNF385A | |
| |
| | Region | | 100-370 | Interaction with HIPK1 | |
| |
| | Region | | 113-236 | Required for interaction with FBXO42 | |
| |
| | Region | | 116-292 | Interaction with AXIN1 | |
| |
| | Region | | 241-248 | Interaction with the 53BP2 SH3 domain | |
| |
| | Region | | 256-294 | Interaction with E4F1 | |
| |
| | Region | | 273-280 | Interaction with DNA | |
| |
| | Compositional bias | | 282-299 | Basic and acidic residues | |
| |
| | Region | | 282-325 | Disordered | |
| |
| | Region | | 300-393 | Interaction with CARM1 | |
| |
| | Motif | | 305-321 | Bipartite nuclear localization signal | |
| |
| | Region | | 319-360 | Interaction with HIPK2 | |
| |
| | Region | | 325-356 | Oligomerization | |
| |
| | Motif | | 339-350 | Nuclear export signal | |
| |
| | Region | | 351-393 | Disordered | |
| |
| | Region | | 359-363 | Interaction with USP7 | |
| |
| | Compositional bias | | 365-384 | Basic residues | |
| |
| | Region | | 368-387 | Basic (repression of DNA-binding) | |
| |
| | Motif | | 370-372 | [KR]-[STA]-K motif | |
| |
| | Region | | 374-393 | Interaction with MORN3 | |
| |
Domain
The N-terminal and C-terminal disordered regions undergo liquid-liquid phase separation (LLPS) following homotetramerization and activation (PubMed:31953488, PubMed:35618207, PubMed:36108750, PubMed:36634798, PubMed:38653238).
Post-translational modifications, such as phosphorylation or lactylation affect the ability to undergo LLPS (PubMed:31953488, PubMed:35618207, PubMed:36634798, PubMed:38653238).
The nuclear export signal acts as a transcriptional repression domain. The TADI and TADII motifs (residues 17 to 25 and 48 to 56) correspond both to 9aaTAD motifs which are transactivation domains present in a large number of yeast and animal transcription factors.
Sequence similarities
Belongs to the p53 family.
Phylogenomic databases
GeneTree
ENSGT00950000183153
InParanoid
P04637
OMA
HKKGEPC
OrthoDB
5915660at2759
PAN-GO
P046373 GO annotations based on evolutionary models
PhylomeDB
P04637
TreeFam
TF106101
eggNOG
ENOG502QVY3Eukaryota
Family and domain databases
View all family and domain features for this entry's canonical sequence in the UniParc Feature Viewer.
CDD
cd08367P53 1 hit
DisProt
DP00086
FunFam
2.60.40.720:FF:000003Cellular tumor antigen p53 1 hit
4.10.170.10:FF:000003Cellular tumor antigen p53 1 hit
Gene3D
2.60.40.7201 hit
6.10.50.201 hit
4.10.170.10p53-like tetramerisation domain 1 hit
IDEAL
IID00015
InterPro
View protein in InterPro
IPR008967p53-like_TF_DNA-bd_sf
IPR012346p53/RUNT-type_TF_DNA-bd_sf
IPR011615p53_DNA-bd
IPR040926p53_TAD2
PANTHER
PTHR11447CELLULAR TUMOR ANTIGEN P53 1 hit
PTHR11447:SF6CELLULAR TUMOR ANTIGEN P53 1 hit
PRINTS
PR00386P53SUPPRESSR
PROSITE
View protein in PROSITE
PS00348P53 1 hit
Pfam
View protein in Pfam
PF00870P53 1 hit
PF08563P53_TAD 1 hit
PF07710P53_tetramer 1 hit
PF18521TAD2 1 hit
SUPFAM
SSF47719p53 tetramerization domain 1 hit
SSF49417p53-like transcription factors 1 hit
MobiDB
Search…
Sequence & Isoforms
Align isoforms (9)
Sequence status
Complete
This entry describes 9 isoforms produced by Alternative promoter usage & Alternative splicing.
P04637-1
This isoform has been chosen as the canonical sequence. All positional information in this entry refers to it. This is also the sequence that appears in the downloadable versions of the entry.
Name
1
Synonyms
p53, p53alpha
See also
sequence in UniParc or sequence clusters in UniRef
Length
393
Mass (Da)
43,653
Last updated
2009-11-24 v4
MD5 Checksum
C133DFCE69F606F20865E9008199F852
MEEPQSDPSVEPPLSQETFSDLWKLLPENNVLSPLPSQAMDDLMLSPDDIEQWFTEDPGPDEAPRMPEAAPPVAPAPAAPTPAAPAPAPSWPLSSSVPSQKTYQGSYGFRLGFLHSGTAKSVTCTYSPALNKMFCQLAKTCPVQLWVDSTPPPGTRVRAMAIYKQSQHMTEVVRRCPHHERCSDSDGLAPPQHLIRVEGNLRVEYLDDRNTFRHSVVVPYEPPEVGSDCTTIHYNYMCNSSCMGGMNRRPILTIITLEDSSGNLLGRNSFEVRVCACPGRDRRTEEENLRKKGEPHHELPPGSTKRALPNNTSSSPQPKKKPLDGEYFTLQIRGRERFEMFRELNEALELKDAQAGKEPGGSRAHSSHLKSKKGQSTSRHKKLMFKTEGPDSD
P04637-2
Name
2
Synonyms
I9RET, p53beta
Note
Expressed in quiescent lymphocytes. Seems to be non-functional. May be produced at very low levels due to a premature stop codon in the mRNA, leading to nonsense-mediated mRNA decay.
See also
sequence in UniParc or sequence clusters in UniRef
Differences from canonical
332-341:
IRGRERFEMF → DQTSFQKENC
342-393: Missing
P04637-3
Name
3
Synonyms
p53gamma
Note
Expressed in quiescent lymphocytes. Seems to be non-functional. May be produced at very low levels due to a premature stop codon in the mRNA, leading to nonsense-mediated mRNA decay.
See also
sequence in UniParc or sequence clusters in UniRef
Differences from canonical
332-346:
IRGRERFEMFRELNE → MLLDLRWCYFLINSS
347-393: Missing
P04637-4
Name
Synonyms
Del40-p53, Del40-p53alpha, p47
See also
sequence in UniParc or sequence clusters in UniRef
Differences from canonical
1-39: Missing
P04637-5
Name
Synonyms
See also
sequence in UniParc or sequence clusters in UniRef
Differences from canonical
1-39: Missing
332-341:
IRGRERFEMF → DQTSFQKENC
342-393: Missing
P04637-6
Name
Synonyms
See also
sequence in UniParc or sequence clusters in UniRef
Differences from canonical
1-39: Missing
332-346:
IRGRERFEMFRELNE → MLLDLRWCYFLINSS
347-393: Missing
P04637-7
Name
Synonyms
Del133-p53, Del133-p53alpha
Note
Produced by alternative promoter usage.
See also
sequence in UniParc or sequence clusters in UniRef
Differences from canonical
1-132: Missing
P04637-8
Name
Synonyms
Note
Produced by alternative promoter usage and alternative splicing.
See also
sequence in UniParc or sequence clusters in UniRef
Differences from canonical
1-132: Missing
332-341:
IRGRERFEMF → DQTSFQKENC
342-393: Missing
P04637-9
Name
Synonyms
Note
Produced by alternative promoter usage and alternative splicing.
See also
sequence in UniParc or sequence clusters in UniRef
Differences from canonical
1-132: Missing
332-346:
IRGRERFEMFRELNE → MLLDLRWCYFLINSS
347-393: Missing
Features
Showing features for alternative sequence, compositional bias.
| | Type | ID | Position(s) | Description | |
--- --- --- |
| | Alternative sequence | VSP_040832 | 1-39 | in isoform 4, isoform 5 and isoform 6 | |
| |
| | Alternative sequence | VSP_040833 | 1-132 | in isoform 7, isoform 8 and isoform 9 | |
| |
| | Compositional bias | | 69-90 | Pro residues | |
| |
| | Compositional bias | | 282-299 | Basic and acidic residues | |
| |
| | Alternative sequence | VSP_006535 | 332-341 | in isoform 2, isoform 5 and isoform 8 | |
| |
| | Alternative sequence | VSP_040560 | 332-346 | in isoform 3, isoform 6 and isoform 9 | |
| |
| | Alternative sequence | VSP_006536 | 342-393 | in isoform 2, isoform 5 and isoform 8 | |
| |
| | Alternative sequence | VSP_040561 | 347-393 | in isoform 3, isoform 6 and isoform 9 | |
| |
| | Compositional bias | | 365-384 | Basic residues | |
| |
Keywords
Coding sequence diversity
#Alternative promoter usage
#Alternative splicing
Technical term
#3D-structure
#Direct protein sequencing
#Proteomics identification
#Reference proteome
Sequence databases
CCDS11118.1 [P04637-1]
CCDS45605.1 [P04637-3]
CCDS45606.1 [P04637-2]
CCDS73966.1 [P04637-7]
CCDS73967.1 [P04637-9]
A25224DNHU53
NP_000537.3NM_000546.5 [P04637-1]
NP_001119584.1NM_001126112.3 [P04637-1]
NP_001119585.1NM_001126113.3 [P04637-3]
NP_001119586.1NM_001126114.3 [P04637-2]
NP_001119587.1NM_001126115.2 [P04637-7]
| | Nucleotide Sequence | Protein Sequence | Molecule Type | Status |
--- ---
| | X02469 EMBL· GenBank· DDBJ | CAA26306.1 EMBL· GenBank· DDBJ | mRNA | |
| | M13121 EMBL· GenBank· DDBJ | AAA59987.1 EMBL· GenBank· DDBJ | Genomic DNA | |
| | M13112 EMBL· GenBank· DDBJ | AAA59987.1 EMBL· GenBank· DDBJ | Genomic DNA | |
| | M13113 EMBL· GenBank· DDBJ | AAA59987.1 EMBL· GenBank· DDBJ | Genomic DNA | |
| | M13114 EMBL· GenBank· DDBJ | AAA59987.1 EMBL· GenBank· DDBJ | Genomic DNA | |
| | M13115 EMBL· GenBank· DDBJ | AAA59987.1 EMBL· GenBank· DDBJ | Genomic DNA | |
| | M13116 EMBL· GenBank· DDBJ | AAA59987.1 EMBL· GenBank· DDBJ | Genomic DNA | |
| | M13117 EMBL· GenBank· DDBJ | AAA59987.1 EMBL· GenBank· DDBJ | Genomic DNA | |
| | M13118 EMBL· GenBank· DDBJ | AAA59987.1 EMBL· GenBank· DDBJ | Genomic DNA | |
| | M13119 EMBL· GenBank· DDBJ | AAA59987.1 EMBL· GenBank· DDBJ | Genomic DNA | |
| | M13120 EMBL· GenBank· DDBJ | AAA59987.1 EMBL· GenBank· DDBJ | Genomic DNA | |
| | K03199 EMBL· GenBank· DDBJ | AAA59989.1 EMBL· GenBank· DDBJ | mRNA | |
| | M14694 EMBL· GenBank· DDBJ | AAA61211.1 EMBL· GenBank· DDBJ | mRNA | |
| | M14695 EMBL· GenBank· DDBJ | AAA61212.1 EMBL· GenBank· DDBJ | mRNA | |
| | M22898 EMBL· GenBank· DDBJ | AAA59988.1 EMBL· GenBank· DDBJ | Genomic DNA | |
| | M22882 EMBL· GenBank· DDBJ | AAA59988.1 EMBL· GenBank· DDBJ | Genomic DNA | |
| | M22883 EMBL· GenBank· DDBJ | AAA59988.1 EMBL· GenBank· DDBJ | Genomic DNA | |
| | M22884 EMBL· GenBank· DDBJ | AAA59988.1 EMBL· GenBank· DDBJ | Genomic DNA | |
| | M22887 EMBL· GenBank· DDBJ | AAA59988.1 EMBL· GenBank· DDBJ | Genomic DNA | |
| | M22888 EMBL· GenBank· DDBJ | AAA59988.1 EMBL· GenBank· DDBJ | Genomic DNA | |
| | M22894 EMBL· GenBank· DDBJ | AAA59988.1 EMBL· GenBank· DDBJ | Genomic DNA | |
| | M22895 EMBL· GenBank· DDBJ | AAA59988.1 EMBL· GenBank· DDBJ | Genomic DNA | |
| | M22896 EMBL· GenBank· DDBJ | AAA59988.1 EMBL· GenBank· DDBJ | Genomic DNA | |
| | M22897 EMBL· GenBank· DDBJ | AAA59988.1 EMBL· GenBank· DDBJ | Genomic DNA | |
| | X01405 EMBL· GenBank· DDBJ | CAA25652.1 EMBL· GenBank· DDBJ | mRNA | |
| | X60011 EMBL· GenBank· DDBJ | CAA42626.1 EMBL· GenBank· DDBJ | mRNA | |
| | X60012 EMBL· GenBank· DDBJ | CAA42627.1 EMBL· GenBank· DDBJ | mRNA | Different termination. |
| | X60013 EMBL· GenBank· DDBJ | CAA42628.1 EMBL· GenBank· DDBJ | mRNA | |
| | X60014 EMBL· GenBank· DDBJ | CAA42629.1 EMBL· GenBank· DDBJ | mRNA | |
| | X60015 EMBL· GenBank· DDBJ | CAA42630.1 EMBL· GenBank· DDBJ | mRNA | |
| | X60016 EMBL· GenBank· DDBJ | CAA42631.1 EMBL· GenBank· DDBJ | mRNA | |
| | X60017 EMBL· GenBank· DDBJ | CAA42632.1 EMBL· GenBank· DDBJ | mRNA | |
| | X60018 EMBL· GenBank· DDBJ | CAA42633.1 EMBL· GenBank· DDBJ | mRNA | |
| | X60019 EMBL· GenBank· DDBJ | CAA42634.1 EMBL· GenBank· DDBJ | mRNA | |
| | X60020 EMBL· GenBank· DDBJ | CAA42635.1 EMBL· GenBank· DDBJ | mRNA | |
| | AF307851 EMBL· GenBank· DDBJ | AAG28785.1 EMBL· GenBank· DDBJ | mRNA | |
| | DQ186648 EMBL· GenBank· DDBJ | ABA29753.1 EMBL· GenBank· DDBJ | mRNA | |
| | DQ186649 EMBL· GenBank· DDBJ | ABA29754.1 EMBL· GenBank· DDBJ | mRNA | |
| | DQ186650 EMBL· GenBank· DDBJ | ABA29755.1 EMBL· GenBank· DDBJ | mRNA | |
| | DQ186651 EMBL· GenBank· DDBJ | ABA29756.1 EMBL· GenBank· DDBJ | mRNA | |
| | DQ186652 EMBL· GenBank· DDBJ | ABA29757.1 EMBL· GenBank· DDBJ | mRNA | |
| | DQ191317 EMBL· GenBank· DDBJ | ABB80262.1 EMBL· GenBank· DDBJ | mRNA | |
| | DQ286964 EMBL· GenBank· DDBJ | ABB80266.1 EMBL· GenBank· DDBJ | mRNA | |
| | X54156 EMBL· GenBank· DDBJ | CAA38095.1 EMBL· GenBank· DDBJ | Genomic DNA | |
| | U94788 EMBL· GenBank· DDBJ | AAC12971.1 EMBL· GenBank· DDBJ | Genomic DNA | |
| | AY838896 EMBL· GenBank· DDBJ | AAV80424.1 EMBL· GenBank· DDBJ | Genomic DNA | |
| | AF135121 EMBL· GenBank· DDBJ | AAD28535.1 EMBL· GenBank· DDBJ | Genomic DNA | |
| | AF135120 EMBL· GenBank· DDBJ | AAD28535.1 EMBL· GenBank· DDBJ | Genomic DNA | |
| | AF136271 EMBL· GenBank· DDBJ | AAD28628.1 EMBL· GenBank· DDBJ | Genomic DNA | |
| | AF136270 EMBL· GenBank· DDBJ | AAD28628.1 EMBL· GenBank· DDBJ | Genomic DNA | |
| | AB082923 EMBL· GenBank· DDBJ | BAC16799.1 EMBL· GenBank· DDBJ | mRNA | |
| | AK312568 EMBL· GenBank· DDBJ | BAG35463.1 EMBL· GenBank· DDBJ | mRNA | |
| | AC007421 EMBL· GenBank· DDBJ Genomic DNA | No translation available. |
| | AC087388 EMBL· GenBank· DDBJ Genomic DNA | No translation available. |
| | CH471108 EMBL· GenBank· DDBJ | EAW90143.1 EMBL· GenBank· DDBJ | Genomic DNA | |
| | CH471108 EMBL· GenBank· DDBJ | EAW90144.1 EMBL· GenBank· DDBJ | Genomic DNA | |
| | BC003596 EMBL· GenBank· DDBJ | AAH03596.1 EMBL· GenBank· DDBJ | mRNA | |
| | AY429684 EMBL· GenBank· DDBJ | AAR10356.1 EMBL· GenBank· DDBJ | mRNA | |
| | AY390341 EMBL· GenBank· DDBJ | AAQ90158.1 EMBL· GenBank· DDBJ | Genomic DNA | |
| | AY359814 EMBL· GenBank· DDBJ | AAR13239.1 EMBL· GenBank· DDBJ | Genomic DNA | |
| | U63714 EMBL· GenBank· DDBJ | AAB39322.1 EMBL· GenBank· DDBJ | Genomic DNA | |
| | AF209136 EMBL· GenBank· DDBJ | AAF36362.1 EMBL· GenBank· DDBJ | Genomic DNA | |
| | AF209128 EMBL· GenBank· DDBJ | AAF36354.1 EMBL· GenBank· DDBJ | Genomic DNA | |
| | AF209129 EMBL· GenBank· DDBJ | AAF36355.1 EMBL· GenBank· DDBJ | Genomic DNA | |
| | AF209130 EMBL· GenBank· DDBJ | AAF36356.1 EMBL· GenBank· DDBJ | Genomic DNA | |
| | AF209131 EMBL· GenBank· DDBJ | AAF36357.1 EMBL· GenBank· DDBJ | Genomic DNA | |
| | AF209132 EMBL· GenBank· DDBJ | AAF36358.1 EMBL· GenBank· DDBJ | Genomic DNA | |
| | AF209133 EMBL· GenBank· DDBJ | AAF36359.1 EMBL· GenBank· DDBJ | Genomic DNA | |
| | AF209134 EMBL· GenBank· DDBJ | AAF36360.1 EMBL· GenBank· DDBJ | Genomic DNA | |
| | AF209135 EMBL· GenBank· DDBJ | AAF36361.1 EMBL· GenBank· DDBJ | Genomic DNA | |
| | AF209148 EMBL· GenBank· DDBJ | AAF36374.1 EMBL· GenBank· DDBJ | Genomic DNA | |
| | AF209149 EMBL· GenBank· DDBJ | AAF36375.1 EMBL· GenBank· DDBJ | Genomic DNA | |
| | AF209150 EMBL· GenBank· DDBJ | AAF36376.1 EMBL· GenBank· DDBJ | Genomic DNA | |
| | AF209151 EMBL· GenBank· DDBJ | AAF36377.1 EMBL· GenBank· DDBJ | Genomic DNA | |
| | AF209152 EMBL· GenBank· DDBJ | AAF36378.1 EMBL· GenBank· DDBJ | Genomic DNA | |
| | AF209153 EMBL· GenBank· DDBJ | AAF36379.1 EMBL· GenBank· DDBJ | Genomic DNA | |
| | AF209154 EMBL· GenBank· DDBJ | AAF36380.1 EMBL· GenBank· DDBJ | Genomic DNA | |
| | AF209155 EMBL· GenBank· DDBJ | AAF36381.1 EMBL· GenBank· DDBJ | Genomic DNA | |
| | AF209156 EMBL· GenBank· DDBJ | AAF36382.1 EMBL· GenBank· DDBJ | Genomic DNA | |
| | AF210309 EMBL· GenBank· DDBJ | AAF63442.1 EMBL· GenBank· DDBJ | Genomic DNA | |
| | AF210308 EMBL· GenBank· DDBJ | AAF63442.1 EMBL· GenBank· DDBJ | Genomic DNA | |
| | AF210310 EMBL· GenBank· DDBJ | AAF63443.1 EMBL· GenBank· DDBJ | Genomic DNA | |
| | AF240684 EMBL· GenBank· DDBJ | AAK76358.1 EMBL· GenBank· DDBJ | Genomic DNA | |
| | AF240685 EMBL· GenBank· DDBJ | AAK76359.1 EMBL· GenBank· DDBJ | Genomic DNA | |
| | AY270155 EMBL· GenBank· DDBJ | AAP30003.1 EMBL· GenBank· DDBJ | Genomic DNA | |
Genome annotation databases
ENST00000269305.9ENSP00000269305.4ENSG00000141510.19 [P04637-1]
ENST00000420246.6ENSP00000391127.2ENSG00000141510.19 [P04637-2]
ENST00000445888.6ENSP00000391478.2ENSG00000141510.19 [P04637-1]
ENST00000455263.6ENSP00000398846.2ENSG00000141510.19 [P04637-3]
ENST00000503591.2ENSP00000426252.2ENSG00000141510.19 [P04637-1]
7157
ENST00000269305.9
ENSP00000269305.4
NM_000546.6
NP_000537.3
uc002gij.4human [P04637-1]
UniRef clusters
Orthologs & paralogs
Disclaimer
Any medical or genetic information present in this entry is provided for research, educational and informational purposes only. It is not in any way intended to be used as a substitute for professional medical advice, diagnosis, treatment or care. Our staff consists of biologists and biochemists that are not trained to give medical advice.
This website requires cookies, and the limited processing of your personal data in order to function. By using the site you are agreeing to this as outlined in our Privacy Notice. |
189033 | https://math.stackexchange.com/questions/4572158/radial-poisson-equation-in-spherical-polar-coordinates | Radial poisson equation in spherical polar coordinates - 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
Radial poisson equation in spherical polar coordinates
Ask Question
Asked 2 years, 10 months ago
Modified2 years, 10 months ago
Viewed 222 times
This question shows research effort; it is useful and clear
1
Save this question.
Show activity on this post.
(This question may belong on Physics. I put in on Mathematics because I saw more similar questions here.)
I have a seperable density function that is expressed in spherical polar coordinates as:
n l m(r)=Y l m(Ω)ρ l m(r)n l m(r)=Y l m(Ω)ρ l m(r)
where Y l m(Ω)Y l m(Ω) is a spherical harmonic, Ω Ω is the angular components of r r, (θ,ϕ)(θ,ϕ) and n l m(r)n l m(r) is the radial component of the density function.
I can use any one of the many ways to solve the Poisson equation
∇2 V l m(r)=−1 ϵ 0 ρ l m(r)(1)(1)∇2 V l m(r)=−1 ϵ 0 ρ l m(r)
to obtain the potential, U(r)U(r) from the density, and indeed I do this in my code, successfully. My question is not about how to solve the Poisson equation, but about the form of the radial Poisson equation. The radial Poisson equation is given in several papers [e.g [Goddard, 1995] ( as:
[d 2 d r 2−l(l+1)r 2]U l m(r)=−4 π r ρ l m(r)[d 2 d r 2−l(l+1)r 2]U l m(r)=−4 π r ρ l m(r)
I want to know what I am missing to derive this form of the radial Poisson equation. Here is my attempt.
I will start with some statements. ϵ 0=0 ϵ 0=0 and I will write the potentials in the form
V l m(r)=Y l m(Ω)U l m(r)(2)(2)V l m(r)=Y l m(Ω)U l m(r)
.
Substituting the definition of ∇2∇2 in spherical polar coordinates,
∇2=1 r 2∂∂r(r 2∂∂r)+1 r 2 sin θ∂∂θ(sin θ∂∂θ)+1 r 2 sin 2 θ∂2∂φ 2∇2=1 r 2∂∂r(r 2∂∂r)+1 r 2 sinθ∂∂θ(sinθ∂∂θ)+1 r 2 sin 2θ∂2∂φ 2
into the Poisson equation (???)(???), together with (???)(???) gives
Y l m(Ω)1 r 2∂∂r(r 2∂∂r)U l m(r)+U l m(r)[1 r 2 sin θ∂∂θ(sin θ∂∂θ Y l m(Ω))+1 r 2 sin 2 θ∂2∂φ 2 Y l m(Ω)]=−Y l m(Ω)ρ l m(r)Y l m(Ω)1 r 2∂∂r(r 2∂∂r)U l m(r)+U l m(r)[1 r 2 sinθ∂∂θ(sinθ∂∂θ Y l m(Ω))+1 r 2 sin 2θ∂2∂φ 2 Y l m(Ω)]=−Y l m(Ω)ρ l m(r)
Then comparing the angular terms with the spherical harmonic differential equation:
[1 sin θ∂∂θ(sin θ∂∂θ)+1 sin 2 θ∂2∂ϕ 2+l(l+1)]Y l m(Ω)=0[1 sinθ∂∂θ(sinθ∂∂θ)+1 sin 2θ∂2∂ϕ 2+l(l+1)]Y l m(Ω)=0
I get
[1 r 2 d d r(r 2 d d r)−l(l+1)r 2]U l m(r)=−ρ l m(r)[1 r 2 d d r(r 2 d d r)−l(l+1)r 2]U l m(r)=−ρ l m(r)
Expanding the derivative gives
[d 2 d r 2+2 r d d r−l(l+1)r 2]U l m(r)=−ρ l m(r)[d 2 d r 2+2 r d d r−l(l+1)r 2]U l m(r)=−ρ l m(r)
So, can someone show me how the extra 2 r d d r 2 r d d r disappears, and where the extra r r on the right hand side comes from?
Thanks.
P.s As a point of interest, while looking for similar questions, I found this question, where the left-hand side is the same as mine (although I would argue that theirs is a Laplace equation, rather than a Poisson equation).
ordinary-differential-equations
spherical-coordinates
poissons-equation
spherical-harmonics
Share
Share a link to this question
Copy linkCC BY-SA 4.0
Cite
Follow
Follow this question to receive notifications
asked Nov 8, 2022 at 17:02
holyjolyholyjoly
61 4 4 bronze badges
2
1 I seems that U m l(r)U m l(r) in Goddard' notation is r⋅U m l(r)r⋅U m l(r) in your notation (d 2 d r 2−l(l+1)r 2)r⋅U m l(r)=r(d 2 d r 2+2 r d d r−l(l+1)r 2)U m l(r)(d 2 d r 2−l(l+1)r 2)r⋅U m l(r)=r(d 2 d r 2+2 r d d r−l(l+1)r 2)U m l(r) Svyatoslav –Svyatoslav 2022-11-08 17:35:45 +00:00 Commented Nov 8, 2022 at 17:35
1 Oh! Thanks! I'll be happy to accept your answer.holyjoly –holyjoly 2022-11-08 18:13:42 +00:00 Commented Nov 8, 2022 at 18:13
Add a comment|
0
Sorted by: Reset to default
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
ordinary-differential-equations
spherical-coordinates
poissons-equation
spherical-harmonics
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
Linked
0Solving the radial part of Poisson's equation via power series method
Related
2Problem with Laplacian while treating polar coordinates as special case of spherical coordinates.
1analytic solution poisson equation spherical coordinates
0Grad in polar coordinates
7Laplacian of spherical coordinates
0What is the error hiding in this trivial integral with spherical polar coordinates and divergence theorem?
1Evaluation of a particular gradient in spherical coordinates.
0A paradox on curl equations in cylindrical and spherical coordinates
Hot Network Questions
How to locate a leak in an irrigation system?
RTC battery and VCC switching circuit
Is existence always locational?
Suspicious of theorem 36.2 in Munkres “Analysis on Manifolds”
Is direct sum of finite spectra cancellative?
alignment in a table with custom separator
Non-degeneracy of wedge product in cohomology
Another way to draw RegionDifference of a cylinder and Cuboid
Is it ok to place components "inside" the PCB
A time-travel short fiction where a graphologist falls in love with a girl for having read letters she has not yet written… to another man
Discussing strategy reduces winning chances of everyone!
Xubuntu 24.04 - Libreoffice
Matthew 24:5 Many will come in my name!
How long would it take for me to get all the items in Bongo Cat?
Storing a session token in localstorage
how do I remove a item from the applications menu
I have a lot of PTO to take, which will make the deadline impossible
Do sum of natural numbers and sum of their squares represent uniquely the summands?
Passengers on a flight vote on the destination, "It's democracy!"
Lingering odor presumably from bad chicken
Does the curvature engine's wake really last forever?
Spectral Leakage & Phase Discontinuites
What is the feature between the Attendant Call and Ground Call push buttons on a B737 overhead panel?
Why are LDS temple garments secret?
Question feed
Subscribe to RSS
Question feed
To subscribe to this RSS feed, copy and paste this URL into your RSS reader.
Why are you flagging this comment?
It contains harassment, bigotry or abuse.
This comment attacks a person or group. Learn more in our Code of Conduct.
It's unfriendly or unkind.
This comment is rude or condescending. Learn more in our Code of Conduct.
Not needed.
This comment is not relevant to the post.
Enter at least 6 characters
Something else.
A problem not listed above. Try to be as specific as possible.
Enter at least 6 characters
Flag comment Cancel
You have 0 flags left today
Mathematics
Tour
Help
Chat
Contact
Feedback
Company
Stack Overflow
Teams
Advertising
Talent
About
Press
Legal
Privacy Policy
Terms of Service
Your Privacy Choices
Cookie Policy
Stack Exchange Network
Technology
Culture & recreation
Life & arts
Science
Professional
Business
API
Data
Blog
Facebook
Twitter
LinkedIn
Instagram
Site design / logo © 2025 Stack Exchange Inc; user contributions licensed under CC BY-SA. rev 2025.9.26.34547
By clicking “Accept all cookies”, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy.
Accept all cookies Necessary cookies only
Customize settings
Cookie Consent Preference Center
When you visit any of our websites, 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 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 manage your preferences. Please note, blocking some types of cookies may impact your experience of the site and the services we are able to offer.
Cookie Policy
Accept all cookies
Manage Consent Preferences
Strictly Necessary Cookies
Always Active
These cookies are necessary for the website to function and cannot be switched off in our systems. They are usually only set in response to actions made by you which amount to a request for services, such as setting your privacy preferences, logging in or filling in forms. You can set your browser to block or alert you about these cookies, but some parts of the site will not then work. These cookies do not store any personally identifiable information.
Cookies Details
Performance Cookies
[x] Performance Cookies
These cookies allow us to count visits and traffic sources so we can measure and improve the performance of our site. They help us to know which pages are the most and least popular and see how visitors move around the site. All information these cookies collect is aggregated and therefore anonymous. If you do not allow these cookies we will not know when you have visited our site, and will not be able to monitor its performance.
Cookies Details
Functional Cookies
[x] Functional Cookies
These cookies enable the website to provide enhanced functionality and personalisation. They may be set by us or by third party providers whose services we have added to our pages. If you do not allow these cookies then some or all of these services may not function properly.
Cookies Details
Targeting Cookies
[x] Targeting Cookies
These cookies are used to make advertising messages more relevant to you and may be set through our site by us or by our advertising partners. They may be used to build a profile of your interests and show you relevant advertising on our site or on other sites. They do not store directly personal information, but are based on uniquely identifying your browser and internet device.
Cookies Details
Cookie List
Clear
[x] checkbox label label
Apply Cancel
Consent Leg.Interest
[x] checkbox label label
[x] checkbox label label
[x] checkbox label label
Necessary cookies only Confirm my choices |
189034 | https://www.cancer.org/content/dam/cancer-org/research/cancer-facts-and-statistics/annual-cancer-facts-and-figures/2023/2023-cff-special-section-lung-cancer.pdf | Cancer Facts & Figures 2023 31 Special Section: Lung Cancer Introduction Lung cancer is the second most commonly diagnosed cancer in both men and women but the most common cause of cancer death, leading to more deaths in 2020 than breast, colorectal, and prostate cancers combined. This burden disproportionately affects people with lower socioeconomic status. Although approximately 80% of lung cancers are caused by cigarette smoking,1 the toll among people who have never smoked is substantial, ranking among the top 10 causes of cancer death when categorized separately.2 Lung cancer is usually fatal because most cases are diagnosed at a late stage and treatment has generally been ineffective. However, over the past two decades, advances in the understanding of tumor biology and the development of targeted treatment, coupled with the introduction of screening, have led to exciting improvements in survival. This special section provides an overview of lung cancer occurrence in the US, including information about risk factors, prevention, and early detection, as well as what the American Cancer Society is doing to reduce the burden.
Lung Anatomy The lungs are a pair of large, spongy, cone-shaped organs in the chest that are part of the respiratory system, which also includes the trachea (windpipe) and the muscles of the chest wall and diaphragm (Figure S1). The lungs are separated from one another by a cavity called the mediastinum, which houses the heart, trachea, esophagus, and many lymph nodes. The right lung is made up of three lobes and is slightly larger than the left lung, which has two lobes. The main function of the lungs is to facilitate respiration by moving air and transferring oxygen into the bloodstream while clearing the body of carbon dioxide and other waste gases. Air enters the lungs through the trachea during inhalation and passes into bronchi (branch-like air passages) that subdivide into tiny bronchioles ending in air sacs called alveoli, where gas transfer occurs. Carbon dioxide and other waste gases are expelled through the trachea during exhalation. The lungs also serve a vital role in the defense of the body by filtering inhaled, potentially harmful particles, such as dust, mold, viruses, and bacteria. However, some inhaled toxins contribute to lung cancer occurrence. What Are the Different Types of Lung Cancer?
Lung cancer encompasses a variety of biologically distinct tumors.3 The two primary types of lung cancer are non-small cell lung cancer (NSCLC), which accounts for 81% of cases, and small cell lung cancer (SCLC), which accounts for 14% of cases. NSCLC is further categorized as adenocarcinoma, which is slightly more common in women, followed by squamous cell carcinoma and large cell carcinoma (Figure S2).4 Squamous cell carcinoma was the most common subtype prior to the 1960s and the introduction of filtered cigarettes, which are thought to contribute to increased incidence of adenocarcinoma via deeper inhalation and increased exposure to smoke toxicants.5 Adenocarcinoma originates in the glands that secrete mucus. It is generally more treatable than other subtypes because it is more likely to be located in the Source: Figure S1. Anatomy of the Respiratory System 32 Cancer Facts & Figures 2023 lung periphery and to have mutations that can be targeted by treatment.6-8 (For more information, see page 36.) Squamous cell carcinoma is more aggressive than adenocarcinoma and originates in cells that line the lung airways.9 Large cell carcinoma can originate in any portion of the lung and is more aggressive than the other two NSCLC subtypes. SCLC is named for the small, round appearance of the cells under a microscope. The most common subtype is small cell carcinoma (oat cell cancer), followed by combined small cell carcinoma. SCLC is slightly more common in women (14%) than in men (13%) and is generally more aggressive than NSCLC. Consequently, SCLC patients are more likely than NSCLC patients to have disease that has spread beyond the lungs at the time of diagnosis (94% versus 70%),10 making treatment difficult.11 Even for those patients who successfully undergo chemotherapy, which is the primary treatment, only a limited number achieve long-term disease control. How Many Lung Cancer Cases and Deaths Are Expected to Occur in 2023?
In 2023, an estimated 238,340 people (117,550 men and 120,790 women) will be diagnosed with lung cancer, and 127,070 people will die from the disease (Table 1). How Many People Alive Today Have a History of Lung Cancer?
As of January 1, 2022, there were 654,620 men and women in the US with a history of lung cancer,12 many of whom were living with metastatic disease.13 About 80% of these individuals were 65 years of age or older, reflecting the advanced median age of diagnosis (71 years).14 More than half (55%) of lung cancer survivors were diagnosed within the past 5 years because of the low survival.12 (See page 36.) What Is the Risk of Developing Lung Cancer?
The lifetime risk of developing lung cancer is approximately 6.2% among men and 5.8% among women, or 1 in 16 men and 1 in 17 women during their lifetime (Table 6). However, these probabilities are based on lung cancer occurrence in the general population so the risk is substantially higher for those with a history of smoking.15 The risk of lung cancer also increases with age, partly because the disease grows for many decades before symptoms develop. More than half (53%) of cases are diagnosed at age 70 or older, and 83% of cases are diagnosed at ages 65 and older. However, the age distribution varies by histologic subtype and race and SCLC: Small cell lung cancer; NSCLC: Non-small cell lung cancer. Percentages for NSCLC subtypes represent the proportion of total NSCLC cases. Percentages may not total 100% because of rounding.
Source: North American Association of Central Cancer Registries, 2022.
©2023, American Cancer Society, Inc., Surveillance and Health Equity Science Figure S2. The Distribution of Lung Cancer by Histologic Subtype (%), US, 2015-2019 Male Female Adenocarcinoma 62% NSCLC 80% NSCLC 82% Squamous Cell 24% Other NSCLC 12% Adenocarcinoma 52% Squamous Cell 35% Other NSCLC 11% Large Cell 1% Large Cell 2% Other 6% Other 5% SCLC 14% SCLC 13% Cancer Facts & Figures 2023 33 ethnicity. For example, NSCLC incidence peaks in ages 80-84 for men but in ages 75-79 for women,16 likely reflecting sex differences in smoking. (See page 37.) For SCLC, incidence peaks in ages 75-79 among both men and women. Incidence among Black men peaks about 5 years earlier than among White men for both NSCLC and SCLC. How Does Lung Cancer Risk Vary Between Different Population Groups? Sex Lung cancer incidence during 2015-2019 was 27% higher among men (64.1 per 100,000) than women (50.3 per 100,000; Figure S3), largely due to historically higher smoking prevalence in men (Figure S4). However, this pattern varies by age and racial and ethnic group. For example, young non-Hispanic White (hereafter White) and Hispanic women have higher lung cancer rates than their male counterparts.17, 18 Notably, this reversal in risk is not fully explained by smoking patterns. The sex gap for mortality is wider than for incidence, with death rates in men (42.2 per 100,000 during 2016-2020) 44% higher than those in women (29.3 per 100,000). This is due in part to differences in the distribution of subtypes and lower survival in men. (See information on lung cancer survival on page 36.) Race & Ethnicity Lung cancer incidence is highest among Black men, whereas mortality is highest among both Black and American Indian and Alaska Native (AIAN) men (Figure S3). AIAN men and women have had the highest smoking prevalence by far since at least the early 1990s, when data first became available (Figure S4). In some regions, including the East, Northern Plains, and Pacific Coast, lung cancer incidence among AIAN women is similar to or higher than among their male counterparts.19 AIAN women have the highest mortality of any racial or ethnic group, 10% higher than White women, who rank second (Figure S3). Lung cancer incidence and mortality among Asian American and Pacific Islander (AAPI) and Hispanic individuals is lower than that among other racial and ethnic groups (Figure S3) due to historically lower smoking prevalence (Figure S4). However, data for broadly defined racial and ethnic groups mask large differences within these heterogeneous populations, of which there are many examples.20 Although lung cancer incidence in AAPI men overall is about 40% lower than in White men, one study found that it is 40% higher among Samoan men.21 Another study reported that among never-smoking women in California and Hawaii, lung cancer was higher in AAPI women than in White women, and was highest in Chinese American AAPI: Asian American and Pacific Islander individuals; AIAN: American Indian and Alaska Native individuals. Age adjusted to the 2000 US standard population. †For AIAN individuals, incidence data are restricted to Purchased/Referred Care Delivery Area counties, and mortality data are adjusted for misclassification on death certificates. All racial groups are exclusive of individuals identifying as Hispanic.
Sources: Incidence, North American Association of Central Cancer Registries 2022; Mortality, National Center for Health Statistics 2022.
©2023, American Cancer Society, Inc., Surveillance and Health Equity Science Cases per 100,000 population Deaths per 100,000 population Figure S3. Lung Cancer Incidence and Mortality Rates by Sex, Race, and Ethnicity, US Incidence, 2015-2019 Mortality, 2016-2020 0 10 20 30 40 50 60 70 80 Female Male 0 10 20 30 40 50 60 70 80 Female Male 64.1 74.8 67.3 66.9 42.1 35.6 50.3 46.9 55.5 57.9 28.3 24.4 42.2 51.0 44.7 51.1 25.6 20.9 29.3 27.8 32.8 36.0 15.4 11.4 All Black White AIAN† AAPI Hispanic 34 Cancer Facts & Figures 2023 women.22 Similarly, among Hispanic individuals in Florida, lung cancer mortality rates among Cuban men were more than 50% higher than in Puerto Rican men and double those in men of other Hispanic origin groups, although rates were still lower than those in Whites.23 Socioeconomic Status The risk of lung cancer is greater in persons with lower socioeconomic status (SES).24 For example, lung cancer death rates in individuals ages 25-74 with ≤12 years of education are nearly 5 times higher in men and 4 times higher in women compared to those in persons with ≥16 years of education.25 This disparity reflects historical differences in smoking prevalence that remain today. In 2021, for example, 21% of individuals without a high school diploma and 31% of individuals with a GED smoked cigarettes compared to 3% of individuals with a graduate degree. (See page 55.) Persons with low SES are also more likely to be diagnosed with advanced-stage disease26 and lack access to high-quality cancer care.27, 28 Place of Residence The states with the highest lung cancer mortality rates are Kentucky, West Virginia, Mississippi, and Arkansas (Table 5), all of which have the highest historical and current smoking prevalence.29 However, there are also pockets of high lung cancer mortality in other states, including in the South and Appalachia (Figure S5). Among AIAN individuals, incidence for those living in the Northern Plains is nearly 5 times higher than for those living in the Southwest, where rates are 64% lower than White individuals living in the region.19 How Has Lung Cancer Occurrence Changed Over Time?
Lung cancer incidence and mortality trends closely mirror the tobacco epidemic with a lag of several decades.30 As a result, lung cancer patterns differ by sex because women started smoking in large numbers later than men and were also much slower to begin quitting.15 Lung cancer mortality rates rose from 3 to 4 per 100,000 in 1930, peaked at 91 per 100,000 men in 1990 and 42 per 100,000 women in 2002 before declining by 58% (38 per 100,000) and 36% (27 per 100,000), respectively, through 2020 (Figure S6). Reductions in mortality began several decades after the release of the first US Surgeon General’s Report on Smoking and Health in 1964, which motivated people to quit smoking.31 Continued reductions in smoking are reflected in steady declines in lung cancer incidence of 2.6% per Figure S4. Trends in Smoking Prevalence by Sex, Race, and Ethnicity, US, 1965-2021 Percent of current smoking adults AAPI: Asian American and Pacific Islander individuals; AIAN: American Indian and Alaska Native individuals. Ever smoked 100 cigarettes in lifetime and now smoke every day or some days. All racial groups are exclusive of individuals identifying as Hispanic beginning in 1990. All estimates are age adjusted. Due to changes in National Health Interview Survey (NHIS) survey design, estimates from 2019 onward are not directly comparable to prior years and are separated from the trend line. Sources: Adult cigarette smoking prevalence 1965-2018, Health United States: 2019; NHIS 1990-2021.
©2023, American Cancer Society, Inc., Surveillance and Health Equity Science Male White White Hispanic Hispanic Black Black AIAN AIAN AAPI AAPI Female All races combined Females Males 0 10 20 30 40 50 60 2020 2015 2010 2005 2000 1995 1990 1985 1980 1975 1970 1965 0 10 20 30 40 50 60 2020 2015 2010 2005 2000 1995 1990 1985 1980 1975 1970 1965 0 10 20 30 40 50 60 2020 2015 2010 2005 2000 1995 1990 1985 1980 1975 1970 1965 Cancer Facts & Figures 2023 35 year in men and 1.1% per year in women since around 2006. However, an increase in the incidence of localized-stage disease of 4.5% per year from 2014 through 201832 suggests that people are being diagnosed earlier. This is probably at least in part due to the uptake of screening, which was first recommended for people at high risk of lung cancer in 2013,33 as well as increased access to care through the Affordable Care Act. At the same time, the decline in mortality has accelerated from 3% per year in men and 2% per year in women during 2005 to 2014 to 5% and 4% per year, respectively, during 2014 to 2020.2 This progress outpaces declines in incidence and likely reflects recent advances in treatment,34, 35 as well as earlier detection.36 Differences by Race and Ethnicity Among men, lung cancer mortality rates have been declining since at least 1990 in each racial and ethnic group, with the steepest drop among Black men (Figure S7). The more favorable trend in Black men likely reflects rapid historic declines in cigarette smoking prevalence (Figure S4) driven by steeper reductions in adolescent cigarette smoking initiation compared to other racial and ethnic groups.37-41 As a result, the disparity in lung cancer mortality among Black versus White men dropped from 40% higher in the early 1990s to 14% higher during 2016-2020 (Figure S3). Among women, lung cancer mortality rates continued to increase until the early or late 2000s in all racial and ethnic groups, with the steepest rise in AIAN women (Figure S7). Consequently, lung cancer mortality in AIAN women surpassed that in Black and White women circa 2000, reflecting their high smoking prevalence (Figure S4). Since the early to late 2000s, mortality has declined among women in all racial and ethnic groups, with the fastest pace among Black women. As a result, the lung cancer mortality rate in Black women was 4% lower than White women in 1990 but 17% lower in 2020. Like Black men, the more favorable lung cancer mortality trend in Black women reflects steeper declines in smoking prevalence.37-41 Figure S5. Lung Cancer Mortality Rates by County, 2016-2020 Age adjusted to the 2000 US standard population.
Source: National Center for Health Statistics, 2022.
Deaths per 100,000 population 5.7-32.2 32.3-39.1 39.2-44.8 44.9-52.4 52.5-119.4 <10 deaths 36 Cancer Facts & Figures 2023 Mortality rates among Hispanic individuals have consistently been lower than any other racial and ethnic group (Figure S7), largely due to their low smoking prevalence (Figure S4) and intensity of smoking. Hispanic smokers are more likely to be light (≤10 cigarettes/day) or intermittent smokers than any other racial or ethnic group.42, 43 Lung Cancer Survival After decades of little improvement, recent advancements in treatment have led to longer survival for lung cancer patients.8, 35 Progress is mostly confined to NSCLC and is more evident in 2-year versus 5-year relative survival rates. For example, among women, 2-year relative survival for NSCLC increased slightly from 32% in 1975-1976 to 36% in 1997-1998, then rose to 54% in 2017-2018 (Figure S8). Increases were similar in men, from 25% to 28% to 43%, respectively. Advances in treatment that have likely contributed to this progress include molecular therapies targeting important mutations, such as in the epidermal growth factor receptor (EGFR) and anaplastic lymphoma kinase (ALK) genes;44 immune checkpoint inhibitors, which boost patient immune response;45, 46 improvements in staging;47 and video-assisted surgery.34 In contrast, 2-year relative survival for SCLC has increased little in absolute terms but doubled in relative terms, from 10% and 8% among women and men, respectively in 1975-1976, to 15% and 13% in 1997-1998 and 19% and 16% in 2017-2018 (Figure S8).
Lung cancer survival rates are higher among women than men (Figure S9), partly reflecting earlier-stage diagnosis and differences in subtype distribution. For example, 28% of women are diagnosed at a localized stage compared to 23% of men (Figure S10). However, 5-year survival rates are higher among women at every stage of diagnosis, albeit mostly confined to NSCLC (Figure S11). Reasons for higher survival in women are unclear but may reflect differences in tumor characteristics and hormones that influence treatment response.48, 49 Female lung cancer patients are also more likely to have tumors with genetic mutations, such as in the EGFR gene, that are amenable to targeted therapies.50 Survival rates also vary by race and ethnicity, ranging from 26% among AAPI individuals to 19% among AIAN individuals (Figure S9), in part because of higher frequency of EGFR mutations among AAPI individuals diagnosed with lung cancer.50 However, White individuals are most likely to be diagnosed with localized-stage disease among both men and women (Figure S10). Lower survival in some groups likely reflects less access to early detection, curative-intent surgery, and new therapies.51-53 Survival Figure S6. Trends in Tobacco Consumption and Lung Cancer Mortality Rates by Sex, 1900-2020 Number of cigarettes per capita Lung cancer death rate per 100,000 persons Age adjusted to the 2000 US standard population. Rates exclude deaths in Puerto Rico and other US territories. Note: Due to changes in ICD coding, numerator information for mortality rates has changed over time.
Source: Death rates: US Mortality Data, 1960-2020, US Mortality Volumes, 1930-1959, National Center for Health Statistics. Cigarette consumption: 1900-1999: US Department of Agriculture. 2000-2015: Consumption of Cigarettes and Combustible Tobacco – United States, 2000-2015. MMWR Weekly Rep 2016; 65(48);1357-1363.
©2023, American Cancer Society, Inc., Surveillance and Health Equity Science 0 500 1000 1500 2000 2500 3000 3500 4000 4500 5000 2020 2010 2000 1990 1980 1970 1960 1950 1940 1930 1920 1910 1900 0 20 40 60 80 100 Cigarette consumption Male mortality Female mortality Free cigarettes for GIs in WWI 1917-19 Free cigarettes for GIs in WWII 1941-45 1964 Surgeon General’s Report Family Smoking Prevention and Tobacco Control Act 2009 Cancer Facts & Figures 2023 37 data for AAPI, Hispanic, and AIAN people should be interpreted with caution due to potential loss to follow-up and racial misclassification.54 What Are the Risk Factors for Lung Cancer?
The primary risk factor for lung cancer is cigarette smoking, which accounts for about 80% of lung cancer cases and deaths (Figure 4).1 Cigarette smoking increases the risk of lung cancer 25-fold in both men and women compared to people who never smoked.15 Other exposures that increase risk include secondhand smoke, radon, asbestos and some other occupational exposures, air pollution, and arsenic in drinking water. Low fruit and vegetable intake may increase risk although evidence is still accumulating.55 Tobacco Use Cigarette Smoking Cigarette smoking began to increase in the early 20th century among men, and after World War II among women,56-58 peaking in the mid-1960s with adult smoking prevalence in 1965 at 51% among men and 34% among women (Figure S4, and Figure S6). While consumption cannot be differentiated by sex before 1965, studies suggest that the decline among men began in the mid-1950s, with increases in consumption among women outweighing the decline among men until the 1960s.30 The pace of the decline in smoking prevalence has been steeper in men than in women because of differences in cessation and transient upticks in female smoking.17 In 2021, adult cigarette smoking was 13% in men and 10% in women,43 a 75% and 70% relative reduction, respectively, since 1965 because of increased awareness about the health hazards of cigarette smoking, increased excise taxes on cigarette products, prohibiting smoking in public places, and counteradvertising among other tobacco control policies. (See Tobacco Control Policies section, page 40.) Historically, smoking prevalence was much higher among Black men than White men, but the gap has narrowed since around 1990 and been eliminated as of 2021 (Figure S4). In contrast, smoking prevalence among Black women was similar to White women prior to 1990 but has since been several percentage points lower. According to data from the National Health Interview Survey, current adult cigarette smoking prevalence in 2021 was 14% in both Black and White men, but 10% and 12% in Black and White women, respectively. The favorable trends in Black individuals are largely due to the sharp decline in smoking initiation among Black adolescents beginning in the late 1970s.37 Additionally, Figure S7. Trends in Lung Cancer Mortality Rates by Sex, Race, and Ethnicity, US, 1990-2020 Deaths per 100,000 population AAPI: Asian American and Pacific Islander individuals; AIAN: American Indian and Alaska Native individuals. Age adjusted to the 2000 US standard population. †Data for AIAN individuals begin with 1997 to include Oklahoma and are adjusted for racial misclassification on death certificates. All racial groups are exclusive of individuals identifying as Hispanic. Source: National Center for Health Statistics, 2022.
©2023, American Cancer Society, Inc., Surveillance and Health Equity Science White White Hispanic Hispanic Black Black AIAN† AIAN† AAPI AAPI Male 0 25 50 75 100 125 2020 2015 2010 2005 2000 1995 1990 0 25 50 75 100 125 2020 2015 2010 2005 2000 1995 1990 Female 38 Cancer Facts & Figures 2023 compared to White people, Black people are more likely to smoke lightly or intermittently.42 A majority (80%) of Black people who smoke consume menthol cigarettes compared to 34% of White people who smoke.59 While it has been hypothesized that menthol cigarette smoking may contribute to elevated lung cancer burden among Black men, research findings have not generally supported this hypothesis.60-62 Research does suggest that Black people are more susceptible to smoking-related lung cancer compared to White people.63 AIAN individuals have the highest smoking prevalence of any racial or ethnic group, with 25% of AIAN adults currently smoking cigarettes in 2020.29 Unlike most racial and ethnic groups, smoking prevalence among AIAN women has been similar to men since the early 1990s, when data first became available (Figure S3). However, smoking varies by region, with the highest reported prevalence in the Northern Plains (42% in both men and women) and the lowest in the Southwest (19% in men and 15% in women).64 Higher prevalence of smoking is largely due to historical and ongoing structural racism that has contributed to lower education and income levels65 and inadequate access to quality health care,66, 67 as well as targeted deceptive tobacco product advertising.68-70 There is also a lack of tailored smoking-cessation programs71 that recognize the cultural importance of traditional tobacco use in some regions.72 As a result, AIAN individuals have fewer quit attempts and slower cessation compared to other racial and ethnic groups.29 Cigarette smoking varies considerably by country of origin and nativity. For example, among the Hispanic population, smoking prevalence is higher among US-born versus foreign-born women and among Puerto Rican individuals (17% in men and 16% in women in 2017-2019) than Cuban individuals (7% and 12%, respectively), though Cuban individuals are more likely to smoke heavily.73 Similarly, cigarette smoking prevalence among AAPI individuals is higher among Filipino (12%) individuals than Chinese (7%) or Asian Indian (6%) individuals.21 Smoking prevalence among Native Hawaiian individuals is higher than any other AAPI group, similar to that among AIAN individuals, with 19.6% of Native Hawaiians living in Hawaii reporting current cigarette smoking during 2018 to 2020.74 Native Hawaiian individuals, along with Black individuals, have a higher smoking-related risk of lung cancer compared to White, Japanese American, or Hispanic individuals.59 Figure S8. Trends in 2-year Relative Survival Rates for Lung Cancer by Sex and Subtype, US, 1975-2018 Percent NSCLC: Non-small cell lung cancer; SCLC: Small cell lung cancer. Survival is based on patients diagnosed during 1975 through 2018, followed through 2019.
Source: Surveillance, Epidemiology, and End Results 8 Registries, 2022.
©2023, American Cancer Society, Inc., Surveillance and Health Equity Science Male, NSCLC Male, SCLC Female, SCLC Female, NSCLC 0 10 20 30 40 50 60 2017-2018 2011-2012 2005-2006 1999-2000 1993-1994 1987-1988 1981-1982 1975-1976 32 25 10 8 36 28 15 13 54 43 19 16 Figure S9. 5-year Relative Survival Rates for Lung Cancer by Sex, Race, and Ethnicity, US, 2012-2018 Percent AAPI: Asian American and Pacific Islander individuals; AIAN: American Indian and Alaska Native individuals. Survival rates are for patients diagnosed during 2012-2018, all followed through 2019. All racial groups exclude individuals identifying as Hispanic.
Source: Surveillance, Epidemiology, and End Results 17 Registries, 2022.
©2023, American Cancer Society, Inc., Surveillance and Health Equity Science 23 23 23 24 22 21 19 19 19 18 17 15 26 27 27 28 25 31 All races AAPI White Hispanic Black AIAN 0 5 10 15 20 25 30 35 Female Male Sexes combined Cancer Facts & Figures 2023 39 Other Combustible Tobacco Products Other forms of combustible tobacco products associated with an increased risk of lung cancer include cigars, pipes, and waterpipes. Current cigar users are more than three times as likely to die from lung cancer compared to individuals who have never smoked any combustible tobacco product.75 Compared to White people, Black people are more than twice as likely to smoke cigars.76 In contrast to patterns of cigarette use, waterpipe use is more prevalent among younger individuals and individuals with higher educational attainment.77 For more information about other combustible tobacco products, see page 55.
Secondhand Smoke Secondhand smoke, or involuntary exposure to tobacco smoke, is the third most common cause of lung cancer in the US,78 with a disproportionate influence on Black individuals and families with lower income.79 Secondhand smoke contains numerous toxic chemicals, including at least 50 known carcinogens,80 and is associated with the greatest risk for small cell lung cancer.81 For more information about secondhand smoke, see page 56. Personal and Family History A history of lung disease, including asthma, chronic bronchitis, COPD, emphysema, pneumonia, and tuberculosis, is associated with increased risk of lung cancer.82, 83 These diseases are thought to influence cancer risk through chronic inflammation of lung tissue.84-87 The estimated excess lung cancer risk ranges from 16% among those with a history of asthma to 2.5-fold among those with a history of COPD.88 In addition, some people are at increased risk because of genetic predisposition.89 Knowledge of inherited lung cancer dates to the 1960s, when excess mortality was noted among relatives of 270 lung cancer patients.90 The International Lung Cancer Consortium estimates that individuals with a first-degree relative with lung cancer are at a 50% increased risk of the disease, with the strongest association for a diagnosed sibling (82%).91 Specific genetic syndromes and mutations that have been associated with excess risk include Li-Fraumeni syndrome (variants in the TP53 tumor-suppressor gene),92 EGFR pathogenic variants,93, 94 and possibly BRCA2, which has been associated with earlier onset of lung cancer by about 12 years95 and risk among women.96 AAPI: Asian American and Pacific Islander individuals; AIAN: American Indian and Alaska Native individuals. Data for AIAN individuals are restricted to Purchased/Referred Care Delivery Area counties. All racial groups are exclusive of individuals identifying as Hispanic.
Source: North American Association of Central Cancer Registries, 2022.
©2023, American Cancer Society, Inc., Surveillance and Health Equity Science Percent Figure S10. Stage at Diagnosis by Sex, Race, and Ethnicity, US, 2015-2019 Males 0 10 20 30 40 50 60 Unknown/Unstaged Distant Regional Localized 0 10 20 30 40 50 60 Unknown/Unstaged Distant Regional Localized 23 22 22 24 23 20 20 20 19 20 21 24 46 46 46 53 50 50 8 8 8 11 10 7 28 29 26 25 26 22 22 26 17 20 23 22 42 43 42 51 45 45 7 7 7 6 9 9 All White Black AAPI AIAN Hispanic Females 40 Cancer Facts & Figures 2023 Environmental Exposures Radon Radon gas is a direct by-product of the radioactive decay of radium-226, part of a lengthy chain of radioactive decay for uranium-238, which is present naturally in rocks and soils.97 Radon is thought to be the second-leading cause of lung cancer after cigarette smoking,98 with the greatest risk for people who smoke.99, 100 Individuals can lower their risk of radon exposure by having their home tested for the gas, regardless of where they live, and taking recommended steps to mitigate exposure when necessary (cdc.gov/ radon/radon-action.html).101, 102 Asbestos and Other Occupational Exposures Occupational exposures associated with increased lung cancer risk include chemical mixtures, such as soot and coal-tar pitch, and compounds such as nickel, chromium, and asbestos.103 In addition, people who work in aluminum production, painting, and steel founding have elevated risk. Work that involves heavy asbestos exposure increases the risk of lung cancer by approximately 70%,104 with exposure to longer and thinner fibers associated with stronger risk.105 Studies of the effectiveness of workplace interventions to limit exposure and mitigate risk remain scarce.
Air Pollution Air pollution is estimated to account for about 1%-2% of lung cancer deaths in the US.106 Outdoor air pollution is made up of a variety of pollutants from many sources, including power generation, transportation, and industrial and agricultural emissions.107, 108 Inhaling particulate matter, a microscopic mixture of solid and liquid pollutants, is linked to an 8%-9% increased risk of lung cancer.108-110 Levels of particulate matter are highest in the Eastern US, but are generally low compared to other parts of the world, such as Asia, North Africa, and the Middle East. Sources of indoor air pollution include coal use in homes,111, 112 burning of biomass for cooking and heating,113 and cooking oil fumes.114, 115 The use of electricity or natural gas for heating and cooking, as well as improved home ventilation, can help prevent increased lung cancer risk due to poor indoor air quality. Arsenic High levels of arsenic in drinking water (at least several hundred micrograms per liter) have been strongly associated with lung cancer in Chile and Taiwan,116-118 but the risk for lower levels, as found in the US, is less clear.116, 119-122 Only a few US counties, mostly in the Southwest, have mean concentrations exceeding 10 µg/L, the Environmental Protection Agency’s maximum concentration limit.122, 123 Tobacco Control Policies Since the publication of the landmark 1964 Surgeon General’s Report on the health hazards of cigarette smoking, tobacco control has led to dramatic declines in cigarette consumption (Figure S6) and, consequently, lung cancer incidence and mortality. Some of the most important tobacco control measures include insurance coverage of tobacco cessation, tobacco excise taxes, laws against smoking in public places, counteradvertising, increasing the tobacco sales age to 21, federal regulation, and funding for evidence-based tobacco control programs.124 Collectively, these measures have contributed to substantial declines in tobacco use in the US.
Figure S11. 5-year Relative Survival Rates for Lung Cancer by Subtype, Stage at Diagnosis, and Sex, US, 2012-2018 Percent NSCLC: Non-small cell lung cancer; SCLC: Small cell lung cancer. Survival rates are for patients diagnosed during 2012-2018, all followed through 2019.
Source: Surveillance, Epidemiology, and End Results 17 Registries, 2022.
©2023, American Cancer Society, Inc., Surveillance and Health Equity Science Male Female 0 10 20 30 40 50 60 70 80 All Stages Distant Regional Local All Stages Distant Regional Local 59 33 43 33 7 11 23 32 29 17 18 70 2 6 8 4 NSCLC SCLC Cancer Facts & Figures 2023 41 Tobacco Excise Taxes Increasing excise taxes on tobacco products regularly and significantly is one of the most effective tobacco control policies. It promotes smoking cessation among adults, discourages initiation in adolescents, and lowers the number of cigarettes smoked among those unable to quit. For each 10% increase in the price of cigarettes, cigarette consumption decreases by an estimated 3%-5%.125 The decrease is nearly double among youths and individuals with lower SES.125-128 The federal cigarette tax has been $1.01 since 2009, and the average state tax was $1.91 as of October 1, 2022,129 ranging from $0.17 in Missouri to $4.50 in the District of Columbia.29 Smoke-free Public Places Comprehensive smoke-free laws (prohibiting smoking in workplaces, restaurants, bars, and gaming facilities) reduce secondhand smoke exposure and youth initiation while increasing cessation, thereby reducing the risk of smoking-related diseases.78, 125 As of October 2022, 62.5% of the US population lived in areas covered by 100% smoke-free laws in the workplace, restaurants, and bars.130 Smoking Cessation Smoking cessation is associated with a reduced risk of lung cancer and is beneficial at any age, with benefits increasing with earlier age at successful cessation.125 Those people who smoke and quit before age 40 reduce their risk of lung cancer by 90% compared to those who continue to smoke throughout their lifetime. In general, people who smoke for their entire adult life lose a decade or more of life compared to people who never smoke because of premature death from lung cancer and other smoking-related diseases.131, 132 Successful cessation usually takes an average of six attempts and increases with the use of FDA-approved cessation medications with counseling.125, 133, 134 Although cessation attempts are highest among Black and AAPI individuals, successful cessation is highest among White individuals,29 perhaps related to the use of cessation aids.135 For more information about reducing tobacco use and exposure, see page 57.
Lung Cancer Screening Lung cancer screening trials in the 1970s using chest radiography (x-rays) with or without sputum (mucus and other matter brought up from the lungs by coughing) examination showed no improvement in patient outcomes.136 In 2011, however, screening high-risk individuals (ages 55 to 74 years with a 30+ pack-year smoking history) with annual low-dose computed tomography (LDCT)137, 138 in the National Lung Screening Trial (NLST) was associated with a 20% reduction in lung cancer mortality compared to chest radiography. More recently, two European trials reported even larger mortality reductions among participants with more moderate disease risk.139, 140 The American Cancer Society and the US Preventive Services Task Force (USPSTF) began recommending lung cancer screening using LDCT for high-risk individuals in 2013,35 and have since expanded the eligibility criteria to people ages 50-80 years with a 20+ pack-year smoking history who currently smoke or have quit within the past 15 years.141 Pack-years is a measurement of smoking history that takes into account duration and quantity of cigarette consumption, both of which determine lung cancer risk. An individual who smokes one pack a day for 20 years and one who smokes two packs a day for 10 years both have a 20 pack-year smoking history. Individuals who do not meet the 20 pack-year threshold, or who do but quit more than 15 years ago, are still at a 10-fold increased risk of lung cancer compared to people who have never smoked.142 There were approximately 8.5 million adults eligible for lung cancer screening in 2020.
Despite evidence that screening in high-risk populations reduces lung cancer mortality, uptake has been low, especially in several Southern states with a high lung cancer burden.143 The exception to this pattern is Kentucky, which although still low, has one of the highest screening rates in the country (13.7% in 2018) as a result of community-engaged programs,144 governmental support,145 expanded Medicaid eligibility, and no preauthorization requirement for lung cancer screening in Medicaid fee-for-service coverage.143, 146 Despite generally low uptake, the proportion of cases diagnosed at a localized stage began to increase 42 Cancer Facts & Figures 2023 following the 2013 USPSTF guideline update, surpassing regional stage in 2016 (Figure S12). In contrast to recent screening rates for some other cancers,147 lung cancer screening rates did not decline in the first year of the COVID-19 pandemic and, in fact, increased in 19 states between 2019 and 2020.148 While this increase is promising, national screening rates among eligible high-risk individuals (about 8.5 million people) remain low (6.5% in 2020). People who are screened are more likely to be older, female, and current smokers,149 with the greatest barriers to screening among Black and socioeconomically disadvantaged individuals.150 Patient and provider education is important for increasing uptake among eligible adults, with patients often placing trust in the decision of their provider.151 The Economic Impact of Lung Cancer Lung cancer causes a substantial loss of earnings in the US, approximately $13 billion in 2019,152 which does not include the costs associated with a cancer diagnosis such as treatment and caregiving. As cancer treatment advances, as it has for lung cancer, the cost of those treatments increases, causing cancer patients and their families to face increasing out-of-pocket costs to receive treatment.153-155 The high cost associated with a lung cancer diagnosis likely exacerbates economic disparities among diagnosed individuals, a disproportionate number of whom are already impoverished due to higher smoking prevalence among those with low SES.24 What Is the American Cancer Society Doing About Lung Cancer?
Research The American Cancer Society, through our Extramural Discovery Science program, funds individual investigators at medical schools, universities, research institutes, and hospitals throughout the US. Currently, this program is funding $28 million in lung cancer research through 70 research grants. Ongoing research includes: • Determining how to deliver high-quality cancer care that maximizes patient quality of life and delivers care that is consistent with patients’ values and preferences • Targeting cancer stem cells to induce anti-tumor immunity in small cell lung cancer Figure S12. Trends in Lung Cancer Incidence Rates by Race and Stage, US, 2004-2019 Cases per 100,000 population Age adjusted to the 2000 US Standard Population and adjusted for delays in case reporting. Rates for White and Black individuals are exclusive of individuals identifying as Hispanic. Source: Surveillance, Epidemiology, and End Results 17 Registries, 2022.
©2023, American Cancer Society, Inc., Surveillance and Health Equity Science Regional Distant Local Unstaged Regional Regional Distant Distant Local Local Unstaged Unstaged All races and ethnicities Black individuals White individuals 0 5 10 15 20 25 30 35 40 45 2018 2016 2014 2012 2010 2008 2006 2004 0 5 10 15 20 25 30 35 40 45 2018 2016 2014 2012 2010 2008 2006 2004 0 5 10 15 20 25 30 35 40 45 2018 2016 2014 2012 2010 2008 2006 2004 Cancer Facts & Figures 2023 43 • Predicting and tracking response to immunotherapy by harnessing information from cell-free DNA in the blood • Developing diagnostic tools for the early detection of lung cancer and the discovery of precision medicine to provide personalized targeted treatment • Utilizing an updated CRISPR approach to identify new tumor-specific vulnerabilities that can be targeted in combination with existing therapies in non-small cell lung cancers National Lung Cancer Roundtable Since its inception in 2017, the American Cancer Society National Lung Cancer Roundtable (NLCRT) has galvanized more than 190 organizations and over 200 leading experts, as well as patient and caregiver advocate representatives, at the national, state, and local levels to collectively partner to problem-solve and achieve enduring systematic change to reduce deaths from lung cancer. The NLCRT engages experts in multidisciplinary problem-solving collaborations, catalyzes action to create, build, and strengthen innovative solutions, and develops and disseminates evidence-based interventions and best practices. The work of the roundtable is guided by its Steering Committee and conducted through the efforts of its 10 strategic priority task groups. The NLCRT engages in public and provider education, targeted research, and health policy initiatives. The roundtable advances lung cancer-related health equity by identifying and working to overcome barriers to equitable access to promote implementation, uptake, and adherence of lung cancer screening and nodule detection and management, promote guideline-concordant staging, and optimize the use of biomarker testing to guide appropriate and timely therapy and care, eliminate the pervasive stigma associated with lung cancer, and strengthen state-based initiatives. The NLCRT was recommended by the 2022 President’s Cancer Panel Report as a priority cancer control model that effectively harnesses the collective power and expertise of the entire lung cancer community to close gaps in cancer screening by connecting people, communities, and systems to improve equity and access. The NLCRT’s mission is to create lung cancer survivors. Visit NLCRT.org for more information. Advocacy Our advocacy affiliate, the American Cancer Society Cancer Action NetworkSM (ACS CAN), is involved in advocacy efforts at both the federal and state levels that reduce the prevalence of tobacco product use and increase access to quality lung cancer screening, treatment, and care. Following are some of the ways ACS CAN is fighting to reduce the impact of tobacco and lung cancer in the US.
• ACS CAN advocates for insurance coverage for comprehensive biomarker testing in state-regulated insurance plans including Medicaid when supported by medical and scientific evidence. • The organization sponsors research seeking to better understand private insurance coverage for comprehensive biomarker testing in lung cancer. • ACS CAN works to improve clinical trial diversity and searchability, including for lung cancer, by improving clinicaltrials.gov, enabling electronic health records to automatically screen patients for trials, and modernizing outdated eligibility criteria for trial entry.
• The organization advocates for evidence-based tobacco control policies to prevent initiation and aid in cessation, including increases in tobacco excise taxes, comprehensive smoke-free laws, insurance coverage for tobacco cessation services, funding for tobacco control programs, and federal regulation of tobacco products, including the prohibition of all flavors in all tobacco products.
• ACS CAN advocates for insurance coverage with no cost sharing of lung cancer screening by all payers, including Medicare, Medicaid, and private insurers.
44 Cancer Facts & Figures 2023 • The organization believes that the American public should be made aware of the known information concerning the potential problem of radon contamination in certain housing areas in the US and how to reduce such a risk. ACS CAN urges federal, state, and local governments to approve legislation that reduces the potential health threat posed by radon by implementing public awareness campaigns and requiring disclosure of radon levels by builders, homeowners, schools, and daycare facilities.
• ACS CAN supports a cap on total out-of-pocket spending for Medicare beneficiaries.
• The organization advocates that Medicare and Medicaid provide coverage of patient navigation services. Patient navigators have shown to help increase cancer screening rates among historically marginalized racial and ethnic populations by providing access to disease prevention education, conducting community outreach, and facilitating public education campaigns. Additionally, given that many cancer screening guidelines are based on family history and personal risk factors, patient navigators can offer individualized advice and help patients assess individual eligibility, improving compliance by increasing a patient’s cancer knowledge and understanding their unique health risks. As such, health care systems should prioritize the use of patient navigators for helping to move cancer patients smoothly and effectively through separate phases of cancer care.
Resources for Lung Cancer Patients and Their Families Lungevity lungevity.org Lung Cancer HELPLine: 1-844-360-5864 Lungevity is the nation’s leading lung cancer-focused nonprofit organization. Their goal is to change outcomes for people with lung cancer through research, education, and support. They have numerous education and supportive resources available to lung cancer patients, including peer-to-peer support and survivorship programs.
CancerCare cancercare.org/diagnosis/lung_cancer CancerCare provides free professional support services for lung cancer patients and their families. These include financial assistance, support groups, and educational materials.
American Lung Association lung.org Lung HelpLine: 1-800-LUNG-USA (1-800-586-4872) The American Lung Association is the nation’s leading organization working to save lives by improving lung health and preventing lung disease through education, advocacy, and research. They provide education on and supportive resources for lung cancer patients and their families, as well as resources to help with smoking cessation.
References 1. Islami F, Goding Sauer A, Miller KD, et al. Proportion and number of cancer cases and deaths attributable to potentially modifiable risk factors in the United States. CA Cancer J Clin. Jan 2018;68(1):31-54. doi:10.3322/caac.21440.
2. Siegel RL, Miller KD, Sandeep-Wagle N, Jemal A. Cancer Statistics, 2023. CA Cancer J Clin. 2023.
3. Nicholson AG, Tsao MS, Beasley MB, et al. The 2021 WHO Classification of Lung Tumors: Impact of advances since 2015. J Thorac Oncol. Nov 19 2021;doi:10.1016/j.jtho.2021.11.003.
4. Wakelee HA, Chang ET, Gomez SL, et al. Lung cancer incidence in never smokers. J Clin Oncol. Feb 10 2007;25(5):472-8. doi:10.1200/ JCO.2006.07.2983.
5. Song MA, Benowitz NL, Berman M, et al. Cigarette Filter Ventilation and its Relationship to Increasing Rates of Lung Adenocarcinoma. J Natl Cancer Inst. Dec 1 2017;109(12)doi:10.1093/ jnci/djx075.
6. Araujo LH, Horn L, Merritt RE, Shilo K, Xu-Welliver M, Carbone DP. Cancer of the lung: non-small cell lung cancer and small cell lung cancer. Abeloff’s clinical oncology. Elsevier; 2020:1108-1158. e16.
7. Chu ZG, Yang ZG, Shao H, et al. Small peripheral lung adenocarcinoma: CT and histopathologic characteristics and prognostic implications. Cancer Imaging. Dec 28 2011;11(1):237-46. doi:10.1102/1470-7330.2011.0033.
8. Thai AA, Solomon BJ, Sequist LV, Gainor JF, Heist RS. Lung cancer. Lancet. Aug 7 2021;398(10299):535-554. doi:10.1016/S0140-6736(21)00312-3.
Cancer Facts & Figures 2023 45 9. Surveillance, Epidemiology, and End Results (SEER) Program (www.seer.cancer.gov) SEERStat Database: Incidence – SEER Research Data, 17 Registries, Nov 2021 Sub (2000-2019) – Linked To County Attributes – Time Dependent (1990-2019) Income/Rurality, 1969-2020 Counties, National Cancer Institute, DCCPS, Surveillance Research Program, released April 2022, based on the November 2021 submission. 10.Data from: SEERStat Database: NAACCR Incidence Data – CiNA Research Data, 1995-2019, with Race/Ethnicity, Standard File, ACS Facts & Figures projection Project (which includes data from CDC’s National Program of Cancer Registries (NPCR), CCR’s Provincial and Territorial Registries, and the NCI’s Surveillance, Epidemiology and End Results (SEER) Registries), certified by the North American Association of Central Cancer Registries (NAACCR) as meeting high-quality incidence data standards for the specified time periods, submitted December 2021. 11. Byers LA, Rudin CM. Small cell lung cancer: where do we go from here? Cancer. Mar 1 2015;121(5):664-72. doi:10.1002/cncr.29098.
12. Miller KD, Nogueira L, Devasia T, et al. Cancer treatment and survivorship statistics, 2022. CA Cancer J Clin. Jun 23 2022;doi:10.3322/caac.21731.
13. Gallicchio L, Devasia TP, Tonorezos E, Mollica MA, Mariotto A. Estimation of the numbers of individuals living with metastatic cancer in the United States. J Natl Cancer Inst. Aug 22 2022;doi:10.1093/jnci/djac158.
14. Howlader N NA, Krapcho M, Miller D, Brest A, Yu M, Ruhl J, Tatalovich Z, Mariotto A, Lewis DR, Chen HS, Feuer EJ, Cronin KA (eds), ed. SEER Cancer Statistics Review, 1975-2018. 2021.
cancer.gov/archive/csr/1975_2018/, based on November 2020 SEER data submission, posted to the SEER web site, April 2021.
15. Thun MJ, Carter BD, Feskanich D, et al. 50-year trends in smoking-related mortality in the United States. N Engl J Med. Jan 24 2013;368(4):351-64. doi:10.1056/NEJMsa1211127.
16. Data from: SEERStat Database: NAACCR Incidence Data - CiNA Research Data, 2015-2019, Delay Adj Factors - ACS Facts & Figures (which includes data from CDC’s National Program of Cancer Registries (NPCR), CCR’s Provincial and Territorial Registries, and the NCI’s Surveillance, Epidemiology and End Results (SEER) Registries), certified by the North American Association of Central Cancer Registries (NAACCR) as meeting high-quality incidence data standards for the specified time periods, submitted December 2021. 17. Jemal A, Miller KD, Ma J, et al. Higher Lung Cancer Incidence in Young Women Than Young Men in the United States. N Engl J Med. May 24 2018;378(21):1999-2009. doi:10.1056/NEJMoa1715907. 18. Data from: SEER 22 Incidence and Mortality, 2000-2019, with Kaposi Sarcoma and Mesothelioma. 19. Kratzer TB, Jemal A, Miller KD, et al. Cancer statistics for American Indian and Alaska Native individuals, 2022: Including increasing disparities in early onset colorectal cancer. CA Cancer J Clin. 2022.
20. Taparra K, Pellegrin K. Data aggregation hides Pacific Islander health disparities. Lancet. Jul 2 2022;400(10345):2-3. doi:10.1016/ S0140-6736(22)01100-X.
21. Torre LA, Sauer AM, Chen MS, Jr., Kagawa-Singer M, Jemal A, Siegel RL. Cancer statistics for Asian Americans, Native Hawaiians, and Pacific Islanders, 2016: Converging incidence in males and females. CA Cancer J Clin. May 2016;66(3):182-202. doi:10.3322/ caac.21335.
22. DeRouen MC, Canchola AJ, Thompson CA, et al. Incidence of Lung Cancer Among Never-Smoking Asian American, Native Hawaiian, and Pacific Islander Females. J Natl Cancer Inst. Jan 11 2022;114(1):78-86. doi:10.1093/jnci/djab143.
23. Pinheiro PS, Callahan KE, Siegel RL, et al. Cancer Mortality in Hispanic Ethnic Groups. Cancer Epidemiol Biomarkers Prev. Mar 2017;26(3):376-382. doi:10.1158/1055-9965.Epi-16-0684.
24. Hiscock R, Bauld L, Amos A, Fidler JA, Munafò M. Socioeconomic status and smoking: a review. Ann. N. Y. Acad. Sci. 2012;1248(1):107-123. 25. Ma J, Jemal A. Temporal Trends in Mortality From Major Cancers by Education in the United States, 2001–2016. JNCI Cancer Spectrum. 2019;3(4)doi:10.1093/jncics/pkz087.
26. Gupta A, Omeogu CH, Islam JY, Joshi AR, Akinyemiju TF. Association of area-level socioeconomic status and non-small cell lung cancer stage by race/ethnicity and health care-level factors: Analysis of the National Cancer Database. Cancer. Jun 20 2022;doi:10.1002/cncr.34327.
27. Moscelli G, Siciliani L, Gutacker N, Cookson R. Socioeconomic inequality of access to healthcare: Does choice explain the gradient? J Health Econ. Jan 2018;57:290-314. doi:10.1016/j.
jhealeco.2017.06.005.
28. Fitzpatrick AL, Powe NR, Cooper LS, Ives DG, Robbins JA. Barriers to health care access among the elderly and who perceives them. Am. J. Public Health. 2004;94(10):1788-1794. 29. American Cancer Society. Cancer Prevention & Early Detection Facts & Figures 2022 Tables and Figures. 2022. 30. Thun M, Peto R, Boreham J, Lopez AD. Stages of the cigarette epidemic on entering its second century. Tob Control. Mar 2012;21(2):96-101. doi:10.1136/tobaccocontrol-2011-050294.
31. Smoking USSGsACo. Smoking and Health: Report of the Advisory Committee to the Surgeon General of the Public Health Service. US Department of Health, Education, and Welfare, Public Health Service; 1964.
32. Siegel RL, Miller KD, Fuchs HE, Jemal A. Cancer statistics, 2022. CA Cancer J Clin. Jan 2022;72(1):7-33. doi:10.3322/caac.21708.
33. Humphrey LL, Deffebach M, Pappas M, et al. Screening for lung cancer with low-dose computed tomography: a systematic review to update the US Preventive Services Task Force recommendation. Ann. Intern. Med. 2013;159(6):411-420. 34. Jones GS, Baldwin DR. Recent advances in the management of lung cancer. Clin Med (Lond). Apr 1 2018;18(Suppl 2):s41-s46. doi:10.7861/clinmedicine.18-2-s41.
35. Howlader N, Forjaz G, Mooradian MJ, et al. The Effect of Advances in Lung-Cancer Treatment on Population Mortality. N Engl J Med. Aug 13 2020;383(7):640-649. doi:10.1056/NEJMoa1916623.
36. Siegel RL, Miller KD, Fuchs HE, Jemal A. Cancer Statistics, 2021. CA Cancer J Clin. 2021;71(1):7-33. doi:
37. Health, United States, 2019. Hyattsville, Maryland. 2021. DOI: 38. Jemal A, Center MM, Ward E. The convergence of lung cancer rates between blacks and whites under the age of 40, United States. Cancer Epidemiol. Biomarkers Prev. 2009;18(12):3349-3352. 39. Fiore MC, Novotny TE, Pierce JP, Hatziandreu EJ, Patel KM, Davis RM. Trends in cigarette smoking in the United States: the changing influence of gender and race. JAMA. 1989;261(1):49-55. 46 Cancer Facts & Figures 2023 40. Anderson C, Burns DM. Patterns of adolescent smoking initiation rates by ethnicity and sex. Tob. Control. 2000;9(suppl 2):ii4-ii8. 41. US Department of Health and Human Services. Tobacco use among US racial/ethnic minority groups—African Americans, American Indians and Alaska Natives, Asian Americans and Pacific Islanders, and Hispanics: A report of the surgeon general. National Center for Chronic Disease Prevention and Health Promotion, Office on Smoking and Health. 1998;90.
42. Reyes-Guzman CM, Pfeiffer RM, Lubin J, et al. Determinants of Light and Intermittent Smoking in the United States: Results from Three Pooled National Health Surveys. Cancer Epidemiol Biomarkers Prev. Feb 2017;26(2):228-239. doi:10.1158/1055-9965.EPI-16-0028.
43. Felicitas-Perkins JQ, Sakuma KLK, Blanco L, et al. Smoking among Hispanic/Latino nationality groups and whites, comparisons between California and the United States. Nicotine Tob Res. 2018;20:1085-1094.
44. Minguet J, Smith KH, Bramlage P. Targeted therapies for treatment of non‐small cell lung cancer—recent advances and future perspectives. Int. J. Cancer. 2016;138(11):2549-2561. 45. Rittmeyer A, Barlesi F, Waterkamp D, et al. Atezolizumab versus docetaxel in patients with previously treated non-small-cell lung cancer (OAK): a phase 3, open-label, multicentre randomised controlled trial. Lancet. 2017;389(10066):255-265. 46. Vokes EE, Ready N, Felip E, et al. Nivolumab versus docetaxel in previously treated advanced non-small-cell lung cancer (CheckMate 017 and CheckMate 057): 3-year update and outcomes in patients with liver metastases. Ann. Oncol. 2018;29(4):959-965. 47. Rami-Porta R, Call S, Dooms C, et al. Lung cancer staging: a concise update. Eur Respir J. 2018;51(5).
48. Chang JW, Asamura H, Kawachi R, Watanabe S. Gender difference in survival of resected non-small cell lung cancer: histology-related phenomenon? J Thorac Cardiovasc Surg. Apr 2009;137(4):807-12. doi:10.1016/j.jtcvs.2008.09.026.
49. Rodriguez-Lara V, Hernandez-Martinez JM, Arrieta O. Influence of estrogen in non-small cell lung cancer and its clinical implications. J Thorac Dis. Jan 2018;10(1):482-497. doi:10.21037/ jtd.2017.12.61.
50. Midha A, Dearden S, McCormack R. EGFR mutation incidence in non-small-cell lung cancer of adenocarcinoma histology: a systematic review and global map by ethnicity (mutMapII). Am. J. Cancer Res. 2015;5(9):2892. 51. Javid SH, Varghese TK, Morris AM, et al. Guideline-concordant cancer care and survival among American Indian/Alaskan Native patients. Cancer. Jul 15 2014;120(14):2183-90. doi:10.1002/cncr.28683.
52. Lovly CM. Expanding Horizons for Treatment of Early-Stage Lung Cancer. N Engl J Med. May 26 2022;386(21):2050-2051. doi:10.1056/NEJMe2203330.
53. Lansdorp-Vogelaar I, Kuntz KM, Knudsen AB, van Ballegooijen M, Zauber AG, Jemal A. Contribution of screening and survival differences to racial disparities in colorectal cancer rates. Cancer Epidemiol Biomarkers Prev. May 2012;21(5):728-36. doi:10.1158/1055-9965.Epi-12-0023.
54. Pinheiro PS, Morris CR, Liu L, Bungum TJ, Altekruse SF. The impact of follow-up type and missed deaths on population-based cancer survival studies for Hispanics and Asians. J Natl Cancer Inst Monogr. Nov 2014;2014(49):210-7. doi:10.1093/jncimonographs/lgu016.
55. Vieira AR, Abar L, Vingeliene S, et al. Fruits, vegetables and lung cancer risk: a systematic review and meta-analysis. Ann Oncol. Jan 2016;27(1):81-96. doi:10.1093/annonc/mdv381.
56. Holford TR, Levy DT, McKay LA, et al. Patterns of birth cohort-specific smoking histories, 1965-2009. Am J Prev Med. Feb 2014;46(2):e31-7. doi:10.1016/j.amepre.2013.10.022.
57. Anderson CM, Burns DM, Dodd KW, Feuer EJ. Chapter 2: Birth-cohort-specific estimates of smoking behaviors for the U.S. population. Risk Anal. Jul 2012;32 Suppl 1:S14-24. doi:10.1111/j.1539-6924.2011.01703.x.
58. Burns D, Lee L, Shen L, et al. Cigarette smoking behavior in the United States: changes in cigarette-related disease risks and their implication for prevention and control. Smoking and tobacco control monograph no. 8. Bethesda, MD: National Cancer Institute, 1997:13-112. (NIH publication no. 97-4213.). 59. U.S. Department of Health and Human Services, Substance Abuse and Mental Health Services Administration, Center for Behavioral Health Statistics and Quality. National Survey on Drug Use and Health, 2020. Accessed August 3, 2022.
samhsa.gov/.
60. Rostron B. Lung cancer mortality risk for U.S. menthol cigarette smokers. Nicotine Tob Res. Oct 2012;14(10):1140-4. doi:10.1093/ntr/ nts014.
61. Brooks DR, Palmer JR, Strom BL, Rosenberg L. Menthol cigarettes and risk of lung cancer. Am J Epidemiol. Oct 1 2003;158(7):609-16; discussion 617-20. doi:10.1093/aje/kwg182.
62. Blot WJ, Cohen SS, Aldrich M, McLaughlin JK, Hargreaves MK, Signorello LB. Lung cancer risk among smokers of menthol cigarettes. J Natl Cancer Inst. May 18 2011;103(10):810-6. doi:10.1093/ jnci/djr102.
63. Haiman CA, Stram DO, Wilkens LR, et al. Ethnic and racial differences in the smoking-related risk of lung cancer. N Engl J Med. 2006;354(4):333-342. 64. Cobb N, Espey D, King J. Health behaviors and risk factors among American Indians and Alaska Natives, 2000-2010. Am J Public Health. 2014;104(suppl 3): S481–S489. doi:10.21 05/A JPH. 2014 .301 879 65. Garrett BE, Dube SR, Trosclair A, et al. Cigarette smoking – United States, 1965-2008. MMWR Suppl. Jan 14 2011;60(1):109-13. 66. Smoking Cessation: A Report of the Surgeon General. 2020. Publications and Reports of the Surgeon General.
67. Cromer KJ, Wofford L, Wyant DK. Barriers to Healthcare Access Facing American Indian and Alaska Natives in Rural America. J Community Health Nurs. Oct-Dec 2019;36(4):165-187. doi:10.1080/0737 0016.2019.1665320.
68. Odani S, Armour BS, Graffunder CM, Garrett BE, Agaku IT. Prevalence and Disparities in Tobacco Product Use Among American Indians/Alaska Natives – United States, 2010-2015. MMWR Morb Mortal Wkly Rep. Dec 22 2017;66(50):1374-1378. doi:10.15585/mmwr.mm6650a2.
69. Services USDoHaH. Tobacco use among U.S. racial/ethnic groups— African Americans, American Indians and Alaska Natives, Asian Americans and Pacific Islanders, and Hispanics: a report of the Surgeon General. 1998. 70. Xu X, Pesko MF, Tynan MA, Gerzoff RB, Malarcher AM, Pechacek TF. Cigarette price-minimization strategies by U.S. smokers. Am J Prev Med. May 2013;44(5):472-6. doi:10.1016/j.
amepre.2013.01.019.
Cancer Facts & Figures 2023 47 71. Choi WS, Beebe LA, Nazir N, et al. All Nations Breath of Life: A Randomized Trial of Smoking Cessation for American Indians. Am J Prev Med. Nov 2016;51(5):743-751. doi:10.1016/j.amepre.2016.05.021.
72. Kunitz SJ. Historical Influences on Contemporary Tobacco Use by Northern Plains and Southwestern American Indians. Am J Public Health. Feb 2016;106(2):246-55. doi:10.2105/AJPH.2015.302909.
73. Miller KD, Ortiz AP, Pinheiro PS, et al. Cancer statistics for the US Hispanic/Latino population, 2021. CA Cancer J Clin. 2021;71(6):466-487. 74. Data from: Hawai’i Health Data Warehouse: Cigarettes – Current Smoking, Adult. 75. Christensen CH, Rostron B, Cosgrove C, et al. Association of cigarette, cigar, and pipe use with mortality risk in the US population. JAMA Intern. Med. 2018;178(4):469-476. 76. Chen-Sankey JC, Mead-Morse EL, Le D, et al. Cigar-Smoking Patterns by Race/Ethnicity and Cigar Type: A Nationally Representative Survey Among U.S. Adults. Am J Prev Med. Jan 2021;60(1):87-94. doi:10.1016/j.amepre.2020.07.005.
77. Park SH, Duncan DT, Shahawy OE, et al. Analysis of State-Specific Prevalence, Regional Differences, and Correlates of Hookah Use in U.S. Adults, 2012-2013. Nicotine Tob Res. 2016;19(11):1365-1374. doi:10.1093/ntr/ntw229.
78. US Department of Health and Human Services. The health consequences of smoking – 50 years of progress: a report of the Surgeon General. Atlanta, GA: US Department of Health and Human Services, Centers for Disease Control and Prevention; 2014.
79. Brody DJ, Faust E, Tsai J. Secondhand smoke exposure among nonsmoking adults: United States, 2015-2018. 2021.
80. US Centers for Disese Control and Prevention, Office on Smoking and Health. How tobacco smoke causes disease: The biology and behavioral basis for smoking-attributable disease: A report of the surgeon general. 2010.
81. Kim CH, Lee YCA, Hung RJ, et al. Exposure to secondhand tobacco smoke and lung cancer by histological type: a pooled analysis of the International Lung Cancer Consortium (ILCCO). Int. J. Cancer. 2014;135(8):1918-1930. 82. Brenner DR, Boffetta P, Duell EJ, et al. Previous Lung Diseases and Lung Cancer Risk: A Pooled Analysis From the International Lung Cancer Consortium. Am. J. Epidemiol. 2012;176(7):573-585. doi:10.1093/aje/kws151.
83. Brenner DR, McLaughlin JR, Hung RJ. Previous lung diseases and lung cancer risk: a systematic review and meta-analysis. PloS one. 2011;6(3):e17479. 84. Rutgers SR, Postma DS, Ten Hacken NH, et al. Ongoing airway inflammation in patients with COPD who do not currently smoke. Thorax. 2000;55(1):12-18. 85. Moldoveanu B, Otmishi P, Jani P, et al. Inflammatory mechanisms in the lung. J. Inflamm. Res. 2009;2:1. 86. Engels EA. Inflammation in the development of lung cancer: epidemiological evidence. Expert Rev. Anticancer Ther. 2008;8(4):605-615. 87. Brody JS, Spira A. State of the art. Chronic obstructive pulmonary disease, inflammation, and lung cancer. Proc. Am. Thorac. Soc. 2006;3(6):535-537. 88. Ang L, Ghosh P, Seow WJ. Association between previous lung diseases and lung cancer risk: a systematic review and meta-analysis. Carcinogenesis. 2021;42(12):1461-1474. doi:10.1093/carcin/ bgab082.
89. Benusiglio PR, Fallet V, Sanchis-Borja M, Coulet F, Cadranel J. Lung cancer is also a hereditary disease. Eur Respir Rev. Dec 31 2021;30(162)doi:10.1183/16000617.0045-2021.
90. Tokuhata GK, Lilienfeld AM. Familial aggregation of lung cancer among hospital patients. Public Health Rep (1896). Apr 1963;78:277-83. 91. Cote ML, Liu M, Bonassi S, et al. Increased risk of lung cancer in individuals with a family history of the disease: a pooled analysis from the International Lung Cancer Consortium. Eur J Cancer. Sep 2012;48(13):1957-68. doi:10.1016/j.ejca.2012.01.038.
92. Bougeard G, Renaux-Petel M, Flaman JM, et al. Revisiting Li-Fraumeni Syndrome From TP53 Mutation Carriers. J Clin Oncol. Jul 20 2015;33(21):2345-52. doi:10.1200/JCO.2014.59.5728.
93. Lou Y, Pecot CV, Tran HT, et al. Germline Mutation of T790M and Dual/Multiple EGFR Mutations in Patients With Lung Adenocarcinoma. Clin Lung Cancer. Mar 2016;17(2):e5-11. doi:10.1016/j.cllc.2015.11.003.
94. Gazdar A, Robinson L, Oliver D, et al. Hereditary lung cancer syndrome targets never smokers with germline EGFR gene T790M mutations. J Thorac Oncol. Apr 2014;9(4):456-63. doi:10.1097/ JTO.0000000000000130.
95. Reckamp KL, Behrendt CE, Slavin TP, et al. Germline mutations and age at onset of lung adenocarcinoma. Cancer. Aug 1 2021;127(15):2801-2806. doi:10.1002/cncr.33573.
96. Li S, Silvestri V, Leslie G, et al. Cancer Risks Associated With BRCA1 and BRCA2 Pathogenic Variants. J Clin Oncol. May 10 2022;40(14):1529-1541. doi:10.1200/JCO.21.02112.
97. IARC Monographs on the Evaluation of the Carcinogenic Risks to Humans: Man-made Mineral Fibres and Radon. 1988. 98. National Research Council (U.S.). Committee on Health Risks of Exposure to Radon. Health effects of exposure to radon. Beir. National Academy Press; 1999:xiv, 500 p.
99. World Health Organization. WHO handbook on indoor radon : a public health perspective. World Health Organization; 2009:xiv, 94 p.
100. Darby S, Hill D, Auvinen A, et al. Radon in homes and risk of lung cancer: collaborative analysis of individual data from 13 European case-control studies. BMJ. Jan 29 2005;330(7485):223. doi:10.1136/bmj.38308.477650.63.
101. A Citizen’s Guide to Radon: The Guide to Protecting Yourself and Your Family from Radon. 2012. Accessed August 12, 2022. sites/default/files/2016-02/documents/2012_a_citizens_guide_to_radon.pdf.
102. Agency UEP. Consumer’s Guide to Radon Reduction. How to Fix Your Home. 2016.
103. Field RW, Withers BL. Occupational and environmental causes of lung cancer. Clin. Chest Med. 2012;33(4):681-703. 104. Järvholm B, Aström E. The risk of lung cancer after cessation of asbestos exposure in construction workers using pleural malignant mesothelioma as a marker of exposure. J Occup Environ Med. Dec 2014;56(12):1297-301. doi:10.1097/jom.0000000000000258.
105. Loomis D, Dement J, Richardson D, Wolf S. Asbestos fibre dimensions and lung cancer mortality among workers exposed to chrysotile. Occup Environ Med. Sep 2010;67(9):580-4. doi:10.1136/ oem.2009.050120.
106. Alberg AJ, Samet JM. Epidemiology of lung cancer. Chest. 2003;123(1):21S-49S. 107. Unger N, Bond TC, Wang JS, et al. Attribution of climate forcing to economic sectors. Proc. Natl. Acad. Sci. U.S.A. 2010;107(8):3382-3387. 48 Cancer Facts & Figures 2023 108. International Agency for Research on Cancer. IARC monographs on the evaluation of carcinogenic risks to humans: vol. 109, outdoor air pollution. Lyon, France: IARC. 2015.
109. Hamra GB, Guha N, Cohen A, et al. Outdoor Particulate Matter Exposure and Lung Cancer: A Systematic Review and Meta-Analysis. Environ. Health Perspect. 2014;122(9):906-911. doi:10.1289/ ehp/1408092.
110. Brauer M, Amann M, Burnett RT, et al. Exposure assessment for estimation of the global burden of disease attributable to outdoor air pollution. Envrion. Sci. Technol. 2012;46(2):652-660. 111. Hosgood III HD, Wei H, Sapkota A, et al. Household coal use and lung cancer: systematic review and meta-analysis of case– control studies, with an emphasis on geographic variation. Int. J. Epidemiol. 2011;40(3):719-728. 112. Lan Q, Chapman RS, Schreinemachers DM, Tian L, He X. Household stove improvement and risk of lung cancer in Xuanwei, China. J. Natl. Cancer Inst. 2002;94(11):826-835. 113. Smith KR, Aggarwal A, Dave R. Air pollution and rural biomass fuels in developing countries: a pilot village study in India and implications for research and policy. Atmos. Environ (1967). 1983;17(11):2343-2362. 114. Chiang T-A, Wu P-F, Ko Y-C. Identification of carcinogens in cooking oil fumes. Environ Res. 1999;81(1):18-22. 115. Shields P, Xu G, Blot W, et al. Mutagens from heated Chinese and US cooking oils. J. Natl. Cancer Inst. 1995;87(11):836-841. 116. Ferreccio C, González C, Milosavjlevic V, Marshall G, Sancha AM, Smith AH. Lung cancer and arsenic concentrations in drinking water in Chile. Epidemiology. 2000:673-679. 117. National Research Council. Arsenic in Drinking Water. 1999. 118. Gallicchio L, Boyd K, Matanoski G, et al. Carotenoids and the risk of developing lung cancer: a systematic review. Am. J. Clin. Nutr. 2008;88(2):372-383. 119. Hinwood AL, Jolley DJ, Sim MR. Cancer incidence and high environmental arsenic concentrations in rural populations: results of an ecological study. Int. J. Environ. Health Res. 1999;9(2):131-141. 120. Lewis DR, Southwick JW, Ouellet-Hellstrom R, Rench J, Calderon RL. Drinking water arsenic in Utah: A cohort mortality study. Environ. Health Perspect. 1999;107(5):359-365. 121. Bates MN, Smith AH, Cantor KP. Case-control study of bladder cancer and arsenic in drinking water. Am. J. Epidemiol. 1995;141(6):523-530. 122. Mendez WM, Eftim S, Cohen J, et al. Relationships between arsenic concentrations in drinking water and lung and bladder cancer incidence in U.S. counties. J. Expo. Sci. Environ. Epidemiol. 2017/05/01 2017;27(3):235-243. doi:10.1038/jes.2016.58.
123. Desimone LA, McMahon PB, Rosen MR. Water Quality in Principal Aquifers of the United States, 1991-2010. US Department of the Interior, US Geological Survey; 2014.
124. Bandi P, Minihan AK, Siegel RL, et al. Updated Review of Major Cancer Risk Factors and Screening Test Use in the United States in 2018 and 2019, with a Focus on Smoking Cessation. Cancer Epidemiol. Biomarkers Prev. 2021;30(7):1287-1299. doi:10.1158/1055-9965.Epi-20-1754 125. Office USPHS. Smoking Cessation: A Report of the Surgeon General. report.pdf. 2020; 126. US Department of Health and Human Services. Reducing tobacco use: a report of the Surgeon General. 2000. US Government Printing Office: Washington, DC. 2000.
127. Chaloupka FJ, Straif K, Leon ME. Effectiveness of tax and price policies in tobacco control. Tob. Control. 2011;20(3):235-238. 128. Wallace RB, Stratton K, Bonnie RJ. Ending the tobacco problem: a blueprint for the nation. National Academies Press; 2007.
129. State Cigarette Excise Tax Rates. Accessed October 21, 2022.
130. Foundation ANR. Overview List – Number of Smokefree and Other Tobacco-Related Laws. 2022. September 12, 2022. https:// no-smoke.org/wp-content/uploads/pdf/mediaordlist.pdf.
131. Jha P, Ramasundarahettige C, Landsman V, et al. 21st-century hazards of smoking and benefits of cessation in the United States. N Engl J Med. 2013;368(4):341-350. 132. Inoue-Choi M, McNeel TS, Hartge P, Caporaso NE, Graubard BI, Freedman ND. Non-Daily Cigarette Smokers: Mortality Risks in the U.S. Am J Prev Med. Jan 2019;56(1):27-37. doi:10.1016/j.
amepre.2018.06.025.
133. Use T, Panel D. Clinical Practice Guideline: Treating Tobacco Use and Dependence-2008 Update. Books Express Publishing; 2008.
134. Krist AH, Davidson KW, Mangione CM, et al. Interventions for tobacco smoking cessation in adults, including pregnant persons: US Preventive Services Task Force recommendation statement. JAMA. 2021;325(3):265-279. 135. Cokkinides VE, Halpern MT, Barbeau EM, Ward E, Thun MJ. Racial and ethnic disparities in smoking-cessation interventions: analysis of the 2005 National Health Interview Survey. Am J Prev Med. 2008;34(5):404-412. 136. Humphrey LL, Teutsch S, Johnson M, Force USPST. Lung cancer screening with sputum cytologic examination, chest radiography, and computed tomography: an update for the U.S. Preventive Services Task Force. Ann Intern Med. May 4 2004;140(9):740-53. doi:10.7326/0003-4819-140-9-200405040-00015.
137. National Lung Screening Trial Research T, Aberle DR, Adams AM, et al. Reduced lung-cancer mortality with low-dose computed tomographic screening. N Engl J Med. Aug 4 2011;365(5):395-409. doi:10.1056/NEJMoa1102873.
138. de Koning HJ, van der Aalst CM, de Jong PA, et al. Reduced Lung-Cancer Mortality with Volume CT Screening in a Randomized Trial. N Engl J Med. Feb 6 2020;382(6):503-513. doi:10.1056/ NEJMoa1911793.
139. de Koning HJ, van der Aalst CM, de Jong PA, et al. Reduced lung-cancer mortality with volume CT screening in a randomized trial. N Engl J Med. 2020;382(6):503-513. 140. Becker N, Motsch E, Trotter A, et al. Lung cancer mortality reduction by LDCT screening—results from the randomized German LUSI trial. International journal of cancer. 2020;146(6):1503-1513. 141. Jonas DE, Reuland DS, Reddy SM, et al. Screening for lung cancer with low-dose computed tomography: updated evidence report and systematic review for the US Preventive Services Task Force. JAMA. 2021;325(10):971-987. 142. Faselis C, Nations JA, Morgan CJ, et al. Assessment of Lung Cancer Risk Among Smokers for Whom Annual Screening Is Not Recommended. JAMA Oncol. 2022;8(10):1428-1437. doi:10.1001/ jamaoncol.2022.2952.
Cancer Facts & Figures 2023 49 143. Fedewa SA, Kazerooni EA, Studts JL, et al. State Variation in Low-Dose Computed Tomography Scanning for Lung Cancer Screening in the United States. J Natl Cancer Inst. Aug 2 2021;113(8):1044-1052. doi:10.1093/jnci/djaa170.
144. Kentucky LEADS Collaborative. Accessed August 25, 2022.
145. Gov. Beshear Signs Bill Establishing Lung Cancer Screening Program. University of Kentucky College of Medicine; July 22, 2022, 2022.
146. Zeliadt SB, Hoffman RM, Birkby G, et al. Challenges implementing lung cancer screening in federally qualified health centers. Am J Prev Med. 2018;54(4):568-575. 147. Fedewa SA, Star J, Bandi P, et al. Changes in Cancer Screening in the US During the COVID-19 Pandemic. JAMA Netw Open. Jun 1 2022;5(6):e2215490. doi:10.1001/jamanetworkopen.2022.15490.
148. Fedewa SA, Bandi P, Smith RA, Silvestri GA, Jemal A. Lung Cancer Screening Rates During the COVID-19 Pandemic. Chest. Feb 2022;161(2):586-589. doi:10.1016/j.chest.2021.07.030.
149. Silvestri GA, Goldman L, Burleson J, et al. Characteristics of Persons Screened for Lung Cancer in the United States: A Cohort Study. Ann Intern Med. 2022.
150. Sosa E, D’Souza G, Akhtar A, et al. Racial and socioeconomic disparities in lung cancer screening in the United States: A systematic review. CA Cancer J Clin. Jul 2021;71(4):299-314. doi:10.3322/caac.21671.
151. Sedani AE, Davis OC, Clifton SC, Campbell JE, Chou AF. Facilitators and Barriers to Implementation of Lung Cancer Screening: A Framework Driven Systematic Review. J Natl Cancer Inst. Aug 22 2022;doi:10.1093/jnci/djac154.
152. Islami F, Marlow EC, Zhao J, et al. Person-years of life lost and lost earnings from cigarette smoking-attributable cancer deaths, United States, 2019. Int J Cancer. Aug 10 2022;doi:10.1002/ijc.34217.
153. Shih Y-CT, Xu Y, Liu L, Smieliauskas F. Rising prices of targeted oral anticancer medications and associated financial burden on Medicare beneficiaries. J. Clin. Onc. 2017;35(22):2482. 154. Dusetzina SB, Huskamp HA, Keating NL. Specialty drug pricing and out-of-pocket spending on orally administered anticancer drugs in Medicare Part D, 2010 to 2019. JAMA. 2019;321(20):2025-2028. 155. Shih Y-CT, Smieliauskas F, Geynisman DM, Kelly RJ, Smith TJ. Trends in the cost and use of targeted cancer therapies for the privately insured nonelderly: 2001 to 2011. J. Clin. Onc. 2015;33(19):2190. |
189035 | https://math.stackexchange.com/questions/2279460/how-to-determine-polar-coordinate-visually | calculus - How to determine polar coordinate visually - Mathematics Stack Exchange
Join Mathematics
By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy.
Sign up with Google
OR
Email
Password
Sign up
Already have an account? Log in
Skip to main content
Stack Exchange Network
Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.
Visit Stack Exchange
Loading…
Tour Start here for a quick overview of the site
Help Center Detailed answers to any questions you might have
Meta Discuss the workings and policies of this site
About Us Learn more about Stack Overflow the company, and our products
current community
Mathematics helpchat
Mathematics Meta
your communities
Sign up or log in to customize your list.
more stack exchange communities
company blog
Log in
Sign up
Home
Questions
Unanswered
AI Assist Labs
Tags
Chat
Users
Teams
Ask questions, find answers and collaborate at work with Stack Overflow for Teams.
Try Teams for freeExplore Teams
3. Teams
4. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams
Teams
Q&A for work
Connect and share knowledge within a single location that is structured and easy to search.
Learn more about Teams
Hang on, you can't upvote just yet.
You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it?
Instead, you can save this post to reference later.
Save this post for later Not now
Thanks for your vote!
You now have 5 free votes weekly.
Free votes
count toward the total vote score
does not give reputation to the author
Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation.
Got it!Go to help center to learn more
How to determine polar coordinate visually
Ask Question
Asked 8 years, 4 months ago
Modified8 years, 4 months ago
Viewed 425 times
This question shows research effort; it is useful and clear
2
Save this question.
Show activity on this post.
I am studying calculating arc length on polar coordinates on khan academy and I encounter this question.
Let R be circular segment that lies inside the circle r=4 c o s(θ)r=4 c o s(θ) and the left of r=s e c(θ)r=s e c(θ).
Calculate the perimeter of the shaded region.
I know how to use formula, and I familiar with polar coordinates. However, when I encountered this I set boundaries for my integral as π/3 π/3 and 5 π/3 5 π/3! Which is wrong. so after struggling a lot and realizing that I am doing things wrong, by solving this equation 4 c o s(θ)=s e c(θ)4 c o s(θ)=s e c(θ) I realized that I must set my boundaries as π/3 π/3 and 2 π/3 2 π/3. The lower boundary is apparent by the diagram, but the other one seems strange to me!
In the hints there is this line:
The entire circle is traced out once from θ=0 θ=0 to θ=π θ=π thus the arc is traced from θ=π/3 θ=π/3 to θ=2 π/3 θ=2 π/3
Unfortunately, I am not even able to understand what this hint even is about! So my questions are: How can I determine boundaries visually from the diagram and what this hint talk about? the circle must be traced out from 0 0 to 2 π 2 π, why it says π π??
calculus
polar-coordinates
Share
Share a link to this question
Copy linkCC BY-SA 3.0
Cite
Follow
Follow this question to receive notifications
edited May 13, 2017 at 17:03
celtschk
44.6k 13 13 gold badges 88 88 silver badges 143 143 bronze badges
asked May 13, 2017 at 16:35
shayanshayan
145 1 1 silver badge 7 7 bronze badges
6
Notice that the way polar coordinates work, the circle r=4∗cos(θ)r=4∗cos(θ) is actually drawn twice during the period 0→2 π 0→2 π. The two circles that are drawn just happen to overlap, resulting in only one that is seen.John Lou –John Lou 2017-05-13 16:43:41 +00:00 Commented May 13, 2017 at 16:43
Imagine you are a lighthouse at the origin. You swing your light from 0 to... whatever it is that you first see the endpoint of the arc. ... there it is. You hit the endpoint of the arc at pi/3. You trace the arc with you light for pi/3 to pi/2. While you do this you distance of the arc goes from 2 down to 0. At pi/2 the distance stops being positive and starts being negative. That's okay,your light bean goes both ways. You continue to trace the arc until you get to the endpoint. That happens at 2pi/3. You are done. If you continue going you hit the endpoint again and you trace it twice.fleablood –fleablood 2017-05-13 16:54:23 +00:00 Commented May 13, 2017 at 16:54
"your light beam goes both ways" for pi/2 to 2pi/3 the arc is behind you. (i.e. r≤0 r≤0. If you were tracing the circle you'd start with theta = 0 => (0,4) at theta = pi/4 => (2,2) and theta = pi/2 => (0,0) that's the top half of the circle. from theta = pi/2 => (0,0) through theta = 3pi/4 => (2,-2) to theta = pi => (0, 4) you have traced the entire circle. If we go from theta = pi to theta = 2pi we will trace the circle TWICE. So in the full theta 0 - 2pi we would trace the arc TWICE. We only want to trace it once. the arc first comes into scope at pi/3 and leaves our scope at 2pi/3.fleablood –fleablood 2017-05-13 17:02:47 +00:00 Commented May 13, 2017 at 17:02
@JohnLou I have studied at khan academy and I am pretty sure I haven't skipped anything but what you say seems strange to me. why circle is drawn twice?shayan –shayan 2017-05-13 17:15:53 +00:00 Commented May 13, 2017 at 17:15
@fleablood thanks. now I have an idea about how to determine boundaries visually, but I have no Idea why my light beam goes both ways. Can you a explain your comment a little more, and write it as an answer?? I am a little confused shayan –shayan 2017-05-13 17:21:44 +00:00 Commented May 13, 2017 at 17:21
|Show 1 more comment
2 Answers 2
Sorted by: Reset to default
This answer is useful
1
Save this answer.
Show activity on this post.
I'm taking a different view here and will determine the perimeter by analysis in the complex plane. We will divide the problem in to two parts, the circular arc and the straight line. These can be readily shown to be described by
r=4 cos θ,θ∈[π/3,2 π/3]r=sec θ,θ∈[−π/3,2 π/3]r=4 cosθ,θ∈[π/3,2 π/3]r=secθ,θ∈[−π/3,2 π/3]
The general equation for the arc length in the complex plane is
s=∫|z˙|d u=∫1+(d y d x)2−−−−−−−−−−√d x s=∫|z˙|d u=∫1+(d y d x)2 d x
We start with the arc...
z=4 cos θ e i θ z˙=(−4 sin θ+i cos θ)e i θ|z˙|=4 sin 2 θ+cos 2 θ−−−−−−−−−−−√=4∫2 π/3 π/3 4 d θ=4 π 3 z=4 cosθ e i θ z˙=(−4 sinθ+i cosθ)e i θ|z˙|=4 sin 2θ+cos 2θ=4∫π/3 2 π/3 4 d θ=4 π 3
The straight line can be much simpler than this, but here we are demonstrating a method. Thus,
z=sec θ e i θ z˙=(tan θ sec θ+i sec θ)e i θ|z˙|=sec θ 1+tan 2 θ−−−−−−−−√=sec 2 θ∫π/3−π/3 sec 2 θ d θ=tan θ∣∣π/3−π/3=2 3–√z=secθ e i θ z˙=(tanθ secθ+i secθ)e i θ|z˙|=secθ 1+tan 2θ=sec 2θ∫−π/3 π/3 sec 2θ d θ=tanθ|−π/3 π/3=2 3
Thus, the total perimeter is given by
s=4 π 3+2 3–√s=4 π 3+2 3
This result has been verified numerically.
Share
Share a link to this answer
Copy linkCC BY-SA 3.0
Cite
Follow
Follow this answer to receive notifications
answered May 19, 2017 at 16:26
Cye WaldmanCye Waldman
8,236 2 2 gold badges 18 18 silver badges 37 37 bronze badges
Add a comment|
This answer is useful
1
Save this answer.
Show activity on this post.
We can calculate the perimeter using regular geometry.
Using the angle markings, we can deduce that the perimeter of the curved part is a third of the perimeter of the circle. Connecting the 1 3 π 1 3 π and the 5 3 π 5 3 π lines to the circle's center, we form two equilateral triangles. Their angles added together must be 2 3 π 2 3 π. Therefore, the perimeter of the curved part is 2 3 π r=4 3 π 2 3 π r=4 3 π. After drawing altitudes from the top to the bottom side on both of the equilateral triangles and using special 30-60-90 triangle rules, we see that half of the line is 3–√3, so the whole line is 2 3–√2 3. Adding these together we get the perimeter as 4 3 π+2 3–√4 3 π+2 3.
Share
Share a link to this answer
Copy linkCC BY-SA 3.0
Cite
Follow
Follow this answer to receive notifications
edited May 19, 2017 at 22:41
answered May 13, 2017 at 17:07
u8y7541u8y7541
738 1 1 gold badge 7 7 silver badges 20 20 bronze badges
3
There is an error here because r=2 r=2 for circle in question Cye Waldman –Cye Waldman 2017-05-19 15:46:25 +00:00 Commented May 19, 2017 at 15:46
@CyeWaldman I fixed the issue, the answer's right now.u8y7541 –u8y7541 2017-05-19 22:42:16 +00:00 Commented May 19, 2017 at 22:42
Thank you but my question is how to determine boundaries.shayan –shayan 2017-05-24 18:22:57 +00:00 Commented May 24, 2017 at 18:22
Add a comment|
You must log in to answer this question.
Start asking to get answers
Find the answer to your question by asking.
Ask question
Explore related questions
calculus
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
1Area inside loop of polar equation, unsolvable problem?
2Find the area using double integral and polar coordinates.
0Polar Coordinate Conversion (Integration)
7Why can't we use the circumference of a circle to find arc lengths of polar curves?
3Calculating distance using polar coordinate metric
0Unit circle method for converting complex number into polar coordinate
0Finding double integrals using polar coordinates.
0Intersection between a line and a circle in polar coordinate system
1Shared area of 2 2 circles (polar integral)
Hot Network Questions
What can be said?
A time-travel short fiction where a graphologist falls in love with a girl for having read letters she has not yet written… to another man
What's the expectation around asking to be invited to invitation-only workshops?
How can blood fuel space travel?
Can I go in the edit mode and by pressing A select all, then press U for Smart UV Project for that table, After PBR texturing is done?
Numbers Interpreted in Smallest Valid Base
Another way to draw RegionDifference of a cylinder and Cuboid
Determine which are P-cores/E-cores (Intel CPU)
Should I let a player go because of their inability to handle setbacks?
how do I remove a item from the applications menu
Does the mind blank spell prevent someone from creating a simulacrum of a creature using wish?
Matthew 24:5 Many will come in my name!
Discussing strategy reduces winning chances of everyone!
What is a "non-reversible filter"?
Analog story - nuclear bombs used to neutralize global warming
Clinical-tone story about Earth making people violent
Direct train from Rotterdam to Lille Europe
How exactly are random assignments of cases to US Federal Judges implemented? Who ensures randomness? Are there laws regulating how it should be done?
What is the meaning of 率 in this report?
Checking model assumptions at cluster level vs global level?
Change default Firefox open file directory
Can a GeoTIFF have 2 separate NoData values?
Who is the target audience of Netanyahu's speech at the United Nations?
Identifying a movie where a man relives the same day
Question feed
Subscribe to RSS
Question feed
To subscribe to this RSS feed, copy and paste this URL into your RSS reader.
Why are you flagging this comment?
It contains harassment, bigotry or abuse.
This comment attacks a person or group. Learn more in our Code of Conduct.
It's unfriendly or unkind.
This comment is rude or condescending. Learn more in our Code of Conduct.
Not needed.
This comment is not relevant to the post.
Enter at least 6 characters
Something else.
A problem not listed above. Try to be as specific as possible.
Enter at least 6 characters
Flag comment Cancel
You have 0 flags left today
Mathematics
Tour
Help
Chat
Contact
Feedback
Company
Stack Overflow
Teams
Advertising
Talent
About
Press
Legal
Privacy Policy
Terms of Service
Your Privacy Choices
Cookie Policy
Stack Exchange Network
Technology
Culture & recreation
Life & arts
Science
Professional
Business
API
Data
Blog
Facebook
Twitter
LinkedIn
Instagram
Site design / logo © 2025 Stack Exchange Inc; user contributions licensed under CC BY-SA. rev 2025.9.26.34547
By clicking “Accept all cookies”, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy.
Accept all cookies Necessary cookies only
Customize settings
Cookie Consent Preference Center
When you visit any of our websites, 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 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 manage your preferences. Please note, blocking some types of cookies may impact your experience of the site and the services we are able to offer.
Cookie Policy
Accept all cookies
Manage Consent Preferences
Strictly Necessary Cookies
Always Active
These cookies are necessary for the website to function and cannot be switched off in our systems. They are usually only set in response to actions made by you which amount to a request for services, such as setting your privacy preferences, logging in or filling in forms. You can set your browser to block or alert you about these cookies, but some parts of the site will not then work. These cookies do not store any personally identifiable information.
Cookies Details
Performance Cookies
[x] Performance Cookies
These cookies allow us to count visits and traffic sources so we can measure and improve the performance of our site. They help us to know which pages are the most and least popular and see how visitors move around the site. All information these cookies collect is aggregated and therefore anonymous. If you do not allow these cookies we will not know when you have visited our site, and will not be able to monitor its performance.
Cookies Details
Functional Cookies
[x] Functional Cookies
These cookies enable the website to provide enhanced functionality and personalisation. They may be set by us or by third party providers whose services we have added to our pages. If you do not allow these cookies then some or all of these services may not function properly.
Cookies Details
Targeting Cookies
[x] Targeting Cookies
These cookies are used to make advertising messages more relevant to you and may be set through our site by us or by our advertising partners. They may be used to build a profile of your interests and show you relevant advertising on our site or on other sites. They do not store directly personal information, but are based on uniquely identifying your browser and internet device.
Cookies Details
Cookie List
Clear
[x] checkbox label label
Apply Cancel
Consent Leg.Interest
[x] checkbox label label
[x] checkbox label label
[x] checkbox label label
Necessary cookies only Confirm my choices |
189036 | https://www.khanacademy.org/math/precalculus/x9e81a4f98389efdf:trig/x9e81a4f98389efdf:inverse-trig/a/inverse-trigonometric-functions-review | Use of cookies
Cookies are small files placed on your device that collect information when you use Khan Academy. Strictly necessary cookies are used to make our site work and are required. Other types of cookies are used to improve your experience, to analyze how Khan Academy is used, and to market our service. You can allow or disallow these other cookies by checking or unchecking the boxes below. You can learn more in our cookie policy
Privacy Preference Center
When you visit any website, it may store or retrieve information on your browser, mostly in the form of cookies. This information might be about you, your preferences or your device and is mostly used to make the site work as you expect it to. The information does not usually directly identify you, but it can give you a more personalized web experience. Because we respect your right to privacy, you can choose not to allow some types of cookies. Click on the different category headings to find out more and change our default settings. However, blocking some types of cookies may impact your experience of the site and the services we are able to offer.
More information
Manage Consent Preferences
Strictly Necessary Cookies
Always Active
Certain cookies and other technologies are essential in order to enable our Service to provide the features you have requested, such as making it possible for you to access our product and information related to your account.
For example, each time you log into our Service, a Strictly Necessary Cookie authenticates that it is you logging in and allows you to use the Service without having to re-enter your password when you visit a new page or new unit during your browsing session.
Functional Cookies
These cookies provide you with a more tailored experience and allow you to make certain selections on our Service. For example, these cookies store information such as your preferred language and website preferences.
Targeting Cookies
These cookies are used on a limited basis, only on pages directed to adults (teachers, donors, or parents). We use these cookies to inform our own digital marketing and help us connect with people who are interested in our Service and our mission.
We do not use cookies to serve third party ads on our Service.
Performance Cookies
These cookies and other technologies allow us to understand how you interact with our Service (e.g., how often you use our Service, where you are accessing the Service from and the content that you’re interacting with). Analytic cookies enable us to support and improve how our Service operates.
For example, we use Google Analytics cookies to help us measure traffic and usage trends for the Service, and to understand more about the demographics of our users.
We also may use web beacons to gauge the effectiveness of certain communications and the effectiveness of our marketing campaigns via HTML emails. |
189037 | https://arxiv.org/pdf/2406.18687 | arXiv:2406.18687v1 [math.QA] 26 Jun 2024
SOME SPECIAL BASES OF THE 2–SWAP ALGEBRAS.
CLAUDIO PROCESI
Abstract. We study the algebra Σ ninduced by the action of the symmetric group Snon V⊗nwhen dim V= 2. Our main result is that the space of symmetric elements of Σ nis linearly spanned by the involutions of Sn.
Introduction
Let V be a vector space of finite dimension d over some field F (in our computations it is convenient to take F = Q the rational numbers or C the complex numbers). In the classical theory of Schur–Weyl a major role is played by the action of the symmetric group Sn on n elements on the nth tensor power V ⊗n by exchanging the tensor factors. The algebra of operators on V ⊗n, generated by these permutations will be denoted by Σ n(d) and called a d–swap algebra .It is the algebra formed by the elements which commute with the diagonal action of GL (V ) or, if V is a Hilbert space, by the corresponding unitary group. The name comes from the use, in the physics literature, to call swap the exchange operator (1 , 2) on V ⊗2.In the literature on quantum information theory the states lying in Σ n(d)are called Werner states and widely used as source of examples, due to fundamental work of the physicist R. F. Werner [ 11 ]. See for instance [ 4], [ 5], [ 6], [ 7] for applications to separability, entangle-ments or the quantum max-cut problem. A classical theorem states that the corresponding algebra homomorphism
F [Sn] → Σn(d) ⊂ End (V ⊗n) is injective if and only if dim V ≥ n.When d = dim V < n the kernel of this map is the two sided ideal of
F [Sn] generated by the antisymmetrizer
Ad+1 := ∑
σ∈Sd+1
ǫσσ, ǫσ the sign of the permutation .
I wish to thank Felix Huber for pointing out the problem and some of the literature.
12CLAUDIO PROCESI
If F has characteristic 0 the algebra F [Sn] decomposes as direct sum of ma-trix algebras indexed by partitions, corresponding to the irreducible rep-resentations of Sn. As for Σ n(d) only the blocks relative to partitions of height ≤ d survive. In the case d = dim V < n an interesting problem is to describe a basis of Σ n(d) formed by permutations. In fact in the physics literature there are several examples of Hamiltonians lying in Σ n(d). Thus it may be convenient to express such Hamiltonian in a given special basis, Moreover in Σ n(d) we have the involution g 7 → g−1, if V is a Hilbert space this coincides with adjuction. So we would like also to describe a basis of Σ n(d)+ the subspace of symmetric elements made by permutations and also a basis of self adjoint operators. In [ 8] I have proved that a possible basis of Σ n(d) is formed by the permutations which are d + 1–good. By definition a permutation g is d + 1–good if and only if it does not contain a decreasing subsequence of length d + 1. By a beautiful Theorem of Schensted this is equivalent to the fact that the pair of tableaux associated to g is of height ≤ d.This of course, by classical theory, is exactly the dimension of Σ n(d) so it is enough to prove that such permutations span Σ n(d) and this, in [ 8] is done by a straightening algorithm deduced from the relations. Since the pair of tableaux associated to g−1 by the Robinson–Schensted correspondence is obtained by exchanging that associated to g, it follows that if g is d + 1–good so is g−1, and this gives also a basis g + g−1, where
g is d + 1–good, for the symmetric elements. We have also the basis g − g−1 for the antisymmetric elements, from which we have a basis over the real numbers for self adjoint operators given by g + g−1and i(g − g−1) , where g is d + 1–good. On the other hand, specially for d = 2, one may want to find a ba-sis formed by simpler type of elements. For this discussion the simplest elements are the elements of order 2 (called involutions ) which are permu-tations with cycles only of order 2,1 and eigenvalues only ±1. We call Σ n(2) the n–swap algebra and denote it simply Σ n. It is known that dim Σ n = Cn the nth Catalan number, see §3.1 for a simple proof. The list of the first 10 Catalan numbers is 1, 2, 5, 14 , 42 , 132 , 429 , 1430 , 4862 , 16796 SOME SPECIAL BASES OF THE 2–SWAP ALGEBRAS. 3
Definition 1.1. The set S of special permutations is formed by the invo-lutions and also by the permutations with cycles only of order 2,1 plus a single cycle of order 3. The 3 cycle can be further normalised to be increasing. Our main Theorem is the following
Theorem 1.2. (1) For each n the algebra Σn has a basis formed by special elements.
(2) Σ +
n
has a basis over C formed by involutions.
(3) The space of real and symmetric elements has a basis over R formed by involutions.
Notice that items (2) and (3) are equivalent and follow from (1). In fact the involutions are symmetric and if a permutation is of the form
g = ab with one 3 cycle a and the rest b is a product 2 or 1 cycles its symmetrization is ( g + g−1) = ( a + a−1)b.If a is a 3 cycle in the algebra Σ 3, by relation ( 1), we have that a + a−1
is the sum of -1 and 3 transpositions. The claim follows. In the same way ( g − g−1) = ( a − a−1)b gives bases for antisymmetric elements, and together for self adjoint operators. The dimensions of the real symmetric elements are, from n = 1 to n = 10 1, 2, 4, 10 , 26 , 76 , 232 , 750 , 2494 , 8524 , · · ·
(see The On-Line Encyclopedia of Integer Sequences A007123 for many interesting informations on this sequence). On the other hand the number I(n) of involutions in Sn from n = 1 to
n = 10 is
I(n) = 1 , 2, 4, 10 , 26 , 76 , 232 , 764 , 2620 , 9496 , · · ·
which is also equal (by the Robinson–Schensted correspondence) to the number of standard Young tableaux with n cells (O.E.I.S A000085). So a curious fact is that these two sequences coincide up to n = 7. We have thus that the involutions are a basis of the real symmetric elements for n ≤ 7 and after that they have linear relations. It would be interesting to understand these relations. The antisymmetric elements are spanned by elements of type ab − a−1b
with a a 3 cycle, which we can assume to be in increasing order, and b
an involution. There are (n
3
) such 3 cycles in Sn and so (n
3
)I(n − 3) such elements. (n
3
)
I(n − 3) = 1 , 4, 20 , 60 , 350 , n = 3 , · · · , 74 CLAUDIO PROCESI
so there appear relations already for n = 5 where the dimension of Σ 5 is 42, while the number of normalised special elements is 46.
Remark 1.3 . The set of special elements has the following compatibility with the partial traces ti : End (V )⊗n → End (V )⊗n−1,
ti : x1 ⊗ · · · xi ⊗ · · · xn 7 → tr (xi)x1 ⊗ · · · xi−1 ⊗ xi+1 ⊗ · · · ⊗ xn.
In fact applied to a permutation decomposed into cycles ti removes i from the cycle, and so a special element is mapped to a special element, and moreover, in case i is a fixed element multiplies by the dimension of the space, in our case 2. Notice that instead the partial trace of a 3–good element may be 3–bad as for instance
t4({3, 4, 1, 2}) = {3, 2, 1}
where by {3, 4, 1, 2} we mean the permutation as string and not as cycle, in cycle form {3, 4, 1, 2} = (1 , 3)(2 , 4) 7 → (1 , 3)(2) = {3, 2, 1}.
Remark 1.4 . Given a basis e1, e 2, e 3, e 4 for the space End (V ) of linear op-erators on V one has the dual basis fi, i = 1 , · · · , 4 for the trace form
tr (ab ). That is 4 operators satisfying tr (eifj ) = δij . Then the operator (1 , 2) : V ⊗ V → V ⊗ V can be written as (1 , 2) = e1 ⊗ f1 + e2 ⊗ f2 + e3 ⊗ f3 + e4 ⊗ f4.
Any involution being product of elements ( i, j ) can the be expressed using this Formula in term of the basis. We have some freedom in the choice of the basis. The most common is the basis by matrix units ei,j in which (1 , 2) = e1,1 ⊗ e1,1 + e1,2 ⊗ e2,1 + e2,1 ⊗ e1,2 + e2,2 ⊗ e2,2.
In particular in Physics are widely used the Pauli matrices .
σ0 :=
∣∣∣∣1 00 1
∣∣∣∣ , σ x :=
∣∣∣∣0 11 0
∣∣∣∣ , σ y :=
∣∣∣∣0 −ii 0
∣∣∣∣ , σ z :=
∣∣∣∣1 00 −1
∣∣∣∣
1
2
∣∣∣∣1 + a b − icb + i c 1 − a
∣∣∣∣ = 1
2 (I + bσ x + cσ y + aσ z).
They are equal to the dual basis up to a scaling by 1
2
so that: (1 , 2) = 1
2 (σ0 ⊗ σ0 + σx ⊗ σx + σy ⊗ σy + σz ⊗ σz ).SOME SPECIAL BASES OF THE 2–SWAP ALGEBRAS. 5
The proof of this Theorem is algorithmic. We give an algorithm which, given as input a permutation σ ∈ Sn produces a linear combination of elements in S which in Σ n equals to σ.2. The algorithm
Usually we write the permutations in their cycle structure. Let us start with the basic antisymmetrizer which vanishes in Σ 3.
A = (1 , 2, 3) + (1 , 3, 2) − (1 , 2) − (1 , 3) − (2 , 3) + 1 (1) (1 , 2, 3) + (1 , 3, 2) = (1 , 2) + (1 , 3) + (2 , 3) − 1.
First remark that in S3 all permutations are special, moreover (2) (1 , 3, 2) = −(1 , 2, 3) + (1 , 2) + (1 , 3) + (2 , 3) − 1.
so a 3–cycle can be normalised. In S4 we have the 4-cycles which are not special and we have to write them as linear combination of special permu-tations in Σ 4. Notice that if we can do this for a single cycle we can do it for all cycles, since permutations of the same cycle structure are conju-gate and clearly the space spanned by special permutations is closed under conjugation. For any n > 3 we have the natural embedding of S3 in Sn as the per-mutations on the first 3 elements. This induces an embedding of A in the algebra of the symmetric group of Sn which we denote by An . This ele-ment vanishes in the swap algebra Σ n. and in C[Sn] generates the ideal of relations for Σ n. Thus in Σ 4 we have the vanishing of (2 , 4) A4 = (4 , 2, 3, 1) + (3 , 4, 2, 1) − (4 , 2, 1) − (3 , 1)(4 , 2) − (3 , 4, 2) + (2 , 4) (3 , 4) A4(2 , 4) = (2 , 3, 1)+(4 , 1)(2 , 3) −(2 , 3, 4, 1) −(4 , 2, 3, 1) −(2 , 3)+(3 , 4, 2)
A4(3 , 4) = (2 , 3, 4, 1) + (3 , 4, 2, 1) − (1 , 2)(3 , 4) − (3 , 4, 1) − (3 , 4, 2) + (3 , 4) .
We then have, in Σ 4, 0 = A4(3 , 4) − (3 , 4) A4(2 , 4) that is (3) −2(2 , 3, 4, 1) = (3 , 4, 2, 1) + (4 , 2, 3, 1) − (1 , 2)(3 , 4) − (3 , 4, 1)
−2(3 , 4, 2) + (3 , 4) − (2 , 3, 1) − (4 , 1)(2 , 3) + (2 , 3) From the vanishing of (2 , 4) A4 we deduce: (4 , 2, 3, 1) + (3 , 4, 2, 1) = (4 , 2, 1) + (3 , 1)(4 , 2) + (3 , 4, 2) − (2 , 4) Substituting in ( 3) we deduce: (4) 2(1 , 2, 3, 4) = 6 CLAUDIO PROCESI
(1 , 4)(2 , 3)+(1 , 2)(3 , 4) −(1 , 3)(4 , 2) −(1 , 4, 2)+(1 , 3, 4)+(2 , 3, 4)+(1 , 2, 3)+(2 , 4) −(3 , 4) −(2 , 3) .
The term (4 , 2, 1) is not normalised, but it can be rewritten using Formula (2).
−(4 , 2, 1) = (1 , 2, 4) − (1 , 2) − (1 , 4) − (2 , 4) + 1 (5) 2(1 , 2, 3, 4) = (1 , 4)(2 , 3)+(1 , 2)(3 , 4) −(1 , 3)(4 , 2)+(1 , 2, 4)+(1 , 3, 4)+(2 , 3, 4)+(1 , 2, 3) −(1 , 2) −(1 , 4) −(3 , 4) −(2 , 3)+1 .
Since all 4 cycles are conjugate we deduce that statement (1) is true for S4.Now notice the following general fact: consider two cycles ( a, A ), (a, B )of lengths h, k respectively where A and B are strings of integers of lengths
h − 1, k − 1 respectively and disjoint. Then their product is the cycle of length h + k − 1: (6) (a, B )( a, A ) = ( a, A, B ), e.g. (1 , 2, 3)(1 , 5, 4, 6) = (1 , 5, 4, 6, 2, 3) .
Thus take a cycle of length p > 4 and, up to conjugacy we may take (7) cp := (1 , 2, 3, 4, 5, . . . , p ) = (1 , 5, . . . , p )(1 , 2, 3, 4) .
In Σ p we have thus that 2 cp equals (1 , 5, . . . , p ) times the expression of Formula ( 5). But then applying again Formula ( 6) we see that the resulting formula is a sum of permutations on p elements which are not full cycles. By iterating then the operation on the cycles of length ℓ with 4 ≤ ℓ ≤ p−1we have a preliminary.
Proposition 2.1. The cycle cp (formula (7)) is a linear combination in
Σp of elements which contain only cycles of length 1,2,3. Hence Σn is spanned by permutations which contain only cycles of length 1,2,3.
Example For p = 5 , 6 we have for 2 cp the formula obtained from Formula ( 5): 2(1 , 2, 3, 4, 5) (7)
= 2(1 , 5)(1 , 2, 3, 4) = (1 , 4, 5)(2 , 3) + (1 , 2, , 5)(3 , 4) − (1 , 3, 5)(4 , 2) − (1 , 4, 2, 5) + (1 , 3, 4, 5) + (3 , 4, 2)(1 , 5) + (1 , 2, 3, 5) + (2 , 4)(1 , 5) − (3 , 4)(1 , 5) − (2 , 3)(1 , 5) In the previous formula appear three 4-cycles, for which we can apply Formula ( 5) (see in the appendix, the expanded Formula ( 10 )). Notice that the final Formula must be invariant under conjugation by powers of the cycle, but this only up to the relations in Σ n.(8) 2(1 , 2, 3, 4, 5, 6) (7)
= 2(1 , 5, 6)(1 , 2, 3, 4) = SOME SPECIAL BASES OF THE 2–SWAP ALGEBRAS. 7
(1 , 4, 5, 6)(2 , 3)+(1 , 2, 5, 6)(3 , 4) −(1 , 3, 5, 6)(4 , 2) −(1 , 4, 2, 5, 6)+(1 , 3, 4, 5, 6) +(3 , 4, 2)(1 , 5, 6)+ (1 , 2, 3, 5, 6) + (2 , 4)(1 , 5, 6) − (3 , 4)(1 , 5, 6) − (2 , 3)(1 , 5, 6) .
Of course in the previous formulas appear 4-cycles, for which we can apply Formula ( 5), and then 5-cycles, for which we can apply the final for-mula developed before. Notice now that in Formula ( 8) all terms are either special or can be expanded into a linear combination of special elements, using the formulas of 4 and 5–cycles, except the term (3 , 4, 2)(1 , 5, 6). In order to prove Theorem 1.2 using Proposition 2.1 it is enough to prove that, in S6, a permutation of type 3 , 3 can be developed as linear combination of special elements, since then we apply recursively this to a product of k disjoint 3-cycles. If k is even we replace them all and if odd we remain with only one 3-cycle which can be normalized if necessary using Formula ( 2). The computation in S6 in principle is similar to that in S4 but now we have to handle a priori many more relations and I had to be assisted by the software ”Mathematica” in order to discover the needed relations. Let me sketch what I did. 2.2. The computation. A set of relations for Σ 6 can be obtained from the antisymmetrizer A6 by multiplication to the left and right by the 720 permutations. Actually it is not necessary to use all permutations since there are 36 × 6 pairs which stabilyze A6 up to sign Finally using these reductions we have 2400 relations each a sum of 6 permutations, of which 3 even and 3 odd. Each 6–cycle c, appearing in these relations, needs to be developed by using the appropriate conjugate of Formula ( 8) by the permutation which has as string the same form of the cycle c and which conjugates the standard 6 cycle into c.So a 6 cycle is replaced, using a conjugate of formula ( 8), by a permu-tation of type 3 , 3 plus a sum of special terms. In this way we obtain 2400 relations which, by inspection contain either 0, 2 or 3 permutations of type 3, 3 and with the remaining terms special. The ones with 0, 2 permutations of type 3 , 3 are linear combinations of special permutations and cannot be used. Remain 360 relations containing 3 permutations of type 3 , 3, arising from relations with two 6–cycles and one permutations of type 3 , 3.
Remark 2.3 . There are 40 permutations of type (3 , 3) but using Formula (1) we can normalise these elements. 8 CLAUDIO PROCESI
If a 3 cycle ( a, b, c ) is not strictly increasing (up to cyclic equivalence) we can replace it by a strictly increasing cycle introducing a sign and adding some special permutations applying a conjugate of formula ( 1). We then are reduced to 10 normalised permutations of type (3 , 3). We have several relations involving these normalised permutations plus special elements and we have to eliminate in one relation all permutations of type (3 , 3) except one, thus obtaining the desired relation. 2.4. The useful relations. Surprisingly in order to obtain the desired relation only the following 2 suffice:
(5 ,6,1)(3 ,4)A6 = (1 ,2,4,3,5,6)+(1 ,4,3,2,5,6) −(2 ,5,6,1)(3 ,4) −(4 ,3,5,6,1) −(5 ,6,1)(4 ,3,2)+(5 ,6,1)(3 ,4) (6 ,1)(4 ,5,3)A6 = (1 ,2,4,5,3,6) + (1 ,4,5,3,2,6) −(2 ,6,1)(4 ,5,3) −(4 ,5,3,6,1) −(6 ,1)(4 ,5,3,2) + (6 ,1)(4 ,5,3)
In Formula ( 8) the contribution to the expansion of 2(1 , 2, 3, 4, 5, 6) of an element of type 3 , 3 is +(3 , 4, 2)(1 , 5, 6). Therefore the contributions of type 3 , 3 of the 4 cycles of length 6 appear-ing in the previous Formulas are obtained by conjugating (3 , 4, 2)(1 , 5, 6) with the permutation which has as string the same form of the cycle. 2 (2 , 4, 3, 5, 6, 1) = (3 , 5, 4)(2 , 6, 1) + · · · , 2 (4 , 3, 2, 5, 6, 1) = (2 , 5, 3)(4 , 6, 1) + · · ·
2 (2 , 4, 5, 3, 6, 1)= (5 , 3, 4)(2 , 6, 1) + · · · , 2 (4 , 5, 3, 2, 6, 1) = (3 , 2, 5)(4 , 6, 1) + · · ·
So, by Remark 2.3 , the previous elements can be written in Σ 6 as 2 (2 , 4, 3, 5, 6, 1) = −(3 , 4, 5)(1 , 2, 6) + · · · , 2 (4 , 3, 2, 5, 6, 1) = −(2 , 3, 5)(1 , 4, 6) + · · ·
2 (2 , 4, 5, 3, 6, 1)= (3 , 4, 5)(1 , 2, 6) + · · · , 2 (4 , 5, 3, 2, 6, 1) = −(2 , 3, 5)(1 , 4, 6) + · · ·
where the · · · represent special elements. Therefore the previous 2 relations multiplied by 2 are of the form (9)
−(3 , 4, 5)(1 , 2, 6) − (2 , 3, 5)(1 , 4, 6) + 2(2 , 3, 4)(1 , 5, 6) + · · ·
(3 , 4, 5)(1 , 2, 6) − (2 , 3, 5)(1 , 4, 6) − 2(3 , 4, 5)(1 , 2, 6) + · · ·
Subtracting the second from the first one has the desired Formula: 0 = 2(2 , 3, 4)(1 , 5, 6) + · · ·
a relation with a single permutation 2(2 , 3, 4)(1 , 5, 6) of type (3 , 3) and the remaining elements special. This gives the desired expression, which is explicited in the Appendix. SOME SPECIAL BASES OF THE 2–SWAP ALGEBRAS. 9
Comments
3.1. The Formula dim Σ n = Cn.
The nth Catalan number is 1
n+1
(2nn
). By the hook Formula it is easily seen that this is the dimension of the irreducible representation of S2n associated to a Young diagram with two rows of length n. This in turn appears as the isotypic component of the SL (2) invariants in V ⊗2n, dim V = 2. By identifying V with V ∗ as SL (2) representations we have an SL (2) linear isomorphism between End (V ⊗n) = V ⊗n ⊗ V ∗⊗ n and V ⊗2n which induces a linear isomorphism with the respective invariants Σn = End SL (V )V ⊗n ≃ (V ⊗2n)SL (V ).
3.2. Several bases. Several bases formed by special elements can be ob-tained from Theorem 1.2 , so a main problem is to describe a best one by combinatorial means. The main advantage of the special elements is that their eigenvalues are ±1 and the two primitive 3-roots of 1. They also are
local in the sense that involve only 2 tensor factors at a time, and at most a single instance of 3 tensor factors. One should compare the complexity of the algorithm to express a per-mutation as linear combination of special ones to that of the algorithm to express a permutation as linear combination of 3-good ones. Some remarks on the algorithm to express a permutation as linear com-bination of 3-good ones. (1) Permutations are ordered lexicographically. (2) The 3-good permutations on n elements are in number Cn the nth
Catalan number. (3) The last 3-good permutation is n, 1, 2, 3, · · · , n − 1. The algorithm takes a permutation σ and checks recursively if there is a string of 3 elements decreasing. If there is not one the permutation is 3-good. Otherwise as soon as one encounters one such sequence, by applying the antisymmetrizer on these elements one obtains that σ is equivalent to a sum with signs of 5 permutations which are lexicographically less than σ.This means that if we have already developed the previous permutations as linear combination of 3-good ones we immediately obtain the developments for σ. Notice that in this development the coefficients are all integers. 3.3. d ≥ 3. One may ask the same question for Σ n(d) and d ≥ 3. The first problem is: Determine the minimum m = m(d) so that Σ m+1 (d) is spanned by the permutations which are NOT m + 1–cycles. 10 CLAUDIO PROCESI
This number m has also other interesting interpretations (see [ 1] page 331 for the interesting history of this question). The same m is the maximum degree of the generators of invariants of
d × d matrices. It is also the minimum degree for which, given an associative algebra R
over a field of characteristic 0, in which every element x satisfies xd = 0 one has Rm(d) = 0. The known estimates for m(d) are the lower bound m(d) ≥ (d+1 2
) due to Kuzmin, see [ 3] or [ 2] and the upper bound m(d) ≤ d2 due to Razmyslov, see [10 ] or [ 1]. Kuzmin conjectures that m(d) = (d+1 2
) which has been verified only for d ≤ 4. 3.4. Transpositions. Let us remark a simple fact
Proposition 3.5. The identity plus all transpositions give linearly inde-pendent operators in all Σn(d).Proof. Of course it is enough to prove this when d = 2, we do it by induc-tion. Assume we have a relation 0 = a · In + ∑
i<j
ai,j (i, j )with In the identity on V ⊗n. Apply the partial trace t1 = t on the first factor as in [ 9]
t : X1 ⊗ X2 ⊗ . . . ⊗ Xn−1 ⊗ Xn 7 → tr (X1)X2 ⊗ . . . ⊗ Xn−1 ⊗ Xn.
As proved in that paper if Sn−1 is the subgroup of Sn fixing 1 we have
t(σ) = 2 σ, ∀σ ∈ Sn−1 t(τ (1 , i )) = τ, ∀τ ∈ Sn−1
So we have in V ⊗n−1.0 = b · In−1 + 2 ∑
1<i<j
ai,j (i, j ), b = 2 a + ∑
1<j
a1,j .
By induction we have b = 0 , a i,j = 0 , ∀1 < i < j .So the relation is, setting aj := a1,j , among In and the n−1 transpositions (1 , i ), i = 2 , n . Apply the partial trace tn on the last factor obtaining in
V ⊗n−1 that a1,j = 0 , ∀j < n and finally a relation among In and (1 , n )which does not exist.
SOME SPECIAL BASES OF THE 2–SWAP ALGEBRAS. 11
3.6. Appendix explicit formulas . In the formula for p = 5 we multiply by 2 4(1 , 2, 3, 4, 5) (7)
= 4(1 , 5)(1 , 2, 3, 4) = 2(1 , 4, 5)(2 , 3) + 2(1 , 2, 5)(3 , 4) − 2(1 , 3, 5)(4 , 2) − 2(1 , 4, 2, 5) + 2(1 , 3, 4, 5) + 2(3 , 4, 2)(1 , 5) + 2(1 , 2, 3, 5) + 2(2 , 4)(1 , 5) − 2(3 , 4)(1 , 5) − 2(2 , 3)(1 , 5) and develop the 4–cycles we obtain
−2(1 , 4, 2, 5) + 2(1 , 3, 4, 5) + 2(1 , 2, 3, 5) = (1 , 2)(3 , 5)+(1 , 2)(4 , 5) −(1 , 3)(2 , 5)+(1 , 3)(4 , 5) −(1 , 4)(2 , 5) −(1 , 4)(3 , 5) −(1 , 5)(2 , 4)+(1 , 5)(3 , 4)+(1 , 5)(2 , 3) +(2 , 4, 5) + (1 , 2, 4) + (1 , 3, 5) + (3 , 4, 5) + (1 , 3, 4) + (1 , 3, 5) + (2 , 3, 5) + (1 , 2, 3)
−(2 , 3) − 2(4 , 5) − (3 , 4) − 2(1 , 2) − (2 , 4) − (1 , 3) − (3 , 5) − (1 , 5) + 3 .
So the formula
(10) 4(1 , 2, 3, 4, 5) (7)
= 4(1 , 5)(1 , 2, 3, 4) = (1 , 2)(3 , 5)+(1 , 2)(4 , 5) −(1 , 3)(2 , 5)+(1 , 3)(4 , 5) −(1 , 4)(2 , 5) −(1 , 4)(3 , 5)+(1 , 5)(2 , 4) −(1 , 5)(3 , 4) −(2 , 3)(1 , 5) +(2 , 4, 5) + (1 , 2, 4) + (1 , 3, 5) + (3 , 4, 5) + (1 , 3, 4) + (1 , 3, 5) + (2 , 3, 5) + (1 , 2, 3)
−(2 , 3) − 2(4 , 5) − (3 , 4) − 2(1 , 2) − (2 , 4) − (1 , 3) − (3 , 5) − (1 , 5) + 3 +2(1 , 4, 5)(2 , 3) + 2(1 , 2, 5)(3 , 4) − 2(1 , 3, 5)(4 , 2) + 2(2 , 3, 4)(1 , 5)
The formula ( 8) for p = 6 (11) 8(1 , 2, 3, 4, 5, 6) (7)
= 8(1 , 5, 6)(1 , 2, 3, 4) = 4(1 , 4, 5, 6)(2 , 3)+4(1 , 2, 5, 6)(3 , 4) −4(1 , 3, 5, 6)(4 , 2) −4(1 , 4, 2, 5, 6)+4(1 , 3, 4, 5, 6) +4(3 , 4, 2)(1 , 5, 6)+4(1 , 2, 3, 5, 6)+4(2 , 4)(1 , 5, 6) −4(3 , 4)(1 , 5, 6) −4(2 , 3)(1 , 5, 6) .
So expanding the terms containing 4 and 5-cycles: we finally have
(12) 8(1 , 2, 3, 4, 5, 6) (7)
= 8(1 , 5, 6)(1 , 2, 3, 4) = 5 − 2(1 , 2) − 3(1 , 3) + (1 , 6) + (2 , 3) − 2(1 , 4)(2 , 3) − 3(1 , 6)(2 , 3) − 3(2 , 4) + 2(1 , 3)(2 , 4) −
(1 , 3)(2 , 6) − (1 , 4)(2 , 6) + (3 , 4) − 2(1 , 2)(3 , 4) − 3(1 , 6)(3 , 4) − 2(2 , 5)(3 , 4) + 2(1 , 6)(2 , 5)(3 , 4) −
2(1 , 5)(2 , 6)(3 , 4) − 2(3 , 5) + 2(2 , 4)(3 , 5) − 2(1 , 6)(2 , 4)(3 , 5) − (3 , 6) + (1 , 2)(3 , 6) − (1 , 4)(3 , 6) −
2(1 , 5)(3 , 6)+2(1 , 5)(2 , 4)(3 , 6)+(1 , 6)(4 , 2) −4(1 , 6)(4 , 5) −2(2 , 3)(4 , 5)+2(1 , 6)(2 , 3)(4 , 5) −2(4 , 6)+ (1 , 2)(4 , 6)+(1 , 3)(4 , 6) −2(1 , 5)(2 , 3)(4 , 6) −2(5 , 6)+2(1 , 3)(5 , 6) −2(2 , 3)(5 , 6)+2(1 , 4)(2 , 3)(5 , 6)+ 2(2 , 4)(5 , 6) −2(1 , 3)(2 , 4)(5 , 6) −2(3 , 4)(5 , 6)+2(1 , 2)(3 , 4)(5 , 6)+(1 , 2, 3)+(1 , 2, 4)+2(3 , 4)(1 , 2, 5) −
2(1 , 2, 6) + 2(3 , 4)(1 , 2, 6) + 2(3 , 5)(1 , 2, 6) + 2(4 , 5)(1 , 2, 6) + (1 , 3, 4) + 2(1 , 3, 5) − 2(2 , 4)(1 , 3, 5) + 2(1 , 3, 6) −2(2 , 4)(1 , 3, 6) −2(2 , 5)(1 , 3, 6)+2(4 , 5)(1 , 3, 6)+2(2 , 3)(1 , 4, 5)+2(1 , 4, 6)+2(2 , 3)(1 , 4, 6) −
2(2 , 5)(1 , 4, 6) −2(3 , 5)(1 , 4, 6)+2(1 , 6)(2 , 3, 5)+(2 , 3, 6)+2(1 , 6)(2 , 4, 5)+(2 , 4, 6)+2(3 , 4)(2 , 5, 6)+ 2(1 , 6)(3 , 4, 5) + (3 , 4, 6) + 2(3 , 5, 6) − 2(2 , 4)(3 , 5, 6) + 2(2 , 3)(4 , 5, 6) + 4(1 , 5, 6)(2 , 3, 4)
Hence the final formula for the permutation of type 3,3 in term of special elements obtained by the method explained in section 2.4 is:
(13) 8(4 , 3, 2)(5 , 6, 1) = 12 CLAUDIO PROCESI (1 , 2) − 2(1 , 3) + (1 , 5) + 4(1 , 6) − 3(2 , 3) + 2(1 , 5)(2 , 3) − 2(1 , 6)(2 , 3) − 3(2 , 4) + 2(1 , 5)(2 , 4) −
2(1 ,6)(2 ,4)+7(2 ,5) −2(1 ,3)(2 ,5) −4(1 ,4)(2 ,5) −6(1 ,6)(2 ,5) −2(2 ,6) −2(1 ,3)(2 ,6)+4(1 ,4)(2 ,6)+ 2(1 ,5)(2 ,6)+(3 ,4) −2(1 ,5)(3 ,4) −2(1 ,6)(3 ,4) −4(2 ,5)(3 ,4) −4(3 ,5)+4(1 ,4)(3 ,5)+4(2 ,4)(3 ,5)+ 4(2 ,6)(3 ,5) −4(1 ,4)(2 ,6)(3 ,5) −4(3 ,6)+4(1 ,2)(3 ,6) −4(1 ,4)(3 ,6) −4(2 ,5)(3 ,6)+4(1 ,4)(2 ,5)(3 ,6) −
2(4 ,5)+2(1 ,2)(4 ,5)+2(1 ,3)(4 ,5)+4(2 ,3)(4 ,5)+4(1 ,3)(2 ,6)(4 ,5)+4(3 ,6)(4 ,5) −4(1 ,2)(3 ,6)(4 ,5) −
2(1 ,2)(4 ,6)+2(1 ,3)(4 ,6) −4(1 ,3)(2 ,5)(4 ,6)+4(1 ,2)(3 ,5)(4 ,6)+6(5 ,6) −4(1 ,2)(5 ,6) −4(4 ,5)(1 ,2,3)+ 4(5 ,6)(1 ,2,3) + (1 ,2,5) −4(3 ,6)(1 ,2,5) −2(1 ,2,6) + 4(4 ,5)(1 ,2,6) + 2(1 ,3,6) + 4(2 ,5)(1 ,3,6) −
4(4 ,5)(1 ,3,6) −4(3 ,5)(1 ,4,2) + 4(5 ,6)(1 ,4,2) + 4(2 ,5)(1 ,4,3) −4(5 ,6)(1 ,4,3) −4(2 ,6)(1 ,4,5) + 4(3 ,6)(1 ,4,5) −(1 ,5,2) −4(1 ,5,6) + 8(3 ,4)(1 ,5,6) + 2(1 ,6,3) −2(1 ,6,5) + (2 ,3,5) + 2(2 ,3,6) −
4(4 ,5)(2 ,3,6) + 4(1 ,6)(2 ,4,3) −(2 ,4,5) + 4(1 ,6)(2 ,4,5) −(2 ,5,3) + 4(1 ,6)(2 ,5,3) + (2 ,5,4) −
2(2 ,5,6) + 2(1 ,4)(2 ,5,6) + 2(2 ,6,4) −4(3 ,5)(2 ,6,4) −2(1 ,4)(2 ,6,5) + (3 ,4,5) −4(1 ,6)(3 ,4,5) + 2(3 ,4,6) −(3 ,5,4)+2(1 ,2)(3 ,5,6) −2(1 ,4)(3 ,5,6)+4(2 ,5)(3 ,6,4) −2(1 ,2)(3 ,6,5)+2(1 ,4)(3 ,6,5) −
2(4 ,5,6) −2(4 ,6,5)
3.7. The corresponding rules of substitution:
In writing an algorithm to transform any given permutation into a sum of special elements one thus uses the following substitutional rules . From: (14) 2( a, b, c, d ) = (a, d )( b, c )+( a, b )( c, d )−(a, c )( d, b )−(a, d, b )+( a, c, d )+( b, c, d )+( a, b, c )+( b, d )−(c, d )−(b, c ).
and: (15) (a, B )( a, A ) = ( a, A, B ), e.g. (1 , 2, 3)(1 , 5, 4, 6) = (1 , 5, 4, 6, 2, 3) for a cycle c := ( a, b, c, d, A ) of length n + 4, where A is of length n > 0, we have (16) c := ( a, b, c, d, A ) = ( a, A )( a, b, c, d ) = (a, d, A )( b, c ) + ( a, b, A )( c, d ) − (a, c, A )( d, b ) − (a, d, b, A )+( a, c, d, A ) + ( b, c, d ) + ( a, b, c, A ) + ( b, d ) − (c, d ) − (b, c ).
This formula now contains only cycles of length < n + 4. By applying recursively these rules we arrive to a linear combination in which no cycles of length > 3 appear. Now the only reduction to be made is if there are pairs of 3-cycles. For these we finally repeat the final rule: (17) 8( d, c, b )( e, f, a ) = (a, b )−2( a, c )+( a, e )+4( a, f )−3( b, c )+2( a, e )( b, c )−2( a, f )( b, c )−3( b, d )+ 2( a, e )( b, d )−2( a, f )( b, d )+7( b, e )−2( a, c )( b, e )−4( a, d )( b, e )−6( a, f )( b, e )−
2( b, f ) − 2( a, c )( b, f ) + 4( a, d )( b, f ) + 2( a, e )( b, f ) + ( c, d ) − 2( a, e )( c, d ) −
2( a, f )( c, d )−4( b, e )( c, d )−4( c, e )+4( a, d )( c, e )+4( b, d )( c, e )+4( b, f )( c, e )−
4( a, d )( b, f )( c, e )−4( c, f )+4( a, b )( c, f )−4( a, d )( c, f )−4( b, e )( c, f )+4( a, d )( b, e )( c, f )−
2( d, e )+2( a, b )( d, e )+2( a, c )( d, e )+4( b, c )( d, e )+4( a, c )( b, f )( d, e )+4( c, f )( d, e )−SOME SPECIAL BASES OF THE 2–SWAP ALGEBRAS. 13
4( a, b )( c, f )( d, e )−2( a, b )( d, f )+2( a, c )( d, f )−4( a, c )( b, e )( d, f )+4( a, b )( c, e )( d, f )+ 6( e, f )−4( a, b )( e, f )−4( d, e )( a, b, c )+4( e, f )( a, b, c )+( a, b, e )−4( c, f )( a, b, e )−
2( a, b, f )+4( d, e )( a, b, f )+2( a, c, f )+4( b, e )( a, c, f )−4( d, e )( a, c, f )−4( c, e )( a, d, b )+ 4( e, f )( a, d, b )+4( b, e )( a, d, c )−4( e, f )( a, d, c )−4( b, f )( a, d, e )+4( c, f )( a, d, e )−
(a, e, b )−4( a, e, f )+8( c, d )( a, e, f )+2( a, f, c )−2( a, f, e )+( b, c, e )+2( b, c, f )−
4( d, e )( b, c, f )+4( a, f )( b, d, c )−(b, d, e )+4( a, f )( b, d, e )−(b, e, c )+4( a, f )( b, e, c )+ (b, e, d )−2( b, e, f )+2( a, d )( b, e, f )+2( b, f, d )−4( c, e )( b, f, d )−2( a, d )( b, f, e )+ (c, d, e )−4( a, f )( c, d, e )+2( c, d, f )−(c, e, d )+2( a, b )( c, e, f )−2( a, d )( c, e, f )+ 4( b, e )( c, f, d ) − 2( a, b )( c, f, e ) + 2( a, d )( c, f, e ) − 2( d, e, f ) − 2( d, f, e )until we arrive at a linear combination of special elements.
References
E. Aljadeff, A. Giambruno, C. Procesi, A. Regev. Rings with polynomial identities and finite dimensional representations of algebras , A.M.S. Colloquium Publications, vol. 66.2020; 630 pp MSC: Primary 16; 15; 14 V. Drensky, E. Formanek, Polynomial identities rings Advanced Course in Mathe-matics, CRM Barcelona Kuzmin, On the Nagata–Higman Thorem, Mathematical structures-Computational mathematics-Mathematical modelling, Sofia 1975, pp. 101–107, Russian 3.3
Jun Takahashi, Chaithanya Rayudu, Cunlu Zhou, Robbie King, Kevin Thompson, Ojas Parekh An SU (2) –symmetric Semidefinite Programming Hierarchy for Quan-tum Max Cut arXiv:quant-ph 2307.15688 Adam Bene Watts, Anirban Chowdhury, Aidan Epperly, J. William Helton, Igor Klep, Relaxations and Exact Solutions to Quantum Max Cut via the Algebraic Struc-ture of Swap Operators arXiv:quant-ph 2307.15661 3.3 3.3
Felix Huber, Igor Klep, Victor Magron and Jurij Volˇ ciˇ c, Dimension–Free Entan-glement Detection in Multipartite Werner States Communications in Mathematical Physics , Volume 396, pages 1051–1070, (2022) Cite this article T. Eggeling and R. F. Werner, Separability properties of tripartite states with U ⊗
U ⊗ U symmetry Phys. Rev. A 63, 042111– 21 March 2001 1
Claudio Procesi, A note on the Formanek Weingarten function . Note di Matematica
v. 17 (2021), no.1. 1
Claudio Procesi, Tensor fundamental theorems of invariant theory , arXiv: 2011.10820. 1
Yu. P. Razmyslov, Identities with trace in full matrix algebras over a field of char-acteristic zero , (Russian) Izv. Akad. Nauk SSSR Ser. Mat. 38 (1974), 723–756. 11
Reinhard F. Werner, Quantum states with Einstein-Podolsky-Rosen correlations admitting a hidden-variable model Phys. Rev. A 40, 4277 Published 1 October 1989
Dipartimento di Matematica, G. Castelnuovo, Universit` a di Roma La Sapienza, piazzale A. Moro, 00185, Roma, Italia
Email address : procesi@mat.uniroma1.it |
189038 | https://www.oxfordlearnersdictionaries.com/us/definition/english/silence_2 | Definition of silence verb from the Oxford Advanced Learner's Dictionary
silence
| | |
--- |
| present simple I / you / we / they silence | /ˈsaɪləns/ /ˈsaɪləns/ |
| he / she / it silences | /ˈsaɪlənsɪz/ /ˈsaɪlənsɪz/ |
| past simple silenced | /ˈsaɪlənst/ /ˈsaɪlənst/ |
| past participle silenced | /ˈsaɪlənst/ /ˈsaɪlənst/ |
| -ing form silencing | /ˈsaɪlənsɪŋ/ /ˈsaɪlənsɪŋ/ |
Join us
Join our community to access the latest language learning and assessment tips from Oxford University Press!
Other results
Nearby words
Oxford Learner's Dictionaries
More from us
Who we are
Oxford University Press is a department of the University of Oxford. It furthers the University's objective of excellence in research, scholarship, and education by publishing worldwide |
189039 | https://brilliant.org/wiki/holomorphic-function/ | Holomorphic Function
Sign up with Facebook
or
Sign up manually
Already have an account?
Log in here.
Sameer Kailasa,
Jake Lai,
and
Jimin Khim
contributed
In complex analysis, a holomorphic function is a complex differentiable function. The condition of complex differentiability is very strong, and leads to an especially elegant theory of calculus for these functions.
Contents
Definitions and Cauchy-Riemann Equations
Analyticity of Holomorphic Functions
Definitions and Cauchy-Riemann Equations
A function f:R→R is said to be differentiable at x∈R if h→0h∈Rlimhf(x+h)−f(x) exists. If the limit exists, its value is called f′(x).
Analogously, a function f:C→C (that is, f takes a complex number and returns a complex number) is said to be complex differentiable at z∈C if h→0h∈Climhf(z+h)−f(z) exists. Again, if the limit exists, its value is called f′(z). If f is complex differentiable at every z∈U⊂C, then f is said to be holomorphic on U.
Although both conditions are notationally the same, the complex differentiability condition is actually much stronger than the differentiability condition for functions R→R.
This is because as h approaches 0 in the complex limit, it must do so from all directions, of which there are infinitely many. (In the real limit, there are only two directions from which h may approach 0: the positive direction and the negative direction.) Thus, if f:C→C is complex differentiable at z∈C, then t→0t∈Rlimthf(z+th)−f(z) exists for all h∈C and equals f′(z).
One can view C as R2 by identifying the complex number x+yi with the pair (x,y). From this perspective, one may write f(z)=f(x,y)=u(x,y)+i⋅v(x,y), where u(x,y)=Re(f(z)) and v(x,y)=Im(f(z)). But then the complex differentiability condition requires
f′(z)=t→0t∈Rlimt⋅1f(z+t⋅1)−f(z)=t→0t∈Rlim(tu(x+t,y)−u(x,y)+i⋅tv(x+t,y)−v(x,y))=∂x∂u(x,y)+i∂x∂v(x,y)
and
f′(z)=t→0t∈Rlimt⋅if(z+t⋅i)−f(z)=−i⋅t→0t∈Rlim(tu(x,y+t)−u(x,y)+i⋅tv(x,y+t)−v(x,y))=−i⋅(∂y∂u+i⋅∂y∂v)=∂y∂v(x,y)−i⋅∂y∂u(x,y).
Setting these equal proves that wherever f is complex differentiable, it satisfies ∂x∂u=∂y∂v,∂y∂u=−∂x∂v. These are the Cauchy-Riemann equations, a system of differential equations that every holomorphic function must satisfy on its domain. Hence, there are severe constraints on how a holomorphic function may behave; namely, it may only behave as these equations prescribe.
Analyticity of Holomorphic Functions
Perhaps the most important property of holomorphic functions is their analyticity. This means that any holomorphic function has an expansion as a power series.
Suppose f:U→C is holomorphic with U an open set. Then there exists a sequence of complex numbers {an}n≥0 such that
f(z)=n=0∑∞an(z−z0)n
within the radius of convergence about z0 (i.e., it converges for all z in some disk centered at z0) and equals f(z) when it converges. Specifically, an=n!f(n)(z0).
Let γ be a closed curve. Using the Cauchy integral formula and geometric series, we have
f(z)=2πi1∫γζ−zf(ζ) dζ=2πi1∫γ1−ζ−z0z−z0f(ζ)/(ζ−z0) dζ=2πi1∫γζ−z0f(ζ)n=0∑∞(ζ−z0z−z0)n dζ=n=0∑∞(2πi1∫γ(ζ−z0)n+1f(ζ) dζ)(z−z0)n=n=0∑∞n!f(n)(z0)(z−z0)n. □
Cite as:
Holomorphic Function.
Brilliant.org.
Retrieved
00:04, August 27, 2025,
from |
189040 | https://e-archivo.uc3m.es/bitstreams/cc097387-074f-4dfe-98ff-f824e4542547/download | UNIVERSIDAD CARLOS III DE MADRID Departamento de Ingenieria de Sistemas y Automatica
APUNTES DE CONTROL DE SISTEMAS II
L. Moreno, S. Garrido, C. Balaguer IV Índice general
1.. Introducción . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.1. Introducción . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2. Concepto de transformada . . . . . . . . . . . . . . . . . . . . . 21.3. Transformada de Fourier . . . . . . . . . . . . . . . . . . . . . . 21.3.1. Transformada inversa de Fourier . . . . . . . . . . . . . 31.4. Transformada de Laplace . . . . . . . . . . . . . . . . . . . . . . 31.4.1. Propiedades de la transformada de Laplace . . . . . . . . 41.4.2. Transformadas de Laplace de algunas funciones . . . . . . 71.4.3. Transformada inversa de Laplace . . . . . . . . . . . . . 71.4.4. Resolución de ecuaciones diferenciales . . . . . . . . . . 11 1.4.5. Operador derivada . . . . . . . . . . . . . . . . . . . . . 12 1.5. Transformada z . . . . . . . . . . . . . . . . . . . . . . . . . . . 13 1.5.1. Propiedades de la transformada en z . . . . . . . . . . . . 13 1.5.2. Transformadas z de algunas funciones . . . . . . . . . . . 15 1.5.3. La transformada z inversa . . . . . . . . . . . . . . . . . 16 1.5.4. Resolución de ecuaciones en diferencias . . . . . . . . . . 18 1.5.5. Operador retardo . . . . . . . . . . . . . . . . . . . . . . 19 1.6. Procesos estocásticos o aleatorios . . . . . . . . . . . . . . . . . 20 1.6.1. Variable aleatoria y proceso aleatorio . . . . . . . . . . . 21 1.6.2. Funciones de distribución y de densidad de probabilidad . 24 1.6.3. Parámetros para describir un proceso estocástico . . . . . 26 1.6.4. Función de autocorrelación . . . . . . . . . . . . . . . . . 29 1.6.5. Función de correlación cruzada . . . . . . . . . . . . . . 31 1.6.6. Función de densidad espectral . . . . . . . . . . . . . . . 33 1.6.7. Descripción espectral de las perturbaciones . . . . . . . . 34 1.6.8. Espectro cruzado . . . . . . . . . . . . . . . . . . . . . . 35 1.6.9. Ruido blanco . . . . . . . . . . . . . . . . . . . . . . . . 35 1.6.10. Descripción de perturbaciones en función del ruido blanco 36
2.. Técnicas clásicas de modelado de sistemas . . . . . . . . . . . 41 2.1. Introducción . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41 2.2. Modelos de entrada/salida . . . . . . . . . . . . . . . . . . . . . 42 2.3. Modelos temporales . . . . . . . . . . . . . . . . . . . . . . . . . 42 2.3.1. Conceptos básicos . . . . . . . . . . . . . . . . . . . . . 42 2.3.2. Modelos en tiempo continuo . . . . . . . . . . . . . . . . 46 2.3.3. Modelos en tiempo discreto . . . . . . . . . . . . . . . . 47 2.4. Modelos frecuenciales . . . . . . . . . . . . . . . . . . . . . . . 69 2.4.1. Respuesta frecuencial de los sistemas muestreados . . . . 71 2.4.2. Espectro de una señal muestrada . . . . . . . . . . . . . . 74 2.4.3. Teorema de muestreo . . . . . . . . . . . . . . . . . . . . 75 2.5. Modelos estocásticos . . . . . . . . . . . . . . . . . . . . . . . . 76 2.5.1. Modelos estocásticos en tiempo discreto . . . . . . . . . . 77
3.. Modelado y análisis de sistemas en el espacio de estados . . 81 3.1. Introducción . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81 3.2. Concepto de estado de un sistema . . . . . . . . . . . . . . . . . 83 3.2.1. Representación matricial de las ecuaciones de estado . . . 84 3.2.2. Función de transferencia y representación en el espacio de estados . . . . . . . . . . . . . . . . . . . . . . . . . . . 91 3.3. Representación de sistemas en el espacio de estados . . . . . . . . 92 3.3.1. Conversión de una ecuación diferencial ordinaria a ecua-ciones de estado . . . . . . . . . . . . . . . . . . . . . . 92 3.3.2. Conversión de una ecuación en diferencias a ecuaciones de estado . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104 3.3.3. Transformaciones entre representaciones . . . . . . . . . 106 3.4. Solución de la ecuación de estado . . . . . . . . . . . . . . . . . 114 3.4.1. Sistemas de tiempo continuo . . . . . . . . . . . . . . . . 115 3.4.2. Obtención de la solución por el método de la transformada de Laplace . . . . . . . . . . . . . . . . . . . . . . . . . 117 3.4.3. Discretización de las ecuaciones de estado en tiempo con-tinuo . . . . . . . . . . . . . . . . . . . . . . . . . . . . 118 3.4.4. Solución de la ecuación de estado en tiempo discreto . . . 122 3.4.5. Obtención de la solución por el método de la transformada en z . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123 3.5. Modelado de las perturbaciones en el espacio de estados . . . . . 125 3.5.1. Perturbaciones en el sistema . . . . . . . . . . . . . . . . 125 3.5.2. Perturbaciones en la medida . . . . . . . . . . . . . . . . 128 3.6. Ejemplos de sistemas físicos . . . . . . . . . . . . . . . . . . . . 131
4.. Técnicas clásicas de control . . . . . . . . . . . . . . . . . . . . 137 4.1. Planteamiento del problema . . . . . . . . . . . . . . . . . . . . 137 4.2. Diseño de controladores clásicos . . . . . . . . . . . . . . . . . . 138 4.3. Especificaciones . . . . . . . . . . . . . . . . . . . . . . . . . . . 139 4.4. Control PID . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141
VI VII
4.4.1. Estructura básica de un controlador PID . . . . . . . . . . 141 4.4.2. Métodos de Ziegler - Nichols . . . . . . . . . . . . . . . 142 4.5. Métodos analíticos de diseño de controladores PID . . . . . . . . 146 4.5.1. Método de asignación de polos . . . . . . . . . . . . . . . 146 4.5.2. Diseño basado en los polos dominantes . . . . . . . . . . 149 4.5.3. Discretización de un controlador PID . . . . . . . . . . . 159 4.5.4. Diseño de controladores PID discretos . . . . . . . . . . . 167 4.5.5. Estructura de un controlador PID discreto real . . . . . . 170 4.6. Diseño de reguladores por síntesis directa . . . . . . . . . . . . . 173 4.6.1. Restricciones de realización física . . . . . . . . . . . . . 175 4.6.2. Conveniencia de simplicidad . . . . . . . . . . . . . . . . 177 4.6.3. Restricciones de estabilidad . . . . . . . . . . . . . . . . 177 4.6.4. Controladores con tiempo de establecimiento mínimo . . . 182
5.. Control de sistemas por realimentación de estado . . . . . . . 191 5.1. Planteamiento del problema . . . . . . . . . . . . . . . . . . . . 191 5.1.1. Modos observables y controlables . . . . . . . . . . . . . 192 5.2. Controlabilidad de un sistema . . . . . . . . . . . . . . . . . . . 193 5.2.1. Controlabilidad de estado . . . . . . . . . . . . . . . . . 193 5.2.2. Controlabilidad de la salida . . . . . . . . . . . . . . . . 198 5.3. Observabilidad de un sistema . . . . . . . . . . . . . . . . . . . . 200 5.3.1. Observabilidad completa de estado . . . . . . . . . . . . 201 5.4. Invarianza de la controlabilidad y observabilidad . . . . . . . . . 203 5.5. Principio de dualidad . . . . . . . . . . . . . . . . . . . . . . . . 205 5.6. Control por realimentación de estado . . . . . . . . . . . . . . . . 206 5.6.1. Sistemas con entrada y salida escalar . . . . . . . . . . . 207 5.6.2. Ajuste de las posiciones de los polos . . . . . . . . . . . . 208 5.6.3. Ajuste de la ganancia . . . . . . . . . . . . . . . . . . . . 215 5.6.4. Modificación del tipo del sistema . . . . . . . . . . . . . 217 5.6.5. Sistemas con entrada vector . . . . . . . . . . . . . . . . 221 5.7. Diseño de observadores de estado . . . . . . . . . . . . . . . . . 222 5.7.1. Observador de orden completo . . . . . . . . . . . . . . . 223 5.7.2. Comportamiento conjunto del sistema realimentado con el observador . . . . . . . . . . . . . . . . . . . . . . . . . 231 5.7.3. Observador de orden reducido . . . . . . . . . . . . . . . 232 5.8. Observador óptimo del estado . . . . . . . . . . . . . . . . . . . 235 5.8.1. Ecuaciones del filtro de Kalman . . . . . . . . . . . . . . 240 VIII 1. INTRODUCCIóN
1.1. Introducción
Uno de los problemas con los que tradicionalmente se tienen que enfrentar aquellos que trabajan en el control o en la identificación de procesos estriba en la existencia de una diversidad de herramientas matemáticas para estudiar, analizar y diseñar dichos sistemas. Por ejemplo, si deseamos analizar y controlar un sistema continuo el sistema podemos modelarlo en cuanto a su respuesta temporal mediante la utilización de la
función de transferencia o mediante la técnica de variables de estado . Además el sistema puede ser controlado en tiempo continuo o en tiempo discreto lo que reque-rirá la utilización de transformadas de Laplace , para el caso de tiempo continuo, o transformadas en Z , para el caso de tiempo discreto. Otras veces nos interesará estudiar la respuesta en frecuencia del sistema para lo que nos basaremos en el es-tudio de la respuesta en régimen permanente del sistema (en base a la función de transferencia). También nos encontramos con frecuencia el caso en el que la utilización de las técnicas anteriores no explica suficientemente el comportamiento del sistema, bien por que este está sometido a perturbaciones no modelables o bien porque nuestro conocimiento del mismo no es suficientemente preciso. En estas situaciones se ha-ce necesario la utilización de técnicas probabilísticas que nos ayuden a caracterizar estas señales o sistemas. Así que es muy frecuente que en el estudio y análisis de un sistema nos encon-tremos con la presencia de diferentes herramientas matemáticas que nos permiten caracterizar, analizar y diseñar un sistema de control de forma que tengamos en cuenta los diferentes efectos presentes en el sistema. El objetivo básico de este capítulo es introducir las herramientas matemáticas que se utilizan normalmente para analizar y sintetizar los sistemas de control de-terministas en el dominio del tiempo, tanto en tiempo discreto como en tiempo continuo, así como en frecuencia. Y se introducirá la herramienta matemática bá-sica para analizar los comportamientos probabilísticos presentes en procesos o en señales. 2 1. Introducción
1.2. Concepto de transformada
A la hora de analizar y diseñar los sistemas de control para una amplia variedad de procesos, nos encontramos con la necesidad de resolver ecuaciones diferencia-les o integro-diferenciales con el fin de conocer la respuesta que el proceso nos va a dar ante determinadas entradas o acciones de control. Estas ecuaciones diferen-ciales pueden resolverse de forma directa o bien se pueden resolver convirtiéndolas en otro tipo de ecuaciones de más fácil resolución. Existen una serie de métodos operacionales que nos permiten transformar problemas de resolución de ecuacio-nes diferenciales lineales o problemas de resolución de ecuaciones en diferencias en problemas de tipo polinómico mucho más fáciles de tratar. Entre las transformaciones más frecuentemente utilizadas para la solución de ecuaciones diferenciales ordinarias de tipo lineal están las transformaciones de tipo integral, y entre estas una muy ampliamente utilizada en control es la denominada
transformada de Laplace .Para el caso de los sistemas con ecuaciones en diferencias se utiliza la denomi-nada transformada z .
1.3. Transformada de Fourier
Dada una función real f (t), se define la transformada de Fourier de la función
f (t) como:
F (ω) = F[f (t)] =
∫ +∞−∞
f (t)e−jωt dt (1.3.1) siendo ω una frecuencia variable, −∞ < ω < +∞, expresada en (rad/s). La función f (t) se dice que tiene transformada de Fourier si la integral 1.3.1 existe para todos los valores de ω. Son condiciones suficientes para que la integral exista que la función f (t) sólo tenga un número finito de discontinuidades, máximos y mínimos a lo largo de un intervalo finito de tiempo, y que la función f (t) sea absolutamente integrable, es decir que
∫ +∞−∞
|f (t)|dt < ∞ (1.3.2) El término exponencial complejo de la transformada 1.3.1 puede expresarse, aplicando la fórmula de Euler, como
e−jωt = cos( ωt ) − j sin( ωt ) (1.3.3) Si sustituimos en la expresión de la transformada de Fourier 1.3.1 el término exponencial por 1.3.3, obtenemos que, 1.4. Transformada de Laplace 3
F (ω) =
∫ +∞−∞
f (t)[cos( ωt ) − j sin( ωt )]d t
=
∫ +∞−∞
f (t) cos( ωt )d t − j
∫ +∞−∞
f (t) sin( ωt )d t (1.3.4) En esta expresión 1.3.4 de la transformada de Fourier se puede apreciar la exis-tencia de dos términos, uno de ellos real y otro imaginario, que denominaremos
R(ω) e I(ω) respectivamente, con lo que F (ω) vendrá dado por
F (ω) = R(ω) + jI (ω) (1.3.5) En la ecuación 1.3.5 se puede apreciar que la transformada de Fourier F (ω) es una función compleja de la variable ω. Para todo valor de la variable ω, tendremos un valor de F (ω) que tiene un módulo, |F (ω)|, y un argumento, ∠F (ω).
|F (ω)| = √R2(ω) + I2(ω)
∠F (ω) = arctan −1
( I(ω)
R(ω)
)
(1.3.6) Si representásemos en una gráfica los valores |F (ω)| para cada ω obtendríamos lo que se denomina el espectro de magnitudes de la señal f (t), y si hiciésemos lo mismo para el argumento ∠F (ω) obtendríamos lo que se conoce como espectro de fase de la señal f (t).
1.3.1. Transformada inversa de Fourier
La transformada inversa es una operación que nos permite obtener la función original f (t) o partir de la transformada de Fourier F (ω). Esta operación viene dada por la siguiente expresión
f (t) = F−1[F (ω)] = 12π
∫ +∞−∞
F (ω)ejωt dω (1.3.7)
1.4. Transformada de Laplace
Dada una función real f (t) tal que f (t) = 0 para t < 0 se define la transfor-mada unilateral de Laplace de la función f (t) como:
F (s) = L[f (t)] =
∫ ∞
0
f (t)e−st dt (1.4.1) 4 1. Introducción
Existe también la transformada bilateral de Laplace, en la que los límites de integración se extienden desde −∞ a +∞, pero dado que no se suele utilizar en ingeniería de control, no se va a tratar. La transformada unilateral de Laplace, que en adelante y por simplificar denominaremos como transformada de Laplace, existe siempre que la integral 1.4.1 converja, para lo que es condición necesaria y sufi-ciente que se verifique que:
∫ ∞
0
|f (t)e−σt |dt < ∞ (1.4.2) para σ finito y real. Dada una transformada de Laplace de una función f (t), al conjunto de todos los números complejos s para los cuales la integral 1.4.1 existe se le denomina
región de convergencia .Dada una función f (t), si existe la transformada de Laplace L(s) y la particu-larizamos para aquellos complejos en los que s = jω obtenemos que
F (s)|s=jω = F (jω ) =
∫ ∞
0
f (t)e−jωt dt (1.4.3) si además la función f (t) = 0 para t < 0, la ecuación 1.4.3 es igual a la transfor-mada de Fourier F (ω) de la función f (t) (compárese con la expresión que veíamos en 1.3.1). Es decir
F (ω) = F (s)|s=jω (1.4.4) Esto nos indica que podemos obtener la transformada de Fourier, F (ω) de una función f (t) (siempre que el valor de esta función para t < 0 sea cero), a partir de su transformada de Laplace, F (s), sin más que sustituir en la expresión de la transformada de Laplace la variable s por jω .
1.4.1. Propiedades de la transformada de Laplace
La transformada de Laplace presenta una serie de propiedades muy útiles que hace que pueda ser considerada como un operador que convierte una función tem-poral f (t) en el dominio del tiempo en una función F (s) de variable compleja s.Veamos estas propiedades: 1. Linealidad
Una propiedad muy importante de la transformada de Laplace es que es un operador lineal. Dadas dos funciones f (t) y g(t) y una constante k se verifica que:
L[f (t) + g(t)] = L[f (t)] + L[g(t)] (1.4.5) 1.4. Transformada de Laplace 5
y
L[kf (t)] = kL[f (t)] (1.4.6) 2. Diferenciación
La transformada de Laplace de la diferencial de una función f (t) se puede expresar de la siguiente manera:
L[ ddt f (t)] = sF (s) − lim
t→0
f (t) = sF (s) − f (0) (1.4.7) siendo F (s) la transformada de la función f (t) y f (0) el límite de la función
f (t) cuando t tiende a cero. En general, para derivadas de orden superior se tiene la siguiente expresión:
L[ dn
dtn f (t)] = snF (s)−sn−1f (0) −sn−2 ddt f (0)+ . . . + dn−1
dtn−1 f (0) (1.4.8) 3. Integración
La transformada de Laplace de la integral de una función f (t) con respecto al tiempo es la transformada de Laplace de la función dividida por s, es decir:
L[
∫ t
0
f (τ )d τ ] = F (s)
s (1.4.9) y si se generaliza para integraciones de órdenes superiores
L[
∫ t1
0
∫ t2
0
. . .
∫ tn
0
f (τ )d τ dτ1 . . . dτn] = F (s)
sn (1.4.10) 4. Desplazamiento en el tiempo
Un desplazamiento en el tiempo de valor a de la función f (t), equivale en el dominio complejo a multiplicar la transformada de Laplace de la función
F (s), por e−sa ; esto es:
L[f (t − a)u0(t − a)] = e−sa F (s) (1.4.11) donde u0(t−a) representa la función escalón unidad desplazada en el tiempo
a unidades. 6 1. Introducción
Teorema del valor inicial
Si la transformada de Laplace de la función f (t) es F (s), entonces lim
t→0
f (t) = lim
s→∞
sF (s) (1.4.12) si existe el límite en el tiempo. 6. Teorema del valor final
Si la transformada de Laplace de la función f (t) es F (s), y si sF (s) es analítica en la parte derecha del plano-s y sobre el eje imaginario, entonces lim
t→∞
f (t) = lim
s→0
sF (s) (1.4.13) Este teorema es de gran utilidad en el análisis y diseño de sistemas de control ya que nos dice el valor final que tomará una función en el dominio del tiempo conociendo el valor que toma su transformada de Laplace para s = 0 .7. Desplazamiento complejo
La transformada de Laplace de una función f (t) multiplicada por e∓αt ,donde
α es una constante, equivale a reemplazar en la transformada de Laplace de la función, F (s), s por s ± a es decir
L[e∓αt f (t]] = F (s ± a) (1.4.14) 8. Convolución real
Sean F1(s) y F2(s) las transformadas de Laplace de las funciones f1(t) y
f2(t), respectivamente, y f1(t) = 0 y f2(t) = 0 para t < 0 entonces
F1(s)F2(s) = L[f1(t) ∗ f2(t)] (1.4.15)
= L[
∫ t
0
f1(τ )f2(t − τ )d τ ] (1.4.16)
= L[
∫ t
0
f2(τ )f1(t − τ )d τ ] (1.4.17) en esta expresión el símbolo ∗ denota la operación convolución en el dominio del tiempo. 1.4. Transformada de Laplace 7
1.4.2. Transformadas de Laplace de algunas funciones
Ejemplo 1.4.1: La transformada de Laplace de la función escalón unitario u0(t)
que se define como
f (t) = u0(t) =
{ 1, t ≥ 00, t < 0 (1.4.18) se obtiene como
F (s) = L[u0(t)] =
∫ ∞
0
u0(t)e−st dt = − 1
s e−st
∣∣∣∞
0
= 1
s (1.4.19)
Ejemplo 1.4.2: La transformada de Laplace de la función exponencial f (t) =
e−αt para t ≥ 0 se obtiene como
F (s) = L[e−αt ] =
∫ ∞
0
e−αt e−st dt = − 1
s + α e−(s+α)t
∣∣∣∞
0
= 1
s + α (1.4.20) En la tabla se pueden ver las transformadas de algunas de las funciones más usuales.
f (t) L(s)
δ(t) 1
u0(t) 1
s
t 1
s2
e−at 1
s+a
te −at 1(s+a)2
tn, t ≥ 0 n!
sn+1
, n = 1 , 2, . . . tne−at , t ≥ 0 n!(s+a)n+1 , n = 1 , 2, . . .
sin( ωt ) ω
(s2+ω2)
cos( ωt ) s
(s2+ω2)
e−at sin ωt ω
(s+a)2+ω2
e−at cos ωt s+a
(s+a)2+ω2
1.4.3. Transformada inversa de Laplace
La operación de obtener la función f (t) a partir de la transformada de Laplace
F (s) se le denomina transformada inversa de Laplace, y se escribe 8 1. Introducción
f (t) = L−1[F (s)] (1.4.21) La transformación integral inversa de Laplace se define como
f (t) = 12πj
∫ c−j∞
c+j∞
F (s)est ds (1.4.22) donde c es una constante real mayor que la parte real de todas las singularidades de F(s) . Esta integral se evalúa en el plano-s complejo. Desde el punto de vista del control de procesos esta forma de obtener la transformación inversa resulta extremadamente costosa y resulta poco práctica. Por ello se acude o a la utilización de tablas de transformadas de Laplace o a la expansión en fracciones parciales.
Expansión en fracciones parciales
En la mayor parte de las aplicaciones de control, no es necesario recurrir a la inversión integral dada por 1.4.22 para obtener la transformada inversa de Laplace. Normalmente tanto las ecuaciones diferenciales como las soluciones de las mismas en los problemas de control suelen tener la forma de funciones racionales en s, del tipo
F (s) = N (s)
D(s) (1.4.23) donde N (s) y D(s) son polinomios en s. Se supone que el grado del polinomio
N (s) del numerador en s es menor o igual que el grado del polinomio D(s) del denominador. Esta expresión puede escribirse en la siguiente forma
F (s) = N (s)
sn + a1sn−1 + . . . + an−1s + an
(1.4.24) donde los coeficientes ai son coeficientes reales. A las raíces del polinomio N (s)
se les denominan ceros y a las raíces del denominador D(s) se les denominan polos
de dicha función F (s).La idea básica del método de la expansión en fracciones parciales consiste en descomponer el polinomio F (s) en una suma de fracciones simples de forma que éstas sean las transformadas de funciones simples y conocidas, de forma que para cada fracción resulte muy fácil calcular la transformada inversa. Como la trans-formada de Laplace es un operador lineal, una vez calculadas las transformadas inversas de las fracciones simples, para obtener la transformada inversa de la fun-ción F (s) basta con sumar las transformadas inversas de las fracciones simples. Veamos los diferentes casos que se pueden presentar. 1.4. Transformada de Laplace 9
Caso de polos simples y reales Si todos los polos de D(s) son simples y reales, entonces la ecuación 1.4.24 toma la siguiente forma
F (s) = N (s)(s + s1)( s + s2) . . . (s + sn) (1.4.25) y en ella ∀i 6 = j, s i 6 = sj . Descomponiendo la expresión 1.4.25 en fracciones parciales obtenemos
F (s) = A1
(s + s1) + A2
(s + s2) + . . . + An
(s + sn) (1.4.26) donde los coeficientes A1, . . . , A n se obtienen como
Ai = [(s + si) N (s)
D(s)
]∣∣∣s=−si
(1.4.27) Obsérvese que cada una de estas fracciones simples son las transformadas de Laplace de funciones de tipo exponencial con exponente −si y coeficiente Ai.
Caso de polos múltiples y reales Si algunos de los polos de D(s) son múl-tiples, entonces la ecuación 1.4.24 toma la siguiente forma
F (s) = N (s)(s + s1)( s + s2) . . . (s + sk−1)( s + sk)n−k+1 (1.4.28) y en ella sk tiene un orden de multiplicidad de n − k + 1 . Descomponiendo la expresión 1.4.28 en fracciones parciales obtenemos
F (s) = A1
(s + s1) + A2
(s + s2) + . . . + Ak−1
(s + sk−1) ++ Ak1
(s + sk) + Ak2
(s + sk)2 + . . . + Akn−k
(s + sk)n−k+1 (1.4.29) donde los coeficientes A1, . . . , A k−1 correspondientes a los polos simples se ob-tienen igual que en el caso anterior, y los correspondientes al polo multiple se obtienen como
Akn−k =
[
(s + sk)n−k+1 N (s)
D(s)
]∣ ∣∣∣s=−sk
(1.4.30)
Akn−k+1 = dds
[
(s + sk)n−k+1 N (s)
D(s)
]∣ ∣∣∣s=−sk
(1.4.31)
. . . Ak1 = 1(n − k)! dn−k
dsn−k
[
(s + sk)n−k+1 N (s)
D(s)
]∣ ∣∣∣s=−sk
(1.4.32) 10 1. Introducción
Caso de polos complejos conjugados La expansión en fracciones parcia-les vista para el caso de polos simples y reales, es válida también para el caso de polos complejos conjugados de la forma s = −σ + jw y s = −σ − jw . En este caso los coeficientes A−σ+jw y A−σ−jw se obtienen de la siguiente forma
A−σ+jw = [(s + σ − jw ) N (s)
D(s)
]∣∣∣s=−σ+jw (1.4.33)
A−σ−jw = [(s + σ + jw ) N (s)
D(s)
]∣∣∣s=−σ−jw (1.4.34)
Ejemplo 1.4.3: Supongamos la siguiente función
F (s) = 2s + 3
s(s + 1)( s + 3) (1.4.35) que si la expandimos en fracciones parciales quedará de la siguiente forma
F (s) = A1
s + A2
(s + 1) + A3
(s + 3) (1.4.36) Los coeficientes vendrán determinados por
A1 = [sF (s)]∣∣∣s=0 = 2s + 3 (s + 1)( s + 3)
∣∣∣s=0 = 1
A2 = [(s + 1) F (s)]∣∣∣s=−1 = 2s + 3
s(s + 3)
∣∣∣s=−1 = − 12
A3 = [(s + 3) F (s)]∣∣∣s=−3 = 2s + 3
s(s + 1)
∣∣∣s=−3 = − 12
con lo que
F (s) = 1
s + − 12
(s + 1) + − 12
(s + 3) (1.4.37) y su transformada inversa será la función
f (t) = 1 − 12 e−t − 12 e−3t (1.4.38)
Ejemplo 1.4.4: Supongamos la siguiente función
F (s) = s2 + 3 s
(s + 1) 3 (1.4.39) que si la expandimos en fracciones parciales quedará de la siguiente forma 1.4. Transformada de Laplace 11
F (s) = A1
(s + 1) + A2
(s + 1) 2 + A3
(s + 1) 3 (1.4.40) Los coeficientes vendrán determinados por
A3 = [(s + 1) 3F (s)]∣∣∣s=−1 = s2 + 3 s
∣∣∣s=−1 = −2
A2 = dds
[(s + 1) 3F (s)]∣∣∣s=−1 = 2 s + 3
∣∣∣s=−1 = 1
A1 = 12!
d2
ds 2
[(s + 1) 3F (s)]∣∣∣s=−1 = 22!
∣∣∣s=−1 = 1
con lo que
F (s) = 1(s + 1) + 1(s + 1) 2 + −2(s + 1) 3 (1.4.41) y su transformada inversa será la función
f (t) = e−t + te −t − t2e−t (1.4.42)
1.4.4. Resolución de ecuaciones diferenciales
Una de las propiedades más útiles de la transformada de Laplace es la simpli-cidad con la que se pueden resolver ecuaciones diferenciales lineales. Veamos un ejemplo.
Ejemplo 1.4.5: Supongamos un cierto sistema, cuya ecuación diferencial es la siguiente
d2x
dt2 + 2 dx
dt + δ = 0 (1.4.43) sabemos además que para dicho sistema los valores de x y ˙x para t = 0 son respectivamente x(0) = 1 y ˙x(0) = 0 , y queremos obtener la salida x(t) que dará el sistema 1.4.43. Aplicando la propiedad de la transformada de Laplace de la derivada tendremos lo siguiente
[s2X(s) − sx (0) − ˙x(0)] + 2[ sX (s) − x(0)] + 1 = 0
si ahora sustituimos x(0) y ˙x(0) por su valor, nos quedará
[s2X(s) − s] + 2[ sX (s) − 1] + 1 = 0
s2X(s) + 2 sX (s) = s + 1
X(s) = s + 1
s(s + 2) 12 1. Introducción
Expandiendo en fracciones parciales esta expresión
X(s) = A1
s + A2
(s + 2) (1.4.44) Los coeficientes vendrán determinados por
A1 = [sX (s)]∣∣∣s=0 = (s + 1) (s + 2)
∣∣∣s=0 = 12
A2 = [(s + 2) X(s)]∣∣∣s=−2 = s + 1
s
∣∣∣s=−2 = 12
con lo que
X(s) =
12
s +
12
(s + 2)
y su transformada inversa será la solución de la ecuación diferencial 1.4.43 para las condiciones iniciales que se han supuesto, es decir
x(t) = 12 (1 + e−2t). (1.4.45)
1.4.5. Operador derivada
Es frecuente la utilización del operador derivada , que se suele representar por el símbolo p o D, aunque aquí se usará el primero de ellos,
py (t) = ddt y(t) (1.4.46) Utilizando este símbolo, una ecuación diferencial de orden ndn
dt n y(t)+ a1
dn−1
dt n−1 y(t) + . . . + an−1
ddt y(t) + any(t) = = b1
dn−1
dt n−1 u(t) + . . . + bn−1
ddt u(t) + bnu(t)
(1.4.47) puede expresarse en forma polinómica
pny(t)+ a1pn−1y(t)+ . . . +an−1py (t)+ any(t) = b1pn−1u(t)+ . . . +bn−1pu (t)+ bnu(t)
(1.4.48) es decir
[pn + a1pn−1 + . . . + an−1p + an]y(t) = [b1pn−1 + . . . + bn−1p + bn]u(t)
A(p)y(t) = B(p)u(t)
y(t) = B(p)
A(p) u(t) (1.4.49) Este operador tiene la ventaja de trabajar directamente en el dominio del tiem-po, lo que en ocasiones resulta útil. Este operador es frecuentemente utilizado en los métodos clásicos de resolución de ecuaciones diferenciales ordinarias. 1.5. Transformada z 13
1.5. Transformada z
La transformada de Laplace puede utilizarse para resolver sistemas de ecua-ciones diferenciales lineales ordinarias en tiempo continuo. Un método operacio-nal equivalente para la resolución de sistemas de ecuaciones en diferencias de tipo lineal en tiempo discreto, es el método basado en la transformada z.Dada una secuencia de números x(k), se define la transformada z de dicha secuencia como
X(z) = Z[x(k)] =
∞
∑
k=0
x(k)z−k (1.5.1) La transformada z definida según la ecuación 1.5.1 se conoce como la trans-formada z unilateral , y en ella se supone que x(k) = 0 para todo k < 0.
1.5.1. Propiedades de la transformada en z
Linealidad .Una propiedad muy importante de la transformada z es que es un operador lineal. Dadas dos funciones f (k) y g(k) y una constante α se verifica que:
Z[f (k) + g(k)] = Z[f (k)] + Z[g(k)] (1.5.2) y
Z[αf (k)] = αZ[f (k)] = αF (z) (1.5.3) 2. Traslación real .La transformada z de una función f (k) desplazada n en el tiempo se puede expresar de la siguiente manera:
Z[f (k − n)] = z−nF (z) (1.5.4) y
Z[f (k + n)] = zn[F (z) −
n−1
∑
k=0
f (k)z−k] (1.5.5) 3. Traslación compleja .Si f (t) tiene la transformada z, F (z), entonces la transformada z de e−at f (t)
se define como F (ze aT ), lo que se conoce como teorema de la traslación compleja. 14 1. Introducción
Z[e−at f (t)] =
n−1
∑
k=0
f (kT )e−akT z−k =
n−1
∑
k=0
f (kT )( ze aT )−k = F (ze aT )
(1.5.6) 4. Teorema del valor inicial .Si la transformada z de la función f (k) es F (z), y si el lim
z→∞
F (z) existe, entonces el valor inicial f (0) de f (k) viene dado por lim
k→0
f (k) = lim
z→∞
F (z) (1.5.7) 5. Teorema del valor final .Si la transformada z de la función f (k), con f (k) = 0 para k < 0, es F (z),y dicha función F (z) tiene todos sus polos dentro del círculo unitario, con la posible excepción de un polo simple en z=1. Entonces el valor final de f (k),puede obtenerse como lim
k→∞
f (k) = lim
z→1
[(1 − z−1)F (z)] (1.5.8) 6. Multiplicación por ak.Si la transformada z de la función f (k) es F (z), entonces la transformada z de akf (k) se obtiene como
Z[akf (k)] = F (a−1z) (1.5.9) 7. Convolución real .Sean F1(z) y F2(z) las transformadas z de las funciones f1(t) y f2(t), res-pectivamente, entonces
F1(z)F2(z) = Z[f1(k) ∗ f2(k)] (1.5.10)
= Z[
N
∑
k=0
f1(k)f2(N − k)] (1.5.11)
= Z[
N
∑
k=0
f2(k)f1(N − k)] (1.5.12) En esta expresión el símbolo ∗ denota la operación convolución en el domi-nio del tiempo (discreto). 1.5. Transformada z 15
Una excepción a este caso se produce si una de las dos funciones es el retardo integral eN T s , ya que en este caso
Z[e−N T s F (s)] = Z[e−N T s ]Z[F (s)] = z−N F (z) (1.5.13)
1.5.2. Transformadas z de algunas funciones
Ejemplo 1.5.1: La transformada z de la función escalón unitario u0(k) que se define como
f (k) = u0(k) =
{ 1, k ≥ 00, k < 0 (1.5.14) se obtiene como
F (z) = Z[u0(k)] =
∞
∑
k=0
1z−k
=
∞
∑
k=0
z−k = 1 + z−1 + z−2 + . . .
= 11 − z−1 = zz − 1 (1.5.15)
Ejemplo 1.5.2: La transformada z de la función exponencial f (k) = e−αk para
k ≥ 0 se obtiene como
F (z) = Z[e−αk ] =
∞
∑
k=0
e−αkT z−k
= 1 + e−αT z−1 + e−α2T z−2 + . . .
= 11 − e−αT z−1
= zz − e−αT (1.5.16) En la tabla se pueden ver las transformadas de algunas de las funciones más usuales. 16 1. Introducción
f (k) F (z)
δ(k) 1
δ(k − n) z−n
u0(k) zz−1
ak zz−a
k z
(z−1) 2
k + 1 z2
(z−1) 2
sin( αk ) (sin α)zz2−(2 cos α)z+1
cos( αk ) z2−(cos α)zz2−(2 cos α)z+1
1.5.3. La transformada z inversa
La transformada z en los sistemas de tiempo discreto juega un papel equivalente al de la transformada de Laplace en los sistemas de tiempo continuo. De forma similar a lo que ocurría en aquella, es necesario después de operar en el plano z
para resolver la ecuación en diferencias convertir dicha solución al dominio del tiempo. En este caso, al pasar de nuevo al dominio del tiempo lo que obtendremos como resultado es la correspondiente secuencia en el tiempo f (k).En el caso de la transformada z inversa, la secuencia que se obtiene sólo está definida en el tiempo en los instantes de muestreo, por lo que resulta una única secuencia f (k) pero que puede corresponder al muestreo de un número infinito de funciones f (t).Un método muy usual de obtener la transformada z inversa es por medio de la expansión en fracciones parciales.
Expansión en fracciones parciales
En la mayor parte de las aplicaciones de control, tanto las ecuaciones en di-ferencias como las soluciones de las mismas suelen tener la forma de funciones racionales en z, del tipo
F (z) = N (z)
D(z) (1.5.17) donde N (z) y D(z) son polinomios en z. Se asume que el grado del polinomio
N (z) del numerador en z es menor o igual que el grado del polinomio D(z) del denominador. 1.5. Transformada z 17
Esta expresión puede escribirse en la siguiente forma
F (z) = N (z)
zn + a1zn−1 + . . . + an−1z + an
(1.5.18) donde los coeficientes ai son coeficientes reales. La idea básica del método de la expansión en fracciones parciales ya se co-mento para el caso de la transformada de Laplace, y consiste en descomponer el polinomio F (z) en una suma de fracciones simples de forma que estas sean las transformadas de funciones simples y conocidas, de forma que para cada fracción resulte muy fácil calcular la transformada inversa. Veamos los diferentes casos que se pueden presentar.
Caso de polos simples y reales Si todos los polos de D(z) son simples y reales, y hay por lo menos un cero en el origen, entonces se expande F (z)
z
de la siguiente forma
F (z)
z = A1
(z − z1) + A2
(z − z2) + . . . + An
(z − zn) (1.5.19) donde los coeficientes A1, . . . , A n se obtienen como
Ai = [(z − zi) F (z)
z
]∣∣∣z=zi
(1.5.20) Se realiza la expansión de la función F (z)/z en vez de la función F (z) para que las fracciones simples de la expansión queden de la misma forma que para las expansiones en fracciones parciales en s.
Caso de polos múltiples y reales Si los polos de D(z) son múltiples, en-tonces la ecuación 1.5.18 toma la siguiente forma
F (z) = N (z)(z − z1)2 (1.5.21) y en ella z1 tiene un orden de multiplicidad de 2. Descomponiendo la expresión 1.5.21 en fracciones parciales obtenemos
F (z)
z = A1
(z − z1)2 + A2
(z − z1) (1.5.22) donde los coeficientes A1 y A2 correspondientes al polo multiple se obtienen como
A1 = [(z − z1)2 F (z)
z
]∣∣∣z=z1
(1.5.23)
A2 = ddz
[(z − z1)2 F (z)
z
]∣∣∣z=z1
(1.5.24) 18 1. Introducción
Ejemplo 1.5.3: Dada la transformada z
F (z) = z(1 − e−αT )(z − 1)( z − e−αT ) (1.5.25) siendo α constante y T el periodo de muestreo, obtener la transformada z inversa por el método de descomposición en fracciones. La expansión en fracciones viene dada por la siguiente expresión
F (z)
z = A1
(z − 1) + A2
(z − e−αT ) (1.5.26) y si aplicamos la expresión 1.5.20 tenemos que
A1 = [(z − 1) F (z)
z
]∣∣∣z=1 = 1
A2 = [(z − e−αT ) F (z)
z
]∣∣∣z=e−αT = 1 (1.5.27) con lo que la expansión en fracciones parciales quedará en la forma
F (z)
z = 1(z − 1) + 1(z − e−αT ) (1.5.28) o lo que es lo mismo
F (z) = 1(1 − z−1) + 1(1 − e−αT z−1) (1.5.29) y si vamos a una tabla de transformadas z veremos que
Z−1[ 1(1 − z−1)
]
= 1
Z−1[ 1(1 − e−αT z−1)
]
= e−αkT (1.5.30) y por tanto su transformada z inversa será
f (kT ) = 1 − e−αkT , k = 0 , 1, 2, . . . (1.5.31)
1.5.4. Resolución de ecuaciones en diferencias
Así como la transformada de Laplace nos permitía resolver ecuaciones dife-renciales lineales, la transformada z nos va a permitir resolver ecuaciones en dife-rencias. Veamos un ejemplo. 1.5. Transformada z 19
Ejemplo 1.5.4: Supongamos un cierto sistema, cuya ecuación en diferencia es la siguiente
x(k + 2) + 2 x(k + 1) + x(k) = 0 (1.5.32) sabemos además que para dicho sistema los valores de x para k = 0 y k = 1 son respectivamente x(0) = 1 y x(1) = 0 , y queremos obtener la salida x(k) que dará el sistema 1.5.32. Aplicando la propiedad de la traslación real de la transformada z tendremos lo siguiente
[z2X(z) − z2x(0) − zx (1)] + 2[ zX (z) − zx (0)] + X(z) = 0
si ahora sustituimos x(0) y x(1) por su valor nos quedará
[z2X(z) − z2] + 2 zX (z) + X(z) = 0
z2X(z) + 2 zX (z) + X(z) = z2
X(z) = z2
z2 + 2 z + 1
X(z) = z2
(z + 1) 2
Expandiendo en fracciones parciales esta expresión
X(z)
z = A1
(z + 1) 2 + A2
(z + 1) (1.5.33) Los coeficientes vendrán determinados por
A1 = [(z + 1) 2 X(z)
z
]∣∣∣z=−1 = z
∣∣∣z=−1 = −1
A2 = ddz
[(z + 1) 2 X(z)
z
]∣∣∣z=−1 = 1
y su transformada inversa será la solución de la ecuación en diferencias 1.5.32 para las condiciones iniciales que se han supuesto, es decir
x(k) = −sin
( π
2 k
)
( −1) k (1.5.34)
1.5.5. Operador retardo
También en el caso de los sistemas en tiempo discreto es frecuente la utilización de un operador de transferencia que nos permita expresar en forma polinómica la relación en el tiempo de la señal de entrada con la de salida. En los sistemas de 20 1. Introducción
tiempo discreto dicho operador es el denominado operador adelanto , que se suele representar por el símbolo q,
qu (k) = u(k + 1) (1.5.35) o su inverso el operador retardo ,
q−1u(k) = u(k − 1) (1.5.36) Utilizando este operador una ecuación en diferencias de orden n quedará en la siguiente forma
y(k)+ a1y(k − 1) + . . . + an−1y(k − (n − 1)) + any(k − n)=b1u(k − 1) + . . . + bn−1u(k − (n − 1)) + bnu(k − n) (1.5.37) puede expresarse en forma polinómica
y(k)+ a1q−1y(k) + . . . + an−1q−(n−1) y(k) + anq−ny(k)=b1q−1u(k) + . . . + bn−1q−(n−1) u(k) + bnq−nnu (k) (1.5.38) es decir
A(q−1)y(k) =
[
b1q−1 + . . . + bn−1q−(n−1) + bnq−n]
u(k)
A(q−1)y(k) = B(q−1)u(k)
y(k) = B(q−1)
A(q−1) u(k)
(1.5.39) El operador retardo es equivalente a z−1 según (1.5.4), aunque en distinto do-minio. Este operador al igual que el operador derivada tiene la ventaja de trabajar directamente en el dominio del tiempo, lo que en ocasiones resulta útil. Este ope-rador es frecuentemente utilizado en las técnicas de identificación de sistemas.
1.6. Procesos estocásticos o aleatorios
Si bien las transformaciones matemáticas vistas en las secciones anteriores nos permiten analizar y resolver con facilidad las ecuaciones diferenciales o en dife-rencias de una amplia gama de sistemas dinámicos, no resultan sin embargo sa-tisfactorias para analizar y explicar una parte importante de la respuesta de una gran parte de sistemas. La parte de los sistemas para la que estas herramientas ma-temáticas no resultan suficientes es aquella ligada a un comportamiento sujeto a una cierta incertidumbre o expresado en otros términos, sujeto a una naturaleza no determinista. Aquella parte de los sistemas cuya naturaleza es no determinista la modelare-mos por medio de técnicas probabilísticas. 1.6. Procesos estocásticos o aleatorios 21 1 1.2 1.4 1.6 1.8 2 2.2 2.4 2.6 2.8 3−5 −4 −3 −2 −1 012345
tx(t)
Fig. 1.1: Función aleatoria
1.6.1. Variable aleatoria y proceso aleatorio
La mayor parte de los sistemas físicos que aparecen en la naturaleza tienden a mostrar una cierta variabilidad con respecto a un cierto valor que podríamos deno-minar de referencia. Supongamos que después del proceso de fabricación de una cierta pieza mecánica procedemos a pesarla, observamos que aunque el peso teóri-co que la pieza debe de tener es de 10 .00 gramos, sin embargo el peso final que se obtiene en todas las piezas no es exactamente ese, sino 9.99; 9 .98; 10 .01; 10 .00; . . . .Si tratásemos de conocer si la próxima pieza que vamos a fabricar pesa exactamen-te 10 .00 , la respuesta la podríamos dar en términos probabilísticos pero no podría-mos asegurar un valor determinado. El peso de esta pieza es lo que se denomina una variable aleatoria .Parece claro que esta variabilidad en el valor de una variable concreta, que hemos denominado variable aleatoria, puede asociarse también a funciones tem-porales que son las que más nos interesan desde el punto de vista del control de procesos. Un posible ejemplo de una función aleatoria en el tiempo es el que se muestra en la figura 1.1. No resulta difícil de intuir que, para un cierto sistema físico, aunque se repita el experimento un número considerable de veces la salida del mismo no va a ser exac-tamente igual. El conjunto de todas las posibles funciones en el tiempo que pueden ser observadas, pertenecen a lo que denominaremos proceso aleatorio o estocás-tico , y que representaremos por {u(t)}. Una función particular de dicho conjunto, la expresaremos como u(t) y la denominaremos función muestra . El valor de una función muestra en un instante de tiempo determinado, t1, es una variable aleatoria y lo representaremos por u(t1). En un proceso estocástico, tenemos por lo tanto, para cada instante de tiempo una variable aleatoria. 22 1. Introducción 0 0.5 1 1.5 2 2.5 3−5 05
tx1(t) 00.5 11.5 22.5 3−5 05
tx2(t) 00.5 11.5 22.5 3−5 05
tx3(t)
Fig. 1.2: Proceso aleatorio
t1 :
[
u1(t1), u 2(t1), . . . , u n(t1)
]
t2 :
[
u1(t2), u 2(t2), . . . , u n(t2)
]
(1.6.1) Como ya se ha indicado para el caso de variables aleatorias, la forma habitual de estudiar magnitudes sometidas a incertidumbre se realiza por medio de la defi-nición y obtención de propiedades estadísticas de dichas variables. Para el caso de procesos estocásticos estas propiedades estadísticas de u(t) deben obtenerse con respecto a un conjunto de datos observados simultáneamente (figura 1.2).
Tipos de procesos estocásticos
Existen diferentes tipos de procesos estocásticos de aparición más o menos frecuente en problemas de ingeniería. Veamos algunos de ellos: 1. Procesos aleatorios continuos .Son aquellos en los que las variables aleatorias u(t1) y u(t2) pueden tomar cualquier valor dentro de un cierto rango de posibles valores. 1.6. Procesos estocásticos o aleatorios 23 0 0.5 1 1.5 2 2.5 3−2 −1.5 −1 −0.5 00.5 11.5 2
tx1(t)
Fig. 1.3: Proceso aleatorio discreto
Al igual que en el caso de las señales para las que definíamos los concep-tos de transformada de Laplace y transformada en z, estos procesos pueden estudiarse en tiempo continuo o discreto. 2. Procesos aleatorios discretos .Son aquellos en los que las variables aleatorias u(t1) y u(t2) sólo toman cier-tos valores (aislados entre sí) dentro de un cierto rango de valores (aunque el número de estos posibles valores puede ser infinito). Al igual que en caso anterior estos procesos pueden estudiarse en tiempo continuo o en tiempo discreto. En el ejemplo de la figura 1.3 se muestra un proceso aleatorio de tipo binario, es decir que el rango de posibles valores es sólo de dos, que aquí son +1 , −1.3. Procesos aleatorios no deterministas .Son procesos aleatorios en los cuales no pueden predecirse los valores fu-turos que tomará el proceso a partir de los valores anteriores que hemos observado de dicho proceso. La mayoría de los procesos aleatorios naturales son no deterministas, bien porque los mecanismos que los generan no son observables o bien porque son extremadamente complejos. En el caso de que los valores futuros de una función pudieran predecirse exactamente a partir de la observación de valores pasados del proceso esta-ríamos ante un proceso determinista. 4. Procesos estacionarios .Para una variable aleatoria de un proceso estocástico, que caracterizamos por medio de parámetros estadísticos, es posible definir una función de densidad de probabilidad. Si las funciones de densidad conjunta y marginales de di-cho proceso no dependen de la elección del origen de tiempos, entonces el proceso se dice que es estacionario. La principal ventaja que presentan estos 24 1. Introducción
procesos es que los parámetros estadísticos del proceso, como por ejemplo el valor medio, permanecen constantes a lo largo del tiempo. Si estos paráme-tros dependen del instante de tiempo que se considere, entonces el proceso será no estacionario. 5. Procesos ergódicos .Algunos procesos estacionarios presentan además la propiedad de que cual-quier conjunto de datos que se tome del proceso presenta las mismas caracte-rísticas estadísticas que todo el proceso. Por ello, en estos procesos se puede determinar la conducta estadística de todo el proceso examinando solamente una muestra típica de la función. Esta propiedad resulta muy difícil de de-mostrar en la mayoría de los casos, si bien es usual el asumirla como cierta por las ventajas prácticas que ofrece.
1.6.2. Funciones de distribución y de densidad de probabilidad
Un aspecto importante para poder caracterizar una variable aleatoria es la for-ma en la que se definen los sucesos asociados con el espacio de probabilidad de-finido para dicha variable aleatoria. Estos sucesos pueden definirse de diferentes formas, si bien aquí utilizaremos un método de uso muy generalizado, como son los conceptos de función de distribución de probabilidad y de densidad de proba-bilidad.
Función de distribución
Sea X una variable aleatoria y x un valor permitido cualesquiera que puede tomar la variable aleatoria X dentro del rango admisible de valores para dicha variable aleatoria. Se define la función de distribución de probabilidad como la probabilidad de que el suceso que la variable aleatoria observada X tome un valor menor o igual que el valor x. Es decir,
F (x) = P r (X ≤ x) (1.6.2) Esta función de distribución de probabilidad es una probabilidad y por tanto debe satisfacer las propiedades básicas exigidas por la definición de probabilidad. Estas restricciones que F (x) debe cumplir pueden resumirse en: 1. 0 ≤ F (x) ≤ 1, −∞ < x < ∞.2. F (−∞ ) = 0 y F (∞) = 1 .3. F (x) es no decreciente para x creciente. 1.6. Procesos estocásticos o aleatorios 25 (a) 1xF(x) 0(b) 1xF(x) 0 x1 x2 x3
Fig. 1.4: Ejemplos de funciones de distribución
P r (x1 < X ≤ x2) = F (x2) − F (x1).En la figura 1.4 se muestran dos posibles funciones de distribución, una en la que la variable aleatoria puede tomar todos los posibles valores dentro del rango
[−∞ , ∞] (a) y otra donde sólo algunos valores son posibles (b).
Función de densidad
Aunque la función de distribución permite describir de forma completa el mo-delo de probabilidad para una variable aleatoria, no es la forma de trabajo más conveniente para la mayoría de los cálculos que se requieren en nuestro caso. Por ello se suele utilizar la derivada de la función de distribución dF (x)
dx
, en vez de uti-lizar F (x) directamente. A esta derivada se le denomina función de densidad de probabilidad , cuando existe, y se define como
f (x) = lim
4x→0
F (x + 4x) − F (x)
4x = dF (x)
dx (1.6.3) El significado de la función de densidad de probabilidad es inmediato a partir de la ecuación 1.6.3,
dF (x) = f (x)dx = P r (x < X ≤ x + dx ) (1.6.4) es decir, la ecuación 1.6.4 nos indica la probabilidad de que el valor observado de la variable aleatoria X este en el rango de valores x y x + dx .Las propiedades de la función de densidad de probabilidad (que no es una probabilidad puesto que puede tomar valores superiores a 1), se pueden resumir en: 1. f (x) ≥ 0, −∞ < x < ∞.2. ∫ ∞−∞ f (x)dx = 1 .26 1. Introducción (a) xf(x) 0(b) xf(x) 0 x1 x2 x3
Fig. 1.5: Ejemplos de funciones de densidad
F (x) = ∫ x
−∞
f (y)dy .4. ∫ x2
x1
f (x)dx = P r (x1 < X ≤ x2).En la figura 1.5 se muestran algunos ejemplos de funciones de densidad de probabilidad, el caso (a) corresponde a una función de densidad continua y el caso (b) a una función de densidad discreta.
1.6.3. Parámetros para describir un proceso estocástico
Como ya se ha comentado, para caracterizar los procesos estocásticos se uti-lizan parámetros estadísticos asociados con las variables aleatorias, u(t), en dife-rentes instantes de tiempo, t. Si el proceso que se estudia es estacionario, estos parámetros permanecen constantes para todo t, por lo que basta con un único con-junto de parámetros para caracterizarlo. Aún suponiendo que el proceso sea estacionario, estos parámetros no resultan en muchas ocasiones fáciles de obtener dado que requieren la repetición del expe-rimento lo que en muchos sistemas no resulta viable dada la lentitud o dificultad para repetirlo en las mismas condiciones. Nos encontramos, en estos casos, con que sólo disponemos de un experimento para poder estimar los parámetros del proceso estocástico. En este tipo de situación la única posibilidad consiste en extraer pro-medios temporales. Estos promedios temporales son razonables si el proceso es ergódico, en cuyo caso la media temporal de un proceso ( en un tiempo infinito) es equivalente a la media temporal de un cierto conjunto muestral. En muchos casos, y salvo que exista una razón clara para impedirlo, se supone que el proceso es ergódico y que por lo tanto el promedio en un tiempo finito resulta una aproximación razonable del parámetro estadístico que nos interesa conocer. Supuesta esta condición de ergodicidad en el proceso, veamos como se pueden obtener algunos de los parámetros estadísticos que nos interesan: 1.6. Procesos estocásticos o aleatorios 27
Media o esperanza
El valor medio o valor esperado de una variable aleatoria X se define a partir de la función de densidad de probabilidad f (x) como
x = E[x] =
∫ ∞−∞
xf (x)d(x) (1.6.5) al símbolo E[] se le denomina valor esperado o esperanza matemática de X. Esta definición es teórica y se supone que la distribución de probabilidad de la variable aleatoria es conocida. Sin embargo con frecuencia la función de distribución de la variable no es conocida y por tanto la función de densidad de probabilidad tampoco se conoce. En estos casos suele utilizarse la media muestral , es decir el valor medio obtenido experimentalmente en una o varias realizaciones de dicha variable aleatoria. Por ejemplo si hemos observado xi(t), i = 1 . . . N realizaciones de una variable aleatoria x a lo largo de un cierto periodo de tiempo, para un instante de tiempo dado t1 el valor medio se define como
x(t1) = 1
N
N
∑
i=1
xi(t1) (1.6.6) Para una cierta función de una variable aleatoria g(x), puede definirse también, y de forma sencilla la esperanza matemática de dicha función:
E[g(x)] =
∫ ∞−∞
g(x)f (x)d(x) (1.6.7)
Momentos
Una función g(x) de importancia considerable es g(x) = xn, ya que nos con-duce al concepto genérico de momento de una variable aleatoria. Se define, el mo-mento de orden n de una variable aleatoria X como
E[xn] =
∫ ∞−∞
xnf (x)d(x) (1.6.8) De todos los posibles momentos definibles para una variable aleatoria, los dos de mayor importancia práctica son los momentos de primer y de segundo orden. El momento de primer orden, n = 1 , es el valor medio que ya hemos comentado anteriormente y el de segundo orden, n = 2 , se denomina valor cuadrático medio .
E[x2] =
∫ ∞−∞
x2f (x)d(x) (1.6.9) Cuando es obtenido experimentalmente para un cierto t1 en un conjunto de realizaciones una variable aleatoria toma la expresión 28 1. Introducción
E[x2(t1)] = 1
N 2
N
∑
i=1
N
∑
j=1
xi(t1)xj (t1) (1.6.10)
Momentos centrados
Estos momentos pueden definirse de forma que sean momentos centrados en torno al valor medio, para ello se definen como:
E[( x − x)n] =
∫ ∞−∞
(x − x)nf (x)d(x) (1.6.11) De todos los momentos centrados, el más conocido y utilizado es el de segundo orden, también conocido como la varianza de una variable aleatoria x. Se define como la esperanza matemática de (x − x)2 y viene dada por:
σ2 = E[( x − x)2] =
∫ ∞−∞
(x − x)2f (x)d(x) (1.6.12) En el caso de que la variable aleatoria sea un vector el concepto se conoce como matriz de covarianza
E[( x − x)( x − x)T ] =
∫ ∞
x1=−∞
. . .
∫ ∞
xn=−∞
(x − x)( x − x)T p(x)d(x1) . . . d (xn)
(1.6.13) En muchos casos suele utilizarse la varianza o covarianza muestral , es decir el valor obtenido experimentalmente en una o varias realizaciones de dicha variable aleatoria. Por ejemplo si hemos observado xi(t), i = 1 . . . N realizaciones de una variable aleatoria x a lo largo de un cierto periodo de tiempo, para dos instantes de tiempo dados t1 y t2 el valor de la covarianza se define como
Cxx (t1, t 2) = cov (x(t1), x (t2)) = 1
N
N
∑
i=1
(xi(t1) − x(t1))( xi(t2) − x(t2)) (1.6.14) a esta expresión se le conoce también como función de autocovarianza .Una propiedad muy útil que presenta la expresión de la varianza es que
E[( x − x)2] = E[x2 − 2xx + x2]= E[x2] − 2E[x]x + x2
= E[x2] − 2x2 + x2
= E[x2] − x2 (1.6.15) 1.6. Procesos estocásticos o aleatorios 29 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5 −1 −0.8 −0.6 −0.4 −0.2 00.2 0.4 0.6 0.8 1
tRxx_1_2
Fig. 1.6: Función de autocorrelación
1.6.4. Función de autocorrelación
Esta función se define como,
Rxx (t1, t 2) = E[x(t1)x(t2)] = 1
N
N
∑
i=1
xi(t1)xi(t2) (1.6.16) y se puede observar que esta expresión y la de la autocovarianza están relacionadas por medio de la siguiente relación
Cxx (t1, t 2) = Rxx (t1, t 2) − x(t1)x(t2) (1.6.17) que en el caso de que el proceso estocástico tenga media cero ambas expresiones coinciden. Si observamos la definición de la función de autocorrelación, esta representa el valor esperado del producto de las variables aleatorias x(t1) y x(t2), que co-rresponden a muestras de un mismo proceso en dos instantes diferentes de tiempo. Este valor esperado dependerá de lo rápidamente que varíe la salida del proceso. Intuitivamente se puede ver que si suponemos dos instantes de tiempo cercanos, la salida habrá variado poco por lo que el valor de la correlación será alto; mientras que si se toman dos instantes de tiempo muy separados entre sí, probablemente el valor de la correlación será bajo. Esta función de autocorrelación tiene una gran importancia práctica en la iden-tificación de sistemas como se verá más adelante. Si la función de autocorrelación Rxx (t1, t 2) es función sólo de la diferencia de tiempo τ = t2 − t1 con independencia del valor del tiempo t1 que se tome, se puede expresar entonces como:
Rxx (t, t + τ ) = Rxx (τ ) (1.6.18) 30 1. Introducción 0 2 4 6 8 10 12 14 16 18 00.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9
nRxx
Fig. 1.7: Función de autocorrelación
Ejemplo 1.6.1: Supongamos una señal simple de tipo triangular, y veamos como se puede obtener las funciones de autocorrelación y autocovarianza mediante el uso de la función xcorr y xcov de Matlab. Definimos el vector x de nueve elementos y aplicamos a continuación la función xcorr tal y como se muestra a continuación,
a
x=[0.1 0.2 0.3 0.4 0.5 0.4 0.3 0.2 0.1]’;
a
Rxx=xcorr(x,x);
a
plot(Rxx)
en la figura 1.7 se muestra la función de autocorrelación generada para dicha señal. Se procede de forma similar para obtener la función de autocovarianza
a
x=[0.1 0.2 0.3 0.4 0.5 0.4 0.3 0.2 0.1]’;
a
Cxx=xcov(x,x);
a
plot(Cxx)
y en la figura 1.8 se muestra la función de autocovarianza generada para dicha señal.
Propiedades de la función de autocorrelación
La función de autocorrelación es una función par de τ , ya que
Rxx (τ ) = Rxx (−τ ) (1.6.19) La función de autocorrelación presenta un valor máximo para τ = 0 , es decir
Rxx (0) ≥ Rxx (τ ) (1.6.20) 1.6. Procesos estocásticos o aleatorios 31 0 2 4 6 8 10 12 14 16 18 −0.5 −0.4 −0.3 −0.2 −0.1 00.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
nCxx
Fig. 1.8: Función de autocovarianza
El valor de la autocorrelación para τ = 0 es igual al valor cuadrático me-dio del proceso, que en el caso de proceso con media cero coincide con la varianza del proceso, es decir
Rxx (0) = x2
= E[x2(t)] = lim
T→∞
12T
∫ T
−T
[x(t)] 2dt (1.6.21)
1.6.5. Función de correlación cruzada
Teniendo en cuenta la definición de la función de autocorrelación es posible definir una función que nos de la correlación entre dos variables aleatorias de dos procesos estocásticos diferentes x(t) e y(t). Esta función se denomina función de
correlación cruzada , Rxy (t, t + τ ), y se define como
Rxy (t, t + τ ) = E[x(t)y(t + τ )] = 1
N
N
∑
i=1
xi(t).y i(t + τ ) (1.6.22) Si en esta definición se supone que ambos procesos son estacionarios conjunta-mente, entonces la función de correlación cruzada dependerá sólo de la diferencia de tiempo τ . Puede ocurrir que los procesos estocásticos sean individualmente es-tacionarios, pero que no lo sean conjuntamente, en cuyo caso la expresión de la función de correlación cruzada dependerá de t además de τ .32 1. Introducción
Si además de estacionarios los procesos estocásticos son ergódicos conjunta-mente, el valor de la función de correlación cruzada puede obtenerse a partir de la correlación de un par de muestras, mediante un promedio temporal, es decir:
Rxy (τ ) = lim
T→∞
12T
∫ T
−T
[x(t)y(t + τ )] dt (1.6.23) La interpretación de la función de correlación cruzada es similar a la de la autocorrelación, en el sentido de que nos da información de cuanto dependen dos variables aleatorias entre sí. De forma análoga a la definición de la función de correlación cruzada entre las variables aleatorias x(t) e y(t), Rxy (t, t + τ ), se puede definir la función de correlación cruzada entre y(t) y x(t), Ryx (t, t + τ ).
Propiedades de la función de correlación cruzada
Las propiedades de la función de correlación cruzada presentan bastantes dife-rencias con las de la función de autocorrelación. Estas pueden ser resumidas como sigue: 1. El valor de las correlaciones cruzadas para τ = 0 coincide, es decir
Rxy (0) = Ryx (0) (1.6.24) aunque no tiene un significado particular y no representan valores cuadráti-cos medios como en el caso de la función de autocorrelación. 2. Las funciones de correlación cruzadas no son por lo general funciones pares de τ . Pero si se verifica que
Rxy (τ ) = Ryx (−τ ) (1.6.25) es decir, que desplazar x(t) en el tiempo en un sentido equivale a desplazar
y(t) en la otra dirección. 3. La función de correlación cruzada a diferencia de la función de autocorrela-ción, no tiene porque tener un máximo de la función para τ = 0 . Pero puede demostrarse que el valor de la correlación cruzada verifica que
|Rxy (τ )| ≤ [Rxx (0) .R yy (0)] 12 (1.6.26) y de forma análoga para Rxy (τ ).1.6. Procesos estocásticos o aleatorios 33
Si los dos procesos estocásticos son estadísticamente independientes, enton-ces se verifica que
Rxy (τ ) = E[x(t)y(t)] = E[x(t)] E[y(t)] = xy = Ryx (τ ) (1.6.27) 5. Si los dos procesos estocásticos además de ser estadísticamente indepen-dientes, alguno de ellos tiene media cero, entonces la función de correlación cruzada se hace cero. Lo contrario, no es necesariamente cierto salvo en va-riables aleatorias normales y distribuidas conjuntamente.
1.6.6. Función de densidad espectral
Se denomina promedio de energía o potencia de un proceso estocástico a
Px = lim
T→∞
12T
∫ T
−T
[x(t)] 2dt (1.6.28) Si expresamos por X(ω) la transformada de Fourier de una señal real x(t),obtenemos la siguiente expresión
X(ω) =
∫ ∞−∞
x(t)e−jωt dt (1.6.29) y si denominamos XT (ω) a la transformada de Fourier de la señal real x(t) trun-cada para valores de tiempo | t |≤ T < ∞ de forma que
xT (t) =
x(t) | t |≤ T < ∞
0 | t |> T
y aplicando el teorema de Parseval que dice que
∫ ∞−∞
f (t).g (t)dt = 12π
∫ ∞−∞
F (ω).G (ω)dω (1.6.30) se deduce que ∫ T
−T
xT (t)2dt = 12π
∫ ∞−∞
|XT (ω)|2dω (1.6.31) y si sustituimos esta expresión en la que utilizábamos para definir la potencia de un proceso estocástico 1.6.28 nos quedará que
Px = lim
T→∞
14πT
∫ ∞−∞
|XT (ω)|2dω (1.6.32) 34 1. Introducción
Se define entonces la densidad espectral de un proceso estocástico {x(t)} co-mo
Sxx (ω) = lim
T→∞
12πT |XT (ω)|2 (1.6.33) con lo que la expresión de la potencia puede reescribirse de la siguiente manera
Px = 12π
∫ ∞−∞
Sxx (ω)dω =
∫ ∞
0
Sxx (ω)dω (1.6.34)
1.6.7. Descripción espectral de las perturbaciones
Las propiedades de las perturbaciones a las que están sometidos la mayoría de los sistemas reales se pueden describir con la ayuda del concepto de espectro o densidad espectral. Este espectro indica, en principio, la distribución de la energía de la señal entre las diferentes frecuencias. De entre las propiedades que presenta el concepto de densidad espectral vea-mos tres que pueden resultar de utilidad: 1. Una señal u(t) m-dimensional puede ser asociada con una función matricial hermítica m × m S uu (ω), a la que se denomina su espectro . ( Una matriz de coeficientes complejos A se dice hermítica si A∗ = A , donde el símbolo expresa la matriz traspuesta conjugada). Estas matrices hermíticas, son matrices simétricas con autovalores reales por lo que siempre pueden ser diagonalizadas. 2. Sea G un sistema lineal y estable. Si u(t) tiene un espectro Suu (ω) y
y(t) = G(p)u(t)
entonces
Syy (ω) = G(jω )Suu (ω)G∗(jω ) (1.6.35) 3. La matriz (con dimensiones m × m)
Pu = 12π
∫ ∞−∞
Suu (ω)dω (1.6.36) es una medida de la energía total de la señal. 1.6. Procesos estocásticos o aleatorios 35
1.6.8. Espectro cruzado
Si consideramos el espectro de una señal separada en dos partes de la forma
z =
yu
(1.6.37) el espectro tendrá la forma
Szz (ω) =
Syy (ω) Syu (ω)
Suy (ω) Suu (ω)
(1.6.38) donde los bloques diagonales Syy (ω) y Suu (ω) corresponden a los espectros de las señales y y u respectivamente. Y los bloques no diagonales se denominan espectro cruzado entre y y u. Como resulta que Szz (ω) es una matriz hermítica tenemos que
Syu (ω) = S∗
uy
(ω) (1.6.39) De aquí se obtiene una definición muy útil que dice que: dos señales son inco-rreladas si su espectro cruzado es idénticamente cero.
1.6.9. Ruido blanco
Como resultado de la definición del concepto de densidad espectral se define un proceso estocástico muy característico y de una gran utilidad en la identificación de procesos, el denominado ruido blanco .Se define como ruido blanco a un proceso estacionario con densidad espectral constante para todas las frecuencias. Es decir
Sxx (ω) = p, −∞ ≤ ω ≤ +∞ (1.6.40) Este proceso, se caracteriza además porque la función de autocorrelación del mismo es una función δ de Dirac multiplicada por el valor de la densidad espectral
p, tal y como se muestra en la figura 1.9.
Rxx (τ ) = pδ (τ ) (1.6.41) El ruido blanco tiene su energía dividida por igual entre todas las frecuencias, esto implica una energía infinita lo que no es real. Es por tanto una señal idealizada. Una característica notable del ruido blanco es que la historia pasada del mismo no contiene información acerca de su valor futuro. Esta función es ideal desde el punto de vista de la identificación de procesos ya que al presentar densidades a todas las frecuencias constituye una señal de ex-citación que generará respuesta por parte del proceso que se trata de identificar a 36 1. Introducción tRxx 0w0pSxx p
Fig. 1.9: Ruido blanco ideal w0pSxx
Fig. 1.10: Ruido blanco con banda limitada
todas esas frecuencias. De esta forma con una única señal de entrada podremos conocer la respuesta del sistema que queremos identificar a todas las frecuencias del espectro. En realidad el ruido blanco idealmente definido no es viable físicamente ya que necesitaríamos una energía infinita para estimular un número infinito de frecuen-cias. Pese a ello, el concepto de ruido blanco resulta de tanta utilidad que se utiliza una versión limitada del mismo, en el que se tiene densidad espectral constante en una cierta banda de frecuencias y densidad cero en el resto.
1.6.10. Descripción de perturbaciones en función del ruido blanco
Supongamos un sistema lineal y estable G(p) al que se le introduce como en-trada una señal del tipo ruido blanco e(t) de intensidad R; la salida del mismo será: 1.6. Procesos estocásticos o aleatorios 37
v(t) = G(p)e(t) (1.6.42) El espectro de esta señal v(t) de acuerdo con las propiedades vistas para el espectro de un sistema lineal es el siguiente
Svv (ω) = G(jω )RG ∗(jω ) (1.6.43) que para el caso escalar se puede expresar como
Svv (ω) = R|G(jω )|2 (1.6.44) Es decir, una señal descrita por un espectro Svv (ω) puede verse como la salida de un sistema (o filtro) sometido a ruido blanco en su entrada. Esto permite mode-lar las perturbaciones en el dominio del tiempo, en aquellos casos en los que las perturbaciones no son ruido blanco, es decir en aquellas con una cierta estructura o en las cuales se puede realizar ciertas predicciones con respecto al futuro a partir de su historia anterior. En muchos casos es posible describir las propiedades de las perturbaciones construyendo un modelo temporal del tipo 1.6.42. Para ello puede ser interesante comenzar a partir del espectro de la perturbación Svv (ω) y tratar de obtener un modelo G estable y una matriz R tal que se verifique la ecuación 1.6.43. Este problema se conoce en teoría de la señal como factorización espectral .Para el caso escalar existe un teorema denominado teorema de la factorización espectral que dice que:
Teorema de la factorización espectral Suponiendo que una función esca-lar, que toma valores reales Svv (ω) ≥ 0 es una función racional de ω2, finita para todo ω. Existe entonces una función racional G(s), con coeficientes reales, y con todos los polos situados estrictamente en el semiplano real negativo, y con todos los ceros situados en el semiplano real negativo o en el eje imaginario,tal que
Svv (ω) = |G(jω )|2 = G(jω )G(−jω ) (1.6.45)
Ejemplo 1.6.2: Supongamos que se desea controlar la temperatura de una habi-tación, dicha temperatura esta sometida a las perturbaciones climáticas debidas a la temperatura del exterior. Una primera posibilidad para tratar de modelar dichas perturbaciones es supo-ner que responden a una perturbación del tipo ruido blanco, pero esta suposición no se corresponde con lo que observamos, ya que es de tipo cíclico. La perturba-ción tiene dos ciclos básicos uno diario con T1 = 24 horas debido a las variacio-nes día-noche y otro largo T2 = 365 .24 horas debido al ciclo verano-invierno. Si consideramos únicamente el ciclo diario, esta señal de perturbación tendría en su espectrograma un pico en ω1 = π
12
en rad hora .38 1. Introducción
Esta señal de perturbación podríamos modelarla como un sistema de segundo orden con dos polos en el eje imaginario situados en ±jω 1, es decir:
ω(t) = 10.1p + ω21
vw(t) (1.6.46) donde vw(t) es un ruido blanco. 1.6. Procesos estocásticos o aleatorios 39
Bibliografía
L. Ljung, System Identification: Theory for the User, Prentice Hall , 1987. L. Ljung and T. Sˆderstrom, Theory and Practice of Recursive Identification,
The MIT Press , 1983. T. Sˆderstrom and P. Stoica, System Identification, Prentice Hall , 1989. J. P. Norton, An Introduction to Identification, Academic Press , 1986. G.R. Cooper and C.D McGuillem, Probabilistic Methods of Signal and Sys-tem Analysis, Oxford University Press , 1999. A. Papoulis, Probabilidad, variables aleatorias y procesos estocásticos, Ed. Universitaria de Barcelona , 1980. B.C. Kuo, Automatic Control Systems, Ed. Prentice Hall , 1991. K. Ogata, Discrete-Time Control Systems, Prentice Hall International Edi-tion , 1995. K. Ogata, Ingeniería de control moderna, Prentice Hall , 1998. E. Andrés Puente, Regulación automática, Servicio de publicaciones ETSIIM-UPM , 1993. 40 1. Introducción 2. TéCNICAS CLáSICAS DE MODELADO DE SISTEMAS
2.1. Introducción
Un problema básico en ingeniería de control y en general en muchas otras ciencias consiste en ser capaces de predecir qué efecto tendrá una cierta acción sobre un sistema físico. Dado que lo que se quiere es predecir una respuesta futura del sistema, resulta necesario utilizar algún tipo de modelo que nos permita el poder hacer esta predicción. Existen diversos tipos de modelos, en algunos de ellos como es el caso de los modelos a escala lo que se busca es reproducir el sistema físico real por medio de alguna maqueta o sistema más manejable sobre el que se pueda experimentar y ex-traer resultados. Estos modelos, aunque interesantes, no son los que más interesan en control, sino aquellos donde las predicciones se realizan sobre un cierto modelo matemático del sistema. Dentro de los modelos matemáticos que se pueden utilizar para analizar los efectos que diferentes acciones van a tener sobre el sistema podemos diferenciar dos grandes grupos de modelos:
Modelos de entrada/salida .En esta clase de modelos se busca una descripción matemática que exprese la relación que existe entre la entrada del sistema y la salida del mismo. Estos modelos no describen el funcionamiento interno del sistema, sino meramente la relación entre la entrada y la salida. Podemos encontrarnos con sistemas diferentes pero que presentan la misma relación entrada/salida por lo que dan lugar al mismo modelo matemático. Estos tipos de modelos son lo que podríamos denominar modelos clásicos.
Modelos basados en el espacio de estados .Esta clase de modelos buscan una descripción más profunda del sistema, ya que no sólo se caracteriza la relación entre la entrada y la salida, sino que además se caracteriza el comportamiento de una serie de variables o magni-tudes internas del sistema. Existen muchos sistemas en los que las magnitu-des internas que se pueden utilizar para caracterizarlo pueden ser escogidas 42 2. Técnicas clásicas de modelado de sistemas
de diferentes formas, por lo que un mismo sistema podrá ser descrito por diferentes modelos matemáticos. Estos tipos de técnicas han dado lugar a lo que se ha denominado teoría moderna de control. En este capítulo abordaremos el estudio de lo que son las técnicas clásicas de modelado de sistemas, y en el capítulo siguiente las técnicas de modelado basadas en el espacio de estados.
2.2. Modelos de entrada/salida
Estos tipos de modelos son lo que podríamos denominar modelos clásicos, y dentro de ellos podemos distinguir varios tipos: modelos temporales, frecuenciales y estocásticos. Los modelos temporales de entrada/salida se caracterizan por per-mitir el estudio de la respuesta temporal del sistema ante una entrada cualesquiera. El conocer la respuesta temporal del sistema nos permitirá analizar efectos tran-sitorios de corta duración en el sistema, como son picos en la respuesta, tiempos de subida o tiempos de establecimiento. Estos modelos temporales pueden ser mo-delos de tiempo continuo basados en ecuaciones diferenciales o bien modelos de tiempo discreto descritos por ecuaciones en diferencias. Los modelos denominados frecuenciales se basan en la caracterización de la relación entrada/salida de un sistema en régimen permanente ante entradas de tipo sinusoidal. Por último los modelos estocásticos más usuales suelen combinar modelos temporales o frecuenciales deterministas y señales de ruido o perturbación que aña-den un comportamiento estocástico al sistema y que requiere ser modelado cuando resulta posible.
2.3. Modelos temporales 2.3.1. Conceptos básicos
Sistemas lineales invariantes en el tiempo
Los modelos de entrada/salida de los sistemas lineales responden a la siguiente forma general,
dn
dtn y(t) + an−1
dn−1
dtn−1 y(t) + . . . + a0y(t) = (2.3.1)
bm
dm
dtm u(t) + bm−1
dm−1
dtn−1 u(t) + . . . + b0u(t) (2.3.2) donde los coeficientes ai, i = 0 , 1, . . . , n − 1 y bj , j = 0 , 1, . . . , m son números reales, u(t) es la entrada al sistema e y(t) es la salida del mismo. Si estos coefi-2.3. Modelos temporales 43
cientes ai y bj no dependen del tiempo, el sistema se dice que es invariante en el tiempo. Estos modelos lineales presentan varias ventajas, entre ellas y quizás la más im-portante es la de la linealidad en la respuesta. Es decir, si α y β son dos constantes arbitrarias y u1(t) y u2(t) son dos entradas al sistema, se verifica que si
u(t) = αu 1(t) + βu 2(t) (2.3.3) entonces la salida del sistema vendrá dada por
y(t) = αy 1(t) + βy 2(t) (2.3.4)
Respuesta impulsional
Dado un sistema lineal invariante en el tiempo, que para una entrada u(t) da una respuesta y(t), se define como respuesta impulsional del sistema la salida
g(t) que daría el sistema cuando la entrada al mismo es un impulso unitario δ(t).Esta respuesta impulsional g(t) contiene toda la información necesaria sobre el sistema, y se puede obtener la salida del mismo, ante cualquier entrada u(t), sin más que realizar la convolución en el dominio del tiempo con esta señal, es decir,
y(t) = g(t) ∗ u(t) (2.3.5)
=
∫ t
0
g(τ )u(t − τ )d τ (2.3.6)
=
∫ t
0
u(τ )g(t − τ )d τ (2.3.7) Si estamos en tiempo discreto tendríamos una secuencia {g(n)} que algunos autores denominan secuencia de ponderación . En este caso la salida del sistema ante una secuencia de entrada cualesquiera se obtendrá como la convolución dis-creta entre la secuencia de respuesta impulsional y la secuencia de entrada,
y(n) =
∞
∑
k=0
g(k)u(n − k) (2.3.8) En términos generales la respuesta impulsional g(t) no resulta demasiado prác-tica a la hora de modelar un cierto sistema ya que resulta mucho más fácil traba-jar en el campo de Laplace, que resolver expresiones integrales que pueden ser complejas. En el caso discreto es algo más fácil pero salvo para los sistemas cu-ya respuesta impulsional tenga un número finito de coeficientes distintos de cero, también resulta laborioso resolver un sumatorio que tiende a ∞.En la figura 2.1 se puede ver que si a una entrada impulsional δ(t), la salida del sistema es g(t), entonces, por linealidad, a una entrada aδ (t), le corresponderá una 44 2. Técnicas clásicas de modelado de sistemas tu(t) y(t) t
δ(t)
g(t) u(t) y(t)
g(t) tu(t) y(t) t
2δ (t) tu(t) y(t) t
aδ(t) ag(t) tu(t) y(t) t
δ(t-t o)g(t-to) to
1.5δ (t-t 1)δ(t-to) to to t1t1
Escalado Desplazamiento Superposición
2g(t)+g(t-to)+1.5g(t-t 1)
Fig. 2.1: Respuesta a un impulso y a una suma de impulsos desplazados
salida ag (t). Como el sistema es invariante en el tiempo, a un impulso desplazado
δ(t − t0) le corresponderá una respuesta desplazada g(t − t0). A una entrada com-puesta de una superposición de impulsos escalados le corresponderá una salida que será la suma de las respuestas correspondientes.
Función de transferencia
Se define como función de transferencia de un sistema lineal e invariante en el tiempo la transformada de Laplace (o la transformada z) de la respuesta impulsio-nal del sistema, supuestas condiciones iniciales nulas. Es decir, en tiempo continuo sería 2.3. Modelos temporales 45
G(s) = L[g(t)] (2.3.9) y en tiempo discreto
G(z) = Z[g(k)] (2.3.10) El concepto de función de transferencia resulta muy útil a la hora de determinar la salida y del sistema ante una entrada determinada u, ya que esta viene dada por la expresión
Y (s) = G(s)U (s) (2.3.11) o en tiempo discreto por
Y (z) = G(z)U (z) (2.3.12) a partir de cuyos valores en el campo complejo, desarrollando en fracciones parcia-les y haciendo las transformadas inversas de cada una de estas fracciones parciales resultaba simple obtener la respuesta analítica que dará el sistema en el dominio del tiempo tal y como se vio en el capítulo anterior. Este concepto de función de transferencia se utilizará mucho en adelante y resulta conveniente el hacer algunas matizaciones al respecto: La función de transferencia se ha definido para sistema lineales e invariantes en el tiempo, por lo que no está definida en el caso de sistemas no lineales. Para obtener la función de transferencia se han supuesto condiciones inicia-les nulas. La función de transferencia expresa la relación que existe entre una señal de entrada y una señal de salida, por lo que sólo permite expresar de forma completa la dinámica de un sistema con una entrada y una salida. Si el siste-ma tuviese más de una entrada o salida, serían necesarias varias funciones de transferencia para expresar todas las posibles relaciones entre cada entrada y cada salida. En este caso estaríamos ante sistemas multivariables MIMO (Multiple Input - Multiple Output). La función de transferencia es independiente de la entrada al sistema. La función de transferencia de un sistema es únicamente función racional de
s o z, y además con un denominador de grado mayor o igual que el numera-dor, para que sea físicamente realizable. 46 2. Técnicas clásicas de modelado de sistemas
2.3.2. Modelos en tiempo continuo
Una parte muy importante de los sistemas que nos interesa controlar corres-ponde a sistemas físicos cuya dinámica viene descrita por medio de ecuaciones diferenciales ordinarias del tipo siguiente
dn
dtn y(t) + an−1
dn−1
dtn−1 y(t) + . . . + a0y(t) =
bm
dm
dtm u(t) + bm−1
dm−1
dtn−1 u(t) + . . . + b0u(t)
(2.3.13) donde u(t) es la entrada al sistema e y(t) la salida del mismo. Si suponemos con-diciones iniciales nulas y tomamos transformadas de Laplace en ambos lados de la ecuación 2.3.13, el resultado es
(sn +an−1sn−1 +. . . +a0)Y (s) = ( bmsm +bm−1sm−1 +. . . +b0)U (s) (2.3.14) Despejando Y (s) en la ec 2.3.14 obtenemos que
Y (s) = sn + an−1sn−1 + . . . + a0
bmsm + bm−1sm−1 + . . . + b0
U (s) (2.3.15) y de aquí que la forma general de la función de transferencia G(s) para un sistema físico descrito por una ecuación diferencial ordinaria de la forma 2.3.13 sea una función racional de s del tipo
G(s) = sn + an−1sn−1 + . . . + a0
bmsm + bm−1sm−1 + . . . + b0
(2.3.16)
Ejemplo 2.3.1: Un automóvil responde a una fuerza aplicada f (t) del motor y a una fuerza debida a la fricción ρv (t), proporcional a la velocidad v(t) del automó-vil. La ecuación diferencial del sistema es
m ˙v(t) = f (t) − ρv (t) (2.3.17) Tomando transformadas de Laplace se obtiene
msV (s) = F (s) − ρV (s) (2.3.18) es decir,
(ms + ρ)V (s) = F (s) (2.3.19) y la función de transferencia de la velocidad respecto de la fuerza es
V (s)
F (s) = 1
ms + ρ (2.3.20) 2.3. Modelos temporales 47
2.3.3. Modelos en tiempo discreto
El desarrollo de los modelos de sistemas en tiempo discreto se basa en las ecuaciones en diferencias. Es un desarrollo muy similar al que se utiliza con las ecuaciones diferenciales ordinarias. De forma similar a aquel caso, sólo se conside-rarán ecuaciones en diferencias lineales y de coeficientes constantes. Una ecuación en diferencias de orden n-ésimo puede expresarse de dos formas: una en función de los términos retardados en el tiempo y(k − 1) , y(k − 2) , u(k − 1) , u(k − 2) , ..., etc; y otra en la que se usan términos avanzados en el tiempo y(k + 1) , y(k + 2) ,
u(k + 1) , u(k + 2) , ..., etc. Ambas formas son útiles. Aquí se utilizará la basada en los términos con retardo, y en función de ellos una ecuación en diferencias de orden n tendrá la siguiente forma
y(k) + an−1y(k − 1) + . . . + a0y(k − n) =
bmu(k) + bm−1u(k − 1) + . . . + b0u(k − m) (2.3.21) La ecuación 2.3.21 puede ser expresada del siguiente modo
y(k) = −an−1y(k − 1) − . . . − a0y(k − n)+bmu(k) + bm−1u(k − 1) + . . . + b0u(k − m) (2.3.22) Esta ecuación 2.3.22 muestra que el valor y(k), es decir la solución de la ecuación en el instante k-ésimo, es función de n + m + 1 datos. Estos datos son: la entrada en ese instante u(k), las m entradas anteriores u(k − 1) , . . . , u (k − m) y las n
salidas anteriores del sistema y(k − 1) , . . . , y (k − n).Si la señal de entrada es causal, es decir si u(k) = 0 para k < 0, entonces
u(−1) = u(−2) = . . . = u(−m) = 0 , por lo que sólo necesitaríamos conocer n
condiciones iniciales correspondientes a los valores de y(−1) , y (−2) , . . . , y (−n)
para poder calcular iterativamente el valor que tomará la salida del sistema en
y(0) , y (1) , y (2) , . . . a medida que se producen las respectivas entradas de control
u(0) , u (1) , u (2) , . . .
Ejemplo 2.3.2: Supongamos el sistema
y(k) + 0 , 2y(k − 1) = u(k) (2.3.23) suponiendo como condiciones iniciales y(−1) = −3, y que la entrada al sistema es un escalón unitario u0(k).La ecuación 2.3.25 puede expresarse como
y(k) = −0, 2y(k − 1) + u(k) (2.3.24) 48 2. Técnicas clásicas de modelado de sistemas
y de aquí que
y(0) = −0, 2y(−1) + u(0) (2.3.25) y como la secuencia de entrada toma valores u(0) = 0 , u(1) = 1 , u(2) = 1, u (3) = 1 , . . . tendremos que
y(0) = −0, 2y(−1) + u(0) = −0, 2.(−3) + 0 = 0 .6
y(1) = −0, 2y(0) + u(1) = −0, 2.(0 .6) + 1 = 0 .88
y(2) = −0, 2y(1) + u(2) = −0, 2.(0 .88) + 1 = 0 .824
y(3) = −0, 2y(2) + u(3) = −0, 2.(0 .824) + 1 = 0 .835
y(4) = −0, 2y(3) + u(4) = −0, 2.(0 , 835) + 1 = 0 .833
. . . . . . . . . . . . . . .
Como se ha mostrado en el ejemplo anterior, si conocemos la expresión en tiempo discreto de un cierto sistema resulta fácilmente calculable por métodos ite-rativos cuál será la salida que dará el sistema en instantes de tiempo sucesivos. Sin embargo no resulta tan fácil el extraer la solución analítica del sistema que sería la que nos permitiese conocer el valor de la salida para un instante de tiempo cuales-quiera sin tener que calcular todos los valores intermedios hasta dicho instante. Si en la expresión 2.3.21 hacemos la transformada z de la misma y suponemos condiciones iniciales nulas obtendremos la función de transferencia G(z) de este sistema de forma similar a lo que ocurría para tiempo continuo.
G(z) = bmz−m + bm−1z−m+1 + . . . + b0
z−n + an−1z−n+1 + . . . + a0
(2.3.26) En el caso del ejemplo 2.3.2, la función de transferencia queda
G(z) = Y (z)
U (z) = 11 + 0 .2z−1
Si reflexionamos un poco sobre las expresiones 2.3.21 y 2.3.26 nos encontra-mos con que el proceso que hemos seguido para obtener la función de transferen-cia en z de un sistema en tiempo continuo es simple y prácticamente inmediato. Sin embargo nos encontramos con que salvo algunos sistemas para los cuales la ecuación en diferencias resulta el modo natural de expresarlo, una gran mayoría de sistemas físicos son descritos por ecuaciones diferenciales ordinarias y no por ecuaciones en diferencias. En estos casos el proceso de obtención del modelo en tiempo discreto del sistema resulta más complejo.
Muestreo de una señal
Una cierta señal continua que es muestreada (es decir observado su valor) cada cierto tiempo T da como resultado de esta observación una cierta secuencia de 2.3. Modelos temporales 49 x(t) x(t)
T
Fig. 2.2: Muestreador
valores {x(0) , x (T ), x (2 T ), x (3 T ), . . . }. Esta secuencia de valores la podemos denominar la señal muestreada y la representaremos como x∗(t) (figura 3.3). Esta señal muestreada x∗(t), puede ser expresada matemáticamente como la suma de una serie infinita de valores de la siguiente forma
x∗(t) =
∞
∑
k=0
x(kT )δ(t − kT ) (2.3.27) donde δ(t − kT ) expresa un impulso unitario en el instante t = kT . Es decir la salida del muestreador corresponde al producto de la señal continua x(t) y el tren de impulsos unitarios δT (t), con
δT (t) =
∞
∑
k=0
δ(t − kT ) (2.3.28) Si hacemos la transformada de Laplace de la señal muestreada 2.3.27 nos queda que
X∗(s) = L[x∗(t)] = L[
∞
∑
k=0
x(kT )δ(t − kT )] =
∞
∑
k=0
L[x(kT )δ(t − kT )] =
∞
∑
k=0
x(kT )L[δ(t − kT )] =
∞
∑
k=0
x(kT )e−kT s (2.3.29) 50 2. Técnicas clásicas de modelado de sistemas x(t)
B0
x(kT)
Fig. 2.3: Bloqueador de orden cero
Si recordamos que en el campo de Laplace, un término de la forma e−as co-rrespondía en el dominio temporal a un desplazamiento en el tiempo de la señal, es decir a un retardo de valor a, y que por otra parte un retardo en el tiempo es lo que denominábamos z, la relación entre ambos conceptos resulta ser
eT s = z (2.3.30) o
s = 1
T ln z (2.3.31) Si sustituimos en la ecuación 2.3.29 el término s por la expresión 2.3.31
X∗(s) |s= 1
Tln z
=
∞
∑
k=0
x(kT )z−k = X(z) (2.3.32)
Bloqueadores de señal
El proceso de bloqueo de señal es el contrario del de muestreo, y tiene por objetivo el obtener una señal de tiempo continuo a partir de una señal de tiempo discreto. Hay que tener en cuenta que las acciones de control que hay que ejercer sobre los sistemas físicos reales normalmente son señales continuas. El modo más simple de obtener una señal continua a partir de una señal de tiempo discreto consiste en suponer que el valor en tiempo continuo de esa señal de tiempo discreto se mantiene hasta el tiempo del proximo ciclo (este tipo de bloqueador se denomina bloqueador de orden cero , ver figura 2.3). Es decir que la salida del bloqueador de orden zero será
y(t) = x(kT ), kT ≤ t < (k + 1) T (2.3.33) 2.3. Modelos temporales 51 u (t-kT)
0 u (t-(k+1)T)
0
x(kT)
Fig. 2.4: Señal entre muestreos
La expresión 2.3.33 podemos reformularla en la siguiente forma, expresando el valor continuo de la señal en un cierto intervalo como la resta de dos señales escalón unitario de tiempo continuo tal y como se muestra en la figura 2.4
y(t) =
∞
∑
k=0
x(kT )[ u0(t − kT ) − u0(t − (k + 1) T )] (2.3.34) Si hacemos la transformada de Laplace de la expresión 2.3.34, ésta da como resultado
L[y(t)] = Y (s) =
∞
∑
k=0
L[x(kT )[ u0(t − kT ) − u0(t − (k + 1) T )]] =
∞
∑
k=0
x(kT )L[u0(t − kT ) − u0(t − (k + 1) T )] =
∞
∑
k=0
x(kT )[ e−kT s
s − e−(k+1) T s
s ]= 1 − e−T s
s
∞
∑
k=0
x(kT )e−kT s (2.3.35) Si comparamos la expresión 2.3.35 con la expresión de X∗(s) que veíamos en la ecuación 2.3.32, se puede deducir que la salida Y (s) del sistema puede ser expresada como
Y (s) = 1 − e−T s
s X∗(s) (2.3.36) 52 2. Técnicas clásicas de modelado de sistemas
Si en la expresión 2.3.36 observamos que X∗(s) es la transformada de Laplace de la señal muestreada de entrada y que Y (s) es la transformada de la salida, se puede deducir que la función de transferencia del bloqueador de orden cero será
GBOC (s) = 1 − e−T s
s (2.3.37) Hemos visto en las secciones anteriores dos conceptos muy importantes. El primero de ellos era que al muestrear un sistema continuo, y por medio de la trans-formación matemática eT s = z obtenemos un modelo que nos permite analizar por medio de la transformada en z en tiempo discreto sistemas físicos continuos cuya transformada de Laplace es fácilmente obtenible y que son muestreados. El segundo aspecto importante que hemos visto es el proceso de bloqueo que nos permite obtener una señal de tiempo continuo utilizable para controlar un sis-tema físico continuo a partir de una señal de tiempo discreto. Este proceso de blo-queo de orden cero responde al modelo matemático (1 − e−T s )/s en el campo de Laplace.
Función de transferencia muestreada
Supongamos ahora un cierto sistema continuo al que se le introduce como en-trada una señal muestreada, tal y como se muestra en la figura 2.5. En este caso, aplicando la transformada de Laplace tendremos que la salida del mismo viene dada por
Y (s) = G(s)X∗(s) (2.3.38) si hacemos la antitransformada de Laplace de la expresión 2.3.38 tendremos que
y(t) = L−1[Y (s)] = L−1[G(s)X∗(s)] =
∫ t
0
g(t − τ )x∗(τ )d τ
=
∫ t
0
g(t − τ )
∞
∑
k=0
x(τ )δ(τ − kT )d τ
=
∞
∑
k=0
∫ t
0
g(t − τ )x(τ )δ(τ − kT )d τ
=
∞
∑
k=0
g(t − kT )x(kT ) (2.3.39) 2.3. Modelos temporales 53 x(t) G (s) y(t) x(t) X(z)
Fig. 2.5: Sistema continuo con entrada muestreada
Si hacemos ahora la transformada z de la expresión obtenida en 2.3.39, tendremos que
Y (z) = Z[y(t)] =
∞
∑
i=0
[
∞
∑
k=0
g(iT − kT )x(kT )] z−i
=
∞
∑
j=0
∞
∑
k=0
g(jT )x(kT )z−(j+k)
=
∞
∑
j=0
∞
∑
k=0
g(jT )x(kT )z−j z−k
=
∞
∑
j=0
g(jT )z−j
∞
∑
k=0
x(kT )z−k
= G(z)X(z) (2.3.40) en el desarrollo de la expresión 2.3.40, se ha realizado el cambio de variable j =
i − k. Por otra parte recordando que la transformada z podía entenderse como la transformada de Laplace particularizada para s = 1
T
ln z. Podríamos expresar 2.3.40 de la siguiente forma
Y ∗(s) = G∗(s)X∗(s) (2.3.41) Una consideración importante a tener en cuenta es el efecto del muestreador a la entrada del sistema. Supongamos ahora que nuestro sistema es el mostrado en la figura 2.6, y supongamos que vamos a muestrear la salida y(t) del mismo
Y (s) = G(s)X(s) (2.3.42) si muestreamos ahora la salida del sistema tenemos que
Y ∗(s) = [ G(s)X(s)] ∗ = [ GX (s)] ∗ (2.3.43) que formulada en términos de transformada z nos conduce a la siguiente expresión 54 2. Técnicas clásicas de modelado de sistemas x(t) G (s) y(t) y(t) Y(z)
Fig. 2.6: Sistema continuo sin entrada muestreada
Y (z) = Z[Y (s)] = Z[G(s)X(s)] = Z[GX (s)] = GX (z) (2.3.44) Es importante resaltar aquí que no es lo mismo la transformada z del producto de las funciones de transferencia G(s)X(s) que el producto de las transformadas z: G(z)X(z) de las G(s) y X(s), es decir
GX (z) = Z[G(s)X(s)] 6 = Z[G(s)] Z[X(s)] = G(z)X(z) (2.3.45) Solamente en el caso de que la entrada al sistema sea una señal impulso mues-treada, se verificará que
G(z) = Z[G(s)] (2.3.46) La conclusión más importante que se extrae de 2.3.45 es que dependiendo del punto donde se muestrea un cierto sistema físico, se pueden obtener diferentes funciones de transferencia en z, es decir diferentes ecuaciones en diferencias.
Ejemplo 2.3.3: Supongamos que deseamos obtener la función de transferencia muestreada G(z) = Z[G(s)] de un sistema físico cuya función de transferencia continua G(s) viene dada por la siguiente expresión
G(s) = 1
s + 2
Este sistema tiene por respuesta impulsional la siguiente función
g(t) = e−2t
si este sistema lo muestreamos con periodo T obtendremos
g(kT ) = e−2kT , k = 0 , 1, 2, . . .
y de esta secuencia podemos obtener la transformada z, es decir
G(z) =
∞
∑
k=0
g(kT )z−k =
∞
∑
k=0
e−2kT z−k =
∞
∑
k=0
(e2T z)−k = 11 − e−2T z−12.3. Modelos temporales 55
Este método de obtención de la transformada z de una función de transferen-cia muestreada resulta bastante laborioso ya que por una parte se hace necesario obtener la respuesta impulsional del sistema para posteriormente hacer la transfor-mada z y obtener la suma de la serie que es la que nos daría la expresión en forma analítica.
Obtención de la función de transferencia z
Habíamos visto que una señal muestreada daba como salida una señal que se podía expresar en la siguiente forma
x∗(t) =
∞
∑
k=0
x(t)δ(t − kT )= x(t)
∞
∑
k=0
δ(t − kT ) (2.3.47) Dado que la transformada de Laplace de un impulso de Dirac era la unidad, que un retardo a de una señal en el tiempo equivale, en el campo de Laplace, a multi-plicar por un término exponencial e−as , y hacemos la transformada de Laplace del sumatorio de la ecuación 2.3.47 obtenemos
L[
∞
∑
k=0
δ(t − kT )] = L[δ(t)] + L[δ(t − T ) + L[δ(t − 2T )] + . . .
= 1 + e−T s + e−2T s + . . .
= 11 − e−T s (2.3.48) Haciendo la transformada de Laplace de la señal muestreada x∗(t) tenemos
X∗(s) = L[x∗(t)] = L[x(t)
∞
∑
k=0
δ(t − kT )] (2.3.49) En esta ecuación se puede observar que X∗(s) es la transformada de Laplace del producto de dos funciones en el dominio del tiempo. Si además recordamos que la transformada de Laplace del producto de dos funciones tiene la siguiente expresión
L[f (t)g(t)] =
∫ ∞
0
f (t)g(t)e−st dt = 12πj
∫ c+j∞
c−j∞
F (p)G(s − p)d p (2.3.50) 56 2. Técnicas clásicas de modelado de sistemas
Aplicando esta expresión a la ecuación 2.3.49 y teniendo en cuenta que las trans-formadas de Laplace de x(t) es X(s) y la del sumatorio de impulsos de Dirac es la de la expresión 2.3.48, resulta la siguiente expresión de la integral de convolución.
X∗(s) = 12πj
∫ c+j∞
c−j∞
X(p) 11 − e−T (s−p) dp (2.3.51) La evaluación de esta integral de convolución a lo largo de la línea c − j∞
a c + j∞ puede realizarse por el método de los residuos formando un contorno cerrado consistente en la propia línea c − j∞ a c + j∞ y un círculo Γ de radio infinito en el semiplano izquierdo que incluya todos los polos de X(p). De esta forma la ecuación 2.3.51 se podrá expresar como
X∗(s) = 12πj
∫ c+j∞
c−j∞
X(p) 11 − e−T (s−p) dp
= 12πj
∮
X(p) 11 − e−T (s−p) dp
− 12πj
∫
Γ
X(p) 11 − e−T (s−p) dp (2.3.52) En el caso de que X(s) este expresado por una expresión polinómica X(s) =
N(s)
D(s)
en la que el grado del denominador sea mayor que el del numerador, el se-gundo término de la expresión 2.3.52 se hace cero. Y de aquí que
X∗(s) = 12πj
∮
X(p) 11 − e−T (s−p) dp (2.3.53) Esta integral se resuelve aplicando la fórmula integral de Cauchy y es igual a la suma de los residuos R en el contorno cerrado para los polos de X(p), es decir
X∗(s) = ∑ R
[
X(p) 11 − e−T (s−p)
]
p=pi
(2.3.54) Las expresiones para el cálculo de los residuos son las siguientes: Polos simples ( p = pi). Para este caso el residuo del polo es
Rpi = lim
p→pi
[
(p − pi) X(p)1 − e−T (s−p)
]
(2.3.55) Polos múltiples ( p = pj , con orden de multiplicidad n). Para este caso el residuo del polo con orden de multiplicidad n es 2.3. Modelos temporales 57
Rpj = 1(n − 1)! lim
p→pj
dn−1
dpn−1
[
(p − pj )n X(p)1 − e−T (s−p)
]
(2.3.56)
Ejemplo 2.3.4: Supongamos un sistema en el que la señal tiene función de trans-ferencia X(s) = 2(s+2) , en este caso tendríamos que:
Rpi=−2 = lim
p→− 2
[
(p + 2)
2(p+2)
1 − e−T (s−p)
]
= lim
p→− 2
21 − e−T (s−p)
= 21 − e−T s e−2T
por tanto tendremos que
X∗(s) = 21 − e−T s e−2T (2.3.57) y de esta expresión teniendo en cuenta que z = eT s obtendríamos la transformada en z,
X(z) = 21 − z−1e−2T (2.3.58) Si suponemos que el periodo de muestreo es de 0.1 segundos la transformada
z de la señal X(s) muestreada tomará el siguiente valor,
X(z) = 21 − 0, 8187 z−1 (2.3.59) Si comparamos las curvas de respuesta en el dominio del tiempo de las señales
X(s) y X(z) se puede observar en la figura 2.7 la correspondencia existente entre ambas.
Función de transferencia discreta equivalente
Supongamos ahora que para un cierto sistema conocemos su función de trans-ferencia continua y que nos interesa conocer su equivalente en tiempo discreto. Este sistema discreto equivalente que buscamos para un cierto proceso o sistema debe tener la entrada discreta y la salida discreta. Para ello, y puesto que el sistema real es continuo, se introduce un bloqueador de señal en la entrada del sistema y un muestreador a la salida del mismo tal y como se ve en la figura 2.8. Veamos que ecuaciones tenemos en el sistema. En primer lugar habíamos visto que la ecuación del bloqueador de orden cero era 58 2. Técnicas clásicas de modelado de sistemas Time (sec.) Amplitude Impulse Response
0 0.5 1 1.5 2 2.5 300.2 0.4 0.6 0.8 11.2 1.4 1.6 1.8 2From: U(1) To: Y(1) Time (sec.) Amplitude Impulse Response
0 0.5 1 1.5 2 2.5 300.2 0.4 0.6 0.8 11.2 1.4 1.6 1.8 2From: U(1) To: Y(1)
Fig. 2.7: Respuesta en tiempo continuo y discreto 2.3. Modelos temporales 59 r(n) BOC G (s) pr(t) y(t) R(z) R(s) Y(s) Y(z) y(n) y(n) r(n) R(z) Y(z) G (z) p
Fig. 2.8: Equivalente en tiempo discreto de un sistema en tiempo continuo
GBOC (s) = 1 − e−T s
s
de aquí tenemos que el conjunto bloqueador-proceso tendrá la siguiente función de transferencia en el campo de Laplace
G(s) = GBOC (s)Gp(s) = 1 − e−T s
s Gp(s) (2.3.60) Por otra parte tenemos que la señal de entrada al sistema es una señal mues-treada R∗(s), por lo que a la salida del proceso tendremos
Y (s) = G(s)R∗(s) (2.3.61) si muestreamos esta señal y(s) para obtener a la salida del sistema una señal dis-creta, la expresión nos queda de la siguiente forma
Y ∗(s) = G∗(s)R∗(s) (2.3.62) que en forma discreta queda como
Y (z) = G(z)R(z) (2.3.63) Por lo tanto estamos en una situación similar al epígrafe anterior donde bus-cábamos obtener X∗(s) discretizando la señal continua, con la diferencia que en este caso lo que buscamos obtener es la función de transferencia muestreada G∗(s)
donde
G(s) = GBOC (s)Gp(s) = 1 − e−T s
s Gp(s)60 2. Técnicas clásicas de modelado de sistemas
operando con esta función de transferencia tenemos que
G(s) = Gp(s)
s − e−T s Gp(s)
s
es decir tenemos la respuesta de Gp(s) ante entrada escalón unitario Gp(s)
s
menos esta misma respuesta retardada T , de lo cual si hacemos la transformada z para esta G(s) tendremos que
Z[G(s)] = Z
[ Gp(s)
s − e−T s Gp(s)
s
]
= Z
[
[ Gp(s)
s ] − Z [ e−T s Gp(s)
s
]
= (1 − z−1)Z
[ Gp(s)
s
]
(2.3.64) y si evaluamos Z[ Gp(s)
s
] por el método de los residuos como hicimos en el epígrafe anterior para la función de transferencia muestreada de una señal, obtendremos la función de transferencia discreta equivalente para dicho sistema continuo.
Ejemplo 2.3.5: Supongamos un sistema cuya función de transferencia en tiempo continuo viene dada por Gp(s) = 3(s+1) , en este caso para obtener el equivalente discreto del sistema suponiendo que se utiliza un bloqueador de orden cero tendre-mos que:
G(z) = (1 − z−1)Z
[ Gp(s)
s
]
veamos en primer lugar la transformada z de Gp(s)
s
para lo que utilizaremos el método de los residuos, en este caso tenemos dos polos para la función por lo que tendremos dos residuos:
Rpi=−1 = lim
p→− 1
[
(p + 1)
3
p(p+1)
1 − e−T (s−p)
]
= lim
p→− 1
−31 − e−T (s−p)
= −31 − e−T s e−T
y2.3. Modelos temporales 61
Rpi=0 = lim
p→0
[
p
3
p(p+1)
1 − e−T (s−p)
]
= lim
p→0
31 − e−T (s−p)
= 31 − e−T s
y sustituyendo eT s por z tendremos que
G(z) = (1 − z−1)
( 31 − z−1 − 31 − z−1e−T
)
(2.3.65) Si suponemos que el periodo de muestreo es de 0.1 segundos la transformada
z del sistema tomará el siguiente valor,
G(z) = (1 − z−1)
( 31 − z−1 − 31 − 0, 3678 z−1
)
(2.3.66)
Función de transferencia muestreada de un ciclo de control típico
Consideremos ahora un bucle o lazo de control típico, tal y como se muestra en la figura 2.9, en el que se muestrea la señal de error. Podemos observar que para este caso
E(s) = R(s) − H(s)Y (s)
Y (s) = G(s)E∗(s)
y de aquí que
E(s) = R(s) − G(s)H(s)E∗(s)
Si muestreamos esta señal de error continua, obtenemos que
E∗(s) = R∗(s) − GH ∗(s)E∗(s)
y despejando E∗(s) tenemos
E∗(s) = R∗(s)1 + GH ∗(s)
De aquí, teniendo en cuenta que Y ∗(s) = G∗(s)E∗(s), se deduce que
Y ∗(s) = G∗(s)R∗(s)1 + GH ∗(s)62 2. Técnicas clásicas de modelado de sistemas r(t) G (s) p y(t) e(t)
+-
H(s) R(s) Y(s) E(s) E(s)
Fig. 2.9: Bucle de control típico muestreado
De esta expresión, y reescribiéndola en términos de la transformada z llegamos a la siguiente expresión
Y (z) = G(z)R(z)1 + GH (z)
Luego la función de transferencia muestreada para un ciclo de control típico será
Y (z)
R(z) = G(z)1 + GH (z) (2.3.67) Veamos sobre este esquema básico de un bucle de control muestreado que se ha mostrado en la figura 2.9 , el que corresponde a un bucle de control por computador donde se incluye el efecto del muestreo de la señal de error y el bloqueo de la salida del controlador digital para hacerla una señal continua (figura 2.10). Veamos que ecuaciones tenemos en el sistema. En primer lugar habíamos visto que la ecuación del bloqueador de orden cero era
GBOC (s) = 1 − e−T s
s
de aquí tenemos que el conjunto bloqueador-proceso tendrá la siguiente función de transferencia en el campo de Laplace
G(s) = GBOC (s)Gp(s) = 1 − e−T s
s Gp(s) (2.3.68) Por otra parte la señal de salida del sistema será
Y (s) = G(s)G∗
c
(s)E∗(s) (2.3.69) o2.3. Modelos temporales 63
Y ∗(s) = G∗(s)G∗
c
(s)E∗(s) (2.3.70) que expresada en términos de transformada z quedará
Y (z) = G(z)Gc(z)E(z) (2.3.71) Además del muestreo de la señal de error tenemos que
E(z) = R(z) − Y (z) (2.3.72) luego
Y (z) = G(z)Gc(z)E(z) = G(z)Gc(z)[ R(z) − Y (z)] (2.3.73) y de aquí podemos deducir que la función de transferencia del lazo de control con muestro de señal de error y bloqueo de la señal de control del regulador es
Y (z)
R(z) = Gc(z)G(z)1 + Gc(z)G(z) (2.3.74) donde G(s) incluye al proceso y al bloqueador de orden cero.
Discretización de un controlador analógico
Resulta evidente que la relación que existe entre la transformada de Laplace de un sistema muestreado y la transformada z, venía dada por la expresión z = eT s .Esta relación no es una expresión racional por lo que con ella no hay una forma fácil e inmediata de obtener el equivalente discreto de un cierto sistema continuo. Es necesario obtener los equivalentes discretos de Gp(s) y de Gc(s) lo cual es un proceso laborioso. En muchas ocasiones además resulta poco práctico puesto que puede ocurrir que el ajuste del regulador se realice de forma empírica, y que no dis-pongamos de la función de transferencia del proceso Gp(s), aunque si tengamos la del controlador analógico que hemos ajustado empíricamente Gc(s). En este tipo de situaciones, en las que lo que nos interesa discretizar es únicamente el con-trolador es frecuente utilizar técnicas aproximadas, aunque estas pueden utilizarse también para discretizar cualquier función de transferencia. En la figura 2.11 se puede observar la situación que se quiere tratar donde un cierto regulador en tiempo continuo (analógico) se quiere sustituir por un conjun-to regulador digital, muestreador y bloqueador cuyo comportamiento global sea equivalente al del controlador analógico existente. Existen diferentes enfoques para realizar esta transformación de una forma rá-pida y simple (aunque no exacta). En general se suelen utilizar técnicas basadas en aproximaciones numéricas de la ecuación diferencial que representa el función de transferencia del controlador en tiempo continuo Gc(s).64 2. Técnicas clásicas de modelado de sistemas Convertidor A/D BOC G (s) G (z) pu(t) e(kT) c(kT) Controlador digital Proceso Convertidor D/A E(s) C(s) U(s) cr(kT) y(t)
+-
Y(s) Computador a) Implementación r(t) BOC G (s) G (z) p y(t) u(t) e(t) e(kT)
+- c(kT) Controlador digital Proceso Convertidor D/A Convertidor A/D R(s) E(s) E(s) C(s) U(s) Y(s) cb) Esquema equivalente
Fig. 2.10: Esquema típico de control por computador 2.3. Modelos temporales 65 BOC G (z) u(t) e(t) e(kT) c(kT) Controlador digital Convertidor D/A Convertidor A/D E(s) E(s) C(s) U(s) cG (s) e(t) u(t) e
Fig. 2.11: Controlador discreto equivalente Ge(z) de un controlador analógico
Gc(s)
Existen dos formas básicas de abordar esta aproximación numérica a la solu-ción de una ecuación diferencial, la integración numérica y la diferenciación nu-mérica. Normalmente se utilizan las técnicas basadas en la integración numérica. Supongamos que el controlador en tiempo continuo tiene la siguiente expresión
Gc(s) = U (s)
E(s) = 1
s (2.3.75) esta expresión corresponde a una integración de la forma
u(t) = u(t0) +
∫ tt0
e(t)d t (2.3.76) Las aproximaciones numéricas a esta expresión se basan en aproximar la integral de la señal de error por un sumatorio de rectángulos o trapezoides, tal y como se muestra en la figura 2.12. Podemos distinguir tres aproximaciones diferentes en función de como aproxi-memos el area de la integral del error durante un periodo de integración:
Método de Euler (I). En este caso el área bajo la integral se aproxima por un rectángulo de base el periodo T y altura el valor de la señal de error en el instantes n.
u(n + 1) = u(n) + T e (n) (2.3.77) haciendo la transformada zde esta ecuación tenemos 66 2. Técnicas clásicas de modelado de sistemas xxxx xxxx xxxx xxxx xxxx xxxx xxxx xxxx xxxx xxxx xxxx xxxx
xxx xxx xxx xxx xxx xxx xxx
xxx xxx xxx xxx xxx
xxx xxx xxx xxx
xxx xxx xxx xxx xxx
xxxx xxxx xxxx xxxx xxxx xxxx
xxx xxx xxx xxx xxx xxx xxx
xxx xxx xxx xxx xxx xxx xxx xxx
xxx xxx xxx xxx xxx xxx xxx xxx xxx
xxxx xxxx xxxx xxxx xxxx xxxx xxxx xxxx xxxx
xxx xxx xxx xxx xxx xxx xxx xxx xxx
te(t) (c)
x
xxxx xxxx xxxx xxxx xxxx xxxx xxxx xxxx xxxx xxxx xxxx xxxx xxxx
xxxx xxxx xxxx xxxx xxxx xxxx xxxx
xxx xxx xxx xxx xxx
xxx xxx xxx xxx xxx
xxx xxx xxx xxx
xxx xxx xxx xxx xxx
xxx xxx xxx xxx xxx xxx
xxx xxx xxx xxx xxx xxx
xxxx xxxx xxxx xxxx xxxx xxxx xxxx xxxx
xxxx xxxx xxxx xxxx xxxx xxxx xxxx xxxx
xxx xxx xxx xxx xxx xxx xxx xxx xxx
xxx xxx xxx xxx xxx xxx xxx xxx xxx
xxx xxx xxx xxx xxx
te(t) (a)
xxxx xxxx xxxx xxxx xxxx xxxx xxxx xxxx xxxx xxxx xxxx xxxx xxxx xxxx
xxx xxx xxx xxx xxx xxx xxx
xxx xxx xxx xxx xxx
xxx xxx xxx xxx xxx
xxx xxx xxx xxx xxx
xxx xxx xxx xxx xxx
xxx xxx xxx xxx xxx
xxxx xxxx xxxx xxxx xxxx xxxx xxxx
xxxx xxxx xxxx xxxx xxxx xxxx xxxx xxxx
xxx xxx xxx xxx xxx xxx xxx xxx xxx
xxx xxx xxx xxx xxx xxx xxx xxx xxx xxx
xxx xxx xxx xxx xxx xxx xxx xxx xxx
te(t) (b)
Fig. 2.12: Aproximaciones a la integración 2.3. Modelos temporales 67
zU (z) = U (z) + T E (z) (2.3.78) y por lo tanto
Ge(z) = U (z)
E(Z) = Tz − 1 (2.3.79) si comparamos esta expresión 2.3.79 con la del controlador analógico 2.3.75, se puede observar que para obtener el equivalente discreto con este método basta con substituir en la función de transferencia del controlador analógico cada s por z−1
T
. Es decir
Ge(z) = Gc(s)|s= z−1
T
(2.3.80)
Método de Euler (II). En este caso el área bajo la integral se aproxima por un rectángulo de base el periodo T y altura el valor de la señal de error en el instante k + 1 .
u(k + 1) = u(k) + T e (k + 1) (2.3.81) haciendo la transformada z de esta ecuación tenemos
zU (z) = U (z) + T zE (z) (2.3.82) y por lo tanto
Ge(z) = U (z)
E(z) = T z z − 1 (2.3.83) si comparamos esta expresión 2.3.83 con la del controlador analógico 2.3.75, se puede observar que para obtener el equivalente discreto con este método basta con substituir en la función de transferencia del controlador analógico cada s por z−1
T z
. Es decir
Ge(z) = Gc(s)|s= z−1
T z
(2.3.84)
Método trapezoidal .También se le conoce como método de Tustin o transformación bilineal. En este caso el área bajo la integral se aproxima promediando los valores de la señal de error en los instantes k y k + 1 y multiplicando por el periodo T .68 2. Técnicas clásicas de modelado de sistemas
u(k + 1) = u(k) + T
2 [e(k) + e(k + 1)] (2.3.85) haciendo la transformada z de esta ecuación tenemos
(z − 1) U (z) = T
2 (z + 1) E(z) (2.3.86) y por lo tanto
Ge(z) = U (z)
E(Z) = T (z + 1) 2( z − 1) (2.3.87) si comparamos esta expresión 2.3.87 con la del controlador analógico 2.3.75, se puede observar que para obtener el equivalente discreto con este método basta con substituir en la función de transferencia del controlador analógico cada s por 2( z−1)
T(z+1)
. Es decir
Ge(z) = Gc(s)|s= 2( z−1)
T(z+1)
(2.3.88)
Ejemplo 2.3.6: Supongamos que se ha diseñado un controlador analógico por métodos clásicos para un cierto sistema, y que la función de transferencia en s de dicho sistema es
Gc(s) = s + 4
s(s + 1)
aplicando la transformación bilineal tendríamos que la Ge(z) del regulador discreto equivalente será
Ge(z) = Gc(s) |s= 2
Tz−1
z+1
= (z + 1)[ z(1 + 2 T ) + (2 T − 1)] T
(z − 1)[ z(T + 2) + ( T − 2)] (2.3.89) 2.4. Modelos frecuenciales 69
Fig. 2.13: Discretización de un controlador analógico
2.4. Modelos frecuenciales
Cuando a un sistema lineal estable se le introduce como entrada una señal sinusoidal de amplitud unitaria y de frecuencia variable u(t) = 1 sin ωt el sistema responde dando a la salida una señal cuya transformada en el campo de Laplace viene dada por
Y (s) = G(s) ws2 + ω2 . (2.4.1) Si esta señal se expande en fracciones parciales aparecen un conjunto de frac-ciones parciales que corresponden a los polos de la función de transferencia G(s)
y los términos que corresponden a la entrada sinusoidal. Es decir
Y (s) = . . . + cs + jω + c∗
s − jω . (2.4.2) Si todos los polos del sistema dan una respuesta estable, su respuesta se amor-tiguará con el tiempo por lo que la respuesta del mismo en régimen permanente sólo retendrá los dos últimos términos del desarrollo 2.4.2. Por lo que la respuesta en régimen permanente en el dominio del tiempo tomará la forma
y(t) = 2 |c| sin( ωt + φ) = A sin( ωt + φ) (2.4.3) donde 70 2. Técnicas clásicas de modelado de sistemas
A = |G(jω )| (2.4.4)
φ = arctan Im [G(jω )]
Re [G(jω )] = ∠G(jω ) (2.4.5) La respuesta frecuencial de un sistema suele representarse por medio de dos curvas, en una de las cuales se muestra la magnitud |G(jω )| y en la otra el desfase
∠G(jω ) para cada valor de ω. La respuesta frecuencial permite determinar la esta-bilidad que tendrá un sistema en bucle cerrado a partir de la respuesta frecuencial en bucle abierto del mismo. Además estas curvas de respuesta pueden obtenerse de forma experimental sin tener conocimiento del modelo matemático del sistema, sin más que excitar el sistema con una señal sinusoidal de frecuencia variable e ir midiendo la amplitud y el desfase de la señal sinusoidal de salida en régimen permanente. Cuando en los diagramas representamos la amplitud escalada en decibelios
20 log |G(jω )| y la frecuencia la representamos en décadas, el diagrama resultante se conoce como diagrama de Bode .Este diagrama está formado por dos curvas de respuesta: una donde se repre-senta el módulo de la respuesta del sistema para diferentes frecuencias de la señal de entrada al mismo y otra en la que se representa el desfase de esa respuesta para diferentes frecuencias de la señal de entrada. Esta forma de caracterizar un sistema presenta algunas propiedades que la ha-cen muy interesante, la primera es la facilidad con la que se pueden obtener expe-rimentalmente estas curvas para un número considerable de sistemas (eléctricos, electrónicos, mecánicos, etc). La segunda es de tipo interpretativo, en el sentido de que nos permite ver o tratar todo sistema físico como si de un filtro se tratase (paso bajo normalmente) que tenderá a amplificar o amortiguar las señales sinusoidales de ciertas frecuencias que se presenten a la entrada del sistema (recordemos que toda señal periódica podía descomponerse en serie de Fourier).
Ejemplo 2.4.1: Sea un sistema cuyo función de transferencia es
G(s) = 5
s2 + s + 5 (2.4.6) veamos su respuesta en frecuencia, para ello utilizaremos la instrucción bode() de MATLAB, tal como se muestra a continuación
num=; den=[1 1 5]; sys=tf(num,den); bode(sys); 2.4. Modelos frecuenciales 71 Frequency (rad/sec) Phase (deg); Magnitude (dB) Bode Diagrams
−30 −25 −20 −15 −10 −5 0510 From: U(1)
10 −1 10 010 1
−200 −150 −100 −50 0To: Y(1)
Fig. 2.14: Diagrama de bode de un sistema de segundo orden
y el resultado se puede ver en la figura 2.14. En el proceso de razonamiento que se ha seguido para determinar la respuesta frecuencial de un sistema hemos partido del modelo matemático en s = σ+jω , y se ha sustituido por jω considerando que se estudia el valor en régimen permanente. Resulta evidente por lo tanto que este modelo G(jω ) matemáticamente deriva del modelo temporal.
2.4.1. Respuesta frecuencial de los sistemas muestreados
Los métodos de análisis de la respuesta frecuencial son aplicables también a los sistemas en tiempo discreto y en términos generales el funcionamiento es similar al caso continuo. Se aplica un señal sinusoidal de frecuencia variable al sistema
g(k), esta señal se muestrea antes de introducirla en el sistema
u(t) = sin ωt
y al muestrearla
u(n) = sin( nωT )
donde T = 2πωT (ωT es la frecuencia de muestreo utilizada en el sistema). Como resultado el sistema generará en su salida una sucesión de impulsos. 72 2. Técnicas clásicas de modelado de sistemas
y(n) =
n
∑
k=0
g(k) sin(( n − k)ωT ) (2.4.7)
= Im {
n
∑
k=0
g(k)ejω (n−k)T } (2.4.8)
= Im {ejωnT n∑
k=0
g(k)e−jωkT } (2.4.9) Si suponemos que el sistema es estable para la frecuencia de muestreo ωT =
2πT
, entonces g(n) → 0 cuando n → ∞ por lo que para valores grandes de n
tendremos que
G∗(s) =
∞
∑
k=0
g(k)e−skT =
n
∑
k=0
g(k)e−skT (2.4.10) con lo que y(n) quedará en la forma
y(n) = Im {ejωnT G∗(jω )} (2.4.11) o expresada en forma sinusoidal
y(n) = |G∗(jω )| sin( ωnT + φ) (2.4.12) donde el término φ es
φ = arg( G∗(jω )) (2.4.13) Si hacemos el cambio de variable z = ejωT en la ecuación 2.4.14, en vez del que proponíamos para obtener la función de transferencia en z (z = esT ) la expresión quedará en la siguiente forma
y(n) = |G(z)|z=ejωT sin( ωnT + φ) (2.4.14) y en esta expresión se puede observar que la salida del sistema es un señal sinusoi-dal que tiene la misma frecuencia que la señal de entrada y esta desfasada respecto a ella, y cuyo módulo depende del |G(z)|z=ejωT . Resulta clara la similitud con la respuesta para el caso continuo que veíamos en la ecuación 2.4.3.
Ejemplo 2.4.2: Sea el sistema del ejemplo anterior cuya función de transferencia es
G(s) = 5
s2 + s + 5 (2.4.15) 2.4. Modelos frecuenciales 73 Frequency (rad/sec) Phase (deg); Magnitude (dB) Bode Diagrams
−80 −60 −40 −20 020 From: U(1)
10 −1 10 010 110 2
−300 −250 −200 −150 −100 −50 050 100 To: Y(1)
Fig. 2.15: Diagrama de bode de un sistema de segundo orden en tiempo discreto
supongamos que dicho sistema se muestrea, y que se obtiene el equivalente dis-creto para periodos de muestreo T = 0 , 1 y T = 1 , para lo que utilizaremos la instrucción c2d() de MATLAB. Obtendremos sus curvas de respuesta en frecuen-cia, para ello utilizaremos de nuevo la instrucción bode() de MATLAB, tal como se muestra a continuación
num=; den=[1 1 5]; sys=tf(num,den); sysd1=c2d(sys,0.1); sysd2=c2d(sys,1); bode(sys,sysd1,sysd2);
y en la figura 2.15 se puede observar las curvas correspondientes al caso de tiempo continuo y a las que se obtienen al muestrear con T = 0 .1 y T = 1 segundos respectivamente. 74 2. Técnicas clásicas de modelado de sistemas
2.4.2. Espectro de una señal muestrada
Consideremos que muestreamos la función f (t) cada T segundos. La función muestreada se podrá expresar como
f ∗(t) = f (kT ), con k = . . . , −2, −1, 0, 1, 2, . . . (2.4.16)
Fig. 2.16: Señales en el dominio temporal y su correspondiente magnitud espectral
Esta función muestreada también se puede expresar como el producto de la función continua y un tren de impulsos
f ∗(t) = f (t)
∞
∑
k=−∞
δ(t − kT ) (2.4.17) Como el tren de impulsos es una función periódica, se puede representar como una serie de Fourier ∞∑
k=−∞
δ(t − kT ) = 1
T
∞
∑
k=−∞
ej(2 πn/T )t (2.4.18) La transformada de Fourier de la señal muestreada es
F ∗(ω) =
∫ ∞−∞
(f (t) 1
T
∞
∑
n=−∞
ej(2 πn/T )t)e−jωt dt (2.4.19) 2.4. Modelos frecuenciales 75
Esta expresión se puede reescribir como
F ∗(ω) = 1
T
∞
∑
n=−∞
∫ ∞−∞
f (t)e−j(ω−2πn/T )tdt (2.4.20) Como la transformada de Fourier de la señal continua es
F (ω) =
∫ ∞−∞
f (t)e−jωt dt (2.4.21) la relación entre las transformadas de Fourier de la señal continua y muestreada es
F ∗(ω) = 1
T
∞
∑
n=−∞
F (ω − 2πn/T ) (2.4.22)
2.4.3. Teorema de muestreo
Cuando una señal continua es muestreada, excepto en el caso de una recons-trucción ideal, parte de la información se pierde. El objetivo, cuando se muestrea, es muestrear a una frecuencia tal que la señal de interés esté bien caracterizada y que la cantidad de información que se pierde sea mínima. La señal continua puede ser, en teoría, reconstruida a partir de la muestreada La salida de un filtro pasa-bajos con un ancho de banda apropiado nos devuelve el espectro continuo original. Según se va disminuyendo la frecuencia de muestreo, la separación entre los sucesivos ciclos del espectro muestreado decrece. Si 2π/T <
2ωm entonces se produce el efecto de aliasing como se puede ver en la figura 2.17 En este caso, no se puede reconstruir la señal original con un filtro paso-bajo. Este nos lleva a que hay un límite a partir del cual no se puede reconstruir la señal:
2πT > 2ωm (2.4.23) donde ωm es la mayor frecuencia de interés en la señal original. Esta ecuación se denomina teorema de muestreo, de Nyquist o de Shannon .76 2. Técnicas clásicas de modelado de sistemas F( ω)
ωs
ωs
ωm
ωm
ωm
ωs> 2 ωm
ωs< 2 ωm
F( ω)F( ω)
Fig. 2.17: Efecto de reducir la frecuencia de muestreo en F ∗(ω).
2.5. Modelos estocásticos
Un modelo genérico de sistema es el formado por un modelo determinista que se quiere controlar sujeto a un cierto comportamiento estocástico en su salida tal y como se muestra en la figura 2.18. Parte determinística U(z) G(z) H(z) E(z) (ruido blanco) V(z) Y(z)
Fig. 2.18: Modelo estocástico del sistema
Hasta el momento nos hemos concentrado en modelos para sistemas lineales de coeficientes constantes deterministas. Sin embargo hay una gran variedad de procesos estocásticos cuya respuesta se modela mejor mediante una combinación de un sistema lineal y un ruido blanco. 2.5. Modelos estocásticos 77
Este tipo de ruidos o perturbaciones es usual en los sistemas reales, donde bajo este término se incluyen una enorme cantidad de efectos físicos existentes y que no pueden ser modelados completamente (no uniformidad de materiales, pequeñas variaciones de parámetros físicos, desconocimiento de otros, etc). Desde un punto de vista práctico nos vamos a concentrar en los modelos es-tocásticos en tiempo discreto que son los que tendrán una mayor utilidad práctica desde el punto de vista de control.
2.5.1. Modelos estocásticos en tiempo discreto
Supongamos que e(n), n = . . . , −1, 0, 1, . . . es un proceso del tipo ruido blan-co discreto. Un grupo de procesos estocásticos muy genérico responde a unos modelos que suelen formularse según unas formas bastante estándar. El ruido o perturbación
v(n) en el sistema podría formularse de la siguiente manera
v(n) = e(n) + h1e(n − 1) + h2e(n − 2) + . . . + hne(n − k)
es decir responde a un modelo de la forma
V (z) = H(z)E(z)
con
H(z) =
∞
∑
i=0
h(i)z−i
De forma similar teníamos que para modelar el sistema lineal de parámetros constantes
y(k) + an−1y(k − 1) + . . . + a0y(k − n) =
bmu(k) + bm−1u(k − 1) + . . . + b0u(k − m) (2.5.1) La ecuación 2.5.1 puede ser expresada del siguiente modo
y(k) = −an−1y(k − 1) − . . . − a0y(k − n)+bmu(k) + bm−1u(k − 1) + . . . + b0u(k − m) (2.5.2) que en términos de la transformada z quedaría de la siguiente forma
Y (z) = G(z)U (z) = A(z)
B(z) U (z)78 2. Técnicas clásicas de modelado de sistemas
con lo que un modelo genérico de proceso estocástico es
Y (z) = G(z)U (z) + H(z)E(z) (2.5.3) Un modelo lineal e invariante en el tiempo está completamente especificado por la función de transferencia pero no ocurre lo mismo en un proceso estocástico, donde para que este completamente especificado se requiere : 1. Su respuesta impulsional {g(n)}∞.2. El espectro Svv (ω) = λ|H(ejω )|2 de la perturbación aditiva, y ,si es posible, la función de densidad de probabilidad de la perturbación e(t) (fe(x)). Un modelo completo, por tanto, vendrá caracterizado por
Y (z) = G(z)U (z) + H(z)E(z) (2.5.4)
fe(x) (2.5.5) con
G(z) =
∞
∑
k=0
g(k)z−k (2.5.6)
H(z) =
∞
∑
k=0
h(k)z−k (2.5.7) Un modelo particular corresponde a la especificación de estas tres funciones
G, H y fe. En la mayoría de los casos resulta poco práctico especificar por enume-ración las secuencias infinitas {g(k)}, {h(k)} junto con fe(x). En vez de lo cual, se toman estructuras de G y H que permitan la especificación en función de un número finito de valores numéricos. Los casos más típicos son las funciones de transferencia racionales. También es frecuente que la función de densidad de probabilidad fe no se espe-cifique como una función, sino que se describe por algunos parámetros caracterís-ticos, los más típicos son el primer y segundo momentos. En el caso de que e(t) sea Gaussiana, la función de densidad de probabilidad queda totalmente especificada por estos dos momentos. 2.5. Modelos estocásticos 79
Bibliografía
E.A. Puente, Regulación Automática I, Sección de Publicaciones ETSIIM ,1980. K. Ogata, Discrete-Time Control Systems, Prentice Hall International Edi-tion , 1995. B. Kuo, Automatic Control Systems, Prentice Hall , 1996. R. Isermann, Digital Control Systems, Springer Verlag , 1989. K.J. Astrˆm, Computer Controlled Systems, Prentice Hall , 1988. 80 2. Técnicas clásicas de modelado de sistemas 3. MODELADO Y ANáLISIS DE SISTEMAS EN EL ESPACIO DE ESTADOS
3.1. Introducción
Los métodos de análisis de sistemas basados en la descripción del sistema por la relación entre su entrada y su salida son de una gran utilidad para tratar sistemas dinámicos con una única entrada y una única salida, ya que son conceptualmente simples y requieren por lo general un número pequeño de cálculos. Estos métodos tienen dificultades para su utilización en sistemas no lineales, excepto en algunos casos muy sencillos, en sistemas con múltiples entradas y salidas, o en sistemas que presentan parámetros que varían con el tiempo. Los métodos de análisis y modelado de sistemas basados en el espacio de esta-dos son los más adecuados para el tratamiento de problemas con entradas y salidas múltiples, así como para el tratamiento de problemas con parámetros variantes en el tiempo e incluso de sistemas no lineales. Las técnicas de modelado de sistemas en el espacio de estados se basan en describir al sistema dinámico por medio de n ecuaciones diferenciales de primer orden para el caso de sistemas descritos en tiempo continuo o por medio de n
ecuaciones en diferencias para el caso de sistemas en tiempo discreto. Estas n
ecuaciones resultantes se describen por medio de una notación matricial, lo que simplifica la representación matemática de las mismas. La idea de estado es un concepto básico en la representación de sistemas en el espacio de estados. Si tratamos de describir intuitivamente lo que significa un estado en un sistema, hemos de recordar que estamos tratando con sistemas de tipo dinámico. Un sistema dinámico genérico se caracteriza porque la salida que da en un cierto instante de tiempo depende de las entradas al sistema en dicho instante y de la situación en la que estaba el sistema como consecuencia de las entradas a las que había sido sometido el sistema anteriormente. La situación en la que se encuentra un sistema en un instante dado puede ser caracterizada por un cierto conjunto de valores numéricos, y con este conjunto de valores numéricos y con el valor de la entrada en el instante actual podemos de-terminar el valor que tomará la salida del sistema en el instante siguiente. A este conjunto de valores numéricos que caracterizan la situación del sistema en un ins-tante dado es lo que denominamos estado del sistema en ese instante de tiempo. 82 3. Modelado y análisis de sistemas en el espacio de estados ue u sRCi
Fig. 3.1: Red RC
Podríamos dar una primera definición, un tanto informal, del concepto de estado.
Definición:
El estado de un sistema dinámico en un instante de tiempo t0 se puede definir como el conjunto más pequeño de valores numéricos que es suficiente para de-terminar la evolución futura del sistema para todo t ≥ t0 conocidos dichos valores numéricos y las entradas al sistema para todo t ≥ t0.
Esta definición, intuitivamente clara, require una formalización mayor, pues no se ha considerado en la misma conceptos tales como unicidad, continuidad y causalidad necesarios para una formalización consistente del concepto de estado. Esta formalización la veremos posteriormente. Otra cuestión importante es: ¿como podemos describir el estado de un cierto sistema físico? Es decir, ¿qué variables del sistema nos permiten caracterizar el estado del mismo? Una primera aproximación al problema consiste en determinar qué variables del sistema almacenan energía , ya que la energía de un sistema suele ser un buen caracterizador del estado del mismo. Veamos un ejemplo.
Ejemplo 3.1 Supongamos el sistema mostrado en la figura 3.1 que está consti-tuido por una resistencia y un condensador.
Solución:
La ecuación diferencial que caracteriza al sistema es la siguiente,
dus
dt = ( ue(t) − us(t)) 1
RC
si tomamos transformadas de Laplace en esta ecuación obtenemos
sU s(s) − us(0) = 1
RC [Ue(s) − Us(s)]
y operando sobre esta expresión obtenemos que
Us(s) = 1
1
RC
s us(0) +
1
RC
1
RC
s Ue(s)3.2. Concepto de estado de un sistema 83
En este sistema se puede observar, que la tensión de salida en un cierto instante de tiempo depende por una parte del valor inicial de la tensión en condensador,
us(0) , y por otra parte de la entrada de control que se le introduce al sistema. Resulta evidente que la variable que mantiene la memoria del estado energético del mismo es la tensión en el condensador us, por tanto, en este sistema el nivel de carga del condensador será la variable que caracteriza el estado del sistema. Si conocemos su valor en un cierto instante de tiempo y la entrada de tensión ue(t)
podemos conocer la salida del sistema. Si tuviésemos como entrada un escalón unitario, la salida del sistema haciendo la antitransformada de Laplace sería,
us(t) = us(0) e −1
RC t
(1 − e −1
RC t
)
3.2. Concepto de estado de un sistema
Aunque el espacio de estados ha sido extensamente utilizado en la descrip-ción de sistemas dinámicos, resulta bastante difícil formular una definición formal satisfactoria de los conceptos de estado y variable de estado. En este sentido la de-finición debida a L.A. Zadeh es una de las más satisfactorias a la hora de expresar las propiedades de ambos conceptos. Consideremos el sistema S, es decir el modelo matemático de un sistema real que tiene como entradas las señales ui(t), i = 1 , 2, . . . , r y como salidas las señales
yj (t), j = 1 , 2, . . . , m , las cuales son función del tiempo. Al conjunto de señales de entrada lo denotaremos por el vector u(t) de dimensiones r × 1 y al de las variables de salida por y(t), de dimensión m × 1.La variable x(t) (ver figura 3.2) puede considerarse como variable de estado del sistema S en el instante t si satisface las siguientes condiciones de consistencia: 1. Los vectores x(t0) y u(t, t 0) determinan de forma unívoca la salida y(t, t 0)
para todos los estados iniciales x(t0) ∈ X, para todos los valores de t ≥ t0
y todo el espacio de las funciones de entrada u de S. Es decir
y(t, t 0) = g(x(t0), u(t, t 0)) (3.2.1) Esta condición implica que el sistema es determinista y causal, por lo que los valores de las señales de salida en un instante de tiempo dado no dependen de salidas posteriores al mismo. 2. Si t1 es un instante de tiempo entre t0 y t, entonces para cada vector de entrada u(t, t 0) y vector de salida que se observa y(t, t 0), considerados desde el instante t1 de forma que dan los segmentos u(t, t 1) e y(t, t 1),existe un subconjunto no vacío de elementos del espacio X, denotado por
S[ x(t0); u(t, t 0)] , cuyos elementos α satisfacen la siguiente relación 84 3. Modelado y análisis de sistemas en el espacio de estados
y(t, t 1) = g(α, u(t, t 0)) (3.2.2) Esta condición asegura que a cada par u(t, t 1), y(t, t 1) le corresponde un estado inicial x(t1) en X.3. Si u(t1, t 0) es invariable y u(t, t 1) varía sobre todas las entradas del espa-cio de funciones de entrada de S, entonces la intersección de los conjuntos
S[ x(t0); u(t, t 0)] , considerados sobre todos los valores de u(t, t 1) es un con-junto no vacío. Esta condición asegura la existencia de al menos un estado en el espacio
X, la cual se refiere a todos los posibles pares de entradas u(t, t 1) y salidas
y(t, t 1), respectivamente. De las condiciones 2 y 3 se tiene que el estado de S en el instante t está deter-minado por el estado inicial x(t0) y el vector de entrada u(t, t 0):
x(t) = f (x(t0), u(t, t 0)) (3.2.3) donde f () es una función unívoca de sus argumentos. x1x(t 0)x(t 1)x2x3
Fig. 3.2: Concepto de estado
3.2.1. Representación matricial de las ecuaciones de estado
Si, tal y como se ha indicado anteriormente, un sistema puede describirse por un conjunto de ecuaciones diferenciales o de ecuaciones en diferencias de primer orden, las n ecuaciones de estado de un sistema dinámico de orden n-ésimo serán:
˙x1(t) = f1(x1(t), . . . , x n(t), u 1(t), . . . , u r(t)) ˙x2(t) = f2(x1(t), . . . , x n(t), u 1(t), . . . , u r(t))
...
˙xn(t) = fn(x1(t), . . . , x n(t), u 1(t), . . . , u r(t))
(3.2.4) 3.2. Concepto de estado de un sistema 85
y las ecuaciones de salida del sistema serán las siguientes
y1(t) = g1(x1(t), . . . , x n(t), u 1(t), . . . , u r(t))
y2(t) = g2(x1(t), . . . , x n(t), u 1(t), . . . , u r(t))
...
ym(t) = gm(x1(t), . . . , x n(t), u 1(t), . . . , u r(t))
(3.2.5) Si expresamos los sistemas de ecuaciones anteriores de forma matricial, donde
x(t) será un vector n × 1, u(t) será un vector r × 1 e y(t) será un vector m × 1,tendremos que
˙x(t) = f (x(t), u(t)) (3.2.6)
y(t) = g(x(t), u(t)) (3.2.7) En el caso de que el sistema sea lineal entonces las ecuaciones de estado podrán escribirse de la forma:
˙x(t) = A(t)x(t) + B(t)u(t) (3.2.8)
y(t) = C(t)x(t) + D(t)u(t) (3.2.9) donde las matrices A(t), B(t), C(t) y D(t) son, en el caso general, matrices variantes en el tiempo de dimensiones A(t) : n × n, B(t) : n × r, C(t) : m × n
y D(t) : m × r.BACDu(t) x(t) y(t) x(t) .++ ++ x
Fig. 3.3: Diagrama de bloques de un sistema en tiempo continuo representado en ecuaciones de estado
En el caso de que el sistema esté definido por ecuaciones en diferencias, las expresiones de las ecuaciones de estado y de salida serán
x(k + 1) = f (x(k), u(k)) (3.2.10)
y(k) = g(x(k), u(k)) (3.2.11) 86 3. Modelado y análisis de sistemas en el espacio de estados ue u sL1C1i1 i2L2C2
Fig. 3.4: Filtro Butterworth de 4 o orden
y en el caso de que el sistema sea lineal o se linealice en torno a algún punto de equilibrio tendremos que
x(n + 1) = G(k)x(k) + H(k)u(k) (3.2.12)
y(k + 1) = C(k)x(k) + D(k)u(k) (3.2.13)
Construcción del modelo de estado de un sistema físico
En términos generales dado un cierto sistema físico de tipo dinámico los pasos a seguir para formular un modelo de estado son los siguientes: 1. Descomponer el sistema en los elementos básicos de forma que podamos identificar sus componentes, variables, entradas, salidas y las relaciones entre ellas. 2. Elección de las variables de estado del sistema, que en una primera aproxi-mación se puede realizar observando los elementos dinámicos del sistema. Nótese que el número de variables de un sistema debe ser igual al orden de las ecuaciones diferenciales que lo describen. 3. Obtención de las ecuaciones de estado para cada variable independiente se-leccionada en el paso anterior. 4. Obtención de la ecuaciones de salida del sistema, a partir de las variables de salida que se desea controlar y de las variables de estado elegidas. Veamos algunos ejemplos de este proceso que se acaba de indicar.
Ejemplo 3.2.1: Supongamos un filtro de Butterworth de 4 o orden construido con elementos pasivos, es decir con bobinas y condensadores, tal y como se muestra en la figura 3.4, obtengamos su modelo de estado. 3.2. Concepto de estado de un sistema 87
Solución:
Las ecuaciones del sistema de acuerdo con las leyes de Kirchoff son:
ue = L1
di1
dt + uC1
uC1 = L2
di2
dt + uC2
C1
duC1
dt = i1 − i2
C2
duC2
dt = i2
us = uC2 (3.2.14) En este ejemplo la elección de las variables de estado es bastante clara, ya que las variables que acumulan energía en el sistema son las tensiones en los con-densadores y las intensidades en las bobinas. Tomando como variables de estado las tensiones en los condensadores y las intensidades que circulan en las bobinas tendremos que
di1
dt = 1
L1
ue − 1
L1
uC1
di2
dt = 1
L2
uC1 − 1
L2
uC2
duC1
dt = 1
C1
i1 − 1
C1
i2
duC2
dt = 1
C2
i2 (3.2.15) es decir
ddt
i1
i2
uC1
uC2
=
0 0 1
L1
00 0 1
L2
1
L2
1
C1
1
C1
0 00 1
C2
0 0
i1
i2
uC1
uC2
+
1
L1
000
ue (3.2.16) y la ecuación de salida sería
us = [0 0 0 1]
i1
i2
uC1
uC2
(3.2.17) 88 3. Modelado y análisis de sistemas en el espacio de estados
Ejemplo 3.2.2: Un modelo simple de operación de un Banco sería el siguiente: El Banco aumenta el dinero en efectivo del que dispone por medio de las imposiciones de los clientes en sus cuentas y por medio de los rendimientos que el Banco obtiene de las inversiones que realiza. El Banco disminuye el dinero efectivo del que dispone por medio de las retiradas de fondos de las cuentas de los clientes, por el pago de los salarios a los empleados del Banco y los diferentes suministros necesarios, y por las inversiones que realiza el Banco. El Banco paga un interés r1 fijo por el dinero que los clientes tienen en sus cuentas, y lo abona mensualmente. A su vez obtiene una rentabilidad r2 por sus inversiones, que le es abonada mensualmente. La política del Banco consiste en invertir cada mes el α % del dinero en efectivo que queda en la caja.
Solución:
Denominemos Rc(k) al total del dinero que los clientes tienen en sus cuentas,
Rb(k) al dinero que el banco tiene invertido, Cb(k) el capital del banco al final del mes, Dc(k) a los nuevos depósitos que realizan los clientes en el mes k-ésimo,
Gc(k) a los gastos o devoluciones de dinero a los clientes de sus cuentas en dicho mes, Gb(k) a los gastos necesarios para el funcionamiento del banco (salarios de los empleados y los diferentes suministros de material) y E(k) al dinero en efectivo en la caja del Banco. El saldo total de las cuentas de los clientes en el mes (k+1) es igual al que tenían en el mes k, más las nuevas imposiciones, menos las retiradas y más los intereses que la Caja les paga por su saldo en ese mes. Es decir:
Rc(k+1) = Rc(k)+ r1Rc(k)+ Dc(k)−Gc(k) = (1 −r1)Rc(k)+ Dc(k)−Gc(k)
El dinero invertido por el banco en el mes k + 1 es el dinero invertido en el mes anterior, más el porcentaje del dinero en efectivo en la caja a final de mes que es invertido.
Rb(k + 1) = Rb(k) + αE (k)
El capital del banco aumenta o disminuye cada mes en una cantidad igual al beneficio que obtiene del total de inversiones menos los gastos del banco 3.2. Concepto de estado de un sistema 89
para obtenerlos (lo que les paga a los clientes de intereses por las cuentas más lo que paga a empleados y suministradores):
Cb(k + 1) = Cb(k) + r2Rb(k) − r1Rc(k) − Gb(k)
El saldo total de las cuentas de los clientes, más el capital del banco debe ser igual al dinero que tiene el banco en sus inversiones más el dinero efectivo en la caja.
E(k) + Rb(k) = Rc(k) + Cb(k)
Tomando como variables de estado Rc(k), R b(k), C b(k) obtenemos que
Rc(k + 1)
Rb(k + 1)
Cb(k + 1)
=
1 + r1 0 0
α 1 − α α
−r1 r2 1
Rc(k)
Rb(k)
Cb(k)
+
1 −1 00 0 00 0 −1
Dc(k)
Gc(k)
Gb(k)
(3.2.18) La salida del sistema en este caso podría ser el propio vector de estado.
Linealización de sistemas en el espacio de estados
Una gran parte de los sistemas físicos que nos encontramos nos conducen a ecuaciones de estado no lineales. Estos tipos de ecuaciones presentan dificultades prácticas para su resolución por lo que cuando resulta posible se trata de obtener aproximaciones lineales a un cierto funcionamiento no lineal en el entorno de un cierto punto de operación. Supuesto un cierto punto de operación del sistema ( x =
x0, u = u0), una aproximación lineal al sistema en el entorno de ese punto de operación se obtiene a partir del desarrollo en serie de Taylor. El desarrollo en serie de Taylor para un sistema multivariable definido por el sistema de ecuaciones
˙x(t) = f (x(t), u(t)) (3.2.19)
y(t) = g(x(t), u(t)) (3.2.20) en el punto de operación ( x = x0, u = u0), se puede expresar como:
ddt ˜x(t) = A ˜x(t) + B ˜u(t) (3.2.21)
˜y(t) = C ˜x(t) + D ˜u(t) (3.2.22) 90 3. Modelado y análisis de sistemas en el espacio de estados
donde ˜x = x − x0, ˜u = u − u0, ˜y = y − y0 y las matrices A, B, C y D son las Jacobianas de f y g para x y u definidas por las siguientes expresiones:
A = ∇f x|x0,u0 (3.2.23)
B = ∇f u|x0,u0 (3.2.24)
C = ∇gx|x0,u0 (3.2.25)
D = ∇gu|x0,u0 (3.2.26) y recordando la expresión de la matriz Jacobiana
∇f x =
∂f 1
∂x 1
. . . ∂f 1
∂x n
... ...
∂f n
∂x 1
. . . ∂f n
∂x n
(3.2.27) podemos obtener las expresiones de A, B, C y D.
Ejemplo 3.4 Un modelo de estado simplificado para un horno de secado viene dado por el siguiente conjunto de ecuaciones de estado:
˙x1 = −a1x2
˙x2 = b1x1 − b2x2 − b3u1
˙x3 = c1x2 + c2u2(T2 − x3) + c3x2x3 (3.2.28) Linealizar el sistema.
Solución:
Sustituyendo las ecuaciones 3.2.28 en las expresiones de las Jacobianas tendremos que
A = ∇f x|x0,u0
=
∂f 1
∂x 1
∂f 1
∂x 2
∂f 1
∂x 3
∂f 2
∂x 1
∂f 2
∂x 2
∂f 2
∂x 3
∂f 3
∂x 1
∂f 3
∂x 2
∂f 3
∂x 3
=
0 −a1 0
b1 −b2 00 (c1 + c3x3,0) (c3x2,0 − c2u2,0)
(3.2.29) y3.2. Concepto de estado de un sistema 91
B = ∇f u|x0,u0
=
∂f 1
∂u 1
∂f 1
∂u 2
∂f 2
∂u 1
∂f 2
∂xu 2
∂f 3
∂u 1
∂f 3
∂u 2
=
0 0
−b1 00 c2(T2 − x3,0)
(3.2.30)
3.2.2. Función de transferencia y representación en el espacio de estados
Tanto la representación en el espacio de estados como la función de transferen-cia de un sistema son dos formas de modelar matemáticamente el comportamiento de un sistema. Resulta evidente que si un sistema puede ser representado mediante dos modelos matemáticos diferentes, existirá alguna relación entre ambos. Supon-gamos que hemos obtenido la representación de estado de un sistema, y que este sistema es lineal e invariante en el tiempo
ddt x(t) = Ax (t) + Bu (t) (3.2.31)
y(t) = Cx (t) + Du (t) (3.2.32) donde las matrices A, B, C y D tienen dimensiones A : n × n, B : n × r,
C : m × n y D : m × r.Si hacemos la transformada de Laplace de las ecuaciones de la representación de estado tenemos que
sX(s) − x(0) = AX (s) + BU (s) (3.2.33)
Y (s) = CX (s) + DU (s) (3.2.34) La función de transferencia nos da la relación que existe entre la entrada y la salida de un sistema. Hemos de relacionar por lo tanto la entrada y la salida del sistema, para ello operamos
sX(s) − AX (s) = BU (s) + x(0) (sI − A)X(s) = BU (s) + x(0)
X(s) = (sI − A)−1BU (s) + ( sI − A)−1x(0) (3.2.35) 92 3. Modelado y análisis de sistemas en el espacio de estados
y sustituyendo en la ecuación de salida del sistema obtenemos
Y (s) = C(sI − A)−1BU (s) + C(sI − A)−1x(0) + DU (s)= [C(sI − A)−1B + D]U (s) + ( sI − A)−1x(0) (3.2.36) En esta expresión se puede observar que la salida del sistema es función por una parte de la entrada de control al mismo y por otra de las condiciones iniciales del sistema. Si suponemos que las condiciones iniciales son nulas, la función de transferencia del sistema será
F (s) = Y (s)
U (s) = C(sI − A)−1B + D (3.2.37) Como
(sI − A)−1 = 1
|sI − A| Adj (sI − A)T
se tiene que el polinomio característico del sistema es |sI − A|, y por tanto, los polos del sistema coinciden con los autovalores de la matriz A.
3.3. Representación de sistemas en el espacio de estados
Tal como se ha indicado en la sección anterior la representación en el espacio de estados de un sistema corresponde a una descripción del mismo por medio de un conjunto de ecuaciones diferenciales o en diferencias de primer orden. Normal-mente no se dispone de las ecuaciones del sistema de forma que pueda obtenerse el modelo en el espacio de estados de forma directa, sino que se tiene un conjunto de situaciones muy variable donde nos podemos encontrar un sistema descrito por una o varias ecuaciones diferenciales, de primer orden o de orden superior, lineales o no lineales. La matriz D en la mayoría de los sistemas no existe, dado que representa un reflejo de la entrada en la salida de forma directa. En el caso de que este efecto se produjera, y dado que la matriz D es de coeficientes constantes, se puede sumar el efecto a la salida. Por ello es necesario desarrollar algún método de conversión de ecuaciones diferenciales ordinarias o de ecuaciones en diferencias a ecuaciones diferenciales o en diferencias de primer orden. Estos métodos los veremos en las secciones si-guientes.
3.3.1. Conversión de una ecuación diferencial ordinaria a ecuaciones de estado
Para un cierto sistema dinámico, la elección de las variables de estado no es única, por lo que no basta con mostrar una única solución puesto que en ocasiones 3.3. Representación de sistemas en el espacio de estados 93
una elección determinada de dichas variables de estado tiene ciertas ventajas frente a otra. Veamos a continuación algunos de los métodos más usuales de elección de las mismas y a que tipo de representaciones dan lugar. Estas representaciones se denominan formas canónicas , y entre ellas tenemos: la forma canónica controlable, la forma canónica observable y la forma canónica de Jordan.
Forma canónica controlable
En este método de representación de un cierto sistema la variable de estado
i-ésima se escoge como la derivada de la variable de estado i − 1. Esta elección es bastante natural en una gran cantidad de sistemas, por ejemplo en los sistemas mecánicos, tanto para el movimiento lineal como para el angular, tenemos que la aceleración es la derivada de la velocidad y la velocidad la derivada de la posición. También se les denomina variables de fase .Supongamos un sistema invariante en el tiempo, con una única señal de entrada
u y una única señal de salida y, cuya dinámica viene determinada por una ecuación diferencial lineal de coeficientes constante
an
dn
dt n y(t) + . . . + a1
ddt y(t) + a0y(t) = bm
dm
dt m u(t) + . . . + b1
ddt u(t) + b0u(t)
(3.3.1) con n ≥ m. Si utilizamos el operador p = ddt la expresión anterior podemos reescribirla de la siguiente forma
anpny(t) + . . . + a1py (t) + a0y(t) = bmpmu(t) + . . . + b1pu (t) + b0u(t) (3.3.2) o bien, si sacamos y(t) y u(t) como factor común en ambos lados de la expresión
(anpn + . . . + a1p + a0)y(t) = ( bmpm + . . . + b1p + b0)u(t) (3.3.3) que podemos expresar de forma más compacta como un cociente de polinomios de la forma
M (p)y(t) = N (p)u(t) (3.3.4) donde
M (p) = anpn + . . . + a1p + a0 (3.3.5)
N (p) = bmpm + . . . + b1p + b0 (3.3.6) De la ecuación 3.3.4 se deduce que
y(t) = N (p)
M (p) u(t) (3.3.7) Si definimos la primera de las variables de estado de forma que
x(t) = 1
M (p) u(t) (3.3.8) 94 3. Modelado y análisis de sistemas en el espacio de estados
se puede observar que la ecuación 3.3.7 ha sido reemplazada por las dos siguientes expresiones
x(t) = 1
M (p) u(t) (3.3.9)
y(t) = N (p)x(t) (3.3.10) Tomando las variables de estado de la siguiente forma
x(t) = x1(t)ddt x(t) = x2(t) = ddt x1(t)
...
dn−1
dtn−1 x(t) = xn(t) = ddt xn−1(t) (3.3.11) donde se ve claramente que se toman como variables de estado las sucesivas deri-vadas de x(t), la ecuación 3.3.9 se puede reescribir de la siguiente forma
dn
dtn x(t) = 1
an
u(t) − an−1
an
xn(t) − . . . − a1
an
x2(t) − a0
an
x1(t) (3.3.12) Las ecuaciones 3.3.11 y 3.3.12 podemos expresarlas de forma matricial como
˙x1(t)˙x2(t)
...
˙xn−1(t)˙xn(t)
=
0 1 0 . . . 00 0 1 . . . 0
... ...
0 0 0 . . . 1
− a0
an
− a1
an
− a2
an
. . . − an−1
an
x1(t)
x2(t)
...
xn−1(t)
xn(t)
+
00
...
0
1
an
u(t)
(3.3.13) Para obtener la ecuación de salida partimos de la ecuación 3.3.10, utilizando la definición de las variables de estado 3.3.11 y suponiendo que n = m obtenemos que
b0x1(t) + b1x2(t) + . . . + bn−1xn(t) + bn
ddt xn(t) = y(t) (3.3.14) y sustituyendo en esta ecuación ddt xn por 3.3.12 nos queda la siguiente expresión
y(t) = ( b0 − bn
a0
an
)x1(t) + ( b1 − bn
a1
an
)x2(t) + . . . . . . + ( bn−1 − bn
an−1
an
)xn(t) + bn
an
u(t) (3.3.15) 3.3. Representación de sistemas en el espacio de estados 95 x
b0
u(t) nxy(t) an
+++ x 1xn-1 x 2x
x
b1 b2 bn
a2
a1
+-+- --+
Fig. 3.5: Representación gráfica de la forma canónica controlable
que formulada en términos matriciales da lugar a
y(t) =
[
b0 − bn a0
an
b1 − bn a1
an
. . . bn−1 − bn an−1
an
]
x1(t)
x2(t)
...
xn(t)
bn
an
u(t)
(3.3.16) Esta representación se denomina forma canónica controlable o también forma canónica en variables de fase . Una simplificación bastante usual se obtiene cuando
an = 1 . En la figura 3.5 se muestra un esquema gráfico de esta forma de representar un sistema en el espacio de estados.
Forma canónica observable
En esta forma canónica de representación las variables de estado son una com-binación lineal de las derivadas de las variables de entrada y de salida. Consideremos un sistema dinámico descrito por una ecuación diferencial de orden n, como en el caso anterior, y cuya ecuación es la mostrada en 3.3.1. Su-pongamos el caso donde n = m y definamos las variables de estado de forma que
xn es función de y(t) y u(t), xn−1 es función de y(t), u(t) y sus derivadas y así sucesivamente, tal y como se muestra a continuación 96 3. Modelado y análisis de sistemas en el espacio de estados
xn(t) = any(t) − bnu(t)
xn−1(t) = an−1y(t) + an
ddt y(t) − bn−1u(t) − bn
ddt u(t)
xn−2(t) = an−2y(t) + an−1
ddt y(t) + an−2
d2
dt2 y(t) −−bn−2u(t) − bn−1
ddt u(t) − bn−2
d2
dt2 u(t)
...
x1(t) = a1y(t) + a2
ddt y(t) + . . . + an
dn−1
dtn−1 y(t) −−b1u(t) − b2
ddt u(t) − . . . − bn
dn−1
dtn−1 u(t) (3.3.17) De la primera ecuación del conjunto de ecuaciones 3.3.17 podemos calcular
y(t), la ecuación de salida del sistema, que da como resultado la siguiente expre-sión
y(t) = 1
an
xn(t) + bn
an
u(t) (3.3.18) Si derivamos las ecuaciones del sistema de ecuaciones 3.3.17 obtenemos
ddt xn(t) = an
d
dt y(t) − bn
ddt u(t)ddt xn−1(t) = an−1
ddt y(t) + an
d2
dt2 y(t) − bn−1
ddt u(t) − bn
d2
dt2 u(t)
...
ddt x2(t) = a2
ddt y(t) + a3
d2
dt2 y(t) + . . . + an
dn−1
dtn−1 y(t) −−b2
ddt u(t) − b3
d2
dt2 u(t) − . . . − bn
dn−1
dtn−1 u(t)ddt x1(t) = a1
ddt y(t) + a2
d2
dt 2 y(t) + . . . + anyn)(t) −−b1
ddt u(t) − b2
d2
dt 2 u(t) − . . . − bn
dn
dt n u(t) (3.3.19) sustituyendo las ecuaciones 3.3.19 en el sistema de ecuaciones 3.3.17 obtenemos 3.3. Representación de sistemas en el espacio de estados 97
que
xn(t) = any(t) − bnu(t)
xn−1(t) = an−1y(t) − bn−1u(t) + ddt xn(t)
xn−2(t) = an−2y(t) − bn−2u(t) + ddt xn−1(t)
...
x1(t) = a1y(t) − b1u(t) + ddt x2(t) (3.3.20) y si aquí sustituimos y(t) por la ecuación de salida 3.3.18 tendremos, n − 1 ecua-ciones de estado, faltaría únicamente la ecuación de estado correspondiente a ˙x1(t)
xn(t) = an−1
an
xn(t) − (bn−1 − bn
an−1
an
)u(t) + ddt xn(t)
xn−1(t) = an−2
an
xn(t) − (bn−2 − bn
an−2
an
)u(t) + ddt xn−1(t)
...
x2(t) = a1
an
xn(t) − (b1 − bn
a1
an
)u(t) + ddt x2(t) (3.3.21) Para obtener la ecuación de estado que nos falta combinamos las ecuaciones 3.3.21 con la ecuación diferencial original expresada en términos de las variables de estado que hemos definido
0 = a0
an
xn(t) − (b0 − bn
a0
an
)u(t) + ddt x1(t) (3.3.22) Escribiendo en forma matricial las ecuaciones 3.3.21 y 3.3.22 obtenemos el conjunto de ecuaciones de estado y de la ecuación 3.3.18 la ecuación de salida, es decir
ddt
x1(t)
x2(t)
...
xn(t)
=
0 0 . . . 0 0 − a0
an
1 0 . . . 0 0 − a1
an
... ... ... ... ...
0 0 . . . 1 0 − an−2
an
0 0 . . . 0 1 − an−1
an
x1(t)
x2(t)
...
xn−1(t)
xn(t)
+
b0 − bn a0
an
b1 − bn a1
an
...
bn−2 − bn an−2
an
bn−1 − bn an−1
an
u(t)
(3.3.23) 98 3. Modelado y análisis de sistemas en el espacio de estados x
b -a b n n 0
u(t)
n
x y(t)
x
++++ x
1
x
n-1
x
2
xan-1 a1
++- --b -a bn-1 n-1 0 b -a b 1 1 0 b0
an
Fig. 3.6: Representación gráfica de la forma canónica observable
y(t) =
[
0 0 . . . 0 0 1
an
]
x1(t)
x2(t)
...
xn−1(t)
xn(t)
bn
an
u(t) (3.3.24) En la figura 3.6 se muestra un esquema gráfico de esta forma de representar un sistema en el espacio de estados.
Forma canónica de Jordan
Este método de elección de las variables de estado se basa en el principio de superposición de sistemas lineales, es decir se busca la descomposición de la res-puesta del sistema lineal en sus componentes individuales. Consideremos en este caso que la función de transferencia del sistema sea
G(s) = Y (s)
U (s) = bmsm + . . . + b1s + b0
ansn + . . . + a1s + a0
(3.3.25) donde m ≤ n.Supongamos en primer lugar que el denominador de dicha función de transfe-rencia no tiene raíces múltiples, con lo que descomponiendo en fracciones parciales 3.3. Representación de sistemas en el espacio de estados 99
se obtiene que
G(s) = bmsm + . . . + b1s + b0
an(s − λ1)( s − λ2) . . . (s − λn)= c1
s − λ1
c2
s − λ2
. . . + cn
s − λn
(3.3.26) en la ecuación 3.3.26 se puede observar que la salida del sistema corresponde a la combinación lineal de n componentes elementales más un término directamente relacionado con la entrada que aparece únicamente en el caso de que n = mY (s) = c1
s − λ1
U (s) + c2
s − λ2
U (s) + . . . + cn
s − λn
U (s) + bn
an
U (s) (3.3.27) Cada una de estas componentes elementales de la ecuación 3.3.27 es la res-puesta de un sistema con función de transferencia
Gi(s) = 1
s − λi
(3.3.28) a la señal de entrada U (s), lo que implica que si cada una de estas componentes básicas la tomamos como una variable de estado Xi(s) que satisface Xi(s) =
Gi(s)U (s), esto implica que la variable de estado xi(t) así definida satisface la siguiente ecuación diferencial
ddt xi(t) = λixi(t) + u(t) (3.3.29) Tenemos, por tanto, n ecuaciones diferenciales como la 3.3.29 que constituyen las ecuaciones de estado y una ecuación de salida que en el dominio del tiempo tiene la forma
y(t) = c1x1(t) + c2x2(t) + . . . + cnxn(t) + bn
an
u(t) (3.3.30) y formulando las ecuaciones 3.3.29 y 3.3.30 en forma matricial tendremos
ddt
x1(t)
x2(t)
...
xn(t)
=
λ1 0 . . . 00 λ2 . . . 0
... ... ...
0 0 . . . λn
x1(t)
x2(t)
...
xn(t)
+
11
...
1
u(t) (3.3.31) 100 3. Modelado y análisis de sistemas en el espacio de estados f+-
1xb1
c1
u(t) y(t) +++++b0
+-
nxbn
cnf
Fig. 3.7: Representación gráfica de la forma canónica de diagonal
y(t) = [ c1 c2 . . . c n]
x1(t)
x2(t)
...
xn(t)
bn
an
u(t) (3.3.32) Si en la forma canónica de Jordan todas las raíces del denominador son simples se la conoce con el nombre de forma canónica diagonal . En la figura 3.7 se muestra una representación gráfica de la forma canónica de diagonal de un sistema. En el caso de que el polinomio característico de la función de transferencia del sistema tenga raíces múltiples, la descomposición en fracciones parciales daría lugar a una expresión del siguiente tipo
G(s) = bmsm + . . . + b1s + b0
an(s − λ1)k(s − λ2) . . . (s − λn)= c1
(s − λ1)k + . . . + ck
s − λ1
ck+1
s − λ2
. . . + cn
s − λn
(3.3.33) Si suponemos que las condiciones iniciales son nulas la respuesta del sistema 3.3. Representación de sistemas en el espacio de estados 101
ante una entrada Y (s) será
Y (s) = c1
(s − λ1)k U (s) + . . . + ck
s − λ1
U (s) + ck+1
s − λ2
U (s) + . . . + cn
s − λn
U (s)
(3.3.34) y tomando como vectores de estado cada una de las fracciones que aparecen en la ecuación 3.3.34 tendremos
X1(s) = c1
(s − λ1)k U (s) = c1
s − λ1
X2(s)
X2(s) = c2
(s − λ1)k−1 U (s) = c1
s − λ1
X3(s)
...
Xk(s) = ck
s − λ1
U (s)
Xk+1 (s) = ck+1
s − λ2
U (s)
...
Xn(s) = cn
s − λn
U (s) (3.3.35) Reescribiendo las ecuaciones anteriores 3.3.35 en el dominio del tiempo nos quedará el sistema expresado como conjunto de ecuaciones diferenciales de primer orden
ddt x1(t) = λ1x1(t) + x2(t)
ddt x2(t) = λ1x2(t) + x3(t)
...
ddt xk(t) = λ1xk(t) + u(t)
ddt xk+1 (t) = λ2xk+1 (t) + u(t)
...
ddt xn(t) = λnxn(t) + u(t) (3.3.36) 102 3. Modelado y análisis de sistemas en el espacio de estados u(t) y(t) s+3 s+2 2s+1 1s
Fig. 3.8: Sistema del ejemplo 3.3.1
y puesto este conjunto de ecuaciones en forma matricial
ddt
x1(t)
x2(t)
x3(t)
...
xk(t)
xk+1 (t)
...
xn(t)
=
λ1 1 0 . . . 0 0 . . . 00 λ1 1 . . . 0 0 . . . 0
... ... ... ...
0 0 0 . . . λ1 0 . . . 00 0 0 . . . 0 λ2 . . . 0
... ... ... ...
0 0 0 . . . 0 0 . . . λn
x1(t)
x2(t)
x3(t)
...
xk(t)
xk+1 (t)
...
xn(t)
+
000
...
11
...
1
u(t)
(3.3.37)
y(t) = [ c1 c2 c3 . . . c k ck+1 . . . c n]
x1(t)
x2(t)
x3(t)
...
xk(t)
xk+1 (t)
...
xn(t)
bn
an
u(t) (3.3.38)
Ejemplo 3.3.1: Dado el sistema mostrado en la figura 3.8, obtengamos las for-mas canónicas de Jordan, controlable y observable.
Solución:
Forma canónica de Jordan. Obtenemos en primer lugar la función de trans-3.3. Representación de sistemas en el espacio de estados 103
ferencia global del sistema,
F (s) = 2( s + 3)
s(s + 1)( s + 2)
y a continuación la expandimos en fracciones parciales
F (s) = −4
s + 1 + 1
s + 2 + 3
s
y de aquí tenemos que tomando como variables de estado las fracciones par-ciales X1(s) = −4
s+1
U (s), X2(s) = 1
s+2
U (s) y X3(s) = 3
s
U (s) obtenemos la siguiente representación de estado
ddt
x1(t)
x2(t)
x3(t)
=
−1 0 00 −2 00 0 0
x1(t)
x2(t)
x3(t)
+
111
u(t)
y(t) = [ −4 1 3]
x1(t)
x2(t)
x3(t)
Forma canónica controlable . Partiendo de la función de transferencia F (s)
del sistema
F (s) = 2( s + 3)
s(s + 1)( s + 2) = 2s + 6
s3 + 3 s2 + 2 s
obtenemos que los coeficientes del polinomio del denominador son a3 =1, a 2 = 3 , a 1 = 2 , a 0 = 0 y los del numerador b1 = 2 , b 0 = 6 . Susti-tuyendo estos coeficientes en la expresión de la forma canónica controlable obtenemos
ddt
x1(t)
x2(t)
x3(t)
=
0 1 00 0 10 −2 −3
x1(t)
x2(t)
x3(t)
+
001
u(t)
y(t) = [6 2 0]
x1(t)
x2(t)
x3(t)
104 3. Modelado y análisis de sistemas en el espacio de estados
Forma canónica observable . Partiendo de los coeficientes de los polinomios del numerador y del denominador de la función de transferencia que hemos obtenido y sustituyendo en la expresión de la forma canónica observable se llega a
ddt
x1(t)
x2(t)
x3(t)
=
0 0 01 0 −20 1 −3
x1(t)
x2(t)
x3(t)
+
620
u(t)
y(t) =
[
0 0 1
]
x1(t)
x2(t)
x3(t)
3.3.2. Conversión de una ecuación en diferencias a ecuaciones de estado
La conversión de ecuaciones en diferencias a ecuaciones de estado es un proce-so similar al que hemos visto para sistemas expresados por ecuaciones diferencia-les, salvo que se trabaja a partir de la transformada z o bien a partir de la expresión en diferencias y utilizando el operador retardo para expresar en forma polinómica las ecuaciones de estado. Supongamos un sistema invariante en el tiempo, con una única señal de entrada
u y una única señal de salida y, cuya dinámica viene determinada por una ecuación en diferencias lineal de coeficientes constantes
any(k −n)+ . . . +a1y(k −1)+ a0y(k) = bmu(k −m)+ . . . +b1u(k −1)+ b0u(k)
(3.3.39) con n ≥ m. Si introducimos el operador retardo q−1 la expresión anterior podemos reescribirla de la siguiente forma
anq−ny(k)+ . . . +a1q−1y(k)+ a0y(k) = bmq−mu(k)+ . . . +b1q−1u(k)+ b0u(k)
(3.3.40) o bien, si sacamos y(t) y u(t) como factor común en ambos lados de la expresión
(anq−n + . . . + a1q−1 + a0)y(k) = ( bmq−m + . . . + b1q−1 + b0)u(k) (3.3.41) que podemos expresar de forma más compacta como un cociente de polinomios de la forma
y(k) = N (q)
M (q) u(k) (3.3.42) 3.3. Representación de sistemas en el espacio de estados 105
A partir de aquí es posible realizar un desarrollo similar al hecho para el caso de sistemas en tiempo continuo, y se llegaría a las mismas expresiones, por lo que no se repetirá el desarrollo.
Ejemplo 3.3.2: Dado un sistema un cuya función de transferencia es
F (z) = z2 + 0 .4z + 3
z3 + 1 .9z2 + 1 .08 z + 0 .18
determinemos las formas canónicas controlable, observable y de Jordan.
Solución:
Forma canónica controlable . Partiendo de la función de transferencia F (z)
del sistema obtenemos que los coeficientes del polinomio del denominador son a3 = 1 , a 2 = 1 .9, a 1 = 1 .08 , a 0 = 0 .18 y los del numerador b2 =1, b 1 = 0 .4, b 0 = 3 y de aquí
x1(k + 1)
x2(k + 1)
x3(k + 1)
=
0 1 00 0 1
−0.18 −1.08 −1.9
x1(k)
x2(k)
x3(k)
+
001
u(k)
y(k) = [2 .82 − 0.68 − 0.9]
x1(k)
x2(k)
x3(k)
Forma canónica observable . Partiendo de los coeficientes de los polinomios del numerador y del denominador de la función de transferencia se obtiene
x1(k + 1)
x2(k + 1)
x3(k + 1)
=
0 0 −0.18 1 0 −1.08 0 1 −1.9
x1(k)
x2(k)
x3(k)
+
2.82
−0.68
−0.9
u(k)
y(k) = [0 0 1]
x1(k)
x2(k)
x3(k)
106 3. Modelado y análisis de sistemas en el espacio de estados
Forma canónica de Jordan. Si obtenemos las raíces del denominador de la función de transferencia del sistema,
F (z) = z2 + 0 .4z + 3 (z + 1)( z + 0 .6)( z + 0 .3)
y a continuación la expandimos en fracciones parciales
F (z) = 12 .85
z + 1 + −26
z + 0 .6 + 14 .14
z + 0 .3
tomando como variables de estado las fracciones parciales X1(z) = 12 .85
z+1
U (z),
X2(z) = 26
z+0 .6
U (s) y X3(z) = 14 .14
z+0 .3
U (z) obtenemos la siguiente represen-tación de estado
x1(k + 1)
x2(k + 1)
x3(k + 1)
=
−1 0 00 −0.6 00 0 −0.3
x1(k)
x2(k)
x3(k)
+
111
u(k)
y(k) = [12 .86 − 26 14 .14]
x1(k)
x2(k)
x3(k)
3.3.3. Transformaciones entre representaciones
Consideremos un sistema definido por una ecuación de estado y de salida en tiempo discreto de la forma
x(k + 1) = Gx (k) + Hu (k)
y(k) = Cx (k) + Du (k) (3.3.43) veamos una serie de técnicas que nos permiten pasar de una representación de estado cualesquiera a alguna otra. Si para el sistema definido por las ecuaciones 3.3.43 definimos un nuevo vector de estado x′(k) de forma que
x(k) = T x ′(k) (3.3.44) donde T es una matriz no singular , es decir, que es invertible, y por tanto las nue-vas variables x′(k) se pueden expresar como combinación lineal de las x(k). Si sustituimos en el sistema de ecuaciones 3.3.43 las variables de estado por las re-cién definidas teniendo en cuenta la relación entre ellas dada por la transformación 3.3.44, obtenemos que
T x ′(k + 1) = GT x ′(k) + Hu (k) (3.3.45) 3.3. Representación de sistemas en el espacio de estados 107
y premultiplicando por la inversa de la matriz de transformación se tiene que
x′(k + 1) = T −1GT x ′(k) + T −1Hu (k) (3.3.46) con lo que si denominamos
G′ = T −1GT (3.3.47)
H′ = T −1H (3.3.48) nos queda que
x′(k + 1) = G′x′(k) + H′u(k) (3.3.49) Procediendo de forma similar para la ecuación de salida del sistema, tendremos que
y(k) = CT x ′(k) + Du (k) (3.3.50) y denominando
C′ = CT (3.3.51)
D′ = D (3.3.52) obtenemos una nueva representación de estado
x′(k + 1) = G′x′(k) + H′u(k) (3.3.53)
y(k) = C′x′(k) + D′u(k) (3.3.54) Dos representaciones de estado son similares si tienen el mismo polinomio característico, y por tanto los mismos valores propios o autovalores. Veamos que esto se verifica para el caso anterior,
|zI − G′| = |zI − T −1GT |
= |zT −1T − T −1GT |
= |T −1(zI − G)T |
= |T −1|| zI − G|| T |
= |zI − G| (3.3.55) Dicho de otra manera, la ecuación característica de la matriz y sus autovalores, no dependen de la base elegida para representar el sistema en el espacio de estados. Puesto que la matriz de transformación sólo requiere que esta sea no singular, existen infinitas representaciones de estado para un sistema dado. Si se desea que el sistema este representado en alguna forma canónica concreta, el problema que se plantea es determinar la matriz de transformación que nos permita pasar de la representación actual a la canónica que nos interese. 108 3. Modelado y análisis de sistemas en el espacio de estados
Transformación a la forma canónica controlable
Dado un sistema expresado en el espacio de estados de la forma
x(k + 1) = Gx (k) + Hu(k) (3.3.56)
y(k) = Cx (k) + Du(k) (3.3.57) puede ser transformado a la forma canónica controlable por medio de la siguiente matriz de transformación
T = M W (3.3.58) donde
M = [ H GH . . . Gn−1H] (3.3.59) y
W =
a1 a2 . . . an−1 1
a2 a3 . . . 1 0
... ... ... ...
an−1 1 . . . 0 01 0 . . . 0 0
(3.3.60) siendo los elementos ai de la matriz W los coeficientes de la ecuación caracterís-tica del sistema
|zI − G| = zn + an−1zn−1 + . . . + a1z + a0 = 0 (3.3.61)
Transformación a la forma canónica observable
Dado un sistema expresado en el espacio de estados de la forma
x(k + 1) = Gx (k) + Hu(k) (3.3.62)
y(k) = Cx (k) + Du(k) (3.3.63) puede ser transformado a la forma canónica observable por medio de la siguiente matriz de transformación
T = ( W N ∗)−1 (3.3.64) donde
N = [ C∗ G∗C∗ . . . (G∗)n−1C∗] (3.3.65) 3.3. Representación de sistemas en el espacio de estados 109
y
W =
an−1 an−2 . . . a1 1
an−2 an−3 . . . 1 0
... ... ... ...
a1 1 . . . 0 01 0 . . . 0 0
(3.3.66) siendo los elementos ai de la matriz W los coeficientes de la ecuación caracterís-tica del sistema indicados en la ecuación 3.3.61
Ejemplo 3.3.3: Sea un sistema un cuya representación de estado es
x1(k + 1)
x2(k + 1)
=
−1 10 −1
x1(k)
x2(k)
+
01
u(k)
y(k) = [1 2]
x1(k)
x2(k)
obtengamos las matrices de transformación T que nos permiten convertir esta representación en las formas canónicas controlable y observable.
Solución:
La ecuación característica de este sistema sistema viene dada por
|zI − G| = z2 + 2 z + 1 = 0
y por tanto los coeficientes del polinomio son a2 = 1 , a 1 = 2 , a 0 = 1 .
Forma canónica observable . La matriz de transformación para pasar esta forma canónica era T = ( W N ∗)−1. Como
N =
[
C∗ G∗C∗
]
=
1 −12 −1
y
W =
a1 11 0
=
2 11 0
con lo que la matriz de transformación será 110 3. Modelado y análisis de sistemas en el espacio de estados
T = ( W N ∗)−1 =
2 11 0
1 2
−1 −1
−1
=
−2 31 −1
T −1 =
1 31 −1
Las matrices G′, H′, C′ del sistema transformado son las siguientes
G′ = T −1GT =
1 31 2
−1 10 −1
−2 31 −1
=
0 −11 −2
H′ = T −1H =
1 31 2
01
=
32
C′ = CT =
[
1 2
] −2 31 −1
=
[
0 1
]
y el sistema una vez transformado quedará
x′(k + 1) = G′x′(k) + H′u(k) =
0 −11 −2
x′(k) +
32
u(k)
y(k) = C′x′(k) = [0 1] x′(k)
Forma canónica controlable . La matriz de transformación para pasar a la forma canónica controlable, era T = M W . Como
M =
[
H GH
]
=
0 11 −1
y
W =
a1 11 0
=
2 11 0
con lo que la matriz de transformación será
T = M W =
0 11 −1
2 11 0
=
1 01 1
3.3. Representación de sistemas en el espacio de estados 111
T −1 =
1 0
−1 1
Las matrices G′, H′, C′ del sistema transformado son las siguientes
G′ = T −1GT =
1 0
−1 1
−1 10 −1
1 01 1
=
0 1
−1 −2
H′ = T −1H =
1 0
−1 1
01
=
01
C′ = CT =
[
1 2
] 1 01 1
=
[
3 2
]
Transformación a la forma canónica de Jordan
En el caso de que la matriz G del sistema tenga valores propios λ1, λ 2, . . . , λ n
diferentes, la matriz G′ de la forma canónica de Jordan será una matriz diagonal, de la forma
G′ = T −1GT =
λ1 0 . . . 00 λ2 . . . 0
... ... ...
0 0 . . . λn
= Λ (3.3.67) La matriz de transformación T puede obtenerse si se tiene en cuenta que los vectores propios vi correspondientes a los valores propios λi cumplen la ecuación característica |λI − G| = 0 , y por tanto se tiene que:
Gv 1 = λ1v1
Gv 2 = λ2v2
... = ...
Gv n = λnvn (3.3.68) La matriz de transformación T , se obtiene a partir de la expresión 3.3.68, re-112 3. Modelado y análisis de sistemas en el espacio de estados
formulándola en forma matricial como:
G[v1 v2 . . . vn] = [ v1 v2 . . . vn]
λ1 0 . . . 00 λ2 . . . 0
... ... ...
0 0 . . . λn
(3.3.69) y llamando T a la matriz cuyas columnas son los vectores propios,
T = [ v1 v2 . . . vn] (3.3.70) el conjunto de igualdades 3.3.69 puede expresarse como
GT = T Λ (3.3.71) donde Λ = G′ es la matriz diagonal que se desea obtener como resultado de la transformación.
Ejemplo 3.3.4: Sea un sistema un cuya representación de estado es
x1(k + 1)
x2(k + 1)
x3(k + 1)
=
1 0 01 −1 00 1 2
x1(k)
x2(k)
x3(k)
+
101
u(k)
y(k) =
[
1 2 1
]
x1(k)
x2(k)
x3(k)
obtengamos las matrices de transformación T que nos permiten convertir esta re-presentación en la forma canónica de Jordan.
Solución:
La ecuación característica de este sistema sistema viene dada por
|λI − G| = ( λ − 1)( λ + 1)( λ − 2) = 0
y por tanto los autovalores son λ1 = 1 , λ 2 = −1, λ 3 = 2 . La matriz de transforma-ción para pasar a la forma canónica diagonal (puesto que las raíces son simples) se obtiene a partir de los vectores propios obtenidos a partir de la siguiente expresión
Gv i = viλi
1 0 01 −1 00 1 2
vi1
vi2
vi3
=
vi1
vi2
vi3
λi (3.3.72) 3.3. Representación de sistemas en el espacio de estados 113
y donde T = [ v1 v2 v3].Sustituyendo en 3.3.72 los λi por su valor obtenemos los siguientes sistemas de ecuaciones. Para λ1 = 1 , el sistema de ecuaciones 3.3.72 queda de la siguiente forma
v11 = v11
v11 − v12 = v12
v12 + 2 v13 = v13
cuya solución nos indica que v11 se puede elegir arbitrariamente, y que v12 = 12 v11
y que v13 = −v12 . Tomando v11 = 2 nos queda que v12 = 1 y v13 = −1 con lo que el vector propio v1 será
v1 =
21
−1
(3.3.73) Para λ2 = −1, el sistema de ecuaciones 3.3.72 queda de la siguiente forma
v21 = −v21
v21 − v22 = −v22
v22 + 2 v23 = −v23
cuya solución nos indica que v22 se puede elegir arbitrariamente, y que v23 =
−13
v22 y que v21 = 0 . Tomando v22 = 3 nos queda que v23 = −1, con lo que el vector propio v2 será
v2 =
03
−1
(3.3.74) Para λ3 = 2 , el sistema de ecuaciones 3.3.72 queda de la siguiente forma
v31 = 2v31
v31 − v32 = 2v32
v32 + 2 v33 = 2v33
cuya solución nos indica que v33 se puede elegir arbitrariamente, y que v31 = 0 yque v32 = 0 . Tomando v33 = 1 nos queda que el vector propio v3 será
v3 =
001
(3.3.75) 114 3. Modelado y análisis de sistemas en el espacio de estados
La matriz de transformación T será
T =
2 0 01 3 0
−1 −1 1
(3.3.76) y su inversa
T −1 =
12
0 0
−1613
0
1313
1
(3.3.77) Las matrices G′, H′, C′ del sistema transformado son las siguientes
G′ = T −1GT =
12
0 0
−1613
0
1313
1
1 0 01 −1 00 1 2
2 0 01 3 0
−1 −1 1
=
1 0 00 −1 00 0 2
H′ = T −1H =
12
0 0
−1613
0
1313
1
101
=
12
− 1643
C′ = CT = [1 2 1]
2 0 01 3 0
−1 −1 1
= [3 5 1] (3.3.78)
3.4. Solución de la ecuación de estado
Hasta este momento se ha estudiado como obtener una representación de esta-do de un sistema, pero para poder analizar el comportamiento de un sistema ante unas ciertas entradas de control es necesario que seamos capaces de determinar la respuesta dinámica del sistema. El cálculo de esta respuesta, dado que la representación es un conjunto de ecua-ciones diferenciales de primer orden o de ecuaciones en diferencias, se denomina solución de la ecuación de estado. Se verá a continuación para el caso de sistemas lineales e invariantes en el tiempo, tanto para el caso de tiempo continuo como para sistemas en tiempo discreto. 3.4. Solución de la ecuación de estado 115
3.4.1. Sistemas de tiempo continuo
Si consideramos un sistema lineal invariante en el tiempo definido por las ecua-ciones de estado
˙x(t) = Ax (t) + Bu (t)
y(t) = Cx (t) + Du (t) (3.4.1) donde las matrices A, B, C y D son matrices de dimensiones A : n×n, B : n×r,
C : m × n y D : m × r. Y los vectores x(t),u(t) e y(t), tienen dimensiones n × 1,
r × 1 y m × 1 respectivamente. Suponiendo que el vector de estado en el instante de tiempo inicial x(t0) es conocido y que se conoce la señal de entrada al sistema
u(t) para t ≥ t0. Se trata de determinar el vector x(t) para t > t 0.La ecuación homogénea correspondiente a la ecuación 3.4.1 es
˙x(t) = Ax (t) (3.4.2) cuya solución viene dada por:
x(t) = eA(t−t0)x(t0) (3.4.3) donde x(t0) es el vector de condiciones iniciales, y la matriz eAt puede obtenerse desarrollando en serie de potencias como
eAt = I + At + A2t2
2! + A3t3
3! + . . . (3.4.4) Si definimos una nueva matriz Φ de la siguiente manera
Φ(t) = eAt (3.4.5) la ecuación 3.4.3 puede reescribirse como
x(t) = Φ(t − t0)x(t0) (3.4.6) La solución de la ecuación no-homogénea 3.4.1 será de la forma
x(t) = Φ(t − t0)C1(t) (3.4.7) Diferenciando la ecuación anterior con respecto a t obtenemos
ddt x(t) = AΦ(t − t0)C1(t) + Φ(t − t0) ddt C1(t)= Ax (t) + Φ(t − t0) ddt C1(t) (3.4.8) y si comparamos la ecuación 3.4.8 con la ecuación 3.4.1 se puede observar que 116 3. Modelado y análisis de sistemas en el espacio de estados
Φ(t − t0) ddt C1(t) = Bu (t) (3.4.9) En la expresión 3.4.9, despejando el término en ddt C1(t) e integrándolo entre
t0 y t obtenemos que
C1(t) =
∫ tt0
Φ−1(τ − t0)Bu (τ )d τ + C2 (3.4.10) con lo que tenemos que la solución de la ecuación 3.4.1 será
x(t) = Φ(t − t0)C1(t)= Φ(t − t0)[
∫ tt0
Φ−1(τ − t0)Bu (τ )d τ + C2]= Φ(t − t0)C2 + Φ(t − t0)
∫ tt0
Φ−1(τ − t0)Bu (τ )d τ (3.4.11) Dado que
Φ(t − t0)Φ−1(τ − t0) = eA(t−t0)e−A(τ −t0) = eA(t−τ ) = Φ(t − τ ) (3.4.12) la ecuación 3.4.11 puede expresarse como
x(t) = Φ(t − t0)C2 +
∫ tt0
Φ(t − τ )Bu (τ )d τ (3.4.13) Nos queda por calcular el vector constante C2. Dicho vector lo calcularemos a partir de las condiciones iniciales en el instante t0. Para ello, si particularizamos la solución de la ecuación de estado que hemos obtenido en 3.4.13 para el instante
t0, y dado que la integral entre t0 y t0 tiene valor nulo, tendremos que
x(t0) = Φ(0) C2 = C2 (3.4.14) por lo tanto la forma resultante de la solución de la ecuación de estado será
x(t) = Φ(t − t0)x(t0) +
∫ tt0
Φ(t − τ )Bu (τ )d τ (3.4.15) La matriz Φ(t) = eAt se conoce como matriz fundamental del sistema o matriz de transición. 3.4. Solución de la ecuación de estado 117
3.4.2. Obtención de la solución por el método de la transformada de Laplace
Un método más eficaz de resolver la ecuación de estado, cuya solución nos con-duce a una expresión de tipo integral, puede obtenerse resolviendo la ecuación de estado en el campo complejo aplicando la transformada de Laplace. Supongamos el sistema definido por
˙x(t) = Ax (t) + Bu (t)
y(t) = Cx (t) + Du (t) (3.4.16) si tomamos transformadas de Laplace en ambas ecuaciones obtenemos que
sX(s) − x(0) = AX (s) + BU (s) (3.4.17)
Y (s) = CX (s) + DU (s) (3.4.18) De la ecuación 3.4.17 se deduce que
X(s) = ( sI − A)−1x(0) + ( sI − A)−1BU (s) (3.4.19) Si ahora hacemos la antitransformada de Laplace de la expresión anterior ten-dremos que
x(t) = L−1[( sI − A)−1]x(0) + L−1[( sI − A)−1BU (s)] (3.4.20) comparando esta expresión con la solución de la ecuación de estado obtenida en 3.4.15 se puede observar que el primer término de la ecuación 3.4.20 es la solu-ción de la ecuación homogénea para u(t) = 0 y el segundo término representa la solución particular correspondiente a la ecuación no homogénea con entrada u(t)
distinta de cero. Por otra parte resulta fácil obtener la expresión de la matriz fundamental Φ(t)
en función de la antitransformada de Laplace ya que igualando los primeros térmi-nos de las ecuaciones 3.4.20 y 3.4.15 resulta
x(t) = Φ(t)x(0) = L−1[( sI − A)−1]x(0) (3.4.21) lo que implica que
Φ(t) = L−1[( sI − A)−1] (3.4.22)
Ejemplo 3.4.1: Supongamos el siguiente sistema
ddt
x1(t)
x2(t)
=
−2 10 −3
x1(t)
x2(t)
+
12
u(t) (3.4.23) 118 3. Modelado y análisis de sistemas en el espacio de estados
sometido a una entrada escalón unitario y con estado inicial nulo. La solución de la ecuación de estado, utilizando la transformada de Laplace, viene dada por la siguiente expresión
x(t) = L−1[( sI − A)−1]x(0) + L−1[( sI − A)−1BU (s)] (3.4.24) en la que el primer término se anula puesto que el sistema tiene condiciones inicia-les nulas. La solución queda entonces reducida a la siguiente expresión
x(t) = L−1[( sI − A)−1BU (s)] (3.4.25) Obtengamos en primer lugar [sI − A] y su inversa,
[
sI − A
]
=
s + 2 −10 s + 3
(3.4.26)
[
sI − A
]−1
=
1
s+2 1(s+2)( s+3)
0 1
s+3
(3.4.27) y a partir de esta expresión tenemos que la respuesta del sistema en el campo de Laplace viene dada por la siguiente expresión
X(s) = [( sI − A)−1BU (s)] =
1
s+2 1(s+2)( s+3)
0 1
s+3
12
1
s =
s+5
s(s+2)( s+3) 2
s(s+3)
(3.4.28) Haciendo la antitransformada de Laplace de esta expresión obtenemos la solución de la ecuación de estado.
x(t) = L−1
s+5
s(s+2)( s+3) 2
s(s+3)
=
56 − 32 e−2t + 23 e−3t
23
− 23 e−3t
(3.4.29)
3.4.3. Discretización de las ecuaciones de estado en tiempo continuo
Aunque desde un punto de vista teórico resulta de gran interés el estudio de los sistemas dinámicos en tiempo continuo, lo cierto es que el control de sistemas basado en técnicas de espacio de estados se realiza en forma discreta. Resulta por lo tanto de gran interés el obtener modelos de estado discretos a partir de los mode-los matemáticos que hemos obtenido de un sistema dado. Dicho de otro modo, si 3.4. Solución de la ecuación de estado 119
hemos obtenido el modelo de estado en tiempo continuo de un sistema expresado por
˙x(t) = Ax (t) + Bu (t)
y(t) = Cx (t) + Du (t) (3.4.30) lo que buscamos es obtener un modelo discretizado de 3.4.30 de la forma
x(k + 1) = Gx (k) + Hu (k)
y(k) = Cx (k) + Du (k). (3.4.31) Esta discretización podemos realizarla de dos formas:
Método Indirecto . Se basa en discretizar de forma convencional la función de transferencia del sistema, para lo cual se obtiene la función de transfe-rencia del sistema, se introducen en él un bloqueador y un muestreador, y a continuación, se obtiene la función de transferencia discreta del sistema para un cierto periodo de muestreo T . En un segundo paso, se convierte la función de transferencia discreta a una representación de estado discreta por medio de alguno de los métodos que hemos visto anteriormente.
Método Directo . En este método buscamos obtener de forma directa las ex-presiones de G y H a partir de las matrices A y B de la representación de estado en tiempo continuo. El primero de los dos métodos no requiere ningún desarrollo adicional, ya que es bien conocida la discretización de una función de transferencia y el paso de esta a una representación de estado ya la hemos estudiado anteriormente. Podríamos utilizar, por ejemplo, el teorema de los residuos para la discretización de un sistema continuo. Concentremos nuestra atención en el segundo de los métodos. Habíamos visto en 3.4.15 que la solución de la ecuación de estado continua era
x(t) = eA(t−t0)x(t0) +
∫ tt0
eA(t−τ )Bu (τ )d τ (3.4.32) Supongamos que las variables de entrada son muestreadas y que posteriormen-te se llevan a un bloqueador de orden cero por lo que permanecen constantes duran-te el intervalo de tiempo T hasta el siguiente muestreo, es decir u(t) = u(t0 + kT )
para t0 + kT ≤ t < t 0 + ( k + 1) T . Entonces, si observamos los valores de la solución de la ecuación de estado en el instante t0 + kT
x(t0 + kT ) = eAkT x(t0) +
∫ t0+kT t0
eA(t0+kT −τ )Bu (τ )d τ (3.4.33) 120 3. Modelado y análisis de sistemas en el espacio de estados
y en t0 + ( k + 1) T
x(t0 + ( k + 1) T ) = eA(k+1) T x(t0) +
∫ t0+( k+1) Tt0
eA(t0+( k+1) T −τ )Bu (τ )d τ
(3.4.34) se puede ver que, si en la expresión 3.4.34 consideramos como estado inicial el correspondiente al instante de tiempo t0 + kT , la expresión nos quedará de la siguiente forma
x(t0 + ( k + 1) T ) = eAT x(t0 + kT ) +
∫ t0+( k+1) Tt0+kT
eA(t0+( k+1) T −τ )Bu (τ )d τ
(3.4.35) Suponiendo por simplicidad que t0 = 0 la expresión anterior se puede expresar como
x(( k + 1) T ) = eAT x(kT ) +
∫ (k+1) TkT
eA(( k+1) T −τ )Bu (τ )d τ (3.4.36) y si en el término integral de la ecuación anterior hacemos el cambio de variable
τ = kT + t, la expresión resultante es
x(( k + 1) T ) = eAT x(kT ) +
∫ T
0
eA(T −t)Bu (kT + t)d t (3.4.37) Teniendo en cuenta que las señales de entrada son muestreadas y se mantienen constantes durante todo el intervalo, tenemos que u(kT + t) = u(kT ), con lo que la expresión 3.4.37 podemos expresarla como
x(( k + 1) T ) = eAT x(kT ) +
∫ T
0
eA(T −t)Bu (kT )d t (3.4.38) y haciendo un segundo cambio de variable, de forma que λ = T − t, la ecuación 3.4.38 se transforma en
x(( k + 1) T ) = eAT x(kT ) +
∫ T
0
eAλBu (kT )d λ (3.4.39) Comparando esta expresión con la ecuación de estado para un sistema de tiempo discreto en el instante k + 1
x(k + 1) = Gx (k) + Hu (k) (3.4.40) se tiene finalmente que
G(T ) = eAT (3.4.41)
H(T ) = (
∫ T
0
eAλdλ)B (3.4.42) 3.4. Solución de la ecuación de estado 121
En el caso de que la matriz A sea no singular , la expresión 3.4.42 puede sim-plificarse, para ello resolvemos la integral, dando como resultado
H(T ) = (
∫ T
0
eAλdλ)B
= A−1[eAT − eA0]B
= A−1[eAT − I]B (3.4.43)
Ejemplo 3.10 Supongamos el mismo sistema que en el ejemplo anterior
ddt
x1(t)
x2(t)
=
−2 10 −3
x1(t)
x2(t)
+
12
u(t)
La matriz fundamental para este sistema, utilizando la transformada de Lapla-ce, viene dada por la siguiente expresión
Φ(t) = L−1[( sI − A)−1]
Obtengamos en primer lugar [sI − A] y su inversa,
[
sI − A
]
=
s + 2 −10 s + 3
[
sI − A
]−1
=
1
s+2 1(s+2)( s+3)
0 1
s+3
la matriz fundamental Φ(t) del sistema es
Φ(t) = L−1[( sI − A)−1] = L−1
1
s+2 1(s+2)( s+3)
0 1
s+3
=
e−2t e−2t − e−3t
0 e−3t
A partir de la matriz fundamental se obtienen las matrices G(T ) y H(T ) del sis-tema en la representación de estado de tiempo discreto con periodo T , resultando
G(T ) = eAT =
e−2T e−2T − e−3T
0 e−3T
122 3. Modelado y análisis de sistemas en el espacio de estados
H(T ) = (
∫ T
0
eAλdλ)B =
∫ T
0
e−2λ e−2λ − e−3λ
0 e−3λ
12
dλ
=
−32 e−2T + 23 e−3T + 32 − 23
−23
e−3T + 23
3.4.4. Solución de la ecuación de estado en tiempo discreto
Dado un sistema representado por las ecuaciones de estado en tiempo discreto
x(k + 1) = Gx (k) + Hu (k) (3.4.44)
y(k) = Cx (k) + Du (k) (3.4.45) la solución de la ecuación de estado puede obtenerse directamente sin más que expresar la ecuación de estado de forma recursiva
x(1) = Gx (0) + Hu (0)
x(2) = Gx (1) + Hu (1) = G2x(0) + GHu (0) + Hu (1)
x(3) = Gx (2) + Hu (2) = G3x(0) + G2Hu (0) + GHu (1) + Hu (2)
...con lo que la expresión recursiva para k quedará de la siguiente forma
x(k) = Gkx(0) +
k−1
∑
j=0
Gk−j−1Hu (j) (3.4.46) En la ecuación 3.4.46, se puede observar que, el estado del sistema en el instante k
tiene dos componentes básicas, una de ellas depende del estado inicial del sistema en el instante 0 y la otra depende de la secuencia de entradas de control que recibe el sistema hasta el instante k. A partir de la expresión anterior es posible expresar la salida del sistema como
y(k) = CG kx(0) + C
k−1
∑
j=0
Gk−j−1Hu (j) + Du (k) (3.4.47) Es posible utilizar las ecuaciones 3.4.46 y 3.4.47 para conocer el valor de las variables de estado y de salida en cualquier instante de tiempo k, pero tiene el inconveniente de no darnos una expresión fácil de calcular de la solución, lo que resulta engorroso, ya que habría que estar calculando los productos de matrices 3.4. Solución de la ecuación de estado 123
correspondientes hasta dicho instante de tiempo y operar con ellos hasta obtener el resultado. Resulta mucho más útil el disponer de una expresión general del estado y de la salida de forma que los valores del estado y de salida puedan obtenerse de una forma más simple.
3.4.5. Obtención de la solución por el método de la transformada en z
Un método más eficaz de resolver las ecuaciones de estado discretas, es el basado en la transformada en z. Dada la ecuación de estado
x(k + 1) = Gx (k) + Hu (k) (3.4.48) si tomamos transformadas en z en ambos lados de la ecuación 3.4.48 resulta
zX(z) − zx(0) = GX (z) + HU (z) (3.4.49) donde X(z) = Z[x(k)] y U (z) = Z[u(k)] . Si agrupamos los términos en X(z)
tendremos que
(zI − G)X(z) = zx(0) + HU (z) (3.4.50) Multiplicando por la izquierda en esta ecuación por la matriz (zI − G)−1 nos queda que
X(z) = ( zI − G)−1zx(0) + ( zI − G)−1HU (z) (3.4.51) y tomando antitransformadas en z
x(k) = Z−1[( zI − G)−1z]x(0) + Z−1[( zI − G)−1HU (z)] (3.4.52) Si comparamos las expresiones 3.4.46 con 3.4.52 se puede deducir que
Gk = Z−1[( zI − G)−1z] (3.4.53) y que k−1∑
j=0
Gk−j−1Hu (j) = Z−1[( zI − G)−1HU (z)] (3.4.54) para k = 1 , 2, 3, . . . .
Ejemplo 3.4.2: Supongamos el sistema
x1(k + 1)
x2(k + 1)
=
0.8 00 0.5
x1(k)
x2(k)
+
12
u(k)124 3. Modelado y análisis de sistemas en el espacio de estados
La matriz Gk para este sistema, utilizando la transformada z, viene dada por la siguiente expresión
GK = Z−1[( zI − G)−1z]
Obtengamos en primer lugar [zI − G] y su inversa,
[
zI − G
]
=
z − 0.8 00 z − 0.5
;
[
zI − G
]−1
=
1
z−0.8
00 1
z−0.5
la matriz fundamental Gk será
Gk = Z−1[( zI − G)−1z] = Z−1
zz−0.8 00 zz−0.5
=
0.8k 00 0.5k
Por otra parte tenemos que el efecto de la entrada sobre el sistema hasta el ins-tante k se obtenía aplicando la transformada z por medio de la siguiente expresión
k−1
∑
j=0
Gk−j−1Hu (j) = Z−1[( zI − G)−1HU (z)]
para el sistema de este ejemplo y suponiendo que la entrada al sistema es un escalón unitario tendremos que
k−1
∑
j=0
Gk−j−1Hu (j) = Z−1[( zI − G)−1HU (z)] = Z−1
1
z−0.8
00 1
z−0.5
12
zz − 1
= Z−1
z
(z−0.8)( z−1)
00 2z
(z−0.5)( z−1)
=
−4 · 0.8k + 5 00 −2 · 0.5k + 4
con lo que la respuesta del sistema para un instante de tiempo k cualesquiera se podrá obtener como:
x1(k)
x2(k)
=
0.8k 00 0.5k
x1(0)
x2(0)
+
−4 · 0.8k + 5 00 −2 · 0.5k + 4
3.5. Modelado de las perturbaciones en el espacio de estados 125
3.5. Modelado de las perturbaciones en el espacio de estados
Si se quiere conseguir que la respuesta del sistema sea la que deseamos a pesar de las influencias que las diferentes perturbaciones tienen sobre el sistema, resulta necesario comprender lo que son, así como describirlas y modelarlas adecuada-mente. Habíamos descrito la relación entre la entrada y la salida de un sistema por medio de la función de transferencia
Y (z) = G(s)U (s) (3.5.1) En la práctica nunca se tiene la suerte de que la señal de salida del sistema corresponda a esta expresión exactamente. Lo más normal es que esta expresión haya que modificarla, dependiendo de las perturbaciones a las que esta sometido el sistema. Una posible expresión de las salidas reales de un sistema podría ser
Y (s) = G(s)U (s) + W (s) (3.5.2) donde la señal W (s) es denominada señal de perturbación o perturbación en la salida, a secas. Indudablemente, esta perturbación es una entrada adicional al siste-ma, pero a diferencia de la entrada de control no podemos utilizarla para controlar el sistema. Las perturbaciones pueden tener diversos orígenes: Variaciones en la carga a la que está sometido el sistema. Errores producidos en las medidas de la salida del sistema debidos al ruido, derivas en los sensores, errores en la transmisión de la señal, etc. Variaciones en la planta y en los actuadores. Simplificaciones y errores en el modelo matemático del sistema que hacen que la salida real no coincida con la modelada. La señal de perturbación la describiremos como un ruido blanco que se añade directamente a la señal del sistema o como la salida de un sistema lineal generada ante una entrada de tipo impulso o ruido blanco, tal y como se muestra en la figura (3.9).
3.5.1. Perturbaciones en el sistema
Entre los diferentes factores que afectan a un sistema tenemos: La señal de control u.126 3. Modelado y análisis de sistemas en el espacio de estados BACw(t) u(t) x(t) y(t) x(t) .++ ++ x
n(t) Sistema Perturbaciones en el sistema Perturbaciones en la salida N
Fig. 3.9: Sistema sometido a perturbaciones
Las perturbaciones en el sistema ω. Dentro de las perturbaciones que afectan al sistema podemos distinguir dos tipos: las que son medibles ωm(por ejem-plo, la temperatura del exterior de un edificio afecta a la climatización del mismo, es una perturbación, puesto que no es controlable, pero es medible) y las no medibles ωm.Las perturbaciones de medida n.Veamos como podemos representarlas matemáticamente en el espacio de esta-dos. Supongamos que las variables de estado del sistema están influenciadas además de por la señal de control por perturbaciones en el sistema. Esto se podría expresar de la siguiente forma
˙x(t) = Ax (t) + Bu (t) + N ω (t) (3.5.3) Las diferentes perturbaciones sobre el sistema ω pueden conocerse en muchas ocasiones, por medio de una serie de experimentos y pruebas sobre el sistema, y en este caso dicha información puede utilizarse para construir un modelo matemático mejorado del sistema. Podemos diferenciar dos situaciones, según que las perturbaciones puedan o no predecirse : 3.5. Modelado de las perturbaciones en el espacio de estados 127 BACu(t) x(t) y(t) x(t) .++ x
Sistema w(t) Nvw(t) Ruido blanco F(s) BACx (t) x (t) .++x
Modelo de las perturbaciones
wwwww
vw(t) w(t) ++Dw
Fig. 3.10: Sistema con perturbaciones de sistema predecibles
Perturbaciones no predecibles . Si las observaciones son no predecibles, esto quiere decir que w debe de ser ruido blanco.
Perturbaciones predecibles . Si las observaciones son predecibles, es decir si w puede predecirse a partir de las observaciones realizadas hasta el ins-tante t, esto quiere decir que tenemos contenida en ω información relevante para la conducta futura del sistema. En este caso w(t) puede representarse por medio de una función de transferencia y un ruido blanco vω
W (s) = F (s)Vw(s) (3.5.4) En el caso de que las perturbaciones sobre el sistema sean predecibles (figura 3.10) estas pueden añadirse a la representación de estado del sistema. En primer lugar formularemos el modelo de estado de las perturbaciones a partir de la función de transferencia entre esta y el ruido blanco vω
˙xω(t) = Aωxω(t) + Bωvω(t) (3.5.5)
ωω(t) = Cωxω(t) + Dωvω(t). (3.5.6) Si combinamos las expresiones 3.5.3 y 3.5.6 en una única ecuación de estado nos queda
˙x = Ax + Bu + N w = Ax + Bu + N C ωxω + N Dvω
. . . . . .
˙xω = Aωxω + Bωvω
(3.5.7) 128 3. Modelado y análisis de sistemas en el espacio de estados
y de aquí, la expresión matricial
˙x
˙xω
=
A N C ω
0 Aω
xxω
+
B
0
u +
N D ω
Bω
vω (3.5.8) Esta expresión 3.5.8 es similar a la expresión 3.5.3 con la salvedad de que aquella información que es predecible en la perturbación la hemos introducido en el modelo como variables de estado y hemos dejado únicamente la parte de perturbaciones no predecibles fuera (es decir el término no incluido en el estado es ruido blanco). Por último si compactamos esta expresión 3.5.8 tenemos
˙xa(t) = Aaxa(t) + Bau(t) + Navω(t) (3.5.9) donde
Aa =
A N C ω
0 Aω
, Ba =
B
0
, Na =
N D ω
Bω
, xa =
x
x¯ω
(3.5.10)
3.5.2. Perturbaciones en la medida
Si tenemos ruido de medida n(t) en el sistema y este es aditivo, es decir si se suma a la variable medida que normalmente es la salida, la ecuación de salida viene dada como
y(t) = Cx (t) + Du (t) + n(t) (3.5.11) Aquí podemos de nuevo tener dos posibles situaciones según sea n:
Ruido blanco . En este caso no hay ninguna información utilizable en el ruido de medida para predecir el comportamiento del sistema, con lo que la expresión 3.5.11 modela adecuadamente el sistema.
El ruido de medida no es blanco . Si el ruido de medida no es blanco (es decir si no tiene un espectro constante) entonces es posible encontrar una cierta relación
n(s) = Fn(s)V2(s) (3.5.12) donde v2(t) es un ruido blanco. 3.5. Modelado de las perturbaciones en el espacio de estados 129 BACu(t) x(t) y(t) x(t) .++ x
Sistema v2(t) Ruido blanco Fn(s) + +n(t) BACx (t) x (t) .++ x
Modelo de las perturbaciones
nnnnn
v2(t) + +Dn
n(t)
Fig. 3.11: Sistema con perturbaciones de salida predecibles 130 3. Modelado y análisis de sistemas en el espacio de estados
En este segundo caso (figura 3.5.2) podemos formular un modelo de estado del ruido de medida a partir de la función de transferencia entre esta y un ruido blanco
v2,
˙xn(t) = Anxn(t) + Bnv2(t) (3.5.13)
n(t) = Cnxn(t) + Dnv2(t) (3.5.14) si combinamos las expresiones 3.5.11 y 3.5.14 en una única ecuación de estado nos queda
˙xa =
Aa 00 An
xa
xn
+
Ba
0
u +
Na 00 Bn
vw
v2
(3.5.15)
y(t) =
[
C Cn
] xxn
+ Du + v2 (3.5.16) y si compactamos esta expresión tendremos
˙x(t) = Ax (t) + Bu (t) + N v 1 (3.5.17)
y(t) = Cx (t) + Du (t) + v2(t) (3.5.18) donde v1 y v2 son ruidos blancos, y podemos considerarla como la representación estándar de sistemas con ruido de sistema y ruido de medida.
Ejemplo 3.5.1: Supongamos un modelo simple de inventario de un almacén. Sea
y(k) el nivel del inventario el día k y u1(k) los pedidos de producto del día k que hace el almacén para reponer sus existencias. Estos pedidos se sirven dos días después. Denominemos u2(k) a las ventas realizadas el día k. Veamos un modelo de variación del inventario del almacén.
Solución:
Supongamos en una primera aproximación que el nivel de ventas se conside-ra simplemente como una variable de entrada al sistema. El nivel del inventario correspondería a la siguiente ecuación
y(k) = y(k − 1) + u1(k − 2) − u2(k − 1) (3.5.19) Si tomamos como entradas u1 y u2 y el estado del sistema lo caracterizamos por medio del nivel de inventario en el almacén y de los pedidos realizados pero aún no recibidos,
u(k) =
u1(k)
u2(k)
x(k) =
y(k)
u1(k − 1)
(3.5.20) 3.6. Ejemplos de sistemas físicos 131
Entonces la representación de estado del sistema será
x(k + 1) =
1 10 0
x(k) +
0 −11 0
u(k)
y(t) =
[
1 0
]
x(k) (3.5.21) Supongamos ahora que queremos refinar más el modelo, para ello podemos considerar las ventas del almacén, como una señal de perturbación, puesto que no tenemos ninguna capacidad de control sobre ellas. Si suponemos que u2(k) = v(k)
y que u1(k) = u(k) el modelo quedará de la siguiente forma
x(k + 1) =
1 10 0
x(k) +
01
u(k) +
−10
v(k)
y(t) =
[
1 0
]
x(k) (3.5.22)
3.6. Ejemplos de sistemas físicos
Ejemplo 3.6.1: Considérese el sistema de la figura 3.12 formado por dos masas
M = 10 Kg y m = 1 Kg unidas por un muelle de coeficiente de elasticidad K =1N/m y de coeficiente de rozamiento viscoso f = 1 N s/m . La entrada al sistema es la fuerza u con la que se desea controlar la posición y. La posición de la primera masa es y, de la segunda masa es d. Obtener el modelo en variables de estado del sistema. M myfKdu
Fig. 3.12: Sistema del ejemplo 3.6.1
Solución: 132 3. Modelado y análisis de sistemas en el espacio de estados
Las dos ecuaciones de equilibrio de fuerzas referida a la primera y segunda masas, respectivamente son
M ¨y + f ( ˙ y − ˙d) + K(y − d) = u (3.6.1)
m ¨d + f ( ˙d − ˙y) + K(d − y) = 0 (3.6.2) Como variables de estado elegimos las posiciones de las masas y sus respecti-vas velocidades
x1 = y; x2 = ˙ y, x 3 = d, x 4 = ˙d
De esta forma, las ecuaciones de equilibrio de fuerzas, en variables de estado quedan como
M ˙x2 + f (x2 − x4) + K(x1 − x3) = um ˙x4 + f (x4 − x2) + K(x3 − x1) = 0
Las ecuaciones de estado de este sistema quedan de la siguiente forma
˙x1 = x2
˙x2 = − 1
M (f x 2 − f x 4 + Kx 1 − Kx 3) + uM
˙x3 = x4
˙x4 = − 1
m (f x 4 − f x 2 + Kx 3 − Kx 1)
y la ecuación de salida es
y = x1
Ejemplo 3.6.2: Sea el sistema hidráulico representado en la figura 3.13, consis-tente en dos depósitos D1 y D2 de superficie transversal A1 y A2, llenos de agua a niveles h1 y h2, y con dos válvulas de paso con resistencia hidráulica R1 y R2. Sea la señal de entrada el caudal qe y la señal de salida el caudal q1. Se desea obtener el modelo del sistema en variables de estado.
Solución: En el sistema existen dos elementos independientes que almacenan ener-gía: los dos depósitos. Por tanto, bastarán dos variables de estado para definir el comportamiento dinámico del sistema. Las ecuaciones físicas del sistema (lineali-zadas) serán:
q1 = 1
R 1
(h1 − h2)
q2 = 1
R 2
h2
qe − q1 = A1 ˙h1
q1 − q2 = A2 ˙h23.6. Ejemplos de sistemas físicos 133 qe
T TT A1h1 h2A2D1 D2 R1 q1 q2R1
Fig. 3.13: Sistema del ejemplo 3.6.1
Eligiendo las variables de estado x=h1, x2 = h2, y siendo la entrada y la salida
u = qe e y = q1 y operando, las ecuaciones de estado del sistema son
˙x1 = − 1
R1A1
x1 + 1
R1A1
x2 + 1
A 1
u
˙x2 = 1
R1A1
x1 − R1 + R2
R1R2A2
x2
y la ecuación de salida es
y = 1
R 1
x1 − 1
R 1
x2
Ejemplo 3.6.3: Considérese el caso de zorros y conejos de Australia. Estudiar la estabilidad de este ecosistema y obtener la matriz de transición de estados.
Solución: El número de conejos, x1 si se dejan solos, crecería indefinidamente, hasta que el suministro de alimentos se agotara. De esta forma
˙x1 = Kx 1
Sin embargo, con los zorros que hay en el continente, tenemos
˙x1 = Kx 1 − ax 2
donde x2 es el número de zorros. Dado que los zorros deben tener conejos para subsistir, su evolución será
˙x2 = −hx 2 + bx 1
Para estudiar la estabilidad, hay que estudiar el signo de los valores propios de la matriz (sI − A)
|sI − A| =
∣∣∣∣∣∣s 00 s
−
K −ab −h
∣∣∣∣∣∣ =
∣∣∣∣∣∣(s − K) a
−b (s + h)
∣∣∣∣∣∣134 3. Modelado y análisis de sistemas en el espacio de estados
= ( s − K)( s + h) + ab = s2 + s(h − K) + ab − Kh
Para que valores propios sean negativos es necesario que los coeficientes de la ecuación de segundo orden sean positivos. Esto implica que h > K y ab > Kh .Suponiendo que los coeficientes sean K = 1 , h = 3 y a = b = 2 la matriz de transición de estados Φ queda
Φ( t) = L−1[( sI − A)−1] =
(2 t + 1) e−t −2te −t
2te −t (−2t + 1) e−t
Bibliografía
S. Domínguez, P. Campoy, J.M. Sebastián, A. Jiménez, Control en el espacio de estados , Prentice Hall, 2002. 1991. T. Glad, L. Ljung Control Theory Taylor and Francis, 2000. K. Ogata, Discrete-Time Control Systems, Prentice Hall International Edi-tion , 1995. K. Ogata, Ingeniería de control moderna, Prentice Hall , 1998. V. Strejc, State Space Theory of Discrete Linear Control , John Wiley and Sons, 1981. 136 Bibliografía 4. TéCNICAS CLáSICAS DE CONTROL
4.1. Planteamiento del problema
El objetivo de todo sistema de control consiste en que dado un cierto siste-ma, del que se dispone de una cierta cantidad de información en base a una serie de medidas de algunas de las señales del mismo, tratar de determinar unas ciertas entradas de control factibles de forma que la variable del sistema que se desea con-trolar siga de forma lo más exacta posible a una cierta señal de referencia, a pesar de la influencia que las posibles perturbaciones, errores de medida y variaciones en la carga del sistema. Un esquema muy general de control del proceso Gp es el que se muestra en la figura 4.1. Gf Gc u yr +-Gp
Fig. 4.1: Diagrama de bloques de un esquema general de control
En este esquema, la señal de control es
U (s) = Gf (s)R(s) − Gc(s)Y (s) (4.1.1) Se puede apreciar una parte formada por un controlador Gc situado en la realimen-tación y un filtro Gf en la entrada del sistema. Este esquema de control tiene como función de transferencia conjunta de todo el sistema la siguiente expresión
Y (s)
R(s) = Gf (s) Gp(s)1 + Gp(s)Gc(s) . (4.1.2) En esta expresión se puede observar que el término de filtrado Gf afecta de forma directa a la ganancia en régimen permanente del sistema realimentado así como a los polos y ceros del mismo al estar situada en la cadena principal antes del lazo de 138 4. Técnicas clásicas de control
realimentación. Sin embargo, no altera las posiciones de los polos y ceros del lazo de realimentación, salvo en el caso de que se produzcan cancelaciones polo-cero entre Gf y la función de transferencia del lazo de realimentación. Por otra parte el término Gc al estar situado en el lazo de realimentación afecta de forma significativa a las posiciones de los polos del sistema en bucle cerrado, alterando de forma importante la dinámica del sistema. Simplificando un poco, po-dríamos pensar que mientras que con el diseño de Gf buscamos ajustar o mejorar la respuesta en régimen permanente del sistema, con Gc se busca ajustar la respuesta transitoria. Esta estructura de control se utilizará de forma extensiva en el control por re-alimentación de estado que se estudiará en el próximo capítulo. Un caso particular de esta estructura de control es cuando Gf = Gc. En este caso se pierde libertad en el diseño ya que sólo se dispone de una función de transferencia. De esta forma, el esquema de control se simplifica tal y como se puede apreciar en el esquema (a) de la figura 4.2, pudiéndose expresar también por medio del esquema equivalente (b) de dicha figura. Gc Gc u yr +-Gp Gc u yr +- Gp (a) (b) e
Fig. 4.2: Controlador en el lazo principal
Este esquema equivalente (b) es el que podríamos denominar esquema clási-co de control, al menos académicamente. Dentro de este grupo de controladores podemos incluir el control PID, que es el que estudiaremos en este capítulo.
4.2. Diseño de controladores clásicos
Dentro de los controladores que responden al esquema de control clásico pode-mos distinguir dos enfoques básicos a la hora de abordar el diseño del controlador
Gc. Estos dos enfoques se diferencian fundamentalmente en la forma de definir la función de transferencia del controlador. 4.3. Especificaciones 139
Controlador con estructura fija . Este grupo de técnicas se caracteriza por utilizar una función de transferencia definida a priori. Dentro de este grupo se incluye el denominado control PID . En este tipo de controladores la fun-ción de transferencia del controlador es fija y en él están presentes: acciones de control proporcionales al error (acción P), acciones de control proporcio-nales a la integral del error (acción I) y acciones de control proporcionales a la derivada del error (acción D). El peso que se le da a cada una de las acciones de control se puede determinar de forma teórica (en el caso de que se conozca la función de transferencia del proceso) o de forma empírica en el caso de que no se conozca dicha función de transferencia. Evidentemente puede ocurrir que uno o varios de los parámetros del controlador anulen alguna de las acciones de control si éstas no fuesen necesarias. 2. Controlador con estructura variable . Este segundo grupo de controlado-res no presenta una estructura de función de transferencia predefinida, sino que ésta se obtiene como resultado de las especificaciones deseadas para el sistema y de la función de transferencia del proceso a controlar. Este es el caso del diseño de controladores mediante la técnica de síntesis directa.
4.3. Especificaciones
Las especificaciones expresan las características que se desean alcanzar en el sistema cuando se introduce un cierto controlador en el mismo. Pueden expresarse de diferentes formas utilizando para ello el dominio de la frecuencia o el dominio del tiempo. En nuestro caso vamos a referirnos a especificaciones en el dominio del tiempo. Las especificaciones respecto a la respuesta en el dominio del tiempo de un sistema se suelen definir con respecto a la respuesta temporal ante una entrada en escalón de un sistema de segundo orden. Para ello se toma como sistema de referencia un sistema con función de transferencia teórica:
G(s) = Kω 2
n
s2 + 2 ξω ns + ω2
n
(4.3.1) cuya respuesta a un escalón es
y(t) = 1 − 1
√1 − ξ2 e−ξω nt sin( ωnt√1 − ξ2 + φ) (4.3.2) con
φ = arctan
√1 − ξ2
ξ (4.3.3) 140 4. Técnicas clásicas de control
Este sistema tiene sus polos situados en
p1,2 = −ξω n ± jω n
√1 − ξ2 (4.3.4) donde ξ es la llamada tasa de amortiguamiento y ωn es la frecuencia natural no amortiguada del sistema. Para un sistema con tasa de amortiguamiento 0 < ξ ≤ 1,(sistema subamortiguado) la curva de respuesta del sistema tiene la forma mostrada en la figura 4.3 d·Mp ty To Ts Mp
ep
Tp
Fig. 4.3: Parámetros de la respuesta temporal
Entre los parámetros que se pueden utilizar para especificar la respuesta tene-mos los siguientes:
Tiempo de pico ,
Tp = πω d
(4.3.5) donde ωd es la frecuencia amortiguada ωd = ωn
√1 − ξ2.
Tasa de decaimiento , d.
d = e−2π ξ√1−ξ2 (4.3.6)
Periodo de la oscilación amortiguada , To.
To = 2πωn
√1 − ξ2 (4.3.7)
Sobreoscilación , Mp.
Mp = e−π ξ√1−ξ2 = √d (4.3.8)
Tiempo de establecimiento , Ts.
Ts ' πξω n
(4.3.9) 4.4. Control PID 141
Por último, se define el error de posición en régimen permanente, ep, como la diferencia entre en valor la referencia deseada y el valor de salida de la varia-ble controlada del sistema (estamos suponiendo por simplicidad que la señal de realimentación es unitaria). Es relativamente fácil obtener, a partir de unas ciertas especificaciones, las po-siciones que deben de tener los polos en cadena cerrada del sistema que cumplen dichas especificaciones.
4.4. Control PID
Los controladores del tipo PID son con mucha diferencia los más frecuente-mente utilizados en la industria de control de procesos, donde más del 95 % de los lazos de control utilizan controladores PID. Su sencillez, versatilidad y capacidad para resolver los problemas básicos que se presentan en los procesos con dinámica favorable y requisitos de funcionamiento modestos hacen de ellos una herramienta imprescindible en el control de procesos. A lo largo de las últimas décadas los controladores PID han sobrevivido a di-ferentes cambios tecnológicos que van desde los primeros controladores desarro-llados a partir de elementos neumáticos hasta los desarrollados con microprocesa-dores pasando por las válvulas electrónicas, los dispositivos analógicos, los tran-sistores y los circuitos integrados. La incorporación de los microprocesadores ha tenido un impacto muy grande ya que ha permitido que los controladores PID se enriquezcan en funciones colaterales sin perder ninguna de sus propiedades, así se les han añadido posibilidades de ajuste automático de los parámetros, posibilidades de ejecución de reglas lógicas o automatismos secuenciales. Por todo ello, y aunque existen técnicas de control más sofisticadas, en el nivel más bajo de control de muchos procesos sigue estando presente o incluso es el mayoritario. En las próximas páginas nos referiremos a él.
4.4.1. Estructura básica de un controlador PID
La versión académica de este controlador viene caracterizada por una ley de control de la forma:
u(t) = K
[
e(t) + 1
Ti
∫ t
0
e(τ )dτ + Td
de (t)
dt
]
(4.4.1) donde u(t) es la variable de control y e(t) es el error expresado como e(t) =
r(t) − y(t) (fig 4.2b). La expresión 4.4.1 muestra los tres tipos de acciones de control: la acción proporcional con peso K, la acción integral con peso KTi y la acción derivativa con peso KT d.142 4. Técnicas clásicas de control
Para ajustar el comportamiento del controlador se dispone de tres parámetros: la ganancia proporcional K, el tiempo integral Ti y el tiempo derivativo Td. Se pue-de observar que si el tiempo integral se hace infinito, la acción integral desaparece y que si el tiempo derivativo se hace cero desaparece la acción derivativa. La función de transferencia del controlador PID es la siguiente
U (s)
E(s) = K[1 + 1
Tis + Tds] = K(1 + sT i + s2TiTd)
sT i
(4.4.2) En este controlador el ajuste busca determinar las posiciones de los dos ceros de la función de transferencia y de la ganancia estática del mismo, para que se cumplan lo mejor posible las especificaciones de diseño deseadas en el sistema. La ecuación 4.4.2 no es físicamente realizable, por lo que se denomina regu-lador PID teórico. Para que un PID sea físicamente realizable habrá que añadir a la acción derivativa un polo, con una influencia limitada. Un regulador PID real tendrá la forma
Gc(s) = K 1 + sT i + s2TiTd
(1 + sT i)(1 + sT d) (4.4.3) El ajuste de los parámetros del regulador se puede hacer de dos formas: 1. Empíricamente . Para lo que se ajusta de forma experimental los valores hasta alcanzar la respuesta deseada. Este sistema puede ser excesivamente lento en muchos sistemas, sobre todo si el tiempo de respuesta de estos es grande. Para evitar este problema, se utilizan los métodos de Ziegler-Nichols que en base a unas medidas muy simples observadas en la respuesta del sistema dan unos valores teóricos de los parámetros del controlador. Valores que se toman como referencia y posteriormente se hace un ajuste más fino a partir de ellos. Estos métodos y otros derivados de ellos se utilizan considerablemente en la industria, tanto para ajuste puramente manual como en controladores PID industriales con sistemas de autoajuste. 2. Teóricamente . Para lo que se determinan de forma analítica los valores del regulador. Se requiere el conocimiento expreso de la función de transferencia del proceso.
4.4.2. Métodos de Ziegler - Nichols
Los métodos de Ziegler - Nichols son dos métodos clásicos de ajuste empírico de los parámetros de un controlador PID. Fueron presentados por dichos autores en 1942. Estos métodos son aún ampliamente utilizados, bien en su forma original o con versiones mejoradas. 4.4. Control PID 143
Ambos métodos se basan en la determinación de algunas características de la respuesta del proceso, temporal o frecuencial, para determinar a partir de dichas características y por medio de unas relaciones matemáticas muy simples los pará-metros del controlador que da una respuesta con tasa de decaimiento de un cuarto entre los valores de la primera y la segunda sobreoscilación.
Método de la respuesta a un escalón
Este primer método se basa en la observación de la respuesta en bucle abierto del sistema ante una entrada en escalón. Observando dicha respuesta, se determinan dos parámetros que se obtienen de la siguiente forma: 1. Se determina el punto de máxima pendiente en la curva de respuesta a un escalón, y en dicho punto se dibuja la tangente a la curva de respuesta en dicho punto. 2. Se determina la intersección de la tangente obtenida anteriormente con los ejes de coordenadas, y se obtienen las distancias a y L. Estos dos parámetros corresponden a la respuesta teórica de un modelo matemático de la forma
G(s) = asL e−sL (4.4.4) que corresponde a un integrador con retardo temporal. Este sistema permite ser caracterizado por dos parámetros a y L tal y como se puede observar en la siguiente figura 4.4. L tya
Fig. 4.4: Respuesta ante entrada escalón de G(s) = asL e−sL
Una vez que se han determinado los parámetros a y L en la respuesta, Ziegler y Nichols proponen como parámetros del controlador PID los indicados en 144 4. Técnicas clásicas de control
Controlador K Ti Td
P 1/a
PI 0.9/a 3L
PID 1.2/a 2L L/ 2
Tab. 4.1: Valores de los parámetros propuestos por Ziegler-Nichols para el método de respuesta a un escalón
la tabla 4.1 obtenidos directamente como función de los parámetros a y L
medidos sobre la respuesta del sistema. Este controlador está diseñado para dar una tasa de decaimiento de un cuarto (d=0.25) entre las magnitudes de la primera y la segunda sobreoscilación, por lo que generalmente presenta una sobreoscilación alta. Tiene la ventaja de que a partir de estos valores es fácil realizar un ajuste más fino para adecuarlos a la respuesta que se desea sin la necesidad de un largo proceso de prueba y error.
Método de la respuesta frecuencial
Este segundo método desarrollado por Ziegler y Nichols se basa en determinar el punto de la curva de Nyquist, para la función de transferencia G(s), en el que dicha curva interseca el eje real negativo (punto de estabilidad relativa). Una vez obtenido este punto, se caracteriza por medio de dos parámetros denominados: ganancia crítica Ku y periodo crítico Tu.Este método se fundamenta en la observación de que muchos sistemas se hacen inestables cuando se hace un control proporcional con ganancia elevada. El método de determinación de los parámetros del regulador PID en los si-guientes pasos: 1. Se cierra el bucle de realimentación con un regulador de tipo proporcional (figura 4.5a). Se va aumentando la ganancia del regulador proporcional has-ta que se obtiene la ganancia crítica Ku en la que se presenta una respuesta oscilatoria como la indicada en la figura 4.5. Para esta ganancia Ku, se de-termina el periodo crítico Tu de la señal de salida oscilatoria. 2. Como en el primer método, se determinan a partir de Ku y Tu los valores de los parámetros del controlador. Los valores propuestos por Ziegler y Nichols se muestran en la tabla 4.2. 4.4. Control PID 145 Ku yr +- G(s) etytyK=Ku, Tu ty
a) b) c) d) KKu
Fig. 4.5: Obtención de la ganancia crítica y del periodo crítico para el segundo método de Ziegler-Nichols
Al igual que en el método anterior el criterio de diseño de los reguladores es conseguir una tasa de decaimiento de un cuarto. Estos reguladores dan buena res-puesta ante variaciones en la carga del sistema, aunque como contrapartida suelen presentar una sobreoscilación alta, como ya se ha comentado anteriormente. En cualquier caso, es fácil realizar un ajuste posterior del regulador partiendo de estos parámetros. 146 4. Técnicas clásicas de control
Controlador K Ti Td
P 0.5Ku
PI 0.4Ku 0.8Tu
PID 0.6Ku 0.5Tu 0.125 Tu
Tab. 4.2: Valores de los parámetros propuestos por Ziegler-Nichols para el método frecuencial
4.5. Métodos analíticos de diseño de controladores PID
Estos métodos buscan determinar de forma analítica los valores más adecuados de los parámetros de un controlador PID de forma que se cumplan unas especifi-caciones determinadas. Veremos dos técnicas analíticas para realizar el cálculo de estos parámetros: la primera de estas técnicas (asignación de polos) es aplicable a sistemas de segundo orden, y la segunda (diseño basado en los polos dominan-tes) es aplicable a sistemas complejos. Ambas técnicas requieren un conocimiento previo de la función de transferencia del proceso que se desea controlar.
4.5.1. Método de asignación de polos
La técnica de asignación de polos trata de encontrar los parámetros de un con-trolador tal que la función de transferencia en cadena cerrada tenga unos polos predeterminados. Los polos en bucle cerrado se determinan de forma que se obten-gan unas ciertas especificaciones en la respuesta del sistema. En términos generales un regulador de tipo P al tener un único parámetro sólo nos permitiría asignar un polo, un regulador PI o uno PD que tienen dos parámetros nos permitirán asignar dos polos y un regulador PID que tiene tres parámetros nos permitiría asignar tres polos. La utilización de uno u otro tipo de regulador dependerá de la función de transferencia del sistema. Supongamos en primer lugar un proceso cuya función de transferencia es de primer orden, con la siguiente parametrización
Gp(s) = Kp
1 + sT (4.5.1) y supongamos que queremos controlar el sistema por medio de un controlador de tipo PI, que responde a la siguiente expresión
Gc(s) = K
(
1 + 11 + sT i
)
(4.5.2) 4.5. Métodos analíticos de diseño de controladores PID 147
el sistema en bucle cerrado tiene la siguiente función de transferencia
G(s) = Gp(s)Gc(s)1 + Gp(s)Gc(s) (4.5.3) La ecuación característica del sistema en bucle cerrado definido por la ecuación 4.5.3 es la siguiente
1 + Gp(s)Gc(s) = 0 (4.5.4) y sustituyendo en 4.5.4 Gp(s) y Gc(s) por las expresiones 4.5.1 y 4.5.2 obtenemos que
1 + K
(
1 + 1
sT i
) Kp
(1 + sT ) = 0
s2 + s
( 1 + KpKT
)
KpKT T i
= 0 (4.5.5) Supongamos que se desea que la respuesta del sistema en bucle cerrado sea la de un sistema de segundo orden con una tasa de amortiguamiento ξ y con una frecuencia ωn. Es decir, se desean situar los polos en cadena cerrada del sistema en las posiciones p1,2 = −ξω n ± ωn
√1 − ξ2. Por lo que la ecuación característica en bucle cerrado tendrá la expresión
s2 + 2 ξω ns + ω2
n
= 0 (4.5.6) Igualando las expresiones 4.5.5 y 4.5.6 obtenemos que
2ξω n = 1 + KpKT (4.5.7)
ω2
n
= KpKT T i
(4.5.8) y despejando en 4.5.7 y 4.5.8 tenemos que
K = 2ξω nT − 1
Kp
(4.5.9)
Ti = 2ξω nT − 1
ω2
n
T (4.5.10) Esta técnica requiere introducir un controlador en el que tengamos dos posibles parámetros de ajuste, un controlador PI, cuando el sistema en cadena abierta sea de primer orden ya que es necesario fijar arbitrariamente las posiciones de dos polos en cadena cerrada. De forma análoga, cuando el sistema es de segundo orden será conveniente por lo general utilizar un controlador PID que tiene tres grados 148 4. Técnicas clásicas de control
de libertad para poder fijar arbitrariamente los tres polos que tendrá el sistema en bucle cerrado. Sea la función de transferencia del sistema de segundo orden de la forma
Gp(s) = Kp
(1 + sT 1)(1 + sT 2) (4.5.11) y supongamos que queremos controlar el sistema por medio de un controlador de tipo PID, que responde a la siguiente expresión
Gc(s) = K (1 + sT i + s2TiTd)
sT i
(4.5.12) con lo que la ecuación característica del sistema en bucle cerrado 1+ Gp(s)Gc(s) = 0, será
s3 + s2
( 1
Ti
1
T2
KpKT d
T1T2
)
s
( 1
T1T2
KpKT1T2
)
KpKT1T2
= 0 (4.5.13) Si deseamos que el sistema tenga en bucle cerrado la siguiente ecuación carac-terística,
(s + α)( s2 + 2 ξω ns + ω2
n
) = 0 (4.5.14) igualando los coeficientes de los polinomios de las expresiones 4.5.13 y 4.5.14 tenemos las siguientes expresiones
ω2
n
α = KpKT1T2
(4.5.15)
ω2
n
2 ξω nα =
( 1
T1T2
KpKT1T2
)
(4.5.16)
2ξω n + α =
( 1
Ti
1
T2
KpKT d
T1T2
)
(4.5.17) y resolviendo este sistema de ecuaciones obtenemos
K = T1T2(2 ξω n + α) + 1
Kp
(4.5.18)
Ti = KpKT1T2ω2
n
α (4.5.19)
Td = T1T2ω3
n
α(ωn + 2 ξα )
T1T2(2 ξω n + α) − 1 + T1ω2
n
α(ω2
n
α + 1) (4.5.20) 4.5. Métodos analíticos de diseño de controladores PID 149
Ejemplo 4.5.1: Supongamos que se desea controlar un sistema cuya función de transferencia es Gp(s) = 1
s(s+2)
de forma que en bucle cerrado tenga una tasa de amortiguamiento ξ = 0 .5 y una frecuencia ωn = 3 .14 rad s .Para controlar este sistema nos basta con un regulador de tipo PD con función de transferencia de la forma
Gc(s) = K(1 + Tds) (4.5.21) ya que el sistema aunque es de segundo orden tiene un polo en el origen en la función de transferencia, por lo que el sistema no requiere añadir un efecto inte-gral en el controlador para anular el error de posición en régimen permanente. Al no incluir efecto integral en el regulador el sistema en cadena cerrada tendrá una ecuación característica de orden 2. El sistema en bucle cerrado tiene una ecuación característica de la forma
1 + K(1 + Tds) 1
s(s + 2) = 0 (4.5.22) y se desea que tenga la ecuación característica 4.5.6 en bucle cerrado. Para ello igualamos 4.5.6 a 4.5.22,
s2 + s(2 + KT d) + K = s2 + 2 ξω ns + ω2
n
= 0 (4.5.23) e igualando los coeficientes de ambos polinomios tenemos que
K = ω2
n
(4.5.24)
Td = 2ξω n − 2
ω2
n
(4.5.25) y sustituyendo por los valores de las especificaciones que se han indicado tenemos que K = 9 .86 y Td = 0 .115 . Con lo que el regulador resultante tiene la expresión:
Gc(s) = 9 .86(1 + 0 .115 s) (4.5.26)
4.5.2. Diseño basado en los polos dominantes
La técnica de asignación de polos trata de ubicar las posiciones de todos los polos en cadena cerrada del sistema. Esto complica considerablemente el método para el caso de sistemas complejos, pudiendo incluso no tener ninguna solución factible el problema. Con lo que, si se quiere controlar con un PID un cierto sistema por medio de esta técnica, los sistemas deben restringirse a los de primer y segundo orden para que esta técnica nos asegure la existencia de una solución. 150 4. Técnicas clásicas de control 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 500.5 11.5 22.5
(a) (b) y(t) t
Fig. 4.6: Respuesta del sistema del ejemplo 7.5.1: (a) en bucle abierto y (b) en bucle cerrado con el controlador diseñado
Si el sistema es complejo, o bien se busca una aproximación del modelo a un primer o segundo orden o bien se utiliza la técnica de diseño basada en los polos dominantes . Esta técnica trata de asignar sólo unos pocos polos del sistema. Esto permite diseñar controladores simples para controlar sistemas complejos, aunque el comportamiento del regulador al instalarlo en el sistema no nos dará de forma exacta las especificaciones que deseamos ya que se ignora en el diseño el efecto de algunos polos. Sin embargo, si la elección de las posiciones es adecuada el comportamiento estará muy próximo a lo que se desea. Esta técnica requiere también que se conozca la función de transferencia del sistema, y la idea de asignar sólo unos pocos polos del sistema en bucle cerrado ha sido muy extensamente utilizada en los primeros diseños de control PID, especial-mente cuando la potencia de los ordenadores no permitía otras posibilidades. Aún hoy, esta técnica tiene un considerable interés ya que una buena parte de los controladores industriales implementan una estructura fija de PID y se utilizan para procesos de complejidad muy variable por lo que resulta necesaria cuando estos son de complejidad alta. Esta asignación basada en los polos dominantes puede realizarse de dos formas: o bien determinando algebraicamente los valores de los parámetros del regulador que sitúan los polos dominantes en las posiciones deseadas o bien mediante el uso del lugar de las raíces. Ambas técnicas nos permiten obtener analíticamente los parámetros de los reguladores PID. 4.5. Métodos analíticos de diseño de controladores PID 151
Determinación algebraica
El problema es similar al de la sección anterior. Si se introduce un regulador P, necesitamos situar un polo para poder determinar el parámetro del regulador. Si se desea introducir un regulador PI o PD es necesario asignar dos polos para calcular los parámetros del regulador y si se introduce un regulador PID es necesario fijar tres polos. Supongamos, para ilustrar el funcionamiento de esta técnica, que se desea di-señar un regulador de tipo PI para controlar un sistema. Dicho regulador tiene dos parámetros por lo que para determinar el valor de los mismos necesitamos situar la posición de dos polos del sistema en bucle cerrado. Supongamos que la ecuación del controlador es de la forma
Gc(s) = K
(
1 + 11 + sT i
)
(4.5.27) el sistema en bucle cerrado tiene la siguiente función de transferencia
G(s) = Gp(s)Gc(s)1 + Gp(s)Gc(s) (4.5.28) con lo que la ecuación característica del sistema en bucle cerrado viene definida por
1 + Gp(s)K
(
1 + 11 + sT i
)
= 0 (4.5.29) Necesitamos fijar las posiciones de dos polos p1,2, por lo que podemos suponer que estos son complejos conjugados de forma que tengan un cierto amortiguamien-to ξ y una cierta frecuencia ωn, es decir
p1,2 = ωn(−ξ ± j√1 − ξ2) (4.5.30) En el caso de que el controlador que se necesita sea de tipo PD estamos en una situación similar, pues hemos de determinar dos parámetros del regulador, cuya función de transferencia es
Gc(s) = K(1 + sT d) (4.5.31) Si lo que se desea es diseñar un regulador de tipo PID para controlar un sistema. Dicho regulador tiene tres parámetros por lo que para determinar el valor de los mismos necesitamos situar la posición de tres polos del sistema en bucle cerrado. Supongamos que la ecuación del controlador es de la forma
Gc(s) = K
(
1 + 11 + sT i
sT d
)
(4.5.32) 152 4. Técnicas clásicas de control
el sistema en bucle cerrado tiene la siguiente función de transferencia
G(s) = Gp(s)Gc(s)1 + Gp(s)Gc(s) (4.5.33) con lo que la ecuación característica del sistema en bucle cerrado viene definida por
1 + Gp(s)K
(
1 + 11 + sT i
sT d
)
= 0 (4.5.34) Necesitamos fijar las posiciones de tres polos p1,2,3, por lo que podemos su-poner que dos de estos son complejos conjugados de forma que tengan un cierto amortiguamiento ξ y una cierta frecuencia ωn, y el otro es un polo situado en −α
es decir
p1,2 = ωn(−ξ ± j√1 − ξ2) (4.5.35)
p3 = −α (4.5.36) La determinación de los valores de los parámetros del controlador PID se haría de forma análoga a lo visto para los controladores con dos parámetros.
Ejemplo 4.5.2: Sea el sistema mostrado en la figura 4.7, cuya función de trans-ferencia es
Gp(s) = 1
s(s + 2)( s + 8) (4.5.37) y supongamos que queremos introducir un controlador en el sistema de forma Gc(s) u yr +-1s(s+2)(s+8) e
Fig. 4.7: Sistema del ejemplo 7.5.2
que el sistema tenga una tasa de amortiguamiento ξ = 0 .5 y una ωn = 3 .14 . Si buscamos fijar sólo los polos dominantes, con estas especificaciones tendríamos que dichos polos deberían estar en
p1,2 = ωn(−ξ ± j√1 − ξ2) = −1.57 ± j2.72 (4.5.38) Dado que el sistema es de tipo 1, con lo que incluye un integrador en cadena abierta, introduciremos un regulador de tipo PD en el sistema K(s + a). Obtenga-mos en primer lugar la ecuación característica del sistema en bucle cerrado 4.5. Métodos analíticos de diseño de controladores PID 153
1 + K(s + a) 1
s(s + 2)( s + 8) = s3 + 10 s2 + (16 + K)s + Ka = 0 (4.5.39) por otra parte, queremos fijar las posiciones de dos de los polos en las posiciones 4.5.38. Como el sistema es de tercer orden el tercer polo estará en una posición
(s + b) con lo que la ecuación característica en bucle cerrado tendrá el aspecto siguiente
(s+b)( s2 +2 ξω ns+ω2
n
) = s3 +(2 ξω n +b)s2 +(2 ξω nb+ω2
n
)s+ω2
n
b = 0 (4.5.40) Al igualar las ecuaciones características 4.5.39 y 4.5.40 obtenemos las ecua-ciones siguientes
10 = 2ξω n + b
16 + K = 2ξω nb + ω2
n
(4.5.41)
Ka = ω2
n
b
y despejando en 4.5.41 tenemos que:
b = 10 − 2ξω n = 6 .86
K = 2ξω nb + ω2
n
− 16 = 15 .4 (4.5.42)
a = ω2
n
bK = 4 .392
con lo que el regulador PD que sitúa los polos dominantes en las posiciones desea-das en cadena cerrada es
Gc(s) = 15 .4( s + 4 .392) (4.5.43) Nótese que la posición del tercer polo en b = 6 .86 está muy alejada de los polos dominantes, por lo que su influencia es muy pequeña.
Determinación basada en el lugar de las raíces
El método del lugar de las raíces , introducido por Evans en 1954, es un método gráfico para determinar las posiciones de los polos en cadena cerrada del sistema a partir de las posiciones de los polos y los ceros del sistema en cadena abierta a medida que alguno de los parámetros (típicamente la ganancia) varía desde cero hasta infinito. 154 4. Técnicas clásicas de control 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 500.2 0.4 0.6 0.8 11.2 1.4
(a) (b) y(t) t (s)
Fig. 4.8: Respuesta del sistema del ejemplo 7.5.2: (a) en bucle abierto y (b) en bucle cerrado con el controlador diseñado
En la práctica, el dibujo del lugar de las raíces de un cierto sistema nos puede indicar si será suficiente variar la ganancia del sistema para alcanzar las especifi-caciones que nos han dado. Es decir, en el caso de que el lugar de las raíces pase por las posiciones deseadas para los polos en cadena cerrada bastará con ajustar al valor adecuado la ganancia del sistema para cumplir con las especificaciones de respuesta transitoria exigidas. Si esto no ocurre, será necesario introducir en el sistema un controlador que modifique el lugar de las raíces en la forma adecuada. Si recordamos el efecto que tenían la adición de polos y ceros al sistema, es relativamente simple determinar que tipo de estructura deberá tener el controlador que introduzcamos en el sistema.
Efecto de la adición de polos . La adición de un polo a la función de trans-ferencia en bucle abierto de un sistema tiene como efecto el añadir una ra-ma más en el lugar de las raíces y como consecuencia las ramas tienden a acercarse o incluso introducirse en el semiplano real positivo. Esto tiende a disminuir la estabilidad relativa del sistema y hace que el tiempo de estable-cimiento de la respuesta sea mayor. Este efecto se puede apreciar en la figura 4.9, donde a medida que se añaden polos aumenta el número de ramas y éstas están cada vez más cerca del semiplano con parte real positiva. Llegando en el caso (c), al introducirse las ramas en el la parte real positiva, para valores altos de la ganancia a inestabilizarse el sistema. 4.5. Métodos analíticos de diseño de controladores PID 155 xx(b) 0jω
σxx(a) 0jω
σxx(c) 0jω
σ
Fig. 4.9: Efecto de la adición de un polo sobre el lugar de las raíces
Efecto de la adición de ceros . La adición de un cero a la función de transfe-rencia en cadena abierta disminuye el número de ramas que se van al infinito por lo que aumenta la estabilidad relativa del sistema al desplazarse las ramas hacia la parte real negativa del plano complejo. Hace al sistema más rápido en su respuesta al introducir un efecto derivativo que anticipa la evolución del sistema. (b) x xx 0jω
σοxxx0jω
σο
(a) (c) xxx0jω
σο
Fig. 4.10: Efecto de la adición de un cero sobre el lugar de las raíces
Este efecto se puede apreciar en la figura 4.10, donde al añadir un cero a un sistema con tres polos como el mostrado en 4.9c sólo dos ramas se van al infinito y ninguna de ellas entra en la parte real positiva, por lo que el sistema no se inestabilizaría para ningún valor de la ganancia.
Ejemplo 4.5.3: Para el sistema de la figura 4.7 se puede observar que si se intro-156 4. Técnicas clásicas de control
duce un regulador proporcional y se varía el valor de la ganancia K
1 + KG p(s) = 0 (4.5.44) los polos de la ecuación característica del sistema se mueven por las ramas del lugar de las raíces, que para este sistema tendría un aspecto como el indicado en la figura 4.11. Si observamos las posiciones de los polos dominantes Pd del sistema se puede apreciar que el lugar de las raíces no pasa por ellos, por lo que no basta con un regulador proporcional. Si deseamos que el lugar de las raíces pase por dichos puntos y que estos sean los polos dominantes en cadena cerrada del sistema, nos basta con introducir un regulador de tipo PD. Este regulador al introducir un cero, si se elige la posición del mismo adecuadamente, puede hacer pasar el lugar de las raíces por las posiciones deseadas. xx x-2 -8 0
ωn
θ
Pd
P' d
Fig. 4.11: Lugar de las raíces para el sistema del ejemplo 7.5.3
En este caso, es necesario determinar en que posición se introducirá el cero y el valor de la ganancia para que los polos se sitúen exactamente en los puntos deseados. Recordando que el lugar de las raíces era el lugar geométrico de los puntos que verificaban la ecuación característica del sistema en bucle cerrado 1 +
G(s) = 0 , y que esta ecuación daba lugar a los criterios del modulo y argumento
|G(s)| = 1 (4.5.45)
∠G(s) = (2 q + 1) π q = 0 , ±1, ±2, . . . (K > 0) (4.5.46) si la función de transferencia G(s) tiene la forma
G(s) = K (s + z1)( s + z2) . . . (s + zm)(s + p1)( s + p2) . . . (s + pn) (4.5.47) 4.5. Métodos analíticos de diseño de controladores PID 157
donde n ≥ m las expresiones 4.5.45 y 4.5.46 quedarán de la siguiente forma
|G(s)| = |K|
∏mi=1 |s + zi|
∏nj=1 |s + pj | = 1 (4.5.48) o lo que es lo mismo
|K| =
∏nj=1 |s + pj |
∏mi=1 |s + zi| =
∏nj=1 distancias desde pj a Pd
∏mi=1 distancias desde zi a Pd
(4.5.49) y
∠G(s) = ∠K +
m
∑
i=1
∠(s + zi) −
n
∑
j=1
∠(s + pj ) = (2 q + 1) π (4.5.50) o lo que es lo mismo
∠K +
m
∑
i=1
∠−−→
ziPd −
n
∑
j=1
∠−−→
pj Pd = (2 q + 1) π (4.5.51) En nuestro caso, dadas las posiciones de los polos del proceso y del regulador, tendremos que si queremos que el lugar de las raíces pase por las posiciones
p1,2 = ωn(−ξ ± j√1 − ξ2) = −1.57 ± j2.72 (4.5.52) deben de cumplirse los criterios del módulo 4.5.48 y del argumento 4.5.50. Los argumentos para los polos cuyas posiciones conocemos son (fig 4.12):
β1 = 120 o (4.5.53)
β2 = arctan 2.72 2 − 1.57 = 81 .01 o (4.5.54)
β3 = arctan 2.72 8 − 1.57 = 22 .93 o (4.5.55) como debe de cumplirse el criterio del argumento
180 o = α1 − (β1 + β2 + β3) (4.5.56) y despejando α1 = 43 .93 o. Conociendo este ángulo podemos determinar la posi-ción donde debe de estar situado el cero para que forme dicho ángulo con el punto donde está situado el polo dominante −1.57 + j2.72 . El resultado es que dicho cero está situado en a = 4 , 39 . Para determinar la ganancia K aplicamos el criterio del módulo
K = n1n2n3
m1
(4.5.57) 158 4. Técnicas clásicas de control xx x-2 -a -8 0
ωn=n 1
α1 β1β2 β3 θ
n2n3 m1 Pdj2.72
Fig. 4.12: Aplicación del criterio del módulo y del argumento para el ejemplo 7.5.3
y sustituyendo por los valores de n1, n 2, n 3, m 1 obtenemos que K = 15 .4. Con lo que el regulador PD que sitúa los polos dominantes en las posiciones deseadas en cadena cerrada es
Gc(s) = 15 .4( s + 4 .39) (4.5.58) Como se puede apreciar los valores que se han obtenido por ambos métodos son similares (Fig.4.13). 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 500.2 0.4 0.6 0.8 11.2 1.4
(a) (b) y(t) t (s)
Fig. 4.13: Respuesta del sistema del ejemplo 7.5.3: (a) en bucle abierto y (b) en bucle cerrado con el controlador diseñado 4.5. Métodos analíticos de diseño de controladores PID 159
4.5.3. Discretización de un controlador PID
Con lo visto hasta este momento sería factible dadas unas ciertas especifica-ciones diseñar un regulador continuo para un cierto proceso. La mayoría de los procesos industriales son de naturaleza continua por lo que resulta natural dise-ñar el controlador en tiempo continuo. Sin embargo, hoy en día, la mayoría de los controladores son de tipo discreto, por lo que resulta necesario discretizar el regu-lador para poderlo implantar. Tal y como se ha visto, la ecuación diferencial de un controlador PID tenía la forma
u(t) = K[e(t) + 1
Ti
∫ t
0
e(τ )dτ + Td
de (t)
dt ] (4.5.59) a la hora de discretizar esta expresión el término derivativo podemos sustituirlo por la diferencia de primer orden y el término integral podríamos aproximarlo por el primero de los métodos de Euler (aproximación rectangular hacia adelante). Esto nos daría como resultado
u(k) = K[e(k) + TTik−1∑
i=0
e(i) + Td
T (e(k) − e(k − 1))] (4.5.60) para obtener una expresión recursiva que evite el sumatorio, se obtiene la expresión para k − 1, que será
u(k − 1) = K[e(k − 1) + TTik−2∑
i=0
e(i) + Td
T (e(k − 1) − e(k − 2))] (4.5.61) y restando de la expresión 4.5.60 la expresión 4.5.61 obtenemos la expresión del controlador PID discretizado
u(k) = u(k − 1) + [ q0e(k) + q1e(k − 1) + q2e(k − 2)] (4.5.62) donde
q0 = K
(
1 + Td
T
)
q1 = −K
(
1 + 2 Td
T − TTi
)
(4.5.63)
q2 = K Td
T
La función de transferencia en z queda de la forma
G(z) = U (z)
E(z) = q0 + q1z−1 + q2z−2
1 − z−1 (4.5.64) 160 4. Técnicas clásicas de control
Si se hubiese aproximado la integral mediante la aproximación trapezoidal, la expresión de u(k) sería
u(k) = K
[
e(k) + TTi
(
e(0) + e(k)2
k−1
∑
i=1
e(i)
)
Td
T (e(k) − e(k − 1))
]
(4.5.65) y el controlador quedaría
u(k) = u(k − 1) + [ q0e(k) + q1e(k − 1) + q2e(k − 2)] (4.5.66) donde
q0 = K
(
1 + Td
T + T
2Ti
)
q1 = −K
(
1 + 2 Td
T − T
2Ti
)
(4.5.67)
q2 = K Td
T
Ejemplo 4.5.4: Supongamos que se ha diseñado un controlador PID continuo con parámetros K = 2 , Td = 5 , Ti = 50 que se desea discretizar con un periodo de muestreo de T = 1 segundos. Utilizando la integración rectangular hacia adelante los parámetros que resultan para el controlador son
q0 = K
(
1 + Td
T
)
= 12
q1 = −K
(
1 + 2 Td
T − TTi
)
= −21 .96 (4.5.68)
q2 = K Td
T = 10
y si se utiliza la integración trapezoidal los parámetros serán:
q0 = K
(
1 + Td
T + T
2Ti
)
= 12 .02
q1 = −K
(
1 + 2 Td
T − T
2Ti
)
= −21 .80 (4.5.69)
q2 = K Td
T = 10
Otra solución a la discretización de reguladores P ID es utilizar el teorema de los residuos con un bloqueador de orden cero y un muestreador de periodo TG(z) = (1 − z−1) ∑
Polos dentro de C
Res
( G(s)
s
11 − epT z−1
)
(4.5.70) 4.5. Métodos analíticos de diseño de controladores PID 161
Uno de los métodos más extendidos de implementación del control por compu-tador, se basa en el diseño del regulador continuo mediante técnicas clásicas en el plano s para su posterior discretización utilizando el teorema de residuos. Nótese que la implementación por computador implica la introducción de un muestreador y de un bloqueador, generalmente de orden cero. La función de transferencia de este bloqueador es 1−e−sT
s
, cuyo numerador equivalente en el plano z es (1 − z−1).En los dos siguientes ejemplos se muestra el método indicado, con la particularidad de aplicar en uno de ellos la discretización a un sistema con polos múltiples.
Ejemplo 4.5.5: Se pretende efectuar un control por computador de un sistema continuo con función de transferencia G(s) = 1
s(s+2)
, con las especificaciones temporales de coeficiente de amortiguamiento ζ = 0 .5 y tiempo de establecimiento
ts = 2 seg .Gc(s) u yr +-1s(s+2) e
Fig. 4.14: Sistema del ejemplo 7.5.5
De las especificaciones temporales se obtiene la posición de los polos domi-nantes del sistema
ζ = cos (θ) → θ = 60 o
ts ≈ πσ → σ = 1 .57
p1,2 = −1.57 ± j2.72
Como se observa en la figura 4.15, para alcanzar los polos dominantes hay que introducir un regulador de tipo P D real con la función de transferencia Gc(s) =
K s+as+b . Para ajustar la posición del polo en −b, utilizaremos el método de cance-lación del cero del regulador en −a con el polo del sistema en −2. Utilizando el criterio del argumento tenemos
∑ α − ∑ β = (2 q + 1) πα1 = 81 o, β 1 = 120 o, β 2 = 81 o
α1 − β1 − β2 − β3 = (2 q + 1) πβ3 = −(2 q + 1) π − 120 o = 60 o162 4. Técnicas clásicas de control xx-a=-2 -b 0
ωn=n 1
β3 β1 β2 θ
n2n3Pd j2.72 xm1=n 2
α1=β2
Fig. 4.15: Distribución de polos y ceros del ejemplo 7.5.5
De esta forma la posición del polo del regulador b es −3.14 . Para la determinación de la ganancia utilizamos el criterio del módulo
K = n1n2n3
m1
= 3.14 × 2.75 × 3.14 2.75 = 9 .86
De esta forma el regulador continuo que cumple las especificaciones tiene la forma
Gc(s) = 9 .86 s + 2
s + 3 .14
Para discretizar este regulador tendremos que elegir en primer lugar el tiempo de muestreo y posteriormente, discretizarlo. Para la elección del tiempo de muestreo
T utilizaremos el método empírico de 1/8 del periodo de la respuesta amortiguada del sistema que es el doble del tiempo de pico Tp (ver apartado siguiente).
To = 2πωn
√1 − ζ2 = 2π
3.14 √1 − 0.52 = 2 .31
T = 2.31 8 = 0 .28 ≈ 0.2
El controlador discreto equivalente está representado en la figura 4.16. Para la obtención de Gc(z) utilizaremos el método de discretización basado en el teorema de los residuos. 4.5. Métodos analíticos de diseño de controladores PID 163 yr +-1s(s+2) e Gc(z) B0u(t) u(k)
Fig. 4.16: Esquema de control por computador del ejemplo 7.5.5
Gc(z) = (1 − z−1) ∑ R
[
Gc(s) 1
s
1(1 − epT z−1)
]
== (1 − z−1) ∑ R
[
9.86 s + 2 (s + 3 .14) 1
s
1(1 − epT z−1)
]
== (1 − z−1)9 .86
[ 23.14 1(1 − z−1) + 1.14 3.14 1(1 − e−3.14 ·0.2z−1)
]
= (1 − z−1)
[
6.277 1(1 − z−1) + 3 .578 1(1 − 0.533 z−1)
]
= 6 .277 + 3 .578 (1 − z−1)(1 − 0.533 z−1) = 9 .855 z − 0.702
z − 0.533
En la figura 4.17 se representan las respuestas del sistema en bucle cerrado, ante entrada escalón, con los reguladores continuo y discretizado. Como se puede observar, ambas respuestas son muy semejantes, siendo la del regulador discreto un poco más oscilatoria, debido a que el tiempo de muestreo elegido es el máxi-mo posible. Si el tiempo de muestreo se reduce, la calidad de la respuesta mejora significativamente. Si, por ejemplo, el tiempo de muestreo es T = 0 .05 seg , el regulador discreto tiene la forma
Gc(z) = 9 .86 z − 0.9073
z − 0.8547
cuya respuesta se puede observar en la figura 4.18. 164 4. Técnicas clásicas de control 0 1 2 3 4 5 600.2 0.4 0.6 0.8 11.2 1.4 tiempo yb) a)
Fig. 4.17: Resultados del sistema del ejemplo 7.5.5 en bucle cerrado con los con-troladores a) continuo y b) discretizado equivalente para T = 0 .20 1 2 3 4 5 600.2 0.4 0.6 0.8 11.2 1.4 tiempo yb) a)
Fig. 4.18: Resultados del sistema del ejemplo 7.5.5 en bucle cerrado con los con-troladores a) continuo y b) discretizado equivalente para T = 0 .05
Ejemplo 4.5.6: Para el sistema y el regulador del ejemplo anterior, supongamos que también se quiere discretizar el sistema continuo a controlar con T = 0 .2 seg .Para ello, el bloqueador de orden cero (según la figura 4.16) cambia de posición situándose en la salida del sistema. Entonces el método de discretización de los residuos se aplicará al regulador y al sistema. 4.5. Métodos analíticos de diseño de controladores PID 165
Gc(z)G(z) = (1 − z−1) ∑ R
[
Gc(s)G(s) 1
s
11 − epT z−1
]
== (1 − z−1) ∑ R
[
9.86 1
s(s + 3 .14) 1
s
11 − epT z−1
]
== 9 .86(1 − z−1)
[
1(2 − 1)!
dds
[
s2 1
s2(s + 3 .14) 1(1 − epT z−1)
]
p=0
1(−3.14) 2
11 − e−3.14 ·0.2z−1
]
Nótese que el residuo correspondiente al polo múltiple en s = 0 solamente contiene la fracción correspondiente al mayor grado s2.
Gc(z)G(z) = 9 .86(1 − z−1)
[[ −(1 − epT z−1 + ( s + 3 .14)( −T )epT z−1)(s + 3 .14) 2(1 − epT z−1)2
]
p=0
++ 13.14 2
1(1 − 0.533 z−1)
]
== (1 − z−1)
[ (1 .628 z−1 − 1) (1 − z−1)2 + 1(1 − 0.533 z−1)
]
== (1 .628 z−1 − 1)(1 − 0.533 z−1) + (1 − z−1)2
(1 − z−1)(1 − 0.533 z−1) == 0 .161 z + 0 .8261 (z − 1)( z − 0.533)
La figura 4.19 representa la comparación entre la respuesta del sistema y regulador continuos con la respuesta del sistema y regulador discretizados.
Determinación de la frecuencia de muestreo
No existe una regla fija en cuanto a la frecuencia de muestreo más adecuada a la hora de implantar un controlador discreto. Lo único que se sabe es que tiene que cumplir el teorema de muestreo (ver apartado 2.4.3). De hecho la elección exacta del periodo de muestreo depende de: las prestaciones que se desean alcanzar, la dinámica del proceso, el espectro de las perturbaciones, el actuador, el equipo de medida disponible y el coste computacional entre otros aspectos. En términos generales, se puede apreciar que, las prestaciones de un cierto sis-tema de control no varían sustancialmente a partir de un cierto periodo de muestreo. Por ello, y dado que muestrear con periodos muy pequeños suele encarecer el coste del controlador, debe de elegirse razonablemente. 166 4. Técnicas clásicas de control 0 1 2 3 4 5 600.2 0.4 0.6 0.8 11.2 1.4 tiempo y
Fig. 4.19: Resultados del sistema del ejemplo 7.5.6 en bucle cerrado con el contro-lador en tiempo continuo y el controlador discreto equivalente
Por otro lado la dinámica del sistema tiene una gran influencia sobre el tiempo de muestreo, tanto por estructura de la función de transferencia como por las cons-tantes de tiempo presentes en ella. En general cuanto mayores son las constantes de tiempo mayores pueden ser los periodos de muestreo. Otro aspecto a considerar es el espectro de las señales de perturbación presentes en el sistema. Por simplificar un poco, dos posibles reglas empíricas de elección podrían ser: Elegir el tiempo de muestreo de forma que este tiempo sea como máximo una décima parte de la menor constante de tiempo presente en el sistema. Si el sistema tiene una respuesta del tipo de un sistema de segundo orden subamortiguada, se puede elegir el periodo de muestreo de forma que sea 8 veces menor que el periodo de la oscilación de la respuesta amortiguada del sistema. Y si la respuesta es de segundo orden sobreamortiguada el periodo se puede tomar 8 veces menor que el tiempo de subida del sistema.
Ejemplo 4.5.7: Si tenemos un sistema cuyos polos dominantes están situados en
p1−2 = −1.57 ± 2.72 j (4.5.71) podríamos considerar como un periodo de muestreo aceptable T según la primera 4.5. Métodos analíticos de diseño de controladores PID 167
de las reglas vistas
T = 1.57 10 = 0 .157 (4.5.72) y según la segunda regla tendríamos que la respuesta amortiguada del sistema ten-dría un periodo to = 2 π/w d, y para el caso que estamos considerando wd = 2 .72
con lo que to = 2 .31 :
T = t0
8 = 2.31 8 = 0 .28 (4.5.73) Si promediamos entre ambos valores, para este sistema un periodo de mues-treo razonable sería de T = 0 .2 segundos. Como se observa en la figura 4.20, las respuestas continua y muestreada son muy similares. 0 1 2 3 4 5 600.2 0.4 0.6 0.8 11.2 1.4
ty(t) (b) (a)
Fig. 4.20: Respuesta del sistema del ejemplo 7.5.7: (a) en tiempo continuo y (b) muestreado con T = 0 .2 segundos
4.5.4. Diseño de controladores PID discretos
Se han estudiado hasta ahora diversos métodos analíticos para el diseño de re-guladores PID continuos, cómo pueden discretizarse posteriormente los controla-dores y cual podría ser un periodo de discretización razonable. Es posible también hacer el diseño del controlador directamente en tiempo discreto. Esto requiere en primer lugar obtener un equivalente discreto del sistema y convertir las especifi-caciones temporales que se deseen para el sistema continuo en especificaciones 168 4. Técnicas clásicas de control Gc(z) u yr +-4(z-0.9)(z-0.6) e
Fig. 4.21: Sistema de control del ejemplo 7.5.6
definidas en términos del plano z discreto (normalmente las posiciones deseadas para los polos en cadena cerrada o al menos las de los polos dominantes). Si suponemos que los polos deseados para el comportamiento del sistema con-tinuo en cadena cerrada están definidos en el plano s, en las posiciones
s = −ξω n ± jω n
√1 − ξ2 (4.5.74) y dado que la relación que existe entre la transformada de Laplace y la transformada
z venía dada por la relación z = eT s , estos puntos pasan al plano z como los puntos
z = exp [ T (−ξω n ± jω n
√1 − ξ2)] (4.5.75) Las posiciones de los polos en el plano z en cadena cerrada, para un cierto periodo de muestreo, tendrán el siguiente módulo y argumento
|z| = e−T ξω n
∠z = ±T ω n
√1 − ξ2 = ±T ω d = ±θ (4.5.76) Las técnicas de obtención de los parámetros de un regulador PID discreto que haga que el sistema cumpla unas ciertas especificaciones son básicamente las mis-mas que las comentadas para el caso continuo. Las reglas de construcción del lugar de las raíces en el plano z son también las mismas. Lo único que cambia es que ahora trabajamos con el círculo unidad estable en el plano z.
Ejemplo 4.5.8: Supongamos que se desea diseñar un controlador discreto para el sistema mostrado en la figura 4.21 de forma que el sistema tenga los polos en cadena cerrada situados en las posiciones p1,2 = 0 .3 ± j0.66 y el error en régimen permanente sea nulo. Si observamos el lugar de las raíces en el plano z, para la ecuación característica del sistema cuando se introduce una ganancia 1+ KG (z) = 0, figura 4.22, se puede observar que no es posible alcanzar las posiciones deseadas,
Pd, para los polos en cadena cerrada del sistema. Por lo que no será suficiente un regulador P. Por otra parte, como se quiere anular el error en régimen permanente y puesto que el sistema no tiene ninguno de sus polos sobre el círculo unidad, será necesario 4.5. Métodos analíticos de diseño de controladores PID 169 xIm x0.6 Re 0.9 0.66 0.3
0.75 Pd
P' d
Fig. 4.22: Lugar de las raíces para el sistema del ejemplo 7.5.6
incluir un integrador en el controlador. Como resulta necesario modificar tanto la dinámica como el tipo del sistema introduciremos un regulador de tipo PID. Dado que la expresión genérica de un PID discreto tenía la forma
Gc(z) = q0 + q1z−1 + q2z−2
1 − z−1 = q0z2 + q1z1 + q2
z(z − 1) (4.5.77) estamos introduciendo en el sistema dos polos en z = 1 y z = 0 y dos ceros cuyas posiciones y ganancia es necesario determinar. Puesto que por las especificaciones sólo se especifican las posiciones de dos polos en cadena cerrada, sólo podremos fijar los valores de dos de los parámetros del PID. Supongamos que escogemos uno de los ceros del controlador de forma que cancela el polo situado en z = 0 .9.Aplicando el criterio del modulo y el argumento a la expresión (se ha eliminado el polo y el cero que se cancelan) 1 + Gc(z)G(z) = 0 , queda:
1 + K 4( z − a)
z(z − 1)( z − 0.6) = 0 (4.5.78) es decir
|Gc(z)G(z)| = |K|
∏mi=1 |z + zi|
∏nj=1 |z + pj | = 1
∠Gc(z)G(z) = ∠K +
m
∑
i=1
∠(z + zi) −
n
∑
j=1
∠(z + pj ) = (2 q + 1) π
podremos determinar los valores de K y de a que hacen que el lugar de las raíces 170 4. Técnicas clásicas de control
pase por los polos p1,2 = 0 .3 ± j0.66 . Los argumentos para los polos son
β1 = 138 .57 o
β2 = 78 .69 o
β3 = 24 .44 o (4.5.79) y como debe de cumplirse el criterio del argumento
α1 = 180 + ( β1 + β2 + β3) = 61 .7o (4.5.80) Conociendo este ángulo podemos determinar la posición donde debe de estar situado el cero para que forme dicho ángulo con el punto donde está situado el polo dominante 0.3 + j0.66 . El resultado es que dicho cero está situado en a = 0 .265 .Para determinar la ganancia K aplicamos el criterio del módulo
4K = n1n2n3
m1
= 0.463 · 0.306 · 0.725 0.496 = 0 .302 (4.5.81) y de aquí que K = 0 .075 . El controlador PID tendrá la siguiente función de trans-ferencia:
Gc(z) = 0.075( z − 0.9)( z − 0.265)
z(z − 1) (4.5.82) Las respuestas del sistema en bucle abierto y cerrado con el controlador P ID
diseñado se muestran en la figura 4.23.
4.5.5. Estructura de un controlador PID discreto real
Se ha comentado anteriormente que la estructura de un controlador PID discre-to venía dada por la siguiente expresión
u(k) = K[e(k) + TTik−1∑
i=0
e(i) + Td
T (e(k) − e(k − 1))] (4.5.83) y en la que tomando transformada z tendremos lo siguiente
U (z) = [ K + KT Ti
11 − z−1 + KT d
T (1 − z−1)] E(z) (4.5.84) si separamos los efectos del controlador PID teórico comentado tal y como se muestra en la figura 4.24, se puede observar varios problemas. El primer problema que se puede apreciar es que el error entre dos instantes consecutivos (es decir la derivada del error) puede tomar valores muy grandes lo que al multiplicar por la ganancia derivativa puede saturar el actuador (esto sucede 4.5. Métodos analíticos de diseño de controladores PID 171 0 10 20 30 40 50 60 010 20 30 40 50 60 70 80 90 100
(a) y(k) Tiempo (ciclos de muestreo) 0510 15 20 25 30 35 40 00.2 0.4 0.6 0.8 11.2 1.4 1.6 1.8 2
y(k) (b) Tiempo ciclos de muestreo)
Fig. 4.23: Respuesta del sistema del ejemplo 7.5.6: (a) en bucle abierto y (b) en bucle cerrado con el controlador diseñado
cuando hay un ruido importante en el sistema o cuando se aplica un escalón en la entrada). Para evitar este problema, muchos controladores PID reales utilizan un esquema de control en el que la ganancia derivativa se sitúa en el lazo de realimen-tación. Esto tiene la ventaja de no experimentar variaciones bruscas que produzcan la saturación del actuador. Un segundo problema denominado saturación integral o en terminología an-glosajona efecto windup se produce debido a que en sistemas lentos la integral del 172 4. Técnicas clásicas de control u(k) Ti 1-z-1 K.T 1 +y(k) (1-z-1) TK.Td r(k) +-e(k) Ky(k) ++
Fig. 4.24: Estructura teórica de un control PID discreto
error puede llegar a tomar valores muy grandes. Esto hace que la acción de control integral tome un valor muy grande también, saturando el actuador durante mucho tiempo. Para evitar este problema se utilizan esquemas de control que permiten la desactivación del efecto integral cuando este alcanza un cierto valor. Un tercer problema se produce cuando en algunos sistemas resulta difícil co-rregir errores muy pequeños del sistema. Esto sucede en ocasiones cuando existen efectos no lineales del tipo huelgo en el sistema, o es costoso en tiempo, o en ener-gía el hacer un ajuste a un valor de referencia con una precisión mayor que un cierto valor. En estos casos se considera una banda muerta o zona de ganancia nula alrededor del valor de regimen permanente deseado para el sistema (por ejemplo de ±1 % ). En esta banda muerta se puede ignorar el error, por lo que en definitiva se está indicando el error admisible con respecto al valor de régimen permanente deseado. En ocasiones esta zona, en vez de utilizarse una ganancia nula puede uti-lizarse una ganancia mayor de lo normal para reducir el error sin correr el riesgo de inestabilizar el sistema con un valor de ganancia alto parar cualquier error. En la figura 4.25 se puede apreciar un esquema real de un controlador PID discreto utilizado en el control de ejes. En este PID se puede observar que se intro-ducen mecanismos para evitar los problemas comentados anteriormente, para ello el efecto derivativo lo tiene situado en el lazo de realimentación, además introduce una banda muerta o banda de tolerancia y el efecto integral puede desconectarse cuando alcanza un cierto valor mediante el interruptor M I realizado por softwa-re. El controlador permite además introducir dos ganancias de prealimentación (en notación anglosajona feedforward): una para compensar los retardos producidos por la inercia del sistema ( Kaf f ) y otra para compensar la fricción en el sistema. 4.6. Diseño de reguladores por síntesis directa 173 u(k) 1-z-1 Ki +Kd(1-z-1) r(k) +-e(k) F. Banda muerta ++ 1+d1z-1+d2z-2 Kp +1+n1z-1+n2z-2 -y(k) ++Kvff(1-z-1) Kaff(1-2z-1+z-2) MI y(k)
Filtro Notch Acción integral Acción derivativa
Fig. 4.25: Estructura de un PID discreto real
Además, estas prealimentaciones mejoran los errores de seguimiento. Está prepa-rado además para realizar un control en cascada con dos lazos de realimentación. Este tipo de control es usual cuando se desea realizar un control de posición en base a un motor donde se controla en el lazo interior la velocidad de giro del motor y en el lazo secundario se hace el control de posición. Es interesante observar que el efecto integral esta situado fuera del lazo interno. El esquema incluye también un filtro Notch para eliminar posibles picos de resonancia que se puedan producir en el sistema.
4.6. Diseño de reguladores por síntesis directa
Los controladores clásicos de tipo PID estaban basados inicialmente en el uso de dispositivos de tipo analógico para conseguir las acciones de control básicas de tipo proporcional, integral y derivativa. Al sustentarse en dispositivos físicos para conseguir las diferentes acciones de control, la estructura de los controladores era rígida lo que condiciona el diseño de los controladores a las acciones de control integradas en ellos. Cuando el sistema es complejo, el diseño de controladores PID puede no ser suficiente para conseguir unas prestaciones adecuadas, pues puede ocurrir que el controlador necesario para ello sea más complejo que el realizable por medio de un controlador PID clásico. El uso de los computadores para realizar el algoritmo de control del sistema ha permitido el poder diseñar e implantar algoritmos de control más complejos y además mantener todos los controladores clásicos. En esta sección se estudiará el diseño analítico de controladores en tiempo discreto que no están restringidos a 174 4. Técnicas clásicas de control
una forma básica de función de transferencia. La técnica podría plantearse también en tiempo continuo, pero su realización sería mucho más compleja, por lo que sólo se analizará para tiempo discreto. Consideremos un sistema de control donde el control está situado en la cadena directa y en el que el controlador es discreto y se realiza por medio de un compu-tador, para lo que se muestrea la salida del sistema y se introduce un bloqueador de orden cero a la salida del controlador para disponer de una acción de control en tiempo continuo sobre el sistema tal y como se muestra en la figura 4.26 (a). Gc(s) u(k) y(t) r(k) +-e(k) Gp(s) B0(s) u(t) y(k) y(k) (a) Gc(z) u(k) r(k) +-e(k) G(z) y(k) (b)
Fig. 4.26: Estructura del control (a) y equivalente discreto del sistema (b).
Si obtenemos el equivalente discreto del sistema nos queda un sistema como el mostrado en la figura 4.26 (b), donde la transformada z del conjunto bloqueador-proceso viene dado por
G(z) = Z[BG p(s)] = Z
[ 1 − e−T s
s Gp(s)
]
= (1 − z−1) ∑
Polos en C
Res
( G(s)
s
11 − epT z−1
) (4.6.1) con lo que la función de transferencia para el sistema equivalente en bucle cerrado viene dada por la siguiente expresión:
F (z) = Y (z)
R(z) = Gc(z)G(z)1 + Gc(z)G(z) (4.6.2) 4.6. Diseño de reguladores por síntesis directa 175
y de aquí, despejando la expresión del controlador Gc(z) tenemos
Gc(z) = F (z)
G(z) − F (z)G(z) = F (z)
G(z)[1 − F (z)] (4.6.3) En la expresión 4.6.3, se puede apreciar que, si conocemos la función de trans-ferencia discreta del sistema G(z) y conocemos la función de transferencia que deseamos que cumpla el sistema en bucle cerrado, F (z), es posible determinar la función de transferencia discreta del controlador. El proceso de diseño del regulador se realizará en tres pasos: Determinar la función de transferencia deseada en cadena cerrada, F (z),para el sistema, de forma que verifique las especificaciones de diseño. Obtener por igualación entre la función de transferencia deseada del sistema en bucle cerrado y la ecuación 4.6.2 la función de transferencia del controla-dor (ecuación 4.6.3). Implantar el controlador obtenido. Este método de apariencia simple requiere tener en cuenta algunos aspectos que de otro modo darían lugar a problemas muy importantes en el controlador. Entre los aspectos a considerar cuidadosamente en el diseño del regulador tenemos tres básicos: El controlador que obtengamos tiene que ser físicamente realizable, por lo que su función de transferencia tiene que ser causal. Es decir, el grado del numerador tiene que ser menor o igual que el del denominador). Es conveniente que el controlador obtenido sea lo más simple posible. Hay que tener en cuenta que pueden existir muchos controladores que den lugar a la función de transferencia deseada, F (z) en cadena cerrada. Es necesario tener en cuenta que aunque matemáticamente es posible realizar la cancelación de un polo o un cero, en la práctica no es posible llevar a cabo una cancelación exacta. Por lo que no resulta conveniente cancelar polos o ceros inestables o críticamente estables del proceso, ya que esto daría lugar a sistemas inestables.
4.6.1. Restricciones de realización física
Para garantizar la realización física del controlador es necesario que la función de transferencia discreta del mismo sea causal, es decir que no dependa de datos futuros. Esto implica que la función de transferencia del controlador debe tener un polinomio en el numerador de grado menor o igual que el del denominador. 176 4. Técnicas clásicas de control
Si suponemos que la función de transferencia del controlador es de la forma
Gc(z) = b0 + b1z−1 + . . . + bmz−m
1 + a1z−1 + . . . + anz−n
= N g c(z)
Dg c(z) (4.6.4) la condición general de realización será la relación entre los grados de los polino-mios:
gr [Dg c(z)] ≥ gr [N g c(z)] (4.6.5) Analicemos un poco más detalladamente esta condición. Para ello observemos que expresando F (z), G(z) y Gc(z) en función de los polinomios de sus numera-dores y denominadores
F (z) = N f (z)
Df (z) (4.6.6)
G(z) = N g (z)
Dg (z) (4.6.7)
Gc(z) = N g c(z)
Dg c(z) (4.6.8) la condición 4.6.5, teniendo en cuenta 4.6.3, se puede expresar de la siguiente for-ma
gr [N f (z)] + gr [Dg (z)] ≤ gr [N g (z)] + gr [Df (z) − N f (z)] (4.6.9) Como además de ser realizable físicamente el controlador, también debe de serlo el sistema en cadena cerrada
gr [N f (z)] ≤ gr [Df (z)] (4.6.10) y teniendo en cuenta esta condición, el último término de la desigualdad 4.6.9 se puede simplificar, con lo que nos queda
gr [N f (z)] + gr [Dg (z)] ≤ gr [N g (z)] + gr [Df (z)] (4.6.11) y reubicando los términos de la desigualdad obtenemos que
gr [Dg (z)] − gr [N g (z)] ≤ gr [Df (z)] − gr [N f (z)] (4.6.12) Esta desigualdad nos dice que, la diferencia entre los grados del denominador y del numerador (en sistemas discretos esto indica el retardo del sistema) en la función de transferencia en bucle cerrado del sistema F (z), debe de ser mayor o igual que el retardo de proceso, para que el controlador sea realizable físicamente. De forma simplificada, se puede decir que el retardo de modelo F (z), tiene que ser igual o mayor que el retardo del proceso a controlar. 4.6. Diseño de reguladores por síntesis directa 177
4.6.2. Conveniencia de simplicidad
Aunque, a primera vista, se podría pensar que como la expresión general de la función de transferencia del controlador
Gc(z) = F (z)
G(z)[1 − F (z)] (4.6.13) es función de F (z), cuanto más sencilla sea F (z) más lo será Gc(z). Sin embargo, esto no es cierto pues hay que tener en cuenta que la función de transferencia del proceso aparece también y esta puede ser muy compleja. Pese a todo, conviene que, en lo posible, el controlador sea lo más sencillo que la función de transferencia del sistema y las especificaciones nos permitan. Para ello, la desigualdad 4.6.12 debe convertirse en la igualdad
gr [Dg (z)] − gr [N g (z)] = gr [Df (z)] − gr [N f (z)] (4.6.14)
4.6.3. Restricciones de estabilidad
Si observamos la figura 4.27 y obtenemos la función de transferencia en cadena cerrada del sistema se tiene que
F (z) = Gc(z)G(z)1 + Gc(z)G(z) (4.6.15) y despejando en 4.6.15 se obtiene la expresión del controlador como función de las funciones de transferencia en cadena cerrada deseada F (z) y del proceso G(z).
Gc(z) = 1
G(z)
F (z)1 − F (z) (4.6.16) Gc(z) u(k) r(k) +-e(k) G(z) y(k)
Fig. 4.27: Estructura típica de un lazo de control
La función de transferencia que se obtiene para el controlador Gc(z), indicado en la ecuación 4.6.16, tiene dos términos claramente diferenciados, un primer tér-mino es la inversa de la función de transferencia del proceso y un segundo término que es función únicamente de la función de transferencia deseada para el sistema 178 4. Técnicas clásicas de control
F (z). Es evidente que el primer término del controlador puede cancelar los polos y ceros del proceso. Matemáticamente sería posible diseñar el regulador de forma que se cancelasen en cadena abierta los polos y ceros inestables del proceso. Sin embargo estas can-celaciones, que matemáticamente no presentan problemas, son sin embargo muy conflictivas en la práctica ya que una cancelación exacta es improbable, lo que ocasiona que el sistema se inestabilice o que no se amortigüe adecuadamente. Es necesario por lo tanto evitar las cancelaciones de aquellos polos o ceros del proceso que estén fuera del círculo unidad o sobre él. Supongamos que en las expresiones del numerador y denominador de la fun-ción de transferencia del proceso indicamos separadamente los polos y ceros si-tuados en el interior del círculo unidad ( Dg i(z), N g i(z)), y los situados sobre el círculo o en el exterior ( Dg e(z), N g e(z)), de forma que
N g (z) = N g e(z)N g i(z)
Dg (z) = Dg e(z)Dg i(z) (4.6.17) Una forma de evitar las cancelaciones indeseadas con los polos y ceros ines-tables (o críticamente estables) de GcG, consiste en hacer que estos polos y ceros inestables aparezcan de forma explícita en F (z) y en 1 − F (z) de forma que can-celen los que aparecen en el término 1/G (z) del controlador. De esta forma, los polos y ceros del proceso no serán cancelados (lo que por otra parte no sería posi-ble realizar de forma exacta con el proceso y sin embargo si es posible realizar en el controlador puesto que en el controlador la G(z) que aparece es la del modelo estimado del sistema y no la real que no se conoce con exactitud). Supongamos entonces que se escoge la función de transferencia F (z) de forma que:
N f (z) = N g e(z)N f ∗(z)[Df (z) − N f (z)] = Dg e(z)[ Df (z) − N f (z)] ∗ (4.6.18) donde el asterisco expresa el resto de la función después de extraer el término correspondiente. Sustituyendo estas expresiones 4.6.17 y 4.6.18 en la 4.6.16 se obtiene que
Gc(z) = Dg (z)N f (z)
N g (z)[ Df (z) − N f (z)] = Dg e(z)Dg i(z)N g e(z)N f ∗(z)
N g e(z)N g i(z)Dg e(z)[ Df (z) − N f (z)] ∗
= Dg i(z)N f ∗(z)
N g i(z)[ Df (z) − N f (z)] ∗ (4.6.19) 4.6. Diseño de reguladores por síntesis directa 179
En la expresión 4.6.19 se aprecia que en el controlador se cancelan los polos y ceros inestables del proceso. Por otra parte, si observamos la función de transfe-rencia del sistema con controlador en cadena abierta, tenemos que
Gc(z)G(z) = Dg i(z)N f ∗(z)
N g i(z)[ Df (z) − N f (z)] ∗
N g e(z)N g i(z)
Dg e(z)Dg i(z)= N f ∗(z)N g e(z)[Df (z) − N f (z)] ∗Dg e(z) (4.6.20) En esta expresión 4.6.20 se puede apreciar que en la función de transferencia en cadena abierta del sistema no se han cancelado los polos y ceros inestables o críti-camente estables del proceso como deseábamos. Podemos resumir estas restricciones diciendo que, para que no se produzcan cancelaciones entre los polos y ceros del regulador con los polos y ceros inestables o críticamente estables del proceso, se tienen que verificar las siguiente restriccio-nes: Los ceros del numerador de la función de transferencia en cadena cerrada deseada, F (z), para el sistema, deben de contener los ceros inestables o crí-ticamente estables de la planta.
N f (z) = N g e(z)N f ∗(z) (4.6.21) La expresión polinómica [Df (z) − N f (z)] debe de contener entre sus ceros los polos inestables o críticamente estables del proceso.
[Df (z) − N f (z)] = Dg e(z)[ Df (z) − N f (z)] ∗ (4.6.22)
Ejemplo 4.6.1: Sea un sistema cuyo equivalente discreto viene dado por la si-guiente función de transferencia:
G(z) = (z − 0.5) (z − 0.2)( z − 0.9) (4.6.23) y supongamos que se quiere diseñar un controlador de forma que los polos en cadena cerrada del sistema estén situados en p1,2 = 0 .3 ± j0.6 y que además tenga un error en régimen permanente ante entrada escalón nulo. De las posiciones de los polos podemos decir que la función de transferencia
F (z) debe de tener la siguiente forma:
F (z) = N f (z)
zm(z − 0.3 + j0.6)( z − 0.3 − j0.6) = N f (z)
zm(z2 − 0.6z + 0 .45)
(4.6.24) 180 4. Técnicas clásicas de control
siendo zm una variable de ajuste, que como sabemos, no influye en la dinámica de
F (z).Por otra parte y para que el sistema tenga error en régimen permanente nulo es necesario introducir un polo en (z − 1) en la función de transferencia en cadena abierta. Es decir
G(z)Gc(z) = F (z)1 − F (z) = N f (z)[Df (z) − N f (z)] = N f (z)(z − 1)[ Df (z) − N f (z)] ∗
(4.6.25) De las ecuaciones 4.6.24 y 4.6.25, tenemos que:
[Df (z) − N f (z)] = zm(z2 − 0.6z + 0 .45) − N f (z) = ( z − 1)[ Df (z) − N f (z)] ∗
(4.6.26) Tenemos que determinar el valor de m y los polinomios N f (z) y [Df (z)−N f (z)] ∗
para obtener la expresión del controlador que satisface las especificaciones de di-seño. La restricción de estabilidad en este caso no afecta puesto que todos los polos y ceros del proceso son estables. Por otra parte la restricción de realización física nos indica que
gr [Dg (z)] − gr [N g (z)] ≤ gr [Df (z)] − gr [N f (z)] (4.6.27) es decir, en nuestro caso
2 − 1 ≤ (m + 2) − gr [N f (z)] (4.6.28) luego gr [N f (z)] ≤ m + 1 . Por simplicidad tomaremos el menor grado m que nos garantice una solución. Supongamos que m = 0 , en este caso N f (z) ≤ 1, es decir
N f (z) = p0+p1z. Por otra parte al hacer m = 0 el grado de gr [Df (z)−N f (z)] = 2 por lo que como [Df (z) − N f (z)] = ( z − 1)[ Df (z) − N f (z)] ∗ el término
[Df (z) − N f (z)] ∗ será un polinomio de grado 1. De acuerdo con esto, la ecuación 4.6.26 nos queda en la siguiente forma
(z2 − 0.6z + 0 .45) − (p0 + p1z) = ( z − 1)( n0 + n1z) (4.6.29) En esta expresión, al ser un polinomio de grado 2, tenemos tres ecuaciones (tantas como coeficientes tiene el polinomio) y cuatro incógnitas ( p0, p 1, n 0, n 1). Es decir tenemos una incógnita más que el número de ecuaciones, luego tenemos 1
grado de libertad adicional. Añadimos una condición más al sistema para igualar el número de ecuaciones con el de incógnitas, esta condición es, por ejemplo, suponer que p1 = 0 . Con esta condición tendremos que
(z2 − 0.6z + 0 .45) − (p0) = ( z − 1)( n0 + n1z) (4.6.30) 4.6. Diseño de reguladores por síntesis directa 181
de esta expresión tenemos que
1 = n1
−0.6 = n0 − n1 (4.6.31)
0.45 − p0 = −n0
y resolviendo obtenemos que la solución al sistema de ecuaciones 4.6.31 es p0 =0.85 , n 0 = 0 .4, n 1 = 1 . Por lo tanto, tendremos que
F (z) = 0.85 (z2 − 0.6z + 0 .45) (4.6.32) y el controlador tendrá la siguiente función de transferencia
Gc(z) = 0.85( z − 0.2)( z − 0.9) (z − 0.5)( z2 − 0.6z − 0.4) (4.6.33) Como se puede observar la función del controlador es más compleja que la de los reguladores PID, incluso para un caso relativamente simple como este. La res-puesta del sistema en bucle abierto y en bucle cerrado con el controlador diseñado se presenta en la figura 4.28. 0 10 20 30 40 50 60 01234567
y(k) Tiempo (ciclos de muestreo) (a) (b)
Fig. 4.28: Respuesta del sistema del ejemplo 7.6.1: (a) en bucle abierto y (b) en bucle cerrado con el controlador diseñado 182 4. Técnicas clásicas de control
4.6.4. Controladores con tiempo de establecimiento mínimo
Un tipo muy interesante de controladores son aquellos sistemas que alcanzan el régimen permanente en un número finito de periodo de muestreo. Es frecuente que el diseño del controlador se realize de forma que el error en régimen perma-nente se haga cero en un número finito de periodos de muestreo. En el caso de que el controlador haga el error cero en el menor número de periodos de muestreo posible, se le denomina controlador de tiempo mínimo ("deadbeat. en denominación anglosajona).
Sistemas sin retardo
Supongamos un proceso general, sin retardo cuya función de transferencia es de la forma
G(z) = b1z−1 + . . . + bmz−m
1 + a1z−1 + . . . + amz−m = B(z−1)
A(z−1) (4.6.34) Para una entrada escalón unitario r(k) = 1 , k = 1 , 2, . . . un sistema sin retardo tendrá una respuesta con tiempo de establecimiento mínimo si (fig. 4.27):
y(k) = r(k) = 1 ∀k ≥ m (4.6.35)
u(k) = u(m) ∀k ≥ m (4.6.36) las transformadas z de estas variables son:
R(z) = 11 − z−1
Y (z) = y(1) z−1 + y(2) z−2 + . . . + 1[ z−m + z−(m+1) + . . . ]
U (z) = u(0) + u(1) z−1 + u(2) z−2 + . . . + u(m)[ z−m + z−(m+1) + . . . ]
(4.6.37) De esta forma, la función de transferencia de un sistema con un tiempo de estable-cimiento mínimo tendrá la siguiente expresión
F (z) = Y (z)
R(z) = p1z−1 + p2z−2 + . . . + pmz−m
= p1zm−1 + p2zm−2 + . . . + pm
zm = P (z)
(4.6.38) donde
p1 = y(1)
p2 = y(2) − y(1) (4.6.39) ... (4.6.40)
pm = 1 − y(m − 1) 4.6. Diseño de reguladores por síntesis directa 183
y
U (z)
R(z) = q0 + q1z−1 + . . . + qmz−m = Q(z) (4.6.41) donde
q0 = u(0)
q1 = u(1) − u(0) (4.6.42) ... (4.6.43)
qm = u(m) − u(m − 1)
La expresión 4.6.38 es equivalente a que el modelo F (z) tenga m polos en el origen. Este modelo corresponde a un sistema con respuesta de tiempo mínimo (dead-beat). Nótese que Y (z)/U (z) y U (z)/R (z) tienen un número finito de términos y retardos, iguales a m, lo que demuestra que son polinomios del mínimo orden posible. Si recordamos que la expresión de la función de transferencia en cadena cerrada tenía la expresión
F (z) = Y (z)
R(z) = Gc(z)G(z)1 + Gc(z)G(z) (4.6.44) y que buscamos determinar el controlador Gc(z) que haga que la respuesta del sistema tenga un tiempo de establecimiento mínimo y sea la descrita por los po-linomios P (z) y Q(z). Tenemos que, al haber supuesto que la entrada al sistema es un escalón unitario y que el error en regimen permanente ante dicha entrada se debe de ser nulo, el valor en regimen permanente de la salida del sistema en lazo cerrado tendrá la siguiente expresión
y(∞) = lim
z→1
(z − 1) R(z)F (z)= lim
z→1
zz − 1
Gc(z)G(z)1 + Gc(z)G(z)= Gc(1) G(1) 1 + Gc(1) G(1) = 1 (4.6.45) De aquí se puede observar que
p1 + p2 + . . . + pm = 1
q0 + q1 + . . . + qm = u(m) = 1
G(1) (4.6.46) donde G(1) es la ganancia estática del proceso. 184 4. Técnicas clásicas de control
Por otra parte de la expresión 4.6.44 obtenemos que
Gc(z) = 1
G(z)
F (z)1 − F (z) (4.6.47) y de esta expresión 4.6.47 y 4.6.38 tenemos que
F (z) = P (z) (4.6.48) Además de las expresiones 4.6.38 y 4.6.41 se deduce que
G(z) = Y (z)
U (z) = P (z)
Q(z) (4.6.49) con lo que la expresión del controlador quedará en la forma
Gc(z) = Q(z)1 − P (z) = q0 + q1z−1 + . . . + qmz−m
1 − p1z−1 − . . . − pmz−m (4.6.50) los parámetros de este controlador se obtienen comparándolos con los coeficientes de la función de transferencia del proceso
G(z) = b0 + b1z−1 + . . . + bmz−m
1 + a1z−1 + . . . + amz−m (4.6.51) con los coeficientes que se obtienen expresando G(z) a partir de las condi-ciones de tiempo mínimo y de error de posición cero en régimen permanente que imponemos, es decir
G(z) = P (z)
Q(z) = p1z−1 + . . . + pmz−m
q0 + q1z−1 + . . . + qmz−m =
p1
q0
z−1 + . . . + pm
q0
z−m
1 + q1
q0
z−1 + . . . + qm
q0
z−m
(4.6.52) con lo que obtenemos
q1 = a1q0 p1 = b1q0
q2 = a2q0 p2 = b2q0
...
qm = amq0 pm = bmq0 (4.6.53) y de la expresión p1 + p2 + . . . + pm = 1 se obtiene que
q0 = 1
b1 + . . . + bm
= u(0) (4.6.54) Como se puede observar de 4.6.53 y 4.6.54 los parámetros del controlador que nos da un tiempo de establecimiento mínimo y error en régimen permanente 4.6. Diseño de reguladores por síntesis directa 185
cero para una entrada escalón se pueden calcular de una forma muy simple. Es interesante observar que el valor inicial de la salida del controlador, u(0), depende sólo de la suma de los parámetros b del proceso. Esta suma tiende a decrecer a medida que el tiempo de muestreo es más pequeño, por lo que el valor de u(0)
aumenta cuanto menor es el periodo de muestreo (hay que tener en consideración este efecto ya que puede llegar a saturar los actuadores). Si introducimos los coeficientes 4.6.53 y 4.6.54 en la expresión del controlador 4.6.50, nos queda que
Gc(z) = Q(z)1 − P (z) = q0A(z−1)1 − q0B(z−1) (4.6.55) Si comparamos esta expresión 4.6.55 con la función de transferencia del pro-ceso 4.6.34, se puede observar que los ceros del controlador se cancelan con los polos del proceso, quedando como función de transferencia del sistema en cadena cerrada la expresión siguiente
F (z) = Gc(z)G(z)1 + Gc(z)G(z)=
q0A(z−1)1−q0B(z−1)
B(z−1)
A(z−1)
1 + q0A(z−1)1−q0B(z−1)
B(z−1)
A(z−1)
= q0B(z−1)= q0B(z)
zm
= p1zm−1 + . . . + pm
zm
= p1z−1 + . . . + pmz−m (4.6.56) que era la que se deseaba obtener. Como se puede observar un sistema con tiempo de establecimiento mínimo tiene todos sus polos en el origen de coordenadas del plano z.En esta sección se ha visto una técnica directa de obtención del regulador para estos sistemas de tiempo mínimo, pero también se podría haber obtenido el regula-dor aplicando el método de síntesis directa visto en la sección anterior, para lo cual habría que tener en cuenta que la función de transferencia en cadena cerrada es de la forma F (z) = P (z) = p1z+p2
z2
y que el controlador debe de incluir un polo en
(z − 1) para anular el error en régimen permanente.
Ejemplo 4.6.2: Supongamos que se tiene un sistema con función de transferencia
G(z) = (z − 0.5) (z − 0.2)( z − 0.9) = z−1 − 0.5z−2
1 − 1.1z−1 + 0 .18 z−2 = B(z−1)
A(z−1) (4.6.57) 186 4. Técnicas clásicas de control
y que se desea diseñar un controlador de forma que la respuesta del sistema sea de tiempo mínimo y con error en régimen permanente nulo. Como el sistema es de orden dos, la respuesta del sistema se alcanzará como mínimo en dos periodos de muestreo, luego la respuesta del sistema será un polinomio del tipo
F (z) = P (z) = p1z−1 + p2z−2 (4.6.58) De acuerdo con 4.6.54 tenemos que:
q0 = 1
b1 + b2
= 2 (4.6.59) y de las expresiones 4.6.53 tenemos que
q1 = a1.q 0 = −2.2
q2 = a2.q 0 = 0 .36
p1 = b1.q 0 = 2
p2 = b2.q 0 = −1 (4.6.60) y de 4.6.59 y 4.6.60 tenemos que la función de transferencia del regulador será
Gc(z) = Q(z)1 − P (z) = 2 − 2.2z−1 + 0 .36 z−2
1 − 2z−1 + z−2 (4.6.61) y la función de transferencia
F (z) = P (z) = 2 z−1 − z−2 (4.6.62) La secuencia de salida de el sistema en cadena cerrada ante entrada escalón uni-tario vendrá dada por la secuencia y(k) = 0 , 2, 1, 1, 1, 1, . . . para k = 0 , 1, 2 . . . .Como se puede observar en la figura 4.29 este tipo de regulador alcanza rápida-mente el valor de régimen permanente pero la sobreoscilación es considerable.
Sistemas con retardo
En el caso de que el proceso tenga retardo, la función de transferencia del proceso tiene una expresión general de la forma
G(z) = z−d b1z−1 + . . . + bmz−m
1 + a1z−1 + . . . + amz−m (4.6.63) esta expresión puede verse como un caso particular de uno más general de la forma
G(z) = b′
1
z−1 + . . . + b′
v
z−v
1 + a1z−1 + . . . + avz−v (4.6.64) 4.6. Diseño de reguladores por síntesis directa 187 0 10 20 30 40 50 60 01234567
(a) (b) y(k) Tiempo (ciclos de muestreo)
Fig. 4.29: Respuesta del sistema del ejemplo 7.6.2: (a) en bucle abierto y (b) en bucle cerrado con el controlador diseñado
donde v = m + d y en el que los coeficientes de la expresión 4.6.64 son los siguientes
am+1 = . . . = av = 0
b′
1
= b′
2
= . . . = b′
d
= 0
b′
d+1
= b1
...
b′
d+m
= bm (4.6.65) Si como en el caso anterior queremos que el sistema tenga un tiempo de esta-blecimiento mínimo. Es necesario que
y(k) = r(k) = 1 ∀k ≥ v = m + d (4.6.66)
u(k) = u(m) ∀k ≥ m (4.6.67) Se puede aplicar directamente lo visto para el caso anterior, con la salvedad de que ahora se anularán los d parámetros finales de Q(z)(qm+1 = . . . = qv = 0 )y los d iniciales de P (z) y los p1 = . . . = pd = 0 con lo que la función de transferencia del controlador quedará
Gc(z) = Q(z)1 − P (z) = q0 + q1z−1 + . . . + qmz−m
1 − pd+1 z−1 − . . . − pm+dz−m+d (4.6.68) 188 4. Técnicas clásicas de control
es decir,
Gc(z) = q0A(z−1)1 − q0B(z−1)z−d (4.6.69) y por tanto la función de transferencia resultante en cadena cerrada quedará de la forma
F (z) = q0B(z−1)z−d
= q0B(z)
zm+d (4.6.70) y su ecuación característica será zm+d = 0 . Es decir, tendrá m + d polos en el origen de coordenadas del plano z.
Ejemplo 4.6.3: Supongamos que se tiene un sistema con función de transferencia
G(z) = z−2 z−1 − 0.5z−2
1 − 1.1z−1 + 0 .18 z−2 = B(z−1)
A(z−1) (4.6.71) y que se desea diseñar un controlador de forma que la respuesta del sistema sea de tiempo mínimo y con error en régimen permanente nulo. Como el sistema es de orden cuatro, la respuesta del sistema se alcanzará como mínimo en cuatro periodos de muestreo, luego la respuesta del sistema será un polinomio del tipo
F (z) = P (z) = p1z−1 + p2z−2 + p3z−3 + p4z−4 (4.6.72) De acuerdo con 4.6.54 tenemos que:
q0 = 1
b1 + b2 + b3 + b4
= 2 (4.6.73) y de las expresiones 4.6.53 tenemos que
q1 = a1.q 0 = −2.2
q2 = a2.q 0 = 0 .36
q3 = a3.q 0 = 0
q4 = a4.q 0 = 0
p1 = b1.q 0 = 0
p2 = b2.q 0 = 0
p3 = b3.q 0 = 2
p4 = b4.q 0 = −1 (4.6.74) y de 4.6.73 y 4.6.74 tenemos que la función de transferencia del regulador será
Gc(z) = Q(z)1 − P (z) = 2 − 2.2z−1 + 0 .36 z−2
1 − 2z−3 + z−4 (4.6.75) 4.6. Diseño de reguladores por síntesis directa 189
y la función de transferencia en cadena cerrada será
F (z) = P (z) = 2 z−3 − z−4 (4.6.76) La secuencia de salida de el sistema en cadena cerrada ante entrada escalón unitario vendrá dada por la secuencia y(k) = 0 , 0, 0, 2, 1, 1, 1, 1, . . . para k =0, 1, 2, . . . (Fig.4.30). 0 10 20 30 40 50 60 70 01234567
(a) (b) y(k) Tiempo (ciclos de muestreo)
Fig. 4.30: Respuesta del sistema del ejemplo 7.6.3: (a) en bucle abierto y (b) en bucle cerrado con el controlador diseñado 190 4. Técnicas clásicas de control
Bibliografía
R. Aracil y A. Jiménez, Sistemas discretos de control, Servicio de publica-ciones ETSIIM-UPM , 1980. K. Aström and T. Hägglund, PID Controllers:Theory, Design, and Tuning,
ISA - The Instrumentation, Systems, and Automation Society , 1994. G.F. Franklin, J.D. Powell and M. Workman, Digital Control of Dynamics Systems, Addison Wesley , 1997. R. Isermann, Digital Control Systems: Volume I Fundamentals, Determinis-tic Control , Springer-Verlag , 1989. K. Ogata, Discrete-Time Control Systems, Prentice Hall International Edi-tion , 1995. K. Ogata, Ingeniería de control moderna, Prentice Hall , 1998. E. Andrés Puente, Regulación automática, Servicio de publicaciones ETSIIM-UPM , 1993. 5. CONTROL DE SISTEMAS POR REALIMENTACIóN DE ESTADO
5.1. Planteamiento del problema
La formulación del modelo de un sistema en el espacio de estados presenta considerables ventajas a la hora de diseñar esquemas de control para sistemas con múltiples entradas de control y salidas del sistema. Aunque es probablemente en este tipo de problemas donde m ·s clara y evidente se muestra esta técnica de re-presentación, an ·lisis y diseño de sistemas de control, no es el único caso ya que también en sistemas con una entrada y una salida puede resultar de considerable utilidad. En este capítulo se van a tratar varios aspectos b ·sicos del diseño de sistemas de control en el espacio de estados. En primer lugar se estudiar · la técnica de con-trol por realimentación del estado (interno) del sistema, suponiendo inicialmente que dicho estado es accesible (medible). A continuación se tratar · el caso en el que las variables de estado del sistema no son accesibles, y se introducir · el concep-to de observador de estado, para posteriormente estudiar el caso de sistemas con variables accesibles y otras que no lo son. Por último se estudiar · el caso de sis-temas sometidos a perturbaciones y como pueden diseñarse observadores que nos permitan estimar de forma óptima el estado del sistema para poder controlarlo con precisión. El esquema b ·sico de control por realimentación de estado se puede ver en la figura 5.1, donde se puede apreciar la diferencia existente entre esta técnica y los reguladores de tipo PID tradicionales. Quiz ·s la diferencia m ·s evidente es que en esta técnica el regulador est · en la realimentación y que esta realimentación no utiliza la salida sino el estado para determinar la acción de control. Esto permite tener un mayor grado de precisión en las acciones ya que es posible diferenciar las causas por las que la salida no toma el valor adecuado lo que no es posible con la estructura cl ·sica de entrada/salida del regulador PID. El diseño de reguladores y de observadores de estado requieren de dos con-ceptos b ·sicos que son la controlabilidad y la observabilidad . Ambos conceptos fueron introducidos por R.E. Kalman. El concepto de controlabilidad es la base de las soluciones al problema de ubicación de los polos del sistema realimentado y el concepto de observabilidad lo 192 5. Control de sistemas por realimentación de estado BACKu x yr + +- + f
Fig. 5.1: Diagrama de bloques de un sistema realimentado en estado
es para el diseño de observadores de estado.
5.1.1. Modos observables y controlables
Supongamos un cierto sistema en tiempo continuo expresado en la forma ca-nónica de Jordan con autovalores distintos. Para este sistema se vio que la matriz de transición tomaba la forma
eAt =
eλ1t 0 . . . 00 eλ2t . . . 0
... ...
0 0 . . . eλnt
(5.1.1) y el componente i-ésimo de la solución de la ecuación de estado x se podía expresar como
xi(t) = eλi(t−t0)xi(t0) +
∫ tt0
eλi(t−τ )Biu(τ )d τ (5.1.2) A este componente de la solución del vector de estado se le denomina modo de respuesta del sistema (modo con autovalor λi). La respuesta total del sistema tiene la forma
y(t) = C1x1(t) + . . . + Cnxn(t) (5.1.3) donde C1, . . . , C n son los coeficientes de la matriz C.Si observamos la expresión 5.1.2 del modo de respuesta xi del sistema, pode-mos apreciar que en el caso de que el vector fila Bi = 0 dicho modo de respuesta 5.2. Controlabilidad de un sistema 193
no se ver · influido por la entrada u(t) al sistema. Como adem ·s en la forma canóni-ca diagonal la variable xi tampoco tiene influencia de las otras variables de estado nos encontramos que no resulta posible influir sobre el valor de dicha variable de estado ni directamente (via entrada de control) ni indirectamente (via la influencia de alguna otra variable de estado). En este caso diremos que es un modo no con-trolable . Si algún modo del sistema es no controlable existir ·n estados del sistema que no son alcanzables y el sistema es no controlable. De forma similar, una columna Ci = 0 en la matriz C da como resultado que dicho modo no sea observable en la salida del sistema. Mientras que si Ci es distinto de cero tendremos un modo observable .Estos conceptos definidos de forma intuitiva para un sistema en la forma canó-nica diagonal, pueden ser generalizados como veremos a continuación.
5.2. Controlabilidad de un sistema
Dado que este tipo de técnicas de control suele realizarse por computador y que el tratamiento matem ·tico del problema es m ·s simple en el caso discreto que en el continuo; se va a suponer que el sistema
˙x(t) = Ax (t) + Bu (t)
y(t) = Cx (t) (5.2.1) es discretizado, de forma que trabajaremos con el siguiente sistema en tiempo dis-creto
x(k + 1) = Gx (k) + Hu (k)
y(k) = Cx (k) (5.2.2) donde
G = eAT
H =
∫ T
0
eAτ Bdτ (5.2.3) siendo T es el periodo con el que se muestrea el sistema. No obstante, la mayoría de los desarrollos de éste capítulo son inmediatamente aplicables a sistemas continuos sin m ·s que realizar el correspondiente cambio de nomenclatura en las matrices y vectores.
5.2.1. Controlabilidad de estado 194 5. Control de sistemas por realimentación de estado
Definición 5.2.1: Se dice que un sistema de control, de dimensión n, es comple-tamente controlable en estado si existe una entrada u definida entre [0 , n − 1]
tal que es posible hacer pasar al sistema desde un estado inicial arbitrario x(0)
a otro estado final arbitrario x(n) en un periodo finito de tiempo ( n periodos de muestreo). De forma intuitiva, podemos pensar que si una variable de estado es indepen-diente de la señal de control, entonces resulta imposible controlar la evolución de esta variable de estado y por lo tanto el sistema no ser · controlable. Veamos cómo se puede analizar si un sistema es o no controlable. Supondremos en primer lugar el caso en que la entrada de control es escalar.
Entrada u(k) escalar
Veamos la controlabilidad para un sistema lineal de tiempo discreto invariante en el tiempo con entrada de tipo escalar. Supongamos que el sistema viene definido por la siguiente ecuación de estado:
x(k + 1) = Gx (k) + Hu (k) (5.2.4) donde G : n × n, H : n × 1, x : n × 1, u : 1 × 1.Recordando que la solución de la ecuación de estado para un sistema lineal de tiempo discreto invariante tenía la forma
x(n) = Gnx(0) +
n−1
∑
j=0
Gn−j−1Hu (j)= Gnx(0) + Gn−1Hu (0) + Gn−2Hu (1) + . . . + Hu (n − 1)
(5.2.5) de donde, puesto en forma matricial, tenemos
x(n) − Gnx(0) =
[
H GH . . . Gn−1H
]
u(n − 1)
u(n − 2)
...
u(0)
(5.2.6) Dado que H es una matriz n × 1 y que G es n × n, la matriz
MC =
[
H GH . . . Gn−1H
]
(5.2.7) tendr · dimensiones n × n. En el sistema de ecuaciones 5.2.6 conocemos los esta-dos iniciales, x(0) , y final, x(n), y podemos calcular la matriz MC . Para poder 5.2. Controlabilidad de un sistema 195
determinar el vector de entradas de control se tiene que verificar que el rango de la matriz MC sea n. A la matriz MC se le denomina matriz de controlabilidad .
Definición 5.2.2 (Condición de controlabilidad del estado): La condición ne-cesaria y suficiente para que un sistema sea completamente controlable en estado es que el rango de la matriz de controlabilidad sea n.
rango (MC ) = n
Ejemplo 5.2.1: Dado el sistema definido por las ecuaciones siguientes
x1(k + 1)
x2(k + 1)
=
−1 00 −2
x1(k)
x2(k)
+
11
u(k) (5.2.8) determinar la secuencia de entradas u(0) , u (1) que hay que introducir al sistema para que este pase del estado [0 0] T en el instante 0 al estado [2 0] T en el ciclo
2.
Solución:
De acuerdo con la ecuación 5.2.6 tenemos que
x1(2)
x2(2)
−
1 00 4
x1(0)
x2(0)
=
1 −11 −2
u(1)
u(0)
(5.2.9) y de aquí sustituyendo x1(0) , x 2(0) , x 1(2) , x 2(2) por los valores deseados del es-tado inicial y final tendremos que
20
−
1 00 4
00
=
1 −11 −2
u(1)
u(0)
(5.2.10)
20
=
1 −11 −2
u(1)
u(0)
(5.2.11) donde
MC =
1 −11 −2
(5.2.12) y dado que el det (Mc) 6 = 0 , el rango de la matriz de controlabilidad es 2 y el siste-ma es completamente controlable en estado. Resolviendo el sistema de ecuaciones 5.2.11 tenemos { 2 = u(1) − u(0) 0 = u(1) − 2u(0) (5.2.13) 196 5. Control de sistemas por realimentación de estado
por tanto,
u(1) = 2 u(0)
y la solución es u(0) = 2 , u (1) = 4 .
Ejemplo 5.2.2: Supongamos un segundo sistema definido por las ecuaciones si-guientes
x1(k + 1)
x2(k + 1)
=
−1 00 −2
x1(k)
x2(k)
+
10
u(k) (5.2.14) determinar la secuencia de entradas (u(0) , u (1)) que hay que introducir al sistema para que este pase del estado [0 0] T en el instante 0 al estado [2 0] T en el ciclo
2.
Solución:
De acuerdo con la ecuación 5.2.6 tenemos que
x1(2)
x2(2)
−
1 00 4
x1(0)
x2(0)
=
1 −10 0
+
u(1)
u(0)
(5.2.15) y de aquí sustituyendo x1(0) , x 2(0) , x 1(2) , x 2(2) por los valores deseados del es-tado inicial y final tendremos que
20
−
1 00 4
00
=
1 −10 0
+
u(1)
u(0)
(5.2.16)
20
=
1 −10 0
u(1)
u(0)
(5.2.17) es decir,
2 = u(1) − u(0) (5.2.18) y para u(1) = 0 , se tiene u(0)=-2. Como
MC =
1 −10 0
(5.2.19) y dado que el rango de la matriz de controlabilidad es 1 el sistema no es com-pletamente controlable en estado. Aunque la condición de controlabilidad no se cumple, si resolvemos el sistema de ecuaciones tenemos que la solución del siste-ma de ecuaciones nos daría 2 = u(1) − u(0) , es decir que si, por ejemplo, fijamos 5.2. Controlabilidad de un sistema 197
u(1) = 0 , entonces u(0) = −2. Aunque el sistema no cumple la condición de con-trolabilidad el estado si es alcanzable desde el punto inicial que se ha considerado. En este ejemplo se observa que si el sistema no es completamente controlable existir ·n estados del sistema que no son alcanzables desde ese estado inicial, pero existir ·n otros que sí son alcanzables desde ese estado. En el ejemplo que hemos visto, el estado [2 0] T es alcanzable desde el estado inicial [0 0] T , sin embargo el estado [2 1] T no sería alcanzable desde el estado [0 0] T puesto que no tene-mos ninguna acción de control que nos permita cambiar el valor de la variable de estado x2.En general, en el caso de sistemas parcialmente controlables, habr · estados que se puedan ubicar libremente en todo el espacio de estados y habr · estados que se ubican en función de los primeros (son linealmente dependientes). øQué estados son libres y cuales son dependientes? Para ello hace falta calcular el rango de los menores de la matriz Mc. En el caso del ejemplo dado por la ecuación 5.2.14, el menor de Mc para el estado x1 es 1, mientras que el menor para el estado x2 es cero. Esto implica que en este ejemplo x1 siempre es libre y x2 siempre es dependiente de x1.
Entrada u(k) vector
Veamos la controlabilidad para un sistema lineal de tiempo discreto invariante en el tiempo con entrada vector. Supongamos que el sistema viene definido por la siguiente ecuación de estado:
x(k + 1) = Gx (k) + Hu (k) (5.2.20) donde G : n × n, H : n × r, x : n × 1, u : r × 1.En este caso la matriz de controlabilidad MC ser · una matriz n × nr , es decir
MC =
[
H GH . . . Gn−1H
]
(5.2.21) y el sistema de ecuaciones esta formado por n ecuaciones y nr incógnitas. Para que el sistema tenga solución el rango de la matriz de controlabilidad debe ser el menor entre n y nr , es decir n. Ahora bien, como el sistema tiene m ·s incógnitas que ecuaciones tenemos n(r − 1) grados de libertad para elegir los posibles valores de las entradas de control que permiten pasar del estado inicial al final. Normalmente se establece algún tipo de restricción adicional entre las variables de entrada o algún criterio para elegir de entre todas las posibles las que nos interesan. 198 5. Control de sistemas por realimentación de estado
5.2.2. Controlabilidad de la salida
En el epígrafe anterior se ha comentado el concepto de controlabilidad de es-tado. El hecho de que un sistema sea completamente controlable en estado no nos garantiza que podamos controlar la salida del sistema, salvo en el caso en el que el sistema tenga una única salida. En general cuando se diseña un sistema de control lo que se busca es controlar la salida del sistema m ·s que controlar el estado del mismo. Por ello, es necesario realizar un an ·lisis similar al anteriormente indicado para la controlabilidad completa del estado, pero en este caso para la salida.
Definición 5.2.3: Se dice que un sistema de control de dimensión n es completa-mente controlable en la salida si existe una entrada u definida entre [0 , n − 1] tal que es posible hacer pasar al sistema desde una salida inicial arbitraria y(0) a otra salida final arbitraria y(n) en un periodo finito de tiempo ( n periodos de muestreo).
Entrada u(k) escalar
Veamos la controlabilidad de salida para un sistema lineal de tiempo discreto invariante en el tiempo con entrada de tipo escalar. Supongamos que el sistema viene definido por las siguientes ecuaciones:
x(k + 1) = Gx (k) + Hu(k) (5.2.22)
y(k) = Cx (k) (5.2.23) donde G : n × n, H : n × 1, C : m × n , x : n × 1, u : 1 × 1, y : m × 1 y m ≤ n.Recordando que la solución de la ecuación de estado para un sistema lineal de tiempo discreto invariante era de la forma
x(n) = Gnx(0) +
n−1
∑
j=0
Gn−j−1Hu (j)= Gnx(0) + Gn−1Hu (0) + Gn−2Hu (1) + . . . + Hu (n − 1)
(5.2.24) de donde tenemos que la salida del sistema tendr · la siguiente expresión
y(n) = Cx (n)= CG nx(0) +
n−1
∑
j=0
CG n−j−1Hu (j) (5.2.25) 5.2. Controlabilidad de un sistema 199
y de aquí que
y(n) − CG nx(0) =
n−1
∑
j=0
CG n−j−1Hu (j)=
[
CH CGH . . . CGn−1H
]
u(n − 1)
u(n − 2)
...
u(0)
(5.2.26) Dado que H es una matriz n×1 y que G es n×n, que C es una matriz m×n, la matriz de controlabilidad de la salida MCS = [CH CGH . . . CGn−1H]
tendr · dimensiones m × n. En el sistema de ecuaciones 5.2.26 conocemos los es-tados inicial y(0) y final y(n), y podemos calcular la matriz MCS . Para poder determinar el vector de entradas de control se tiene que verificar que el rango de la matriz MCS es m. A la matriz MCS se le denomina matriz de controlabilidad de la salida .
Definición 5.2.4 (Condición de controlabilidad de la salida): La condición ne-cesaria y suficiente para que un sistema sea completamente controlable en la salida es que el rango de la matriz de controlabilidad de la salida sea m.
rango (MCS ) = m
Definición 5.2.5: La controlabilidad completa de estado implica controlabilidad completa de la salida si y sólo si las m filas de la matriz C son linealmente inde-pendientes.
Ejemplo 5.2.3: Dado el sistema definido por las ecuaciones siguientes
x1(k + 1)
x2(k + 1)
=
−1 00 −2
x1(k)
x2(k)
+
11
u(k)
y1(k)
y2(k)
=
−1 11 −1
x1(k)
x2(k)
determinemos si el sistema es completamente controlable en salida.
Solución:
Si obtenemos la matriz de controlabilidad de la salida de este sistema tenemos que:
MCS =
[
CH CGH
]
=
0 −10 1
(5.2.27) 200 5. Control de sistemas por realimentación de estado
cuyo rango es 1. Por lo que el sistema no es completamente controlable en la salida, aunque el sistema es controlable en estado ya que
Mc = [ H GH ] =
1 −11 −2
y, por tanto rg (Mc) = 2 . Esto se podía haber detectado ya que las filas de la matriz C no son linealmente independientes. En este caso, los dos menores de Mc
son distintos de cero lo que implica que podemos controlar cualquier estado. Sin embargo, en la matriz de controlabilidad de salida Mcs , el menor de y1 es −2 y el de la salida y2 es cero. Por lo que la salida y1 ser · controlable e y2 ser · linealmente dependiente de la primera. Esto significa que podremos situar libremente una salida (y1), pero la otra ( y2) estar · en función de la primera.
5.3. Observabilidad de un sistema
Un aspecto muy importante en los sistemas en el espacio de estados es la posi-bilidad de estimar u observar el valor que toman las variables de estado que no son medibles directamente. Dicha estimación o reconstrucción se realiza a partir de la medida de la salida del sistema durante un cierto número mínimo de periodos de muestreo y de los valores de la señal de control que se le introducen al sistema. Se considera que la salida siempre es medible. BACSistema u x yObservador del estado del sistema x^++ x
Fig. 5.2: Concepto de observador de estado 5.3. Observabilidad de un sistema 201
Supongamos un sistema definido por las siguientes ecuaciones:
x(k + 1) = Gx (k) + Hu(k) (5.3.1)
y(k) = Cx (k) (5.3.2) donde G : n × n, H : n × 1, C : m × n , x : n × 1, u : 1 × 1, y : m × 1 y m ≤ n.La solución de la ecuación de estado para un sistema lineal de tiempo discreto e invariante era
x(n) = Gnx(0) +
n−1
∑
j=0
Gn−j−1Hu (j) (5.3.3) y la salida del sistema tenía la siguiente expresión
y(n) = CG nx(0) +
n−1
∑
j=0
CG n−j−1Hu (j) (5.3.4) En la ecuación 5.3.4 conocemos G, H y C y conocemos también la secuencia de entradas que se le ha introducido al sistema u(0) , . . . , u(n − 1) por lo que el segundo término de la parte derecha de dicha expresión nos es completamente conocido. Puesto que este término siempre es conocido, podemos simplificar un poco el problema de determinar la observabilidad de un sistema limit ·ndonos a analizar el sistema no forzado , es decir con entrada de control cero. Dicho de otro modo, si podemos determinar para el sistema no forzado cual era el estado inicial de partida del sistema a partir de la observación de las salidas del sistema, entonces podremos determinarlo también para el sistema sujeto a una secuencia de entradas de control.
5.3.1. Observabilidad completa de estado
Definición 5.3.1: Un sistema se dice que es completamente observable si todo estado inicial x(0) puede ser determinado a partir de la observación de la salida
y(k) del sistema en un número finito de intervalos de muestreo. Considerando el sistema no forzado, las ecuaciones de este quedan reducidas a las siguientes ecuaciones
x(n) = Gx (n − 1)
y(n − 1) = Cx (n − 1) (5.3.5) y aplicando recursivamente la expresión de la ecuación de estado obtenemos que la salida del sistema ser ·
y(n − 1) = CG n−1x(0) (5.3.6) 202 5. Control de sistemas por realimentación de estado
De acuerdo con la definición de observabilidad, dada la secuencia de salidas
y(0) , y(1) , . . . , y(n − 1) un sistema ser · observable si es posible determinar a partir de ellas el estado inicial x(0) . Si expresamos las salidas en función del estado inicial tenemos
y(0) = Cx (0)
y(1) = CGx (0)
...
y(n − 1) = CG n−1x(0) (5.3.7) y en forma matricial
y(0)
y(1)
...
y(n − 1)
=
CCG
...
CG n−1
x(0) (5.3.8) A la matriz [C CG . . . CG n−1]T se le denomina matriz de observa-bilidad , MO. Como G es una matriz n × n y C es m × n, la matriz MO =
[C CG . . . CG n−1]T tiene dimensión mn × n. Por lo tanto, para que el sistema de ecuaciones tenga solución dicha matriz MO tiene que tener rango n.
Definición 5.3.2 ( Condición de observabilidad): La condición necesaria y su-ficiente para que un sistema sea observable es que el rango de la matriz de obser-vabilidad sea n.
rango (MO) = n
Una situación típica en la que un sistema no es completamente observable se produce en los sistemas en los que existen cancelaciones polo-cero. Si observa-mos en los polos y ceros del sistema la existencia de estas cancelaciones, esto nos indicar · que el sistema no ser · completamente controlable u observable. En el ca-so de un sistema con una entrada y una salida el sistema ser · a la vez controlable y observable si no se pueden efectuar cancelaciones polo-cero en la función de transferencia. 5.4. Invarianza de la controlabilidad y observabilidad 203
Ejemplo 5.3.1: Dado el sistema definido por las ecuaciones siguientes
x1(k + 1)
x2(k + 1)
=
−1 00 −2
x1(k)
x2(k)
+
11
u(k)
y(k) =
[
0.5 1
] x1(k)
x2(k)
determinar la observabilidad del mismo.
Solución:
Si obtenemos la matriz de observabilidad de estado del sistema tenemos que:
MO =
CCG
=
0.5 1
−0.5 −2
(5.3.9) cuyo rango es 2 por lo que el sistema ser · completamente observable.
5.4. Invarianza de la controlabilidad y observabilidad
Se ha visto anteriormente que un sistema admite múltiples representaciones en el espacio de estados, recordemos las formas canónicas por limitarnos a las que hemos comentado m ·s extensamente. Una pregunta que surge de forma autom ·tica es, si las condiciones de controlabilidad y observabilidad se mantienen ante un cambio de base en la representación de estado utilizada, es decir, al pasar de una forma canónica a otra. Para responder a esta cuestión, supongamos que realizamos un cambio de base, caracterizado por la matriz de transformación T , no singular, en la representación de estado
x(k + 1) = Gx (k) + Hu (k)
y(k) = Cx (k)
x(k) = T x ′(k) (5.4.1) con lo que tenemos una nueva representación de estado
x′(k + 1) = T −1GT x ′(k) + T −1Hu (k)
y(k) = CT x ′(k) (5.4.2) donde las nuevas matrices G′, H′, C′ ser ·n
G′(k + 1) = T −1GT H′(k) = T −1HC′(k) = CT (5.4.3) 204 5. Control de sistemas por realimentación de estado
y para esta nueva representación de estado la matriz de controlabilidad ser ·
M ′C =
[
T −1H T −1GT T −1H . . . T −1Gn−1T T −1H
]
=
[
T −1H T −1GH . . . T −1Gn−1H
]
= T −1 [
H GH . . . Gn−1H
]
= T −1MC (5.4.4) y puesto que T es una matriz no singular, el rango de la matriz de controlabilidad
M ′C y de la matriz de controlabilidad del sistema original MC coinciden. Por lo que si el sistema 5.4.1 es controlable, el sistema 5.4.2 en la nueva base también ser ·
controlable. Si hacemos una an ·lisis similar para la observabilidad, tendremos que la matriz de observabilidad del sistema 5.4.2 tendr · la siguiente expresión
M ′O =
CT CT T −1GT
...
CT T −1Gn−1T
=
CT CGT
...
CG n−1T
=
CCG
...
CG n−1
T
= MOT (5.4.5) y dado que se ha tomado una matriz de transformación T no singular, y de forma similar al caso de la controlabilidad, los rangos de las matrices de observabilidad
M ′O y MO son iguales. Por tanto, aunque el sistema cambie de base, si el sistema era observable seguir · siendo observable. Es importante observar que lo contrario no tiene porque verificarse, es decir, si el sistema es no observable, esto no implica que no exista ningún cambio de base que nos conduzca a otra representación de estado observable. De hecho, si el 5.5. Principio de dualidad 205
sistema es no observable (pero controlable) siempre nos es posible encontrar una nueva representación de estado que sea observable. Lo mismo podemos decir para la controlabilidad.
5.5. Principio de dualidad
Veamos qué relación existe entre la controlabilidad y la observabilidad. Supon-gamos un sistema S1 lineal e invariante en el tiempo definido por
x(k + 1) = Gx (k) + Hu (k)
y(k) = Cx (k) (5.5.1) donde las diferentes matrices tienen dimensión: x : n × 1, u : r × 1, y : m × 1, G :
n × n, H : n × r, C : m × n. Y un sistema S2, que denominaremos sistema dual
del S1, definido por las siguientes ecuaciones
x′(k + 1) = GT x′(k) + CT u′(k)
y′(k) = HT x′(k) (5.5.2) donde: x′ : n × 1, u′ : m × 1, y′ : r × 1. El principio de dualidad debido a Kalman dice lo siguiente.
Teorema 5.5.1 (Principio de dualidad): El sistema S1 es completamente con-trolable en estado sí y sólo si el sistema S2 es completamente observable, y S1 es completamente observable sí y sólo si el sistema S2 es completamente controlable en estado. Si observamos las matrices de controlabilidad y observabilidad del sistema S1
tenemos que
MC =
[
H GH . . . Gn−1H
]
MO =
CCG
...
CG n−1
(5.5.3) 206 5. Control de sistemas por realimentación de estado
y si ahora observamos las matrices de controlabilidad y observabilidad del sistema
S2 resulta lo siguiente
M ′C =
[
CT GT CT . . . (GT )n−1CT
]
=
CCG
...
CG n−1
= MO
M ′O =
HT
HT GT
...
HT (Gn−1)T
=
[
H GH . . . Gn−1H
]
= MC (5.5.4) y como se puede observar M ′O = MC y M ′C = MO tal y como indica el principio de dualidad. Este principio de dualidad se ha utilizado aunque sin nombrarlo explícitamente cuando se obtuvieron las representaciones canónicas de un sistema a partir de la función de transferencia, donde la forma canónica controlable y la observable son sistemas duales.
5.6. Control por realimentación de estado
Supongamos que queremos controlar un cierto sistema, en el que todas las va-riables de estado est ·n disponibles (medibles o accesibles) para ser utilizadas en la realimentación del estado del sistema.
Teorema 5.6.1: Si un sistema es completamente controlable en estado, entonces los polos en bucle cerrado del sistema pueden ser libremente situados en las po-siciones p1, p 2, . . . , p n que se deseen mediante una realimentación de estado con una matriz K de ganancias adecuada. Este esquema de control se muestra en la figura 5.3 har · que la ecuación carac-terística del sistema realimentado tome la forma
(z − p1)( z − p2) . . . (z − pn) = 0 (5.6.1) 5.6. Control por realimentación de estado 207
La elección de las posiciones deseadas de los polos se hace de forma que se cumplan unas ciertas especificaciones de diseño. Esta técnica es aplicable sólo a sistemas lineales invariantes en el tiempo. Veamos cómo podemos determinar el valor de la matriz de ganancias que nos permite situar los polos del sistema reali-mentado en el lugar deseado. HGCKu x yr + +- + z .I
-1
Realimentación del estado
Fig. 5.3: Realimentación del estado
Esta técnica de control nos va a permitir situar los polos del sistema reali-mentado en las posiciones que deseemos para que se cumplan ciertos requisitos en cuanto a la respuesta din ·mica del sistema. El mero hecho de poder situar los polos en las posiciones que deseemos no nos garantiza que el sistema cumpla las especificaciones de régimen permanente ya que esta técnica no permite modificar las posiciones de los ceros del sistema por lo que posteriormente veremos como ajustar la respuesta en régimen permanente del sistema.
5.6.1. Sistemas con entrada y salida escalar
Supongamos el sistema mostrado en la figura 5.3, las ecuaciones del mismo son
x(k + 1) = Gx (k) + Hu(k)
y(k) = Cx (k) (5.6.2) donde G : n × n, H : n × 1, C : 1 × n , x : n × 1, u : 1 × 1 e y : 1 × 1.Adem ·s, la señal de control al sistema viene dada por la siguiente expresión
u(k) = r(k) − Kx (k) (5.6.3) De las ecuaciones 5.6.2 y 5.6.3 tenemos que (siendo K de dimensión 1 × n)208 5. Control de sistemas por realimentación de estado
x(k + 1) = Gx (k) + H[−Kx (k) + r(k)] (5.6.4)
= [G − HK ]x(k) + Hr(k) (5.6.5) Tomando transformada z, se tiene
zX(z) = ( G − HK )X(z) + HR(z) (5.6.6) por tanto,
(zI − G + HK )X(z) = HR(z) (5.6.7)
X(z) = ( zI − G + HK )−1HR(z) (5.6.8) la salida tendr · la expresión
Y (z) = C[zI − G + HK ]−1HR(z) (5.6.9) de donde se tiene que la función de transferencia del sistema realimentado tendr ·
la siguiente forma
F (z) = Y (z)
R(z) = C[zI − G + HK ]−1H (5.6.10) en la ecuación 5.6.10 se puede observar que el denominador de la función de transferencia viene dado por el determinante de |zI − (G − HK )|. Por tanto, si deseamos un cierto comportamiento din ·mico del sistema, nos basta con ajustar las posiciones de los polos del sistema en cadena cerrada. Para ello, es necesario determinar en primer lugar las posiciones p1, p 2, . . . , p n deseadas para los polos, con ello determinamos la ecuación característica deseada y de aquí
|zI − G + HK | = ( z − p1)( z − p2) . . . (z − pn) (5.6.11) En la expresión 5.6.11 lo único que no se conoce son los valores de los coe-ficientes de la matriz K, pero igualando los coeficientes del término del lado iz-quierdo a los de la parte derecha de la expresión es posible resolver el sistema. La condición necesaria y suficiente para poder resolver este sistema es que el rango de la matriz de controlabilidad de estado sea n, es decir, que el sistema sea totalmente controlable. Adem ·s, es f ·cilmente demostrable que si el sistema es totalmente controlable la matriz (zI − G + HK ) es no singular.
5.6.2. Ajuste de las posiciones de los polos
Dado un sistema
x(k + 1) = Gx (k) + Hu(k) (5.6.12) 5.6. Control por realimentación de estado 209
completamente controlable en estado, si realimentamos el estado de forma que
u(k) = r(k) − Kx (k), para que los polos en bucle cerrado del sistema se sitúen en las posiciones z = p1, z = p2, . . . , z = pn, basta con igualar las expresiones características, es decir
|zI−G+HK | = ( z−p1)( z−p2) . . . (z−pn) = zn+a1zn−1+. . . +an−1z+an = 0
(5.6.13) En la expresión de la izquierda tenemos un polinomio en z que ser · función de los coeficientes de la matriz de realimentación de estado. Igualando dichos coefi-cientes a los de la ecuación característica deseada, obtendremos un sistema de n
ecuaciones con n incógnitas que resolviéndolo nos permite obtener los coeficientes de la matriz de realimentación.
Ejemplo 5.6.1: Supongamos el sistema definido por las siguientes ecuaciones
x1(k + 1)
x2(k + 1)
=
0 1
−1 −0.5
+
01
u(k) (5.6.14)
y(k) = [1 0]
x1(k)
x2(k)
(5.6.15) y se desea determinar los valores de las ganancias de la matriz K de realimentación de estado para que los polos en cadena cerrada del sistema estén en p1,2 = −0.5 ±
j0.5.
Solución:
Comprobemos en primer lugar que el sistema sea completamente controlable en estado, para ello obtendremos la matriz de controlabilidad
MC =
[
H GH
]
=
0 11 −0.5
(5.6.16) que tiene rango 2, por lo que el sistema es completamente controlable en estado. Dado que el sistema es controlable es factible encontrar una realimentación de estado que cumpla nuestras exigencias. De acuerdo con la ecuación 5.6.11 tenemos que
|zI − G + HK | = ( z − p1)( z − p2) . . . (z − pn) = 0 (5.6.17) que para nuestro caso concreto da la siguiente expresión
|zI − G + HK | = z2 − z + 0 .5 = 0 (5.6.18) 210 5. Control de sistemas por realimentación de estado
y sustituyendo en el primer término las matrices por sus valores y operando
z2 + z(0 .5 + k2) + (1 + k1) = z2 − z + 0 .5 = 0 (5.6.19) en la que igualando los coeficientes de ambos polinomios obtenemos
k1 = −0.5
k2 = −1.5
y la matriz de realimentación de estado ser · K = [ −0.5−1.5] . La respuesta del sistema en cadena abierta y del sistema con la realimentación de estado diseñada se muestra en la figura 5.4. 0 5 10 15 20 25 30 35 40 45 50 −0.5 00.5 11.5 22.5
y(k) Tiempo (ciclos de muestreo) (a) (b)
Fig. 5.4: Respuesta del sistema del ejemplo 8.4: (a) sin realimentación y (b) siste-ma realimentado
Determinación de las posiciones de los polos por transformación
En el caso en el que el sistema este representado en una forma que no sea la for-ma canónica controlable el método de comparación de coeficientes puede dar lugar a c ·lculos complejos para la obtención de las ganancias de realimentación, especial-mente para sistemas de orden elevado. En este caso existe posibilidad transformar el sistema a la forma canónica controlable y diseñar en esta nueva representación el regulador y después convertirlo de nuevo a la representación del sistema que teníamos inicialmente. 5.6. Control por realimentación de estado 211
Veamos como se realiza este proceso. Para ello, supongamos que se tiene un sistema con la siguiente representación de estado,
x(k + 1) = Gx (k) + Hu(k) (5.6.20)
y(k) = Cx (k) (5.6.21) cuya matriz de controlabilidad viene dada por la siguiente expresión
MC =
[
H GH . . . Gn−1H
]
(5.6.22) y supongamos que aplicamos una transformación T que transforma al sistema a la forma canónica controlable
x(k) = T x ′(k) (5.6.23) con lo que el nuevo sistema quedaría en la forma
x′(k + 1) = T −1GT x ′(k) + T −1Hu(k) (5.6.24)
y(k) = CT x ′(k) (5.6.25) Esta nueva representación tiene la siguiente matriz de controlabilidad, tal y como se vio en 5.4.4,
MC′ = T −1 [
H GH . . . Gn−1H
]
= T −1MC (5.6.26) y de aquí se puede deducir que
T = MC M −1
C′
(5.6.27) pudiéndose obtener la matriz de transformación T a partir de las dos matrices de controlabilidad. Una vez que el sistema est · en la forma canónica controlable y verifica la condición de controlabilidad completa de estado, se diseña el regulador por realimentación de estado. En esta representación tendremos que
|zI − G′ + H′K′| = ( z − p1)( z − p2) . . . (z − pn) = 0 (5.6.28) e igualando los coeficientes podremos determinar la matriz de ganancias del re-gulador K′. El sistema realimentado en la forma canónica controlable tendr · las siguientes ecuaciones
x′(k + 1) = T −1GT x ′(k) + T −1Hu(k) (5.6.29)
u(k) = −K′x′(k) + r(k) (5.6.30)
y(k) = CT x ′(k) (5.6.31) 212 5. Control de sistemas por realimentación de estado
y de aquí
x′(k + 1) = T −1GT x ′(k) − T −1HK ′x′(k) + T −1Hr(k) (5.6.32)
= (T −1GT − T −1HK ′)x′(k) + T −1Hr(k) (5.6.33)
y(k) = CT x ′(k) (5.6.34) Si ahora deshacemos la transformación, volviendo al sistema en la representación inicial, para lo cual aplicamos x′ = T −1x, tendremos que
x(k + 1) = Gx (k) − HK ′T −1x(k) + Hr(k) (5.6.35)
= (G − HK ′T −1)x(k) + Hr(k) (5.6.36)
y(k) = Cx (k) (5.6.37) y comparando con la ecuación característica de un sistema realimentado en la re-presentación original
|zI − G + HK | = 0 (5.6.38) tenemos que la matriz de ganancia de la realimentación expresada en el sistema de representación original tendr · la forma
K = K′T −1 (5.6.39) Una limitación importante a este método se produce en el caso de que el sistema sea no-controlable y se pueda transformar a la forma canónica controlable. En este caso, ser · posible determinar la matriz de ganancias K′ para controlar el sistema en la forma canónica controlable, pero la matriz de transformación T para pasar a dicha forma canónica es singular por lo que no es posible obtener la inversa de la transformación.
Ejemplo 5.6.2: Supongamos el sistema definido por las siguientes ecuaciones
x1(k + 1)
x2(k + 1)
=
−0.5 00 −1
x1(k)
x2(k)
+
01
u(k) (5.6.40)
y(k) = [1 1]
x1(k)
x2(k)
(5.6.41) y se desea determinar los valores de las ganancias de la matriz K de realimentación de estado para que los polos en cadena cerrada del sistema estén en p1,2 = z−0.5±
j0.5.5.6. Control por realimentación de estado 213 K' HGCu x yr + +- + z I
-1
Realimentación del estado Tx'
-1 u=-K'x'=-(KT)x'=-Kx
Sistema
Fig. 5.5: Transformación de un sistema a otra representación de estado para con-trolarlo.
Solución:
Comprobemos en primer lugar que el sistema sea completamente controlable en estado, para ello obtendremos la matriz de controlabilidad
MC =
[
H GH
]
=
0 01 −1
(5.6.42) que tiene rango 1, por lo que el sistema es no es completamente controlable en estado. Dado que el sistema es no controlable en estado, si se quiere encontrar una realimentación de estado que cumpla nuestras exigencias ser · necesario obtener una representación de estado que sea controlable. En este caso, obteniendo la función de transferencia del sistema tenemos que
F (z) = C[zI − G]−1H = z + 0 .5
z2 + 1 .5z + 0 .5214 5. Control de sistemas por realimentación de estado
y pasando de aquí a la forma canónica controlable tenemos que
x′
1
(k + 1)
x′
2
(k + 1)
=
0 1
−0.5 −1.5
+
01
u(k) (5.6.43)
y(k) = [1 0.5]
x′
1
(k)
x′
2
(k)
(5.6.44) esta nueva representación tiene la siguiente matriz de controlabilidad
M ′C =
[
H′ G′H′
]
=
0 11 −1.5
(5.6.45) que tiene rango 2 luego ser · controlable. Diseñaremos en esta nueva representación de estado las ganancias de la matriz de realimentación K′ que sitúan los polos del sistema en cadena cerrada en las posiciones que se desean. De acuerdo con la ecuación 5.6.11 tenemos que
|zI − G + HK | = ( z − p1)( z − p2) . . . (z − pn) = 0 (5.6.46) que para nuestro caso concreto da la siguiente expresión
|zI − G′ + H′K′| = z2 − z + 0 .5 = 0 (5.6.47) y sustituyendo en el primer término las matrices por sus valores y operando
z2 + z(1 .5 + k′
2
) + (0 .5 + k′
1
) = z2 − z + 0 .5 = 0 (5.6.48) en la que igualando los coeficientes de ambos polinomios obtenemos
k′
1
= 0
k′
2
= −2.5
y la matriz de realimentación de estado ser · K′ = [0 − 2.5] . En este caso no es posible aplicar la inversa de la transformación T para obtener la matriz de ganan-cias en la representación de estado inicial. Comprobémoslo, para ello obtenemos la matriz de transformación que nos permite pasar de una representación a otra,
T = MC M ′
C
−1
=
0 01 −1
1.5 11 0
=
0 00.5 1
(5.6.49) y como el rango de T es 1 la matriz no es invertible y el sistema no es controlable. 5.6. Control por realimentación de estado 215 HGCKu x yr + +- + z .I
-1
k0
Planta
Fig. 5.6: Realimentación de estado con ajuste de ganancia
5.6.3. Ajuste de la ganancia
Supongamos el sistema mostrado en la figura 5.6, en el que se ha incluido una ganancia k0 con el fin de ajustar la ganancia est ·tica del sistema de forma que tengamos la posibilidad de ajustar el error que comete el sistema en régimen permanente. Las ecuaciones del sistema en este caso son
x(k + 1) = Gx (k) + Hu(k) (5.6.50)
y(k) = Cx (k)
donde G : n × n, H : n × 1, C : 1 × n , x : n × 1, u : 1 × 1 e y : 1 × 1.La señal de control u(k) que se introduce al sistema viene dada por la siguiente expresión
u(k) = k0r(k) − Kx (k) (5.6.51) De las ecuaciones 5.6.50 y 5.6.51 tenemos que
x(k + 1) = Gx (k) + H[−Kx (k) + r(k)] = [G − HK]x(k) + Hk0r(k) (5.6.52) Con lo que tendremos que la función de transferencia del sistema en este caso tendr · la siguiente forma
F (z) = Y (z)
R(z) = C[zI − G + HK ]−1Hk0 (5.6.53) En la ecuación 5.6.53 se puede observar que el denominador de la función de transferencia viene dado por el determinante de la matriz [zI − (G − HK )] y216 5. Control de sistemas por realimentación de estado
en el numerador nos aparece la ganancia k0 que se ha introducido en la cadena principal. Por tanto, si deseamos un cierto comportamiento din ·mico del sistema nos basta con ajustar las posiciones de los polos del sistema en cadena cerrada y para ajustar el error en régimen permanente, una vez fijados los polos se ajusta el valor de k0.Se especifican en primer lugar las posiciones p1, p 2, . . . , p n deseadas para los polos, y con ello determinamos la ecuación característica deseada
|zI − G + HK | = ( z − p1)( z − p2) . . . (z − pn) (5.6.54) y de aquí se deducen como en el caso anterior los valores de los coeficientes de la matriz K. Una vez que se determina la matriz de realimentación de estado queda por ajustar k0.Si suponemos que la entrada al sistema es R(z), la respuesta en régimen per-manente tendr · la siguiente expresión lim
k→∞
y(k) = lim
z→1
(1 − z−1)Y (z) = lim
z→1
(1 − z−1)R(z)F (z) (5.6.55) En la ecuación 5.6.55 si se conoce el valor deseado de la salida en régimen permanente para la entrada R(z) introducida, es posible determinar el valor de la ganancia k0 de la parte derecha de la expresión, con lo que tendríamos ajustado también el régimen permanente.
Ejemplo 5.6.3: Supongamos el mismo sistema del ejemplo anterior (8.6.2), en dicho ejemplo se ha determinado los valores de la ganancia de realimentación de estado parar situar los polos en unas posiciones determinadas. Tratemos ahora de conseguir que el error en régimen permanente ante entrada escalón sea cero.
Solución:
Veamos primero si el sistema cumple la especificación respecto al error en régimen permanente. La función de transferencia del sistema en bucle cerrado
F (z) = Y (z)
R(z) = C[zI − G + HK ]−1H (5.6.56) desarrollando esta expresión tenemos
F (z) =
[
1 0
] z −10.5 z − 1
−1
01
= 1
z2 − z + 0 .5 (5.6.57) Si al sistema se le introduce en la entrada un escalón unitario, el valor de la salida en régimen permanente ser ·5.6. Control por realimentación de estado 217
lim
k→∞
y(k) = lim
z→1
(1 − z−1)R(z)F (z)= lim
z→1
(1 − z−1) zz − 11
z2 − z + 0 .5= 2 (5.6.58) Es decir, para una entrada unitaria la salida en régimen permanente del sistema nos da el valor 2. Si queremos que la salida del sistema en régimen permanente tenga valor 1, introducimos una ganancia k0 en el sistema con lo que la función de transferencia en bucle cerrado quedar · como
F (z) = Y (z)
R(z) = C[zI − G + HK ]−1Hk0 (5.6.59) y la respuesta en régimen permanente ser ·
lim
k→∞
y(k) = lim
z→1
(1 − z−1)R(z)F (z)= lim
z→1
(1 − z−1) zz − 1
k0
z2 − z + 0 .5= k0
0.5 (5.6.60) luego para que el valor de la respuesta del sistema en régimen permanente sea 1
implica que k0 = 0 .5.
5.6.4. Modificación del tipo del sistema
En determinadas situaciones no es posible ajustar satisfactoriamente la res-puesta din ·mica y la respuesta en régimen permanente del sistema a la vez con estas dos técnicas que se han comentado. Esto pasa especialmente en sistemas de orden bajo, donde los grados de libertad para elegir las posiciones de los polos en cadena cerrada del sistema son limitados. Un caso típico donde se desea que la salida del sistema siga a la entrada del mismo son los servosistemas. Si suponemos un servosistema, en el que el sistema es de orden dos, únicamente disponemos de dos posiciones de polos para ajustar. Si se ajustan las posiciones de los polos en cadena cerrada para que la din ·mica sea razonablemente r ·pida (como un sistema de segundo orden con una pequeña sobreoscilación), elegiremos un par de polos complejos conjugados. En este caso el tipo del sistema en cadena cerrada ser · cero, por lo que probablemente no anular · el error en régimen permanente del sistema ante entrada escalón y adem ·s ante entrada rampa se har · infinito. Si situ ·semos un polo en el punto +1 del plano z (equivale a un integrador) en este caso anularíamos el error en régimen permanente pues el sistema sería de tipo uno, pero entonces el sistema se haría excesivamente lento lo que puede no ser deseable. 218 5. Control de sistemas por realimentación de estado
Si el sistema fuese de tercer orden, se podrían fijar las posiciones de los tres polos de forma que: el tipo sea uno (con lo que el error en régimen permanente ante una entrada en rampa se anularía) y tenga dos polos complejos conjugados (que diesen una respuesta din ·mica buena). En el caso de que el orden del sistema no nos permita satisfacer las dos condi-ciones de diseño simult ·neamente, es posible elevar el tipo del sistema añadiendo un integrador en la cadena principal del sistema (figura 5.7). Al añadir un integra-dor al sistema, este aumenta el orden y el tipo del sistema. En definitiva hemos añadido un grado de libertad m ·s al sistema (una variable de estado adicional) que hemos utilizado para poner un polo en el círculo unidad. ++ z I
-1
rControl Integral vHCKuy+-kI
Planta Gx++zI
-1 +-eRealimentación de estado
Fig. 5.7: Realimentación de estado con integrador adicional
El sistema de la figura 5.7 tiene las siguientes ecuaciones,
x(k + 1) = Gx (k) − HKx (k) + Hkiv(k)
v(k + 1) = v(k) + e(k)
e(k) = r(k) − y(k)
y(k) = Cx (k) (5.6.61) En esta expresión podemos reescribir v(k + 1) como
v(k + 1) = v(k) + r(k) − Cx (k) (5.6.62) Si consideramos ahora un nuevo vector de estado aumentado con v(k), tendre-mos la siguiente expresión 5.6. Control por realimentación de estado 219
x(k + 1)
v(k + 1)
=
G − HK Hki
−C 1
x(k)
v(k)
+
01
r(k)
y(k) =
[
C 0
] x(k)
v(k)
(5.6.63) En la figura 5.8 se puede ver un esquema equivalente al de la figura 5.7, en el que se observa que la introducción del integrador en la cadena principal y la reali-mentación de la salida del sistema constituyen en el fondo una realimentación de estado pero a la que hemos añadido una variable de estado adicional que acumula el error cometido entre la entrada y la salida del sistema. En este esquema, la for-ma de diseño fuerza a que este polo adicional que se añade sea un integrador para aumentar el tipo del sistema. En el caso de que el sistema tuviese un número sufi-ciente de polos que pudiésemos asignar, se podría diseñar la matriz de ganancias de realimentación del estado de forma que uno de ellos fuese un integrador, lo que haría innecesario introducir un control integral adicional. ++ z I
-1
rControl Integral vHCKuy+-kI
Planta Gx++zI
-1 +-eCRealimentación de estado
Fig. 5.8: Realimentación de estado con integrador adicional (2)
Ejemplo 5.6.4: Supongamos el mismo sistema del ejemplo (8.6.2) y (8.6.3). En dichos ejemplos se ha determinado los valores de la ganancia de realimentación de estado parar situar los polos en unas posiciones determinadas así como la ganancia necesaria para que el error en régimen permanente ante entrada escalón fuese ce-ro. Tratemos ahora de conseguir que el error en régimen permanente ante entrada rampa sea también cero. 220 5. Control de sistemas por realimentación de estado
Solución: Veamos en primer lugar cual sería el error que se comete en régimen permanente ante una entrada de tipo rampa para el sistema sin introducir el control integral, y si es posible conseguir algún valor de ganancia k0 que anule dicho error. Si definimos el error en régimen permanente como la diferencia entre la sali-da deseada (en este caso se busca que la salida siga a la entrada) y la salida real tendremos que:
e(k) = r(k) − y(k) (5.6.64) y tomando transformadas en z, tendremos que
E(z) = R(z) − Y (z) = R(z) − F (z)R(z) = (1 − F (z)) R(z) (5.6.65) Para ver el error en régimen permanente hacemos el límite de esta expresión del error para t → ∞ y se obtiene el siguiente resultado lim
k→∞
e(k) = lim
z→1
(1 − z−1)R(z)(1 − F (z)) = lim
z→1
(1 − z−1) T z −1
(1 − z−1)2
(
1 − k0
z2 − z + 0 .5
)
= T (0 .5 − k0)0 (5.6.66) En esta expresión se puede apreciar que salvo para el caso k0 = 0 .5 donde el error quedaría indeterminado, en el resto de los casos el error en régimen perma-nente tiende hacia infinito (en la pr ·ctica las posibilidades de ajustar k0 al valor 0.5 son remotas por lo que las pequeñas desviaciones har ·n que en la pr ·ctica el error tienda a infinito siempre). Supongamos ahora que se introduce un control integral en el sistema, man-teniendo los valores que habíamos calculado para la matriz de ganancia K de la realimentación de estado. Las matrices del sistema con control integral son
G′ =
G − HK Hki
−C 1
=
0 1 0
−1 −0.5 ki
−1 0 1
H′ =
001
, C′ =
[
1 0 0
]5.6. Control por realimentación de estado 221
La función de transferencia del sistema quedar · de la siguiente forma
F ′(z) = C′[zI − G′]−1H′
=
[
1 0 0
]
z −1 01 z + 0 .5 −ki
1 0 z − 1
−1
001
= ki
(z − 1)( z2 + 0 .5z − 1) + ki
y al igual que antes, haciendo el límite para t → ∞ en la expresión del error en régimen permanente se obtiene que lim
k→∞
e(k) = lim
z→1
(1 − z−1)R(z)(1 − F ′(z)) = lim
z→1
(1 − z−1) T z −1
(1 − z−1)2
(
1 − ki
(z − 1)( z2 − z + 0 .5) + ki
)
= 0.5Tki
(5.6.67) en este caso se aprecia que el error si bien no se anula completamente se puede hacer muy pequeño muestreando el sistema con un periodo T pequeño y haciendo grande el valor de ki.
5.6.5. Sistemas con entrada vector
Hasta este momento hemos considerado el problema de diseñar una realimen-tación de estado de forma que los polos en cadena cerrada del sistema se sitúen en unas posiciones determinadas previamente para el caso de sistemas con entrada de control de tipo escalar. En el caso de que la entrada de control al sistema sea un vector tenemos mucha m ·s libertad para elegir las señales de control para controlar el sistema. Recordemos que, mientras que la matriz de controlabilidad tenía dimen-siones n × n para sistemas con entrada escalar, para el caso de entrada vectorial las dimensiones eran n × nr (suponiendo que el estado tiene dimensiones n × 1 y la entrada de control r × 1). Veamos un ejemplo muy simple para ilustrar esta situación.
Ejemplo 5.6.5: Supongamos el sistema definido por las siguientes ecuaciones
x1(k + 1)
x2(k + 1)
=
0 1
−0.5 −0.5
x1(k)
x2(k)
+
0 01 0.25
u1(k)
u2(k)
(5.6.68) 222 5. Control de sistemas por realimentación de estado
y se desean determinar los valores de las ganancias de la matriz K (de dimensión
2×2) de realimentación de estado para que los polos en cadena cerrada del sistema estén en p1 = 0 .5 ± j0.5.
Solución: Comprobemos en primer lugar que el sistema sea completamente con-trolable en estado, para ello obtendremos la matriz de controlabilidad
MC = [ H|GH ] =
0 0 1 0.25 1 0.25 −0.5 −0.125
(5.6.69) que tiene rango 2 por lo que el sistema es completamente controlable en estado. Puesto que se desea que los polos del sistema en cadena cerrada estén en p1 =0.5 − j0.5 y p1 = 0 .5 + j0.5, tendremos que igualar la ecuación característica del sistema en cadena cerrada a la definida por los polos p1 y p2. Dando como resultado da la siguiente expresión
|zI − G + HK | = z2 − z + 0 .5 = 0 (5.6.70) y sustituyendo en el primer término las matrices por sus valores y operando
z2 + z(0 .5 + k12 − 0.25 k22 ) + 0 .5 + k11 + 0 .25 k21 = z2 − z + 0 .5 = 0 (5.6.71) en la que igualando los coeficientes de ambos polinomios obtenemos
k12 + 0 .25 k22 = −1.5
k11 + 0 .25 k21 = 0 (5.6.72) como se puede observar tenemos dos ecuaciones y cuatro incógnitas, luego te-nemos dos grados de libertad a la hora de elegir los coeficientes de la matriz de ganancias de la realimentación. Si tomamos, por ejemplo, k12 = k21 = 1 entonces tendremos que k22 = −10 y k11 = −0.25 . Con lo que la matriz de realimentación de estado ser ·
K =
−0.25 11 −10
5.7. Diseño de observadores de estado
Ocurre con una cierta frecuencia que las variables de estado de un sistema o una parte de ellas no son medibles (o accesibles). De hecho es frecuente que únicamente la variable de salida del sistema sea medible, o que el sensor necesario para medir una cierta variable tiene que desecharse por su alto coste económico, o por otro tipo de razones. En todos estos casos resulta necesario estimar aquellas 5.7. Diseño de observadores de estado 223 HGCSistema u(k) x(k) y(k) Observador del estado del sistema x(k) ^++ z I
-1
Fig. 5.9: Sistema con observador del estado
variables de estado que no son medibles directamente. La estructura genérica de un observador de estado responde al esquema mostrado en la figura 5.9. Puesto que se est · suponiendo que las variables de estado no son accesibles o medibles, la única información que podemos utilizar para determinar el estado en el que est · el sistema son las entradas que se le suministran y las salidas con las que responde. En el caso de que ninguna de las variables de estado sea medible nece-sitaremos que el observador de estado estime todas ellas, a este tipo de observador se le denomina observador de orden completo .
5.7.1. Observador de orden completo
Una primera idea para el diseño de un observador del estado de un sistema po-dría consistir en incluir en el observador un modelo matem ·tico (lo m ·s completo posible) del sistema e introducir como entradas al observador las entradas de con-trol que se le van suministrando al sistema real (fig. 5.10). Este primer esquema tiene tres defectos b ·sicos: Para que este esquema funcionase necesitaríamos conocer el estado inicial en el que est · el sistema, ya que de otro modo evolucionar · de forma distinta. El segundo problema del esquema de observador propuesto se debe a que las matrices G, H y C del sistema físico real no se conocen con exactitud, sino que suelen ser estimaciones de las mismas obtenidas por algún método 224 5. Control de sistemas por realimentación de estado
de identificación o de modelado del sistema. Esto implica que el observador diseñado tendr · pequeñas desviaciones con respecto al sistema real. Estas desviaciones se ir ·n acumulando en el tiempo, haciendo que pasado un cier-to periodo de tiempo, el estado estimado no se corresponda con el estado verdadero del sistema. El tercer problema se debe a las perturbaciones a las que el sistema real est ·
sometido y que no son medibles. En esta situación y aunque las matrices de sistema que incluimos en el observador sean las verdaderas, el efecto de las perturbaciones haría que el estado que estima el observador y el real no coincidan al cabo de un cierto tiempo. HGCu(k) y(k) ^Observador del estado del sistema x(k) ^++ z I
-1
Fig. 5.10: Observador de orden completo del estado
Debemos tener en cuenta en el diseño algún mecanismo que nos permita mo-dificar el estado estimado por el observador cuando el estado del observador no coincida con el del sistema real. Puesto que el estado del sistema real no lo pode-mos medir para compararlo con el estimado por el observador, lo que se hace es comparar la salida estimada del sistema con la salida del sistema real y el error se utiliza para modificar el estado tal y como se muestra en la figura 5.11. Es decir introducimos una realimentación del error de estimación en el observador. Para que sea posible diseñar un observador, un requisito previo imprescindible es que se pueda estimar matem ·ticamente el estado en un cierto instante de tiempo. Tal y como se vio cuando se introdujo el concepto de observabilidad de un sistema, este es un requisito previo para que podamos diseñar un observador. 5.7. Diseño de observadores de estado 225
Definición 5.7.1: La condición necesaria y suficiente para poder observar las va-riables de estado es que se verifique la condición de observabilidad del sistema.
Fig. 5.11: Observador del estado
Función de transferencia del observador
El observador de la figura 5.11 tiene la siguiente din ·mica
ˆx(k + 1) = G ˆx(k) + Kee(k) + Hu (k)ˆy(k) = C ˆx(k) (5.7.1)
e(k) = y(k) − ˆy(k)
tomando transformadas z en las expresiones anteriores tenemos que
z ˆX(z) = G ˆX(z) + KeE(z) + HU (z) (5.7.2)
ˆY (z) = C ˆX(z) (5.7.3)
E(z) = Y (z) − ˆY (z) (5.7.4) y de aquí tenemos
z ˆX(z) = ( G − KeC) ˆX(z) + HU (z) + KeY (z) (5.7.5) o bien
ˆX(z) = [ zI − G + KeC]−1HU (z) + [ zI − G + KeC]−1KeY (z) (5.7.6) 226 5. Control de sistemas por realimentación de estado HGCu(k) y(k) ^Observador x(k) ^++ z I
-1
e(k) y(k) Ke
+++-x(k) ^HGCSistema u(k) x(k) y(k) ++zI
-1
r(k) +K-
Fig. 5.12: Sistema con observador completo y realimentación del estado
De la ecuación 5.7.6 se deduce que el observador tiene como función de trans-ferencia la siguiente expresión:
ˆX(z) =
[
[zI − G + KeC]−1H [zI − G + KeC]−1Ke
] U (z)
Y (z)
(5.7.7) y la ecuación característica del observador viene dada por la siguiente ecuación
|zI − G + KeC| = 0 (5.7.8) De la ecuación 5.7.8 se deduce que es necesario diseñar la ganancia de reali-mentación en el observador Ke de forma que la din ·mica del mismo sea adecuada para el sistema que estamos observando. Tendremos que diseñar esta ganancia del observador de forma que la din ·mica del mismo sea m ·s r ·pida que la del sistema. Si no se hace así, se tendr · un error muy importante entre los valores estimados y los reales, lo que har · que la regulación posterior del sistema funcione mal. 5.7. Diseño de observadores de estado 227
Error cometido por el observador
Las ecuaciones 5.7.1 nos muestran la din ·mica del observador. Sin embargo, m·s que la din ·mica del observador, lo que nos interesa es el error cometido por el observador. Este error se define como la diferencia entre el valor estimado del estado que nos da el observador y el valor verdadero del estado,
˜x(k) = x(k) − ˆx(k) (5.7.9) si formulamos la expresión del error de estimación de estado en el instante k + 1
obtenemos que
˜x(k + 1) = x(k + 1) − ˆx(k + 1) = Gx (k) + Hu (k) − [G − KeC] ˆ x(k) − Hu (k) − KeCx (k)= (G − KeC)[ x(k) − ˆx(k)] = (G − KeC) ˜ x(k) (5.7.10) En la expresión 5.7.10 se aprecia que la din ·mica del error viene determinada por los autovalores de la matriz G−KeC. Resulta conveniente que los autovalores de esta matriz estén dentro de la zona estable, que para los sistemas de tiempo discreto es el círculo de radio unidad. Adem ·s es conveniente que la din ·mica del error sea muy r ·pida, de forma que ante una variación brusca en el estado (por una perturbación por ejemplo) el error se anule lo antes posible. De esta forma el observador seguir · fielmente el estado del sistema. Sin embargo, es necesario tener en cuenta que, cuando se conectan el observa-dor y la realimentación de estado, la matriz Ke aparece también en el numerador de la función de transferencia. Esto implica que, si el sistema esta sometido a per-turbaciones en la salida y/o ruido en la medida, este ruido es amplificado. Desde el punto de vista de diseño del observador, la elección de Ke es un compromiso entre la rapidez de adaptación del observador a los cambios de estado y la sensibilidad al ruido del sistema y de medida.
Diseño de la ganancia del observador por comparación de coeficientes
El diseño de la ganancia del observador se hace de forma similar al visto para la realimentación de estado, es decir se determina la ganancia del observador Ke de forma que la ecuación característica de del mismo tenga los polos en las posiciones deseadas p1, p 2, . . . , p n.
|zI − G + KeC| = ( z − p1)( z − p2) . . . (z − pn) = 0 (5.7.11) 228 5. Control de sistemas por realimentación de estado
Ejemplo 5.7.1: Supongamos el sistema definido por las siguientes ecuaciones
x1(k + 1)
x2(k + 1)
=
0 1
−1 −0.5
x1(k)
x2(k)
+
01
u(k) (5.7.12)
y(k) =
[
1 0
] x1(k)
x2(k)
(5.7.13) y se desea diseñar un observador de estado para este sistema, teniendo en cuenta que se han ajustado los valores de las ganancias de la matriz K de realimentación de estado para que los polos en cadena cerrada del sistema estén en p1 = z − 0.5 −
j0.5 y p1 = z − 0.5 + j0.5.
Solución:
Comprobemos en primer lugar que el sistema sea completamente observable en estado, para ello obtendremos la matriz de observabilidad
MO =
CCG
=
1 00 1
(5.7.14) que tiene rango 2, luego el sistema es observable. Dado que nos interesa que la din ·mica del error cometido por el observador sea lo m ·s r ·pida posible, haremos que los polos del observador estén situados en el origen (esta respuesta se conoce como respuesta de tiempo mínimo, o en anglosajón dead-beat ). Por lo tanto
|zI − G + KeC| = z2 = 0 (5.7.15) es decir
∣∣∣∣∣∣z 00 z
−
0 1
−1 −0.5
+
ke1
ke2
[
1 0
]∣∣∣∣∣∣ = z2 = 0 (5.7.16) y de aquí obtenemos
z2 + z(ke1 + 0 .5) + (0 .5ke1 + ke2 + 1) = z2 = 0 (5.7.17) e igualando los coeficientes obtendremos que ke1 = −0.5 y ke2 = −0.75
Diseño de la ganancia del observador por transformación
En el caso en el que el sistema este representado en una forma que no sea la for-ma canónica observable el método de comparación de coeficientes puede dar lugar 5.7. Diseño de observadores de estado 229
a c ·lculos complejos para la obtención de las ganancias de realimentación, especial-mente para sistemas de orden elevado. En este caso existe posibilidad transformar el sistema a la forma canónica observable y diseñar en esta nueva representación el observador y después convertirlo de nuevo a la representación del sistema que teníamos inicialmente. Para ello, supongamos que se tiene un sistema con la siguiente representación de estado,
x(k + 1) = Gx (k) + Hu(k) (5.7.18)
y(k) = Cx (k) (5.7.19) y el observador de orden completo para esta representación viene expresado por
ˆx(k + 1) = G ˆx(k) + Hu (k) + Ke(y(k) − ˆy(k)) ˆy(k) = C ˆx(k) (5.7.20) y transformando la ecuación queda
ˆx(k + 1) = ( G − KeC) ˆ x(k) + Hu (k) + Key(k) (5.7.21) que es la expresión definitiva del observador de orden completo. De la expresión 5.7.21, la relación de ˜x(k + 1) y ˜x(k) y considerando ˜x(k) = x(k) − ˆx(k) se tiene
˜x(k + 1) = (G − KeC) ˜ x(k) (5.7.22) cuya ecuación característica es
|zI − G + KeC| = 0 (5.7.23) Al igual que para el caso del diseño de la realimentación de estado, es posible transformar el sistema a otra forma canónica, normalmente a la observable, diseñar aquí el observador y convertirlo de nuevo a la representación original. Para el sistema en la representación original la matriz de observabilidad es
MO =
CCG
...
CG n−1
(5.7.24) y supongamos que aplicamos una transformación T que transforma al sistema a la forma canónica observable
x(k) = T x ′(k) (5.7.25) 230 5. Control de sistemas por realimentación de estado
con lo que el nuevo sistema quedaría en la forma
x′(k + 1) = T −1GT x ′(k) + T −1Hu(k) (5.7.26)
y(k) = CT x ′(k) (5.7.27) La relación entre las matrices de observabilidad de ambas representaciones venía dada por la expresión 5.4.5
M ′O = MOT (5.7.28) y de aquí que
T = M ′− 1
O
MO (5.7.29) pudiéndose obtener la matriz de transformación T a partir de las dos matrices de observabilidad. Una vez que el sistema esta en la forma canónica observable y verifica la condición de observabilidad completa de estado, se diseña el observador de estado. En esta nueva representación tendremos que
|zI − G′ + K′eC′| = ( z − p1)( z − p2) . . . (z − pn) = 0 (5.7.30) e igualando los coeficientes podremos determinar la matriz de ganancias del ob-servador K′e. Si aplicamos la transformación inversa T −1 al sistema y ˜x(k) =
x(k) − ˆx(k) se tiene
˜x′(k + 1) = (G′ − K′eC′) ˜x′(k)
y(k) − ˆy(k) = C′ ˜x′(k)
para volver a la representación original aplicamos x′ = T −1x, y recordando que
G′ = T −1GT y C′ = CT , obtenemos que
˜x(k + 1) = (G − T K ′eC) ˜ x(k)
y(k) − ˆy(k) = C ˜x(k)
Y comparando ahora con la expresión del modelo de error de estimación en el estado en el sistema original tenemos que la matriz de ganancia del observador expresada en el sistema de representación original tendr · la forma
Ke = T K ′e (5.7.31) Una limitación importante a este método se produce en el caso de que el sistema sea no-observable y se pueda transformar a la forma canónica observable. En este caso, ser · posible determinar la matriz de ganancias K′e para observar el sistema en la forma canónica observable, pero la matriz de transformación T para pasar a dicha forma canónica es singular por lo que no es posible obtener la inversa de la transformación. 5.7. Diseño de observadores de estado 231
5.7.2. Comportamiento conjunto del sistema realimentado con el observador
Las ecuaciones 5.7.1 nos muestran la din ·mica del error que comete el obser-vador. Desde un punto de vista pr ·ctico nos interesa conocer como se comporta la din ·mica del sistema al introducir el observador en el lazo de realimentación. Recordando que las ecuaciones del sistema y la señal de realimentación son,
x(k + 1) = Gx (k) + Hu (k)
y(k) = Cx (k)
u(k) = −K ˆx(k) (5.7.32) introduciendo la expresión de la señal de realimentación en la ecuación de estado, y sumando y restando el término HKx (k) en dicha expresión obtenemos que
x(k + 1) = Gx (k) − HK ˆx(k)= [G − HK ]x(k) + HK [x(k) − ˆx(k)] (5.7.33) sustituyendo x(k) − ˆx(k) por ˜x(k) nos queda lo siguiente
x(k + 1) = [ G − HK ]x(k) + HK ˜x(k) (5.7.34) teniendo en cuenta ahora que la din ·mica del error de estimación de estado come-tido por el observador venía dada por
˜x(k + 1) = [ G − KeC] ˜ x(k) (5.7.35) combinando las expresiones de la ecuación de estado del sistema realimentado 5.7.34 y la ecuación de estado del error de estimación de estado 5.7.35 en un único modelo de estado, tendremos
x(k + 1) ˜x(k + 1)
=
G − HK HK
0 G − KeC
x(k)˜x(k)
(5.7.36) La ecuación característica del sistema con realimentación y observador se ob-tiene de 5.7.36 y vendr · dada por la siguiente expresión
∣∣∣∣∣∣
zI − G + HK −HK
0 zI − G + KeC
∣∣∣∣∣∣ = 0 (5.7.37) de esta última expresión se puede observar que la ecuación característica del sistema se puede escribir en la forma
|zI − G + HK | . |zI − G + KeC| = 0 (5.7.38) 232 5. Control de sistemas por realimentación de estado
La expresión 5.7.38 nos dice que la ecuación característica del sistema con realimentación de estado y observador es el producto de las ecuaciones caracterís-ticas del sistema realimentado y del observador de estado. Dicho en otras palabras, que son independientes, lo que permite que se diseñen de forma independiente y posteriormente sean combinados. Hay que tener en cuenta que la din ·mica global del sistema se complica puesto que ahora adem ·s de los polos del sistema realimen-tado tendremos los que añade el observador de estado. Para que estos no influyan en la din ·mica del sistema deben amortig¸arse r ·pidamente, es decir deben de estar situados próximos al origen o en él si se desea que el observador elimine el error de estimación r ·pidamente de forma que no influya de forma sustancial en la din ·mica global del sistema. Se diseñar · en primer lugar la ganancia de realimentación de estado y en fun-ción de donde se hayan situado los polos del sistema realimentado se determinar ·nunas posiciones adecuadas para el observador. Por tanto el diseño se realizar · en dos pasos.
5.7.3. Observador de orden reducido
Los estimadores de estado que hemos diseñado anteriormente tenían como ob-jetivo la estimación o reconstrucción de todas las variables de estado del sistema. Sin embargo, existen una serie de problemas donde algunas de las variables de es-tado pueden ser medidas con precisión. En estos casos, donde podemos medir las variables directamente, cabe preguntarse si es necesario hacer una reconstrucción completa del estado. Este tipo de problemas no requiere hacer una reconstrucción completa del estado, sino que basta con estimar las variables de estado que no son medibles. Esto ocurre sobre todo cuando parte de las variables de estado del sistema son variables de salida. A estos tipos de estimadores u observadores se les denomina observadores de orden reducido , y si el orden del observador es el mínimo posible se le denomina
observador de orden mínimo (Fig.5.13). Supongamos que en el modelo del sistema tenemos que una parte de las varia-bles de estado son medibles xm y otra parte de las variables no xn. Si dividimos el vector de estado x en dos submatrices, el modelo de estado del sistema nos queda de la siguiente forma
xm(k + 1)
xn(k + 1)
=
Gmm Gmn
Gnm Gnn
xm(k)
xn(k)
+
Hm
Hn
u(k)
y(k) =
[
I O
] xm(k)
xn(k)
(5.7.39) si esta expresión matricial la ponemos en forma de ecuaciones tenemos 5.7. Diseño de observadores de estado 233 nx (k) ^u(k) x(k) ^HGCSistema u(k) x(k) y(k)= ++ z I
-1
Kr(k) +Observador reducido x (k)
m
x (k)
m
x (k)
m
-
Fig. 5.13: Observador de orden reducido
xm(k + 1) = Gmm xm(k) + Gmn xn(k) + Hmu(k) (5.7.40)
xn(k + 1) = Gnm xm(k) + Gnn xn(k) + Hnu(k) (5.7.41) De la ecuación 5.7.40 podemos medir o conocer todos los términos salvo el término en xn(k) de la parte derecha de la expresión. Si separamos este término la ecuación nos queda
xm(k + 1) − Gmm xm(k) − Hmu(k) = Gmn xn(k) (5.7.42) y la ecuación 5.7.41 podemos reescribirla agrupando entre corchetes los términos medibles
xn(k + 1) = Gnn xn(k) + [ Gnm xm(k) + Hnu(k)] (5.7.43) Si comparamos las ecuaciones 5.7.42 y 5.7.43 con las de un sistema ficticio genérico de la forma
x′(k + 1) = G′x′(k) + H′u′(k) (5.7.44)
y′(k) = C′x′(k) (5.7.45) 234 5. Control de sistemas por realimentación de estado
de forma que
x′(k) = xn(k)
y′(k) = xm(k + 1) − Gmm xm(k) − Hmu(k) (5.7.46)
G′ = Gnn
H′u′(k) = Gnm xm(k) + Hnu(k)
C′ = Gmn
Si para el sistema 5.7.44 un observador de estado tiene la forma 5.7.21, enton-ces ˆx′(k + 1) = ( G′ − KeC′) ˆx′(k) + H′u′(k) + Key′(k) (5.7.47) sustituyendo aquí las matrices y vectores por las expresiones completas 5.7.46 nos queda que el observador para la parte no medible del sistema tiene la siguiente ecuación
ˆxn(k + 1) = (Gnn − KeGmn ) ˆ xn(k) + Gnm xm(k) + Hnu(k)+Ke[xm(k + 1) − Gmm xm(k) − Hmu(k)] (5.7.48) y en el diseño este observador de orden mínimo, tendremos que ajustar la matriz
Ke de forma que los polos de la ecuación característica estén en unas posiciones determinadas por
|zI − (Gnn − KeGmn )| = 0 (5.7.49)
Ejemplo 5.7.2: Supongamos el sistema formado por dos integradores sucesivos, con un periodo de muestreo de 0.1 segundos, definido por las siguientes ecuaciones
x1(k + 1)
x2(k + 1)
=
1 0.10 1
x1(k)
x2(k)
+
0.01 0.1
u(k) (5.7.50)
y(k) =
[
1 0
] x1(k)
x2(k)
(5.7.51) dado que la variable de salida es medible y coincide con una de las variables de estado, se desea diseñar un observador de estado para estimar el valor de la otra variable.
Solución:
Comprobemos en primer lugar que el sistema sea completamente observable, para ello obtendremos la matriz de observabilidad
MO =
CCG
=
1 01 0.1
(5.7.52) 5.8. Observador óptimo del estado 235
que tiene rango 2, luego es observable. Como se ha indicado que hay una variable de estado medible y otra no, las dimensiones de las submatrices del estado no medible y de los medibles son: Gnn : 1 × 1 y Gmm : 1 × 1. Estas submatrices son
Gmm = 1 , G mn = 0 .1, G nm = 0 , G nn = 1
Por lo tanto, la ecuación característica del observador de orden reducido tendr ·
orden 1. Y si buscamos que la respuesta del observador sea lo m ·s r ·pida posible, haremos que el polo de esta ecuación característica esté en el origen.
|zI − (Gnn − KeGmn )| = z = 0 (5.7.53) y sustituyendo en la expresión por las submatrices correspondientes, tendremos
z − (1 − 0, 1ke) = z = 0 (5.7.54) de donde se tiene que el valor de ke = 10 .Nótese que este desarrollo es cierto si la matriz de salida C = [ IO ]. Si no fuera así, habría que efectuar una transformación para poner la matriz C en la forma adecuada, realizar el c ·lculo de Ke según lo expuesto, y posteriormente devolver el sistema a su estado inicial.
5.8. Observador óptimo del estado
Supongamos que tenemos un sistema sometido a ruido en el sistema y a rui-do de medida, representado en el espacio de estados en tiempo discreto por las siguientes ecuaciones
x(k + 1) = Gx (k) + Hu (k) + v1(k) (5.8.1)
y(k) = Cx (k) + v2(k) (5.8.2) donde:
v1(k) es un ruido blanco con una distribución de probabilidad normal con media cero y covarianza conocida R1, es decir v1 ∼ N (0; R1). Es decir,
E[v1] = 0 y E[v1vT
1
] = R1.
v2(k) es un ruido blanco con una distribución de probabilidad normal con media cero y covarianza conocida R2, es decir v2 ∼ N (0; R2). Es decir,
E[v2] = 0 y E[v2vT
2
] = R2.Supondremos adem ·s que E[v1vT
2
] = 0 . Es decir, los ruidos son indepen-dientes.
R1 y R2 son matrices semidefinidas positivas. 236 5. Control de sistemas por realimentación de estado HGCu(k) y(k) ^Observador x(k) ^++ z I
-1
e(k) y(k) Ke
+++-x(k) ^HGCSistema u(k) x(k) y(k) ++zI
-1
+v (k)
1
++v (k)
2
Fig. 5.14: Observador de estado de un sistema sometido a perturbaciones
Supondremos adem ·s que el modelo de estado es controlable y observable. Debido al efecto del ruido de medida y del ruido del sistema (figura 5.14) el observador de estado que hemos visto hasta el momento podría no dar unas pres-taciones adecuadas. Resulta por ello deseable que el observador de estado que di-señemos optimice algún tipo de criterio de forma que se mejore el funcionamiento del mismo, para lo que se propone que Ke varíe con el tiempo k.La estructura que supondremos para el estimador de estado es la misma que para el observador de estado completo (, es decir 5.7.20)
ˆx(k + 1) = G ˆx(k) + Hu (k) + Ke(k)[ y(k) − ˆy(k)] = G ˆx(k) + Hu (k) + Ke(k)[ y(k) − C ˆx(k)] (5.8.3) Si denominamos ˜x(k) = x(k) − ˆx(k) al error de reconstrucción, entonces la din ·mica del error de reconstrucción es la siguiente 5.8. Observador óptimo del estado 237
˜x(k + 1) = x(k + 1) − ˆx(k + 1) = Gx (k) + Hu (k) + v1(k) − G ˆx(k) − Hu (k) − Ke[y(k) − C ˆx(k)] = G ˜x(k) + v1(k) − Ke(k)[ y(k) − C ˆx(k)] = G ˜x(k) + v1(k) − Ke(k)[ x(k) + v2(k) − C ˆx(k)] = (G − Ke(k)C) ˜ x(k) + v1(k) − Ke(k)v2(k) (5.8.4) Si calculamos la esperanza del error de reconstrucción E[ ˜ x(k + 1)] obtenido a partir de la expresión 5.8.4 tendremos
E[ ˜ x(k + 1)] = E[( G − Ke(k)C) ˜ x(k) + v1(k) − Ke(k)v2(k)] (5.8.5) Y si al término de ruido lo denominamos w(k) = v1(k) − Ke(k)v2(k), se puede observar f ·cilmente que su valor medio es cero, puesto que los ruidos del sistema y de medida se habían supuesto como ruidos blancos de media cero, luego
E[w(k)] = E[v1(k) − Ke(k)v2(k)] = E[v1(k)] − Ke(k)E[v2(k)] = 0 (5.8.6) y su covarianza tiene la forma
E[w(k + 1) wT (k + 1)] = E[( v1(k) − Ke(k)v2(k))( v1(k) − Ke(k)v2(k)) T ]= Ke(k)E[v2(k)vT
2
(k)] KTe (k) − E[v1(k)vT
2
(k)KTe (k)]
−E[Ke(k)v2(k)vT
1
(k)] + E[v1(k)vT
1
(k)] = Ke(k)E[v2(k)vT
2
(k)] KTe (k) + E[v1(k)vT
1
(k)] = Ke(k)R2KTe (k) + R1 (5.8.7) El objetivo que perseguimos es determinar una ganancia de realimentación Ke
en el observador de forma que se minimice la varianza del error de estimación, la cual denominaremos P (k) y que corresponde a la siguiente expresión
P (k) = E[( ˜ x(k) − E[ ˜ x(k)])( ˜ x(k) − E[ ˜ x(k)]) T ]= E[ ˜ x(k) ˜ xT (k)] − E[ ˜ x(k)E[ ˜ xT (k)]] −
E[E[ ˜ xT (k)] ˜ x(k)] + E[ ˜ x(k)] E[ ˜ xT (k)] = E[ ˜ x(k) ˜ xT (k)] − E[ ˜ x(k)] E[ ˜ xT (k)] (5.8.8) 238 5. Control de sistemas por realimentación de estado
Si aplicamos 5.8.6 a la expresión 5.8.5 nos queda que el valor medio del error de reconstrucción tiene la siguiente expresión
E[ ˜ x(k + 1)] = E[( G − Ke(k)C) ˜ x(k) + w(k)] = (G − Ke(k)C)E[ ˜ x(k)] (5.8.9) y en esta expresión 5.8.9 se puede apreciar que si E[x(0)] = m0 y se toma como valor inicial del estimador en el instante cero, es decir ˆx(0) = m0, entonces esta expresión del valor medio del error de reconstrucción tomar · valor cero. Dicho de otro modo, el error de estimación ser · cero para todo k ≥ 0, independientemente del valor de Ke(k) que se elija. Por otra parte, la expresión 5.8.8 de la matriz de covarianza P (k), (puesto que el valor medio del error de reconstrucción es cero para todo k, E[ ˜ x(k)] = 0 ), se simplifica y nos queda
P (k) = E[ ˜ x(k) ˜ xT (k)] (5.8.10) Si relacionamos la covarianza del error de reconstrucción en un instante k + 1
con la del instante anterior k, tendremos que
P (k + 1) = E[ ˜ x(k + 1) ˜ x(k + 1) T ]= E[(( G − Ke(k)C) ˜ x(k) + w(k))(( G − Ke(k)C) ˜ x(k) + w(k)) T ]= E[(( G − Ke(k)C) ˜ x(k))(( G − Ke(k)C) ˜ x(k)) T ]+E[(( G − Ke(k)C) ˜ x(k)) w(k)T ]+E[w(k)(( G − Ke(k)C) ˜ x(k)) T ] + E[w(k)w(k)T ]= (G − Ke(k)C)P (k)( G − Ke(k)C)T + R1 + Ke(k)R2KTe (k)
(5.8.11) Donde se ha tenido en cuenta que el estado y los ruidos de sistema y medi-da son independientes por lo que sus matrices de covarianza cruzada son nulas
E[ω(k)˜ x(k)T ] = 0 y E[x(k)˜ ω(k)T ] = 0 , y adem ·s se ha sustituido la covarianza del ruido w(k) por el valor obtenido en la expresión 5.8.7. Buscamos, según se ha dicho, un observador de estado que minimice la cova-rianza del error de reconstrucción, es decir que minimice P (k). Esta covarianza para el estado k + 1 , como función de la del instante k y de las covarianzas de los ruidos del sistema, la tenemos expresada en 5.8.11.
Observación:
Decir que P (k) es mínimo quiere decir que la forma cuadr ·tica αT P (k)α (valor escalar) es mínima, donde α es un vector n × 1 arbitrario. 5.8. Observador óptimo del estado 239
De esta observación y de la expresión 5.8.11 tenemos
αT P (k + 1) α = αT [GP (k)GT + R1 − Ke(k)CP (k)GT
−GP (k)CT KTe (k) + Ke(k)( R2 + CP (k)CT )KTe (k)] α
(5.8.12) la ganancia Ke puede determinarse a partir de 5.8.12 agrupando en dos términos la expresión
αT P (k + 1) α = αT [GP (k)GT + R1 + Ke(k)( R2 + CP (k)CT )KTe (k)]
−Ke(k)CP (k)GT − GP (k)CT KTe (k)] α
= αT [GP (k)GT + R1
+[ Ke(k) − GP (k)CT (R2 + CP (k)CT )−1][R2 + CP (k)CT ][ Ke(k) − GP (k)CT (R2 + CP (k)CT )−1]T
−GP (k)CT (R2 + CP (k)CT )−1CP (k)GT ]α
(5.8.13) En la expresión 5.8.13 se puede observar que el primer término y el último no dependen de Ke(k) y el segundo término no es negativo ya que la matriz R2 +
CP (k)CT es definida positiva. Por lo tanto el mínimo de esta expresión se obtiene cuando Ke(k) se elige para que haga cero el segundo término. Es decir, cuando
Ke(k) = GP (k)CT (R2 + CP (k)CT )−1 (5.8.14) y en este caso la expresión de la matriz de covarianzas P (k + 1) queda de la siguiente forma
P (k + 1) = GP (k)GT + R1
− GP (k)CT × (R2 + CP (k)CT )−1CP (k)GT (5.8.15) Esta expresión 5.8.15 es una ecuación de Riccati . Como se puede observar el valor de la ganancia Ke(k) que minimiza la varianza del error de reconstrucción no depende de α. La matriz de covarianzas del error de reconstrucción en el instante de tiempo cero (dado que el error de reconstrucción en el instante inicial era cero por la forma en que habíamos elegido la estimación inicial) ser ·
P0 = E[ ˜ x(0) − ˜x(0) T ]= E[( x(0) − m0)( x(0) − m0)T ]= P0 (5.8.16) A las expresiones 5.8.3, 5.8.14 y 5.8.15 se las conoce como filtro de Kalman .240 5. Control de sistemas por realimentación de estado
5.8.1. Ecuaciones del filtro de Kalman
Resumiendo, dado un sistema
x(k + 1) = Gx (k) + Hu (k) + v1(k) (5.8.17)
y(k) = Cx (k) + v2(k) (5.8.18) las ecuaciones del observador óptimo del estado (filtro de Kalman) son:
ˆx(k + 1) = G ˆx(k) + Hu (k) + Ke(k)[ y(k) − C ˆx(k)]
Ke(k) = GP (k)CT (R2 + CP (k)CT )−1 (ganancia de Kalman)
P (k + 1) = GP (k)GT + R1
− GP (k)CT × (R2 + CP (k)CT )−1CP (k)GT (5.8.19) El problema de la reconstrucción se ha resuelto como un problema de optimi-zación paramétrica a partir del modelo del observador de estado completo. Pero el filtro de Kalman puede interpretarse también como la media condicional del estado en k + 1 dadas las observaciones acumuladas hasta el instante k de la salida del sistema
ˆx(k + 1) = E[x(k)|Y k] (5.8.20) este tratamiento puede encontrarse en una gran parte de los textos dedicados al filtro de Kalman. 5.8. Observador óptimo del estado 241
Bibliografía
S. Barnett y R.G. Cameron, Introduction to Mathematical Control Theory
Oxford University Press , 1992. S. Domínguez, P. Campoy, J.M Sebasti ·n y A. Jiménez, Control en el Espacio de Estado Prentice Hall, Serie Autom ·tica-Robótica , 2001. B. C. Kuo, Automatic Control Systems, Ed. Prentice Hall , 1991. N. Nise, Sistemas de Control para Ingeniería, Ed. CECSA , 2002. K. Ogata, Ingeniería de control moderna, Ed. Prentice Hall , 2002. K. Ogata, Discrete-Time Control Systems, Prentice Hall International Edi-tion , 1995. E. Umez-Eronini, Din ·mica de Sistemas y Control Thomson Learning , 2001. Índice alfabético
función de densidad de probabilidad, 24 modo controlable, 193 operador adelanto, 20 asignación de polos, 146 autocovarianza, 28 bloqueadores de señal, 50 control PID, 141 control por posicionamiento de polos, 208 control por realimentación de estado, 206 controlabilidad, 193 controlabilidad de estado, 193 controlabilidad de la salida, 198 controlador de tiempo mínimo, 181 controlador deadbeat, 181 controladores con tiempo de estable-cimiento mínimo, 181 covarianza, 28 densidad espectral, 34 diagrama de Bode, 70 discretización de las ecuaciones de es-tado, 118 discretización de un controlador analó-gico, 63 discretización de un controlador PID, 159 diseño de controladores PID discre-tos, 167 diseño de observadores de estado, 222 diseño de reguladores por síntesis di-recta, 173 espectro cruzado, 35 espectro de una señal muestrada, 74 esperanza matemática, 27 estado, 82 filtro de Kalman, 239 forma canónica controlable, 93 forma canónica de Jordan, 98 forma canónica diagonal, 100 forma canónica observable , 95 frecuencia natural no amortiguada, 140 función de autocorrelación, 29 función de correlación cruzada, 31 función de densidad espectral, 33 función de distribución, 24 función de transferencia, 44 función de transferencia discreta equi-valente, 57 función de transferencia muestreada, 52 ganancia de Kalman, 240 Kalman, 239 linealización de sistemas, 89 método basado en determinación al-gebraica para el ajuste de PID, 151 método basado en el lugar de las raí-ces para el ajuste de PID, 153 método basado en los polos dominan-tes para el ajuste de PID, 149 242 Índice alfabético 243
método de asignación de polos para ajuste de PID, 146 método de Euler, 65 método de la respuesta a un escalón para ajuste de PID, 143 método de la respuesta frecuencial pa-ra ajuste de PID, 144 método trapezoidal, 67 métodos de Ziegler y Nichols, 142 modelos de entrada/salida, 42 modelos temporales, 42 modo observable, 193 momento, 27 momentos centrados, 28 muestreo de una señal, 48 observabilidad, 200 observador óptimo del estado, 235 observador de orden completo, 223 observador de orden reducido, 232 observadores de estado, 222 operador retardo, 20 periodo de la oscilación amortiguada, 140 perturbaciones en el sistema, 125 principio de dualidad, 205 proceso aleatorio, 21 procesos ergódicos, 24 procesos estacionarios, 23 procesos estocásticos, 20 respuesta frecuencial de los sistemas muestreados, 71 respuesta impulsional, 43 ruido blanco, 35 sobreoscilación, 140 solución de la Ecuación de Estado, 114 solución de la ecuación de estado en tiempo discreto, 122 tasa de amortiguamiento, 140 tasa de decaimiento, 140 teorema de la factorización espectral, 37 teorema de muestreo, 75 teorema del valor final, 6, 14 teorema del valor inicial, 6, 14 tiempo de establecimiento, 140 tiempo de pico, 140 transformada de Fourier, 2 transformada de Laplace, 3 transformada inversa de Fourier, 3 transformada inversa de Laplace, 7 transformada z, 13 variable aleatoria, 21 variables de estado, 86 varianza, 28 Ziegler-Nichols, 142 |
189041 | https://www.chemguide.co.uk/14to16/organic/alkanes.html | | | | | | | | | | | | | | |
--- --- --- --- --- ---
| Chemguide: Core Chemistry 14 - 16 This page introduces alkanes and the important term "homologous series". It looks at the combustion of alkanes and their reactions with chlorine and bromine. I am assuming that you have already read the pages about organic formulae and names and about isomerism. Homologous series There are huge numbers of organic compounds, but they fall into families known as homologous series. All the members of a homologous series have the same type of bonding. The homologous series of the alkanes only has carbon-hydrogen bonds and carbon-carbon single bonds apart from methane, CH4, the first member of the series which only has carbon-hydrogen bonds. All members of a homologous series . . . have similar chemical properties; have physical properties which change in a regular way; can be given a general formula. For example: The alkanes have a very limited chemistry. The only simple things they do are burn or react with chlorine or bromine. All alkanes have the same reactions because they all have the same sort of bonds. Reactions happen when you break bonds. The boiling points of the alkanes increase in a regular way. The first four alkanes (up to butane, C4H10) are gases at room temperature. Then you get liquids as the boiling points increase, and eventually solids when they get really big. You really need to remember that the first four are gases. A general formula lets you work out the formula of any member of a homologous series if you know the number of carbon atoms. For alkanes, the general formula is CnH2n+2 So, for example, methane with one carbon atom has (2 x 1) + 2 carbon atoms so the formula is CH4. Octane with 8 carbon atoms has (2 x 8) + 2 carbon atoms so the formula is C8H18. Notice that as you go from one alkane to the next, the formula increases by CH2. The alkanes The alkanes are saturated hydrocarbons. A saturated hydrocarbon is one which has the maximum possible number of hydrogens for a given carbon chain. For example, consider propane (an alkane) and propene (an alkene). Both have 3 carbon atoms but propane is C3H8, whereas propene is C3H6. That is because propene contains a carbon-carbon double bond. propane: CH3CH2CH3 propene: CH3CH=CH2 Propene is known as an unsaturated hydrocarbon because you could fit another two hydrogens around the carbon chain if you replaced the double carbon-carbon bond by a single one. The first 6 alkanes are: | | | --- | | methane | CH4 | | ethane | C2H6 | | propane | C3H8 | | butane | C4H10 | | pentane | C5H12 | | hexane | C6H14 | Structural isomerism Alkanes from butane on have structural isomers. For example, in butane you can arrange the 4 carbon atoms and 10 hydrogens in two different ways: CH3CH2CH2CH3 which is just called butane; and which is called 2-methylpropane. |
| Note:You need to check your syllabus to find out how much, if anything, you need to know about structural isomerism. If you do need to know about it, you should make sure that you at least know about the examples on the page about isomerism. |
| Chemically the isomers of butane behave in exactly the same way. That is because they both contain only carbon-carbon bonds and carbon-hydrogen bonds. The reactions that organic compounds undergo is determined by what sorts of bond are present. Combustion of alkanes Alkanes burn in air or oxygen. What you get depends on how much oxygen is available. Complete combustion In a plentiful supply of air, they burn to form carbon dioxide and water, and you get lots of heat evolved. For example, with methane or butane: CH4 + 2O2 CO2 + 2H2O 2C4H10 + 13O2 8CO2 + 10H2O |
| Note:You could be asked for the equation for almost any given alkane, so there is no point in learning these equations. You have to be able to work them out. Equations involving alkanes with an odd number of carbons are much easier than those with even numbers. If you aren't sure how to handle these, read the equations page. |
| Incomplete combustion If there isn't enough oxygen, the hydrogen always gets what is available first. You will always get water, but you will also get either carbon monoxide or carbon (as soot) formed as well as carbon dioxide. Carbon monoxide is colourless, odourless and poisonous, and so potentially very dangerous. It interferes with the way oxygen is carried around the body. Carbon monoxide combines irreversibly with a molecule in the blood called haemoglobin. Normally oxygen bonds to the haemoglobin, and gets carried around the body in the blood stream to where it is needed. The carbon monoxide binding with the haemoglobin prevents that happening. Reactions of alkanes with chlorine or bromine A mixture of methane and chlorine reacts on exposure to sunlight. A substitution reaction happens. A substitution reaction is one in which an atom in a molecule is replaced by a different atom. In this case, a hydrogen atom in the methane is replaced by a chlorine atom. CH4 + Cl2 CH3Cl + HCl The organic product, CH3Cl, is called chloromethane and is a colourless gas at room temperature, and you also get steamy fumes of hydrogen chloride gas. Similarly a mixture of methane and bromine vapour forms bromomethane and hydrogen bromide on exposure to sunlight. Bromomethane is also a gas at room temperature, and hydrogen bromide also forms steamy fumes. In both cases, it is the UV in the sunlight which triggers the reaction by splitting the chlorine or bromine molecules into individual atoms. Ethane, CH3CH3, will behave similarly forming CH3CH2Cl or CH3CH2Br. Things are a bit more complicated if you do the similar reactions with propane (or any other bigger alkane). With propane, CH3CH2CH3, the hydrogen being replaced could either be in the middle of the chain or at the end. That means that you will get both of these two different molecules formed if you were using chlorine: 1-chloropropane: CH3CH2CH2 Cl 2-chloropropane: CH3CH ClCH3 |
| Note:This is all a massive simplification! In fact, if you do these reactions it is impossible to stop at the stage where just one hydrogen is replaced. You get a mixture of products that, in the methane and chlorine case, would include CH3Cl, CH2Cl2, CHCl3 and CCl4 as well, of course, as HCl. This isn't a useful reaction! I suspect it is only introduced at this level so that you meet the term "substitution reaction". |
| Where would you like to go now? To the organic chemistry menu . . . To the Chemistry 14-16 menu . . . To Chemguide Main Menu . . . © Jim Clark 2021 | |
189042 | https://study.com/academy/lesson/using-additive-properties-of-definite-integrals.html | Using Additive Properties of Definite Integrals | 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
Math Courses / Saxon Calculus Homeschool: Online Textbook Help Course
Using Additive Properties of Definite Integrals
Instructor Carmen AndertShow bio
Carmen has two master's degrees in mathematics has has taught mathematics classes at the college level for the past 9 years.
Cite this lesson
In this lesson, we will introduce the three additive properties of definite integrals and discuss how they may be used in solving homework problems. We will attempt to highlight the situations in which a calculus student is most likely to encounter a need for these properties.
Create an account
Table of Contents
Area Under a Curve
Two More Additive Properties
Lesson Summary
Show
Area Under a Curve ------------------
A contractor is building a house with a pitched roof. In order to determine the amount of insulation needed, he must find the area of a cross section of the attic. The roof can be described by the graph of the function y=f(x) as shown below, where the x-axis is the floor of the attic.
Graph of y=f(x)
When f(x) is positive for x between a and b and a is less than b, the definite integral of f(x)dx from a to b is simply the area of the region between the graph of y=f(x) and the x-axis between x=a and x=b. Therefore, the contractor needs to evaluate the definite integral of f(x)dx from 0 to 9. How can he do this? Using simple geometry, we see that the area in question is composed of two triangles, which are shaded in green and purple in the graph shown below.
Definite integral of f(x)dx from 0 to 9
Now, we know that the area of the green triangle is 1/233 = 4.5 and the area of the purple triangle is (1/2)63=9. Therefore, the definite integral of f(x)dx from 0 to 9 is 4.5 + 9 = 13.5. In other words, the definite integral of f(x)dx from 0 to 9 is the sum of the definite integrals of f(x)dx from 0 to 3 (green triangle) and of f(x)dx from 3 to 9 (purple triangle). This is an illustration of the additive property of definite integrals:
Additive Property 1
Example
You are asked to find the definite integral from -2 to 6 of the piecewise defined function f defined by x² if x<3 and x-1 if x>=3. Here is how to solve:
Using the additive property of definite integrals, the definite integral of f(x)dx from -2 to 6 is equal to the definite integral of f(x)dx from -2 to 3 plus the definite integral of f(x)dx from 3 to 6.
Each of these integrals can be easily solved by replacing f(x) with its definition on that interval.
The definition of the function from -2 to 3 is x². Using this definition, we can use our calculators to calculate the the definite integral of x² dx from -2 to 3, which gives 35/3
The definition of the function from 3 to 6 is x-1. Using our calculators, we can calculate the definite integral of x-1dx from 3 to 6, which gives 21/2.
To complete the problem, simply add 35/3 + 21/2 to get 133/6.
A student is most likely to need the additive property listed above when he or she is asked to find a definite integral of a piecewise defined function. It should also be mentioned that although the property as stated above is most often applied when b is between a and c, it is true regardless of the order of a, b, and c.
To unlock this lesson you must be a Study.com Member.
Create your account
Lesson
Quiz
Course
6.9K views
Two More Additive Properties ----------------------------
In addition to the main additive property previously discussed, there are two more additive properties of definite integrals. The second additive property is that the definite integral of f(x)dx from b to a is the opposite of the definite integral of f(x)dx from a to b:
Additive Property 2
This property is used to switch the upper and lower bounds of integration. Specifically, it is often desirable to have the upper bound higher than the lower bound; using this property, we can switch the bounds at the price of a negative sign.
The third additive property is that the definite integral from a to a of f(x)dx is zero:
Additive Property 3
Example
You are asked to evaluate the definite integral of g(x)dx from 0 to -4 given the graph of y=g(x) shown below.
Graph of y=g(x)
Can we find an antiderivative of g(x)? No! Because we are given only the graph of g(x), we must use geometry to solve this problem. Using the second additive property, the definite integral of g(x)dx from 0 to -4 is the opposite of the definite integral of g(x)dx from -4 to 0. From the graph, we can see that the area bounded between the graph of y=g(x) and the x-axis between x=-4 and x=0 is a quarter of a circle having radius 4. Therefore, the definite integral of g(x)dx from -4 to 0 is 1/4π4², or 4π. Finally, the definite integral of g(x)dx from 0 to -4 is -4π.
Because we needed to use geometry to solve this example, it was imperative to have the upper bound of integration greater than the lower bound. However, this ordering is not always necessary; algebraic problems requiring the use of an antiderivative to evaluate a definite integral can be solved using the Fundamental Theorem of Calculus regardless of the ordering of the upper and lower bounds.
Example
You are asked to evaluate the definite integral of cos(x²-x)dx from π to π.
Are you thinking of complicated techniques you could use to find an antiderivative of cos(x²-x)? Although this train of thought is tempting, it will only lead to a headache and a lot of wasted time! Instead, simply notice that the upper and lower bounds of integration are the same. Therefore, by the third additive property, the answer to this problem is zero.
To unlock this lesson you must be a Study.com Member.
Create your account
Lesson Summary --------------
It is important to note that the three additive properties stated above are true for any integrable function f(x) and any real numbers a, b, and c. The first property says you can split a definite integral into two (or more) definite integrals over disjoint intervals, and it is most often applied when integrating piecewise defined functions. The second property says that the upper and lower bounds of integration may be switched at the price of a negative sign. This property is most often applied when it is important to have the bounds in a particular order. Finally, the third property says that the definite integral is 0 anytime the two bounds of integration are the same.
To unlock this lesson you must be a Study.com Member.
Create your account
Register to view this lesson
Are you a student or a teacher?
I am a student I am a teacher
Unlock Your Education
See for yourself why 30 million people use Study.com
Become a Study.com member and start learning now.
Become a Member
Already a member? Log In
Back
Resources created by teachers for teachers
Over 30,000 video lessons& teaching resources‐all in one place.
Video lessons
Quizzes & Worksheets
Classroom Integration
Lesson Plans
I would definitely recommend Study.com to my colleagues. It’s like a teacher waved a magic wand and did the work for me. I feel like it’s a lifeline.
Jennifer B.
Teacher
Try it now
Back
Start today. Try it now
Saxon Calculus Homeschool: Online Textbook Help
30 chapters | 247 lessons
Ch 1. Saxon Calculus: Real Numbers
Ch 2. Saxon Calculus: Algebra
Ch 3. Saxon Calculus: Algebra Theorems
Ch 4. Saxon Calculus: Geometry
Ch 5. Saxon Calculus: Logic
Ch 6. Saxon Calculus: Trigonometry
Ch 7. Saxon Calculus: Graphing...
Ch 8. Saxon Calculus: Basics of...
Ch 9. Saxon Calculus: Graphing Functions &...
Ch 10. Saxon Calculus: Analysis of...
Ch 11. Saxon Calculus: Limits of...
Ch 12. Saxon Calculus: Asymptotic & Unbounded...
Ch 13. Saxon Calculus: Continuity as a...
Ch 14. Saxon Calculus: Parametric, Polar &...
Ch 15. Saxon Calculus: Concept of the...
Ch 16. Saxon Calculus: Derivative at a...
Ch 17. Saxon Calculus: Derivative as a...
Ch 18. Saxon Calculus: Second...
Ch 19. Saxon Calculus: Applications of the...
Ch 20. Saxon Calculus: Computation of...
Ch 21. Saxon Calculus: Riemann Sums
Ch 22. Saxon Calculus: Interpretations & Properties of...
Definite Integrals: Definition 6:49
How to Use Riemann Sums to Calculate Integrals 7:21
Calculating Integrals of Simple Shapes 7:50
Definite Integrals: Rate of Change
Using Additive Properties of Definite Integrals
7:38 Next Lesson Linear Properties of Definite Integrals
Ch 23. Saxon Calculus: Applications of...
Ch 24. Saxon Calculus: Fundamental Theorem of...
Ch 25. Saxon Calculus: Techniques of...
Ch 26. Saxon Calculus: Applications of...
Ch 27. Saxon Calculus: Numerical Approximation...
Ch 28. Saxon Calculus: Concept of...
Ch 29. Saxon Calculus: Series of...
Ch 30. Saxon Calculus: Taylor Series
Using Additive Properties of Definite Integrals Related Study Materials
Related Topics
Browse by Courses
Math 104: Calculus
AP Calculus AB & BC: Exam Prep
LSAT Study Guide and Test Prep
SAT Subject Test Mathematics Level 2: Practice and Study Guide
Linear Algebra: Help & Tutorials
Geometry Proofs: Help & Tutorials
Prentice Hall Pre-Algebra: Online Textbook Help
Basic Geometry: Help & Review
Math 103: Precalculus
GED Math: Quantitative, Arithmetic & Algebraic Problem Solving
Glencoe Geometry: Online Textbook Help
CAHSEE Math Exam: Test Prep & Study Guide
TExES Mathematics 7-12 (235) Study Guide and Test Prep
CLEP College Algebra Study Guide and Exam Prep
CLEP College Mathematics Study Guide and Exam Prep
Browse by Lessons
Anti-Derivatives: Calculating Indefinite Integrals of Polynomials
Finding the Integral of e^x
Indefinite Integral Overview, Rules & Examples
Finding Antiderivatives | Rules & Examples
How to Antidifferentiate Sums
Antiderivatives of Products of Constants & Functions
Solving the Integral of Cos(x)
How to Antidifferentiate Logarithmic Functions
Indefinite Integrals as Anti Derivatives
Finding Antiderivatives Using Initial Conditions
Integration of Piecewise Function | Calculation, Steps & Examples
Integral of Csc(X) | Overview, Steps & Antiderivative
Finding the Antiderivative of 1/cos(x)
How to Use Riemann Sums for Functions and Graphs
Riemann Sum to Integral | Overview, Formula & Examples
Create an account to start this course today
Used by over 30 million students worldwide
Create an account
Like this lessonShare
Explore our library of over 88,000 lessons
Search
Browse
Browse by subject
College Courses
Business
English
Foreign Language
History
Humanities
Math
Science
Social Science
See All College Courses
High School Courses
AP
Common Core
GED
High School
See All High School Courses
Other Courses
College & Career Guidance Courses
College Placement Exams
Entrance Exams
General Test Prep
K-8 Courses
Skills Courses
Teacher Certification Exams
See All Other Courses
Upgrade to enroll×
Upgrade to Premium to enroll in Saxon Calculus Homeschool: Online Textbook Help
Enrolling in a course lets you earn progress by passing quizzes and exams.
Track course progress
Take quizzes and exams
Earn certificates of completion
You will also be able to:
Create a Goal
Create custom courses
Get your questions answered
Upgrade to Premium to add all these features to your account!
Upgrade Now
Upgrade to Premium to add all these features to your account!
Download the app
Plans
Student Solutions
Teacher Solutions
Study.com for Schools
Working Scholars® Solutions
Online tutoring
About Us
Blog
Careers
Teach For Us
Press Center
Ambassador
Scholarships
Support
FAQ
Site Feedback
Download the app
Working Scholars® Bringing Tuition-Free College to the Community
© Copyright 2025 Study.com. All other trademarks and copyrights are the property of their respective owners. All rights reserved.
Terms of UsePrivacy PolicyDMCA NoticeADA ComplianceHonor Code For Students
× |
189043 | https://www.inchcalculator.com/circumference-calculator/ | Circumference Calculator
Calculate the circumference of a circle using the radius, diameter, or area with the calculator below.
On this page:
Joe is the creator of Inch Calculator and has over 20 years of experience in engineering and construction. He holds several degrees and certifications.
Ethan has a PhD in astrophysics and is currently a satellite imaging scientist. He specializes in math, science, and astrophysics.
Cite This Page
How to Find the Circumference of a Circle
A circle is a round, two-dimensional shape with no corners or straight edges where each point along its edge is equidistant from its center point. The circumference is the distance around the outside of a circle, also known as the perimeter of a circle.
If the path around the circle were flattened out into a straight line, the circumference would equal the length of the line. A circle is also defined by its radius, which is the length of a line from the center point of a circle to the outer edge.
Another defining feature is its diameter, which is the length of a line from an edge of a circle to the opposite edge passing through the center of the circle. The diameter is equal to twice the length of the radius.
Circumference Formula
You can find a circle’s circumference using its radius and the constant π using the formula:
C = 2πr
The circumference C is equal to 2 times pi times the radius r of the circle r.
Pi is a mathematical constant with an approximate value of 3.141592… Pi can also be approximated as the fraction 22/7, but because it’s an irrational number, it cannot be expressed precisely as a fraction.
Diameter to Circumference Formula
Since the diameter is twice the length of the radius, the formula to find the circumference given the diameter is:
C = πd
The circumference C is equal to pi times the diameter of the circle d.
Area to Circumference Formula
The area is the amount of space the circle occupies. The formula to calculate the circumference given the area of a circle is:
C = 2 × π × A
The circumference C is equal to 2 times the square root of pi times the area A of the circle.
How to Find the Diameter From Circumference
It is possible to solve for the diameter of a circle using its circumference. Since we already know how to find the circumference from the diameter, simply solve the above equation for diameter:
d = C / π
Thus, the diameter d of a circle is equal to the circumference C divided by pi.
This is the same formula that our circumference to diameter calculator uses to solve this problem.
Frequently Asked Questions
Why is the circumference 2πr?
The formula to find circumference is defined as πd since π is a ratio of a circle’s circumference to its diameter. The diameter of a circle is equal to twice the length of the radius, so the formula 2πr is equivalent to πd for determining the circumference.
Is circumference 2πr or πd?
The formulas 2πr and πd are equivalent ways to find the circumference of a circle. Since a circle is most frequently defined in terms of its radius, the formula 2πr is the most commonly used circumference formula.
Is the radius half the circumference?
Not quite. The radius of a circle is equal to half of its diameter, but it is not equal to half the circumference. The diameter is equal to the circumference divided by π, so the radius is equal to the circumference divided by π divided by 2.
This can be simplified to the following formula:
r = C ÷ 2π
The radius r is equal to the circumference C divided by 2 times pi.
Are circumference and perimeter the same thing?
Yes! The circumference is essentially the perimeter of a circle. The circumference is more specifically defined as the boundary of a curved geometric figure, so when working with circles, the term circumference is used to express its perimeter.
Is there a formula for the circumference of an ellipse?
Unfortunately, the circumference of an ellipse cannot be precisely defined, but there are several formulas that can calculate it with good approximation.
One formula to approximate the circumference of an ellipse is:
C = π( 3(a + b) – √[(3a + b)(a + 3b)] )
The circumference C of an ellipse is approximately equal to pi times 3 times the semi-major axis a plus the semi-minor axis b, minus the square root of 3 times a plus b, times a plus 3 times b.
How do you find the circumference of a sphere?
The circumference of a sphere is equal to the distance around a plane that passes through its center point, called a great circle. Since the two-dimensional slice of the sphere is a circle, the formula to find the circumference of a sphere is the same as the formula for a circle.
C = 2πr
The circumference of a sphere is equal to 2 times π times the sphere’s radius r.
If you want to calculate the surface area of the sphere, use the formula:
SA = 4πr
Where SA is the surface area and r is the radius of the sphere.
Shapes Icon Similar Geometry Calculators
Have Feedback or a Suggestion?
About |
189044 | http://www.acted.co.uk/forums/index.php?threads/geometric-distribution-memoryless-property.3378/ | New posts
Search forums
Menu
Log in
Register
Install the app
How to install the app on iOS
Follow along with the video below to see how to install our site as a web app on your home screen.
Note: This feature may not be available in some browsers.
We are pleased to announce that the winner of our Feedback Prize Draw for the Winter 2024-25 session and winning £150 of gift vouchers is Zhao Liang Tay. Congratulations to Zhao Liang. If you fancy winning £150 worth of gift vouchers (from a major UK store) for the Summer 2025 exam sitting for just a few minutes of your time throughout the session, please see our website at for more information on how you can make sure your name is included in the draw at the end of the session.
Please be advised that the SP1, SP5 and SP7 X1 deadline is the 14th July and not the 17th June as first stated. Please accept our apologies for any confusion caused.
Forums
Legacy Subjects
CT Subjects
CT3
You are using an out of date browser. It may not display this or other websites correctly.You should upgrade or use an alternative browser.
Geometric distribution - memoryless property
Thread starter Simon C
Start date
S
Simon C
Member
#1
The geometric distribution is said to have the memoryless property P(X > x + n|X > n) = P(X > x).The proof for the Type 1 Geometric distribution is shown in the ActEd notes Chapter 4 page 7. I assumed this could also be proved for the Type 2 distribution but, on attempting it, I get P(X > x + n|X > n) = P(X > x – 1). For example, under the Type 2 basis I calculate that P(X > 2 + 4|X > 4) = P(X > 1).I am fairly sure that my algebra is correct here, especially as the Type 2 distribution is "one out" from the Type 1 distribution. However I am struggling to interpret the meaning of this result or understand why P(X > x + n|X > n) = P(X > x) does not still apply in the Type 2 situation.I calculate P(X <= x) to be 1 - q^x for the Type 1 distribution and 1 - q^(x + 1) for the Type 2 distribution. This gives P(X > x) as q^x for the Type 1 distribution and q^(x + 1) for the Type 2 distribution.Please could someone let me know where my calculations have gone wrong or alternatively help me to interpret why the Type 2 result differs.Thanks in advance for any help provided.
J
Julie Lewis
Member
#2
Suppose that X is Type 2 Geometric, so that P(X=x) = pq^x . Then:P(X>=x+k | X>=k) = P(X>=x+k) / P(X>=k)= {pq^(x+k) + pq^(x+k+1) + ...} / {pq^(k) + pq^(k+1) + ...}Dividing the numerator and denominator by pq^k gives{q^x + q^(x+1) + ...} / { 1 + q + ... }The denominator is equal to 1/(1-q) = 1/p So:P(X>=x+k | X>=k) = p{q^x + q^(x+1) + ... } = P(X>=x)
S
Simon C
Member
#3
Thanks very much Julie. I've followed your proof through and it makes sense. However one aspect that is still confusing me is that under Type 1 we have proved P(X > x + n | X > n) = P(X > x) whereas under Type 2 we have proved P(X >= x + k | X >= k) = P(X >= x).I don't think P(X > x + n | X > n) = P(X > x) holds under Type 2. I'm struggling to gain an intuitive understanding of why we are looking at slightly different criteria under the Type 1 and Type 2 scenarios. What is the reason for this?Thanks againSimon
J
Julie Lewis
Member
#4
Ah, I see your point.The difference between Type 1 and Type 2 is that, with Type 1 we are counting the number of trials until the first success, and with Type 2 we are counting the number of failures before the first success. So if X is Type 1 Geom, then Y = X-1 is Type 2.So we have:P(X > x + n | X > n) = P(X > x)or equivalently:P(X-1 > x + n-1 | X-1 > n-1) = P(X-1 > x-1)Now replacing X-1 by Y, this is:P(Y > x + n -1 | Y > n-1) = P(Y > x-1)or:P(Y >= x+n | Y>=n) = P(Y>= x)and that's it!
S
Simon C
Member
#5
Thanks for that - am glad to see I was on the right track.Thinking about it further, the different symbols in the formulae make intuitive sense if we consider the following reasoning. For the Type 1 memoryless property, we're considering the probability of there being more than a certain number of trials before the first success. If x equals a particular value under Type 1, then success has occurred at that value so this value of x does not meet the criteria that we are interested in. For a Type 2 variable to meet the memoryless property, we need to make sure that it is considered in a consistent manner to a Type 1 variable. If we are interested in there being more than x trials before the first success under Type 1, this is equivalent to there being more than or equal to x failures under Type 2. Therefore, when considering the memoryless property, it is appropriate and consistent to use > for Type 1 and >= for Type 2.
You must log in or register to reply here.
Share:
Facebook X (Twitter) Reddit Pinterest Tumblr WhatsApp Email Share Link
Forums
Legacy Subjects
CT Subjects
CT3 |
189045 | https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Thermodynamics/Ideal_Systems/Ideal_Gas_Processes | Ideal Gas Processes - Chemistry 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
Ideal Systems
Thermodynamics
{ }
{ Entropy_of_Mixing : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", Ideal_Gas_Processes : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", Thermodynamics_of_Mixing : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1" }
{ Advanced_Thermodynamics : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "Basics_Thermodynamics_(General_Chemistry)" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", Calorimetry : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", Chemical_Energetics : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", Energies_and_Potentials : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", Fundamentals_of_Thermodynamics : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", Ideal_Systems : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", Path_Functions : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "Real_(Non-Ideal)_Systems" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", Thermochemistry : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", Thermodynamic_Cycles : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", The_Four_Laws_of_Thermodynamics : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1" }
Wed, 01 Jan 2025 22:42:05 GMT
Ideal Gas Processes
1915
1915
Delmar Larsen
{ }
Anonymous
Anonymous User
2
false
false
[ "article:topic", "showtoc:no", "license:ccbyncsa", "licenseversion:40", "author@Joanne Chan" ]
[ "article:topic", "showtoc:no", "license:ccbyncsa", "licenseversion:40", "author@Joanne Chan" ]
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. Physical & Theoretical Chemistry
4. Supplemental Modules (Physical and Theoretical Chemistry)
5. Thermodynamics
6. Ideal Systems
7. Ideal Gas Processes
Expand/collapse global location
Ideal Gas Processes
Last updated Jan 1, 2025
Save as PDF
Entropy of Mixing
Thermodynamics of Mixing
Page ID 1915
( \newcommand{\kernel}{\mathrm{null}\,})
Table of contents
1. Introduction
2. A Quick Recap on Thermodynamics
3. Thermodynamic and Ideal Gases
1. A Monatomic Ideal Gas
2. A Diatomic Ideal Gas
3. Work in Ideal Gases
4. The Heat Capacity and State Functions
Types of Ideal Gas Processes
Isobaric
Isochoric
Isothermal
Adiabatic
Example 1
Solution
Example 2
Solution
Example 3
Solution
Example 4
Solution
Example 5
Solution A
Solution B
References
In this section we will talk about the relationship between ideal gases in relations to thermodynamics. We will see how by using thermodynamics we will get a better understanding of ideal gases.
Introduction
In the realm of Chemistry we often see many relations between the former and its relations to Physics. By utilizing both Chemistry and Physics we can get a better understanding for the both mentioned. We will use what we know about Ideal Gases and Thermodynamics to try to understand specific processes that occur in a system.
A Quick Recap on Thermodynamics
Before we discuss any further, let’s do a very quick recap on the important aspects of thermodynamics that are important to know for ideal gas processes. Some of this will be a quick review and some will be relatively new unless you have seen it in your Physics class.
So first off let’s state the First law of thermodynamics:
ΔU=Q+W
The whole point of stating this equation is to remind us that energy within any given system is conserved. What that means is that no energy is EVER created or destroyed, but it is simply converted from one form to another, such as heat to work and vice versa.
In case you may be confused with some of these symbols, here is a short explanation in a table.
ΔU This is the total change in the internal energy of the gas.
Q This is the total heat flow of the gas
• When Q is negative (-), heat is being removed from the system
• When Q is positive (+), heat is being added to the system
W This is the total work done on or by the gas
• When W is negative (-), work is being done by the system
• When W is positive (+), work is being done on system
Thermodynamic and Ideal Gases
Below are two equations that describe the relationship between the internal energy of the system of a monatomic gas and a diatomic gas. In a monatomic (mono-: one) gas, since it only has one molecule, the ways for it have energy will be less than a diatomic gas (di-: two) since a diatomic gas has more ways to have energy (Hence, diatomic gas has a 5/2 factor while a monatomic gas has a 3/2).
Looking at these two equations we have also conclude that the internal energy (ΔU) only has an effect on the kinetic energy of the gas molecules (movement). Nowhere in these two equations do we see that the potential energy being affected.
A Monatomic Ideal Gas
A Monatomic Ideal Gas Equation:
ΔU=3 2nRΔT
In a monatomic gas, it has a total of three translational kinetic energy modes (hence, the 3/2).
A Diatomic Ideal Gas
A Diatomic Ideal Gas Equation:
ΔU=5 2nRΔT
In a diatomic gas, it has a total of three translational kinetic energy modes and two rotational energy modes (hence, the 5/2).
Work in Ideal Gases
In relations to the first law of thermodynamics, we can see that by adding heat (Q) or work (W) the internal energy of the gaseous system can be increased. Also, that during compression of the system, the volume of the gas will decrease and response its temperature will increase and thus the internal energy of the system will also increase since temperature is related to energy. And this is true except in an isothermal system (which we’ll talk more about later). That is why when a gas is compressed the work is positive and when it is being compressed, it is negative.
It may also be good to know that the area under the curve is work. If you have taken Calculus, you may remember the integral as it is used to find the area under the curve (or graph) as shown below.
(1)W=−∫P d V
OR
W = - (area under curve)
In this case, you can literally take the area of the triangle or work with integrals.
Work = Area = (1/2)base x height
or
Work = ∫F(x) dx
The Heat Capacity and State Functions
When certain state functions (P, V, T) are held constant, the specific heat of the gas is affected. Below is the universal formula for a gas molecule when its pressure is held constant:
c p=c v+R
When this formula is rearranged we get the heat capacity of the gas when its volume is held constant:
c v=R−c p
Types of Ideal Gas Processes
There are four types of thermodynamics processes. What this basically means is that in a system, one or more variable is held constant. To keep things simple, below are examples as to how keeping a certain variable in a system constant can lead to.
Isobaric
This is a process where the pressure of the system is kept constant.
ΔP=0
An example of this would be when water is boiling in a pot over a burner. In this case, heat is being exchanged between the burner and pot but the pressure stays constant.
To derive this process we start off by using what we know, and that is the first law of thermodynamics:
ΔU=Q+W
Rearranging this equation a bit we get:
Q=ΔU+W
Next, since pressure is equal to WΔV, it can be denoted as:
Q=ΔU+pΔV
Now, the ideal gas law can be applied (PV=nRΔT) and since pressure is constant:
Q=ΔU+nRΔT
For the next step, we will assume that this number of moles of gas stays constant throughout this process:
Q=n c V ΔT+nRΔT
Simplifying the equation some more by taking out the nΔT from both equations we get:
Q=n(c V+R)ΔT
Knowing that c p = c v + R we can substitute for cp:
Q=n c P ΔT
Now we got the equation for an isobaric process!
Figure: Isobaric Process in Graphical Form
Isochoric
This is a process where the volume of the system is kept constant.
ΔV=0
An example of this would be when you have Helium gas sealed up in a container and there is an object (like a piston) pushing down the container (exerting pressure). But, gas molecules is neither entering nor exiting out of the system.
Let’s find the equation for this process, as before let’s start off with the first law of thermodynamics:
ΔU=Q+W
Rearranging this equation a bit we get:
Q=ΔU+W
In this case, since volume is constant, ΔV = 0:
Q=ΔU
Since the internal energy of the system equals to the amount of heat transferred we can replace ΔU with the ideal gas equation for heat:
Q=nC VΔT
Above is the ideal gas equation for an isochoric process!
Figure: Isochoric Process in Graphical Form
Isothermal
•This is a process where the temperature of the system is kept constant.
ΔU = 0, ΔT = 0
•When volume increases, the pressure will decrease, and vice versa.
ΔT = 0 then:ΔV ↑and P ↓ORΔV↓ and P ↑ (inverse relationship)
•As an example, gas molecules are sealed up in a container but an object on top of the container (such as a piston) pushes down on the container in a very slow fashion that there is not enough to change its temperature.
Figure: Isothermal Process in Graphical Form
To derive the equation for an isothermal process we must first write out the first law of thermodynamics:
ΔU=Q+W
Rearranging this equation a bit we get:
Q=ΔU+W
Since ΔT = 0. Therefore we are only left with work:
Q=W
As such we get:
W=−pΔV
Making this equation into an ideal gas equation we get:
W=nRt V
In order to get to the next step we need to use some calculus:
Q=nRTlnV f V i
And there you go! The equation for an isothermal process.
Adiabatic
•This is a process where no heat is being added or removed from the system.
•Or can be simply stated as: no heat transfer (or heat flow) happening in a system.
•In freshman chemistry, only the basic idea of this process is needed and that is when there is no heat transfer, Q = 0.
Figure: Adiabatic Process in Graphical Form
Example 1
The volume of a gas in a container expanded from 1L to 3L upon releasing the piston upward. From following graph, find the amount work associated with the expansion of the gas.
f(x)=x+3 from [1, 3]
Solution
As stated above, we know that the amount of work is the same as saying the area under the curve. In this case we can look for shapes we can easily find the area of.
Here, we have a shape that is similar to a rectangle.
Work= Area = Length x Width
= (3)(3-1)
=(3)(2)
=6
Work = 6 Joules
Example 2
Calvin is observing an unknown monatomic gas molecule (sealed inside a container) in his Freshman Chemistry Lab. He has been told by his lab instructor that there are four moles of this unidentified gas in the container. The laboratory room’s temperature was initially set at room temperature when he started the lab, but by the time he was almost finished with the lab the temperature has gone up 10°C. What is the total internal energy of this unknown gaseous substance by the time Calvin’s lab session ended?
Solution
From the problem, we know that the unknown gaseous substance is monatomic, so we would have to use the equation for a monatomic gas:
ΔU=3 2nRΔT
Also from the problem, we know that the substance is at room temperature with is 25°C, but that was just extra information given and we didn’t really need to know that to solve the problem. Since we know that the temperature change from start to finish was +10°.
ΔT=+10K
The problem also given us the number of moles that was in the container:
n=4mol
For the R-value, we can choose any R constant, but to make this problem a little easier we will choose a R-constant that will cancel out any other units except for joules.
R=8.3145J molK
Plugging in all these values into the equation we have:
ΔU=3 2(4mol)(8.3145J molK)(10K)
=498J
Example 3
In an isochoric system, three moles of hydrogen gas is trapped inside an enclosed container with a piston on top of it. The total amount of internal energy of the gaseous system is 65 Joules, and the temperature of the system decreased from 25°C to 19°C. What is the specific heat of the gas molecules?
Solution
Since the system is an isochoric system, there will be zero change in volume, therefore:
ΔV=0
And the equation for an isochoric system will be:
Q=nC VΔT
From the equation we know the following:
n=3mol
Q=ΔT=(25−19)K=6K
(Since we are only trying to find the difference between the starting point and the final point, we do not need to do conversions to degrees Kelvin)
Q=65J
Before plugging the values in, let’s rearrange the isochoric equation and set it to what we are trying to find:
C V=Q nΔT
Now, we can plug the values in:
C V=65J(3mol)(6K)
=3.6J molk
Example 4
While looking over some of her lab data, a chemistry student notices she forgot to record the numbers of moles of the gas molecule she looked at. The experiment she did that day had kept the pressure constant and the temperature went down two degrees in the process of the experiment. Assuming all variables are also ideal, how many moles was she dealing with?
Solution
From the problem we are given the following:
ΔT=+2K
ΔU=Q=110J
Since are given that the pressure of the gas was kept constant:
C p=14J Kmol
We have enough evidence to conclude that this is an isobaric process:
Q=n\c P ΔT
Rearranging the equation to fit what we are trying to find we get:
n=Q\c P ΔT
Now we can plug in our values:
n=110J 14J Kmol2K
=4molesofgas
Example 5
A chemistry student is looking at 5.00 grams of the monatomic Helium gas that is put into a container that expands from 10 L to 13 L. The container is confined at a constant temperature of 30°C in an enclosed system.
What is the total energy of the system?
Is the pressure of the system increasing, decreasing, or not affected?
Solution A
First, let’s list what is given in the problem:
V i=10L
V f=13L
Converting temperature from Celsius to Kelvin we have:
T=(30+273.15)K=303.15K
And since temperature is constant, this is an isothermal process:
Q=nRTlnV f V i
In order to use this equation we will need to convert Helium gas to moles:
n=grams grams mole
=5.00gHe 4.00gHe molHe
=1.25molHe
We will choose a universal R-constant that will easily cancel out all the units that will help us get the answer without any other conversions:
=8.31451J Kmol
Plugging in all these values into the isothermal equation we get:
Q=(1.25mol)(8.31451J Kmol)(303.15K)ln13L 10L
=827J
Solution B
It is a rule that in an isothermal process that volume and pressure have an inverse relationship. Meaning in one goes up, the other must go down. In this case, volume went up so the pressure must have decreased.
References
Waite, Boyd A. "A Gas Kinetic Explanation of Simple Thermodynamic Processes." Journal of Chemical Education 62.3 (1985): 224-27. Print.
Gislason, Eric A., and Norman C. Craig. "General Definitions of Work and Heat in Thermodynamic Processes." Journal of Chemical Education 64.8 (1987): 660-68. Print.
Snider, Eric H. Ideal Gas Law, Enthalpy, Heat Capacity, Heats of Solution and Mixing. New York: American Institute of Chemical Engineers, 1984. Print.
Redfern, Francis. "Physics Book - Chapter 13." Texarkana College. 2001. Web. 28 Feb. 2011. .
Young, Hugh D., Roger A. Freedman, Francis Weston Sears, and Hugh D. Young. Sears and Zemansky's University Physics. San Francisco: Pearson/Addison Wesley, 2004. 748-51. Print.
Ideal Gas Processes is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Joanne Chan.
Back to top
Entropy of Mixing
Thermodynamics of Mixing
Was this article helpful?
Yes
No
Recommended articles
CalorimetryCalorimetry is the process of measuring the amount of heat released or absorbed during a chemical reaction. By knowing the change in heat, it can be d...
Path Functions
A System and Its SurroundingsA primary goal of the study of thermochemistry is to determine the quantity of heat exchanged between a system and its surroundings. The system is the...
Constant Pressure CalorimetryBecause calorimetry is used to measure the heat of a reaction, it is a crucial part of thermodynamics. In order to measure the heat of a reaction, the...
2nd Law of ThermodynamicsThe Second Law of Thermodynamics states that the state of entropy of the entire universe, as an isolated system, will always increase over time. The s...
Article typeSection or PageLicenseCC BY-NC-SALicense Version4.0Show Page TOCno on page
Tags
author@Joanne Chan
© Copyright 2025 Chemistry 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× |
189046 | https://literacyideas.com/anchor-charts/ | Effective Anchor Charts for Engaging Classroom Writing
Skip to content
TEACHING RESOURCES
Search for:
Writing
Foundations of Writing
Text Types and Writing Genres
The Author’s Purpose
The Writing Process
Superb Sentence Writing
Perfect Paragraph Writing
Editing & Proofreading Skills
Excellent Essay Writing
How to Start an Essay with Strong Hooks and Leads
How to write convincing Conclusions
5 Paragraph (Hamburger) Essay Structure
Writing to Entertain
Fables
Fairy Tales
The Narrative
Narrative Overview
Short Story Writing
Writing engaging Characters and Settings
How to Write a Great Plot
Poetry
Introduction to Poetry
Elements of Poetry
Fun Poetry for Kids
Simile Poems
Writing to Inform
Article Writing
Autobiographies
Biographies
Compare and Contrast Essays
Descriptive Texts
Explanatory Texts
Expository Essays
Formal Letter Writing
Historical Recounts
Informational Texts
Personal Narratives
Procedural Texts
Recount Writing
Hypothesis Writing
Transactional Writing
Writing to Persuade
Advertisements
Arguments & Discussions
Book Reviews
Debating Speeches
Opinion Writing Essentials
Persuasive Writing
Text Response
Writing Prompts
Essay Writing
Lessons & Strategies
Premium Writing Resources
Reading
The Foundations of Reading
Cause & Effect
Compare & Contrast
Facts & Opinions
The Main Idea
Making Inferences
Phonics
Point of View & Perspective
Comprehension Strategies
Sequencing Events
Reading Literature
Elements of Literature
The Elements of a Story
Completing a Novel Study
Elements of Poetry
Reading Informational Texts
Anchor Charts
Critical Thinking & Reading
Fake News Activities
Fake News & Misinformation
Graphic Organizers
Research Strategies
Guided Reading Strategies
Lessons & Strategies
Language
Language Conventions
Parts of Speech
Sentence Structure
A Complete Guide to Grammar
A complete guide to Punctuation
Literary Devices & Figures of Speech
Figurative Language
13 Literary Devices to Supercharge your Writing Skills
Hyperbole: A Complete Guide for Students and Teachers
A guide to Onomatopoeia
A Guide to Personification
Sarcasm, Satire & Parody
Glossary of literary terms
Lessons & Strategies
Premium Reading Resources
Multiliteracies
Media Literacy
TEACHING RESOURCES
Anchor Charts to Improve Writing Skills
ByShane Mac DonnchaidhJuly 23, 2021 March 18, 2024 March 18, 2024
Table of Contents
WHAT ARE ANCHOR CHARTS?
WHAT ARE THE BENEFITS OF ANCHOR CHARTS FOR WRITING AND READING?
Anchor Charts Provide Increased Student Engagement
Anchor Charts Deepen Comprehension
Anchor Charts Supports Independent Work
WHAT DO ANCHOR CHARTS LOOK LIKE?
HOW ARE ANCHOR CHARTS CREATED IN THE CLASSROOM?
Anchor Charts as Writing Tools – Examples
1. Point of View Anchor Chart
2. Instructional Writing Anchor Chart
USEFUL VIDEOS TO EXPLAIN ANCHOR CHARTS
OTHER GREAT ARTICLES RELATED TO ANCHOR CHARTS FOR WRITING
WHAT ARE ANCHOR CHARTS?
Anchor charts are tools that support learning in the classroom. They can be used to support everything from classroom management strategies to the teaching of writing.
Essentially, they are visual prompts that provide students with information regarding their prior learning on a given topic. These visual prompts are used to provide a scaffold to support the students during guided practice and independent work.
WHAT ARE THE BENEFITS OF ANCHOR CHARTS FOR WRITING AND READING?
Anchor charts are becoming increasingly popular in classrooms for some very good reasons. They offer a wealth of benefits for students and teachers alike.
Here are just a few of the great benefits of using anchor charts as writing tools in the classroom.
Anchor Charts Provide Increased Student Engagement
Anchor charts are an effective way of encouraging student engagement. Not only do they increase student confidence when engaged in a writing task, but they help to keep students on task by offering support in the form of visual prompts that help unstick the stuck! These anchor chart examples below provide students with a great visual point of reference to learn from.
101 DIGITAL & PRINT GRAPHIC ORGANIZERS FOR ALL CURRICULUM AREAS
Introduce your students to 21st-century learning with this GROWING BUNDLE OF 101 EDITABLE & PRINTABLE GRAPHIC ORGANIZERS. ✌NO PREP REQUIRED!!!✌ Go paperless, and let your students express their knowledge and creativity through the power of technology and collaboration inside and outside the classroom with ease.
Whilst you don’t have to have a 1:1 or BYOD classroom to benefit from this bundle, it has been purpose-built to deliver through platforms such as ✔GOOGLE CLASSROOM, ✔ OFFICE 365, ✔ or any CLOUD-BASED LEARNING PLATFORM.
DOWNLOAD NOW
Anchor Charts Deepen Comprehension
Often, students get involved in the actual production of the anchor charts themselves. When helping to produce the anchor charts, students will have opportunities to reconstruct their learning and thereby deepen their comprehension of the material in the process.
As they construct their charts, students begin to make new connections between the various aspects of their learning as they organize these aspects in a visually comprehensible manner.
Anchor Charts Supports Independent Work
Learning to write well can be one of the most challenging things a student learns to do at school.
As students learn to navigate the demands of various writing genres, seemingly endless questions arise at word, sentence, and whole text level.
This, in turn, makes heavy demands on the teacher’s time as individual students struggle with the various challenges of a given task.
Fortunately, anchor charts can help alleviate some of that burden by providing a visual resource and reference point that help students to answer many of the more commonly asked questions for themselves. This frees up the teacher from having to repeatedly answer the same questions throughout the course of a lesson, making more time to offer support where it’s most needed.
WHAT DO ANCHOR CHARTS LOOK LIKE?
Anchor charts come in all shapes and sizes and can be commercially bought or produced collaboratively by students and teachers in class.
Commercially bought Anchor Charts are great for permanent displays within the classroom.
Usually well-presented in bold lettering using dynamic colors, professionally-produced charts work well for topics that recur throughout the course of the year and are complex enough to require ongoing reinforcement.
When anchor charts are self-produced, they are usually handwritten in large print and displayed in a prominent position in the classroom for easy reference.
Usually, a co-creation between the student and the teacher, the charts should contain only the essential information regarding the topic.
When deciding what to include on a chart, think about the concepts, strategies, and prior learning that will most help students to work independently when engaged in their work.
HOW ARE ANCHOR CHARTS CREATED IN THE CLASSROOM?
To produce an anchor chart in the classroom requires very little in the way of resources other than some chart paper and some colored markers. Other than these, and defining a clear purpose and focus for your anchor chart, there is no specific preparation required.
However, there are a number of common elements to consider when producing anchor charts for use in the classroom. Some of these include:
●Paper: Decide whether you are using adhesive paper, lined paper, blank paper, colored paper etc
●Font-Size:This should be large enough to see from the various working areas of the classroom
●Collaboration: Is it teacher-produced or a collaboration? What is the level of student involvement?
Where Anchor Charts are to be co-created with students, generally, they will be produced in collaboration with the students as you teach the lesson.
The chart will include the most important content and relevant strategies. In the case of the various writing genres, a list of the main criteria that must be included works well.
The anchor charts can then be used by the students as a checklist to refer to as the writing is produced. They can also serve for a final check when the work has been completed.
Here are some general tips to help ensure you get the most out of Anchor Charts in your classroom:
● Keep things simple
● Be sure the writing is well organised and easy to read
● Use headings and bullet points to help display the main points
● Use different colors for headings, bullet points etc
● Use simple pictures, graphs, illustrations etc to help reinforce points
● Don’t fill with lots of distracting details or graphics
Anchor Charts as Writing Tools – Examples
Anchor charts can be used very effectively to break down many of the more complex aspects of writing.
From punctuation use to the specific criteria for various writing genres, Anchor charts are a fantastic way to visually reinforce student understanding of these diverse processes.
The content of each chart will be dependent on their focus. But, let’s look at the possible content of two examples of Anchor Charts to help serve as models for what might be contained in an anchor chart produced in your classroom.
1. Point of View Anchor Chart
The Point of View Anchor Chart can be used both to help guide students in identifying the point of view in a text as well as to help in the creation of the student’s own texts. We have an excellent guide on point of view that can be found here.
Looking out for keywords is an effective way to determine the point of view in a piece of writing. Point-of-view keywords are generally centred around the pronouns and the level of insight and perspective we are offered.
Let’s look at some of the more common points of view used, first in a little detail and then at how they might appear on an Anchor Chart.
First Person – a character is telling the story (narrator), and we often gain insight into the characters’ thoughts. Clues that indicate a first person will be the use of pronouns such as I,my,me,mine, we,us etc.
As bullet points, this might look like this:
●First Person
○ Character narrates the story
○ Narrator is in the story
○ Narrator’s thoughts are revealed
○ Uses pronouns: I,my,me,mine, we, us etc.
Third Person Limited– the narrator is outside the story and telling the story. In third-person limited, the writer sticks closely to the point of view of a single character, so we are usually only privy to that one character’s thoughts and experiences. The narrator does not know everything about the events that occur in the story. Indications that the third person is being used may be the use of characters’ names and pronouns, such as he, him, his, himself, she, her, hers, herself,it,its, itself, they, them, their,theirs, and themselves.
In brief:
●Third Person Limited
○ Narrator is outside the story
○ Narrator tells the story primarily from one character’s POV
○ Only the main character’s thoughts and feelings are revealed
○ Narrator has limited knowledge of events
○ Uses pronouns: he, him, his, himself, she, her, hers, herself,it,its, itself, they, them, their,theirs, and themselves.
Third Person Omniscient – the narrator tells the story and is privy to everything. Not only all the details of every event in the story, but the interior life of each character, such as emotions, opinions, and feelings, whether expressed or not. The usual third-person pronouns are used along with the character names.
In brief:
●Third Person Omniscient
○ Narrator tells the story from ‘above.’
○ Narrator knows the thoughts and feelings of every character
○ Narrator knows everything that happens
○ Uses pronouns: he, him, his, himself, she, her, hers, herself,it,its, itself, they, them, their,theirs, and themselves.
2. Instructional Writing Anchor Chart
The criteria for writing clear instructions can be very handily displayed as an anchor chart.
When writing a set of instructions, students can refer to the chart to help organize their writing. The same chart will also serve as a checklist for self-assessment at the end.
An instructional writing anchor chart may include information such as:
Instructional Writing:
● Includes an explanatory title, e.g. ‘How to…’
● Laid out in bullet points or numbered instructions
● Uses time connectives to organise, e.g. ‘first’, ‘then’, ‘finally’ etc.
● Uses imperatives to instruct the reader
● Use straightforward, functional language
● Supported by illustrations or diagrams
● Diagrams and illustrations contain captions
In Conclusion
As we can see, anchor charts can serve as useful writing tools that support the development of student writing skills in the classroom.
When displayed prominently in the classroom, they can help students efficiently bridge the gap between being emergent writers lacking in self-confidence to becoming self-assured, independent writers.
It is important to remember, too, that though anchor charts are great tools that support students, ultimately, the intent is for the students to internalize the knowledge and information they contain. So don’t allow them to become a permanent crutch!
USEFUL VIDEOS TO EXPLAIN ANCHOR CHARTS
OTHER GREAT ARTICLES RELATED TO ANCHOR CHARTS FOR WRITING
13 Literary Devices to Supercharge your Writing Skills
The Writing Process
Graphic Organizers for Writing and Reading
Elements of Literature
Multiliteracies
Essay Writing: A complete guide for students and teachers
© 2025 Literacy IdeasAbout Us
WritingToggle child menu
Foundations of WritingToggle child menu
Text Types and Writing Genres
The Author’s Purpose
The Writing Process
Superb Sentence Writing
Perfect Paragraph Writing
Editing & Proofreading Skills
Excellent Essay Writing
How to Start an Essay with Strong Hooks and Leads
How to write convincing Conclusions
5 Paragraph (Hamburger) Essay Structure
Writing to EntertainToggle child menu
Fables
Fairy Tales
The NarrativeToggle child menu
Narrative Overview
Short Story Writing
Writing engaging Characters and Settings
How to Write a Great Plot
PoetryToggle child menu
Introduction to Poetry
Elements of Poetry
Fun Poetry for Kids
Simile Poems
Writing to InformToggle child menu
Article Writing
Autobiographies
Biographies
Compare and Contrast Essays
Descriptive Texts
Explanatory Texts
Expository Essays
Formal Letter Writing
Historical Recounts
Informational Texts
Personal Narratives
Procedural Texts
Recount Writing
Hypothesis Writing
Transactional Writing
Writing to PersuadeToggle child menu
Advertisements
Arguments & Discussions
Book Reviews
Debating Speeches
Opinion Writing Essentials
Persuasive Writing
Text Response
Writing Prompts
Essay Writing
Lessons & Strategies
Premium Writing Resources
ReadingToggle child menu
The Foundations of ReadingToggle child menu
Cause & Effect
Compare & Contrast
Facts & Opinions
The Main Idea
Making Inferences
Phonics
Point of View & Perspective
Comprehension Strategies
Sequencing Events
Reading LiteratureToggle child menu
Elements of Literature
The Elements of a Story
Completing a Novel Study
Elements of Poetry
Reading Informational TextsToggle child menu
Anchor Charts
Critical Thinking & Reading
Fake News Activities
Fake News & Misinformation
Graphic Organizers
Research Strategies
Guided Reading Strategies
Lessons & Strategies
LanguageToggle child menu
Language ConventionsToggle child menu
Parts of Speech
Sentence Structure
A Complete Guide to Grammar
A complete guide to Punctuation
Literary Devices & Figures of SpeechToggle child menu
Figurative Language
13 Literary Devices to Supercharge your Writing Skills
Hyperbole: A Complete Guide for Students and Teachers
A guide to Onomatopoeia
A Guide to Personification
Sarcasm, Satire & Parody
Glossary of literary terms
Lessons & Strategies
Premium Reading Resources
MultiliteraciesToggle child menu
Media Literacy
Search for: |
189047 | https://study.com/skill/learn/adding-rational-expressions-with-common-denominators-quadratic-factoring-explanation.html | Adding Rational Expressions with Common Denominators and Quadratic Factoring | Algebra | 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
Adding Rational Expressions with Common Denominators and Quadratic Factoring
Algebra 2 Skills Practice
Click for sound
4:37
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
00:04 Adding rational…
02:03 Adding rational…
Jump to a specific example
Speed Normal
0.5x
Normal
1.25x
1.5x
1.75x
2x
Speed
Kathryn Boddie
Instructors Kathryn Boddie
Kathryn has taught high school or university mathematics for over 10 years. She has a Ph.D. in Applied Mathematics from the University of Wisconsin-Milwaukee, an M.S. in Mathematics from Florida State University, and a B.S. in Mathematics from the University of Wisconsin-Madison.
View bio
Example SolutionsPractice Questions
Adding Rational Expressions with Common Denominators and Quadratic Factoring
Step 1: Add the rational expression by adding their numerators together.
Step 2: Combine any like terms in the numerator.
Step 3: Fully factor the numerator.
Step 4: Cancel any common factors between the numerator and the denominator.
Adding Rational Expressions with Common Denominators and Quadratic Factoring - Vocabulary
Rational Expression: A rational expression is a ratio of expressions. This means the numerator and the denominator may have multiple terms.
Common Denominator: The denominator of a rational expression is the bottom of the fraction. To add rational expressions, they must have the same denominator - called a common denominator.
We will use these steps and definitions to add rational expressions with common denominators and quadratic factoring in the following two examples.
Example Problem 1: Adding Rational Expressions with Common Denominators and Quadratic Factoring
Simplify the following:
x 2 x+5+3 x−10 x+5
Step 1: Add the rational expression by adding their numerators together.
The two expressions have the same denominator, so we can write as one rational expression by adding the numerators together.
x 2 x+5+3 x−10 x+5=x 2+3 x−10 x+5
Step 2: Combine any like terms in the numerator.
There are no like terms to combine in the numerator.
Step 3: Fully factor the numerator.
To factor the numerator, x 2+3 x−10, we need two numbers that multiply to -10 and add to 3. These numbers are 5 and -2. So, the numerator factors as:
x 2+3 x−10=(x+5)(x−2)
This turns our expression into:
x 2+3 x−10 x+5=(x+5)(x−2)x+5
Step 4: Cancel any common factors between the numerator and the denominator.
The numerator and denominator both have a factor of (x+5). Canceling this common factor, we have:
(x+5)(x−2)(x+5)=x−2
Therefore,
x 2 x+5+3 x−10 x+5=x−2
Example Problem 2: Adding Rational Expressions with Common Denominators and Quadratic Factoring
Simplify the following:
x 2+2 x−10 x−3+2 x−11 x−3
Step 1: Add the rational expression by adding their numerators together.
x 2+2 x−10 x−3+2 x−11 x−3=x 2+2 x−10+2 x−11 x−3
Step 2: Combine any like terms in the numerator.
Combining like terms,
x 2+2 x−10+2 x−11 x−3=x 2+4 x−21 x−3
Step 3: Fully factor the numerator.
x 2+4 x−21 x−3=(x+7)(x−3)x−3
Step 4: Cancel any common factors between the numerator and the denominator.
(x+7)(x−3)(x−3)=x+7
Therefore,
x 2+2 x−10 x−3+2 x−11 x−3=x+7
Get access to thousands of practice questions and explanations!
Create an account
Table of Contents
Adding Rational Expressions with Common Denominators and Quadratic Factoring
Vocabulary
Example Problem 1
Example Problem 2
Test your current knowledge
Practice
Adding Rational Expressions with Common Denominators & Quadratic Factoring
Related Courses
High School Algebra II: Homework Help Resource
College Preparatory Mathematics: Help and Review
General Math Lessons
Topic Summary Lessons
6th-8th Grade Math: Practice & Review
Related Lessons
Equation in Math | Definition, Parts & Examples
Relations and Functions | Meaning, Relationship & Types
Advanced Algebraic Concepts
Algebraic Structures
Geometric Algebra
Recently updated on Study.com
Videos
Courses
Lessons
Articles
Quizzes
Concepts
Teacher Resources
Electromagnetic Induction: Conductor to Conductor &...
Ethnic Groups in Indonesia | Demographics & People
Gods of the Winter Solstice
Holes by Louis Sachar | Themes, Quotes & Analysis
Aurangzeb | Empire, Achievements & Failures
Libya Ethnic Groups | Demographics, Population & Cultures
Cold War Lesson for Kids: Facts & Timeline
The Chronicles of Narnia Series by C.S. Lewis | Overview...
Isotherms Definition, Maps & Types
Abstinence | Definition & Types
Project Cost Management: Planning & Estimation
General Studies Literature: Help & Review
GACE 713 Study Guide - English (6-12) Exam Prep
GACE 711 Study Guide - Mathematics (6-12) Exam Prep
GACE 706 Study Guide - Middle Grades Language Arts (4-8)...
GACE 704 Study Guide - Middle Grades Math (4-8) Exam Prep
GACE 735 Study Guide - Special Education General...
Physics 112L: Physics II with Lab
Science 101L: Intro to Life Sciences with Lab
CLEP Introductory Business Law Study Guide and Exam Prep
TASC Reading: Prep and Practice
ScienceFusion The Dynamic Earth: Online Textbook Help
Common Core History & Social Studies Grades 11-12:...
TExMaT Master Science Teacher 8-12 (092): Practice &...
Psychology 102: Educational Psychology
CSET Math Subtest I (211) Study Guide and Test Prep
Glencoe Physical Science: Online Textbook Help
To Kill a Mockingbird Study Guide
McDougal Littell The Americans: Online Textbook Help
Rock in Performing Arts | History, Characteristics & Facts
Reproductive Endocrinology Meaning & Specialization
Drug Rehabilitation Types, Effects & Importance
Cardiology Definition, Tests & Specializations
Neurology | Definition, Importance & Disorders
Anaerobic Bacteria Definition, Types & Examples
Seasonal Festivals | List & Examples
Clinical Assessment Definition, Importance & Types
The August Revolution of 1945 | History, Causes & Aftermath
Controlled Drinking: Strategies & Treatment
Wat Tyler & the Peasants' Revolt in 1381 | Facts & Rebellion
Self-Fulfillment Definition, History & Importance
Reducing Cognitive Dissonance
Dissociative Amnesia: Treatment, Symptoms, Causes &...
What is a Behavior Assessment in Sport Psychology?
Intrinsic Motivators: Examples & Overview
Aerial Perspective in Psychology | Definition & Examples
Lithium Treatment: Types, Uses & Definition
How to Use & Cite AI Tools in College Saver Course...
Understanding Generative AI as a Student: Uses, Benefits...
WEST Prep Product Comparison
What Are the Features of My Institutional Student Account...
How to Pass the SLLA Exam
How to Pass the TExES Exams
504 Plans in Ohio
How to Learn French Quickly
History Books for Kids
Transition Words Lesson Plan
School Closures in Virginia Due to Coronavirus: Learning...
Quiz & Worksheet - Aslan in The Lion, the Witch & the...
Quiz & Worksheet - Barbados Ethnic Groups
Quiz & Worksheet - The Mule in Their Eyes Were Watching God
Quiz & Worksheet - Who Dies in Bridge to Terabithia?
Quiz & Worksheet - Periodic Functions
Quiz & Worksheet - Legacy of the Neo-Babylonian Empire
Quiz & Worksheet - SEO Definition & Examples
Math
Social Sciences
Science
Business
Humanities
Education
Art and Design
History
Tech and Engineering
Health and Medicine
Plans
Study help
Test prep
College credit
Teacher resources
Working Scholars®
School group plans
Online tutoring
About us
Blog
Careers
Teach for us
Press Center
Ambassador
Scholarships
Support
Contact support
FAQ
Site feedback
Resources and Guides
Download the app
Study.com on Facebook Study.com on YouTube Study.com on Instagram Study.com on Twitter Study.com on LinkedIn
© Copyright 2025 Study.com. All other trademarks and copyrights are the property of their respective owners. All rights reserved.
Contact us by phone at (877)266-4919, or by mail at 100 View Street#202, Mountain View, CA 94041.
About Us
Terms of Use
Privacy Policy
DMCA Notice
ADA Compliance
Honor Code For Students
Support
× |
189048 | https://www.sciencedirect.com/topics/neuroscience/glasgow-coma-scale | Skip to Main content
Sign in
Chapters and Articles
You might find these chapters and articles relevant to this topic.
Coma Scales
Glasgow Coma Scale
The Glasgow Coma Scale (GCS) (Table 1) is almost universally used in emergency rooms and intensive care units (ICUs), and is by far the most common coma scale cited in the neurosurgical literature. However, it has not been used consistently in different hospitals, and the later versions of the GCS have not been adequately tested for reliability. The GCS was designed for the initial assessment of patients with head injury. Problems with the use of the GCS arise when patients are intubated and cannot respond verbally or if the eyes are swollen shut, preventing ocular assessment. A theoretical disadvantage is the three-dimensional assessment: The total score is obtained by adding the values for three motor activities – eye opening, best motor response, and best verbal response. These are assumed to be independent variables but they are not. Because they covary, their addition may not be valid. Furthermore, to achieve a score of 6–12, there are more than 10 simple combinations of variables, each with very different clinical profiles. It seems unlikely that all patients with specific scores ranging from 6 to 10 will be equivalent in disease severity. Additionally, there is little difference in outcome over several different score values (e.g., between 10 and 15). The GCS is often insufficiently sensitive for the detection of changes in the level of consciousness in patients following head injury or with masses and risk of herniation when they are in lighter stages of impaired consciousness. Furthermore, in the application of the GCS to patients who have been in the ICU for an extended period of time, eye opening does not equal conscious awareness because patients with persistent vegetative state (VS) may show this and patients with seizures show spontaneous eye opening.
Table 1. Glasgow Coma Scale
| Item | Factor | Score |
---
| Best motor response | Obeys | 6 |
| Localizes | 5 |
| Withdraws (flexion) | 4 |
| Abnormal flexion | 3 |
| Extensor response | 2 |
| Nil | 1 |
| Verbal response | Oriented | 5 |
| Confused conversation | 4 |
| Inappropriate words | 3 |
| Incomprehensible sounds | 2 |
| Nil | 1 |
| Eye opening | Spontaneous | 4 |
| To speech | 3 |
| To pain | 2 |
| Nil | 1 |
However, the GCS has been the standard scoring assessment throughout the world for 20 years. It seems unlikely that it will be easily replaced, even by potentially superior scoring systems. The Innsbruck and the Edinburgh-2 Coma Scales have some of the same problems as the GCS but the Reaction Level Scale (RLS 85) has a number of advantages over the others.
View chapterExplore book
Read full chapter
URL:
Reference work2014, Encyclopedia of the Neurological Sciences (Second Edition)T.E. Gofton, G.B. Young
Chapter
First contact management
2011, Managing Sports Injuries (Fourth Edition)Christopher M. Norris PhD MSc MCSP
The Glasgow Coma Scale
The Glasgow Coma Scale (Teasdale and Jennett, 1974) is a series of tests that are given a numerical value which can then be used to objectify an athlete's state of consciousness (Table 6.11). The scale also forms domain 3 of the SCAT2 evaluation. The first test relates to the eyes, and determines whether the athlete opens the eyes spontaneously or in response to sound (verbal command) or pain. Opening the eyes to verbal command merely means that the person has registered sound; it does not imply that they necessarily understand the command. The second test is of verbal response, and assesses the athlete's reaction to simple questions such as ‘where are you’ or ‘what is your name’. The test assesses whether the athlete is aware of him/herself and the environment. The third test is of motor response. The maximum score is 6 if the athlete is able to perform actions correctly to verbal commands such as ‘move your arm’. If the athlete fails to respond, a painful stimulus is applied by the practitioner pressing their knuckles into the athlete's sternum, or pressing the athlete's fingers together around a pen. Painful stimuli to the face or palm of the hands should be avoided as these can give reflex eye closing and hand closing respectively (Magee, 2002). Where reflex responses alone result, flexion of the arms and hands together with adduction of the upper limb and extension of the lower limb with plantarflexion of the feet (decorticate posturing) indicates a lesion above the red nucleus. Extension of the arms with pronation of the forearm (decerebrate posturing) indicates a lesion of the brainstem. The time of the test should be noted and the test repeated every 15–30 minutes to note any degeneration of results.
Where the score is between 3 and 8 on the coma scale, emergency care is required immediately as a severe head injury is present. Those with scores of 9–11 are considered to have a moderate head injury, and those with a score of 12 or higher are considered to have a mild head injury.
Caution must always be exercised with head injuries. Unfortunately, the practitioner or coach who has to decide whether to allow an athlete to continue playing has no way of knowing if secondary brain damage is going to develop. At the time of injury, bleeding may have occurred which could accumulate and give rise to a subdural haematoma.
An athlete who remains unconscious should be placed in the recovery position until an ambulance is available to take him or her to hospital. If there is bleeding or discharge from an ear, the athlete should be turned so that the affected ear is dependent. Nothing should be given by mouth, and the athlete should not be left unattended. Testing for responses should continue regularly (every 10 minutes or more frequently) and any changes in the athlete's condition should be recorded.
View chapterExplore book
Read full chapter
URL:
Book2011, Managing Sports Injuries (Fourth Edition)Christopher M. Norris PhD MSc MCSP
Chapter
Maxillofacial trauma
2007, Oral and Maxillofacial Surgery (Second Edition)J.P. Shepherd
The Glasgow Coma Scale
First described by two Glasgow neuroscientists, the Glasgow Coma Scale (GCS) is an internationally recognized method for measuring coma. It cannot discriminate between causes of coma, such as brain injury or alcohol intoxication, but it provides an excellent means of assessing the need for hospital admission and recovery (Table 13.1).
View chapterExplore book
Read full chapter
URL:
Book2007, Oral and Maxillofacial Surgery (Second Edition)J.P. Shepherd
Chapter
Traumatic Brain Injury, Part I
2015, Handbook of Clinical NeurologyScott A. Goldberg, ... Andrew Jagoda
The Glasgow Coma Scale
The Glasgow Coma Scale (GCS) was developed in 1974 as a measure of the depth of impaired consciousness from a TBI. It was intended to provide an easy to use tool to facilitate communication between care providers and was intended for serial use to monitor a patient's neurologic function over time (Teasdale and Jennett, 1974). It has since been adapted for widespread use based on its relative simplicity and association with prognosis and has been further modified for use in the pediatric population (Holmes et al., 2005). The GCS consists of three dimensions: motor responsiveness, verbal performance, and eye opening (Table 23.1).
Table 23.1. The Glasgow Coma Scale
| | | | | | | |
--- --- ---
| Empty Cell | 1 | 2 | 3 | 4 | 5 | 6 |
| Motor response | None | Extensor | Flexor | Localizes | Command | Spontaneous |
| Verbal response | None | Unintelligible | Inappropriate | Confusion | Spontaneous | |
| Eye opening | None | To pain | To command | Spontaneous | |
(Adapted from Teasdale and Jennett, 1974.)
Low GCS has been shown to correlate with poor outcomes, with mortality rates as high as 76% for patients with a post-resuscitative GCS of 3 (Baxt and Moody, 1987; Marshall et al., 1991). As the GCS was developed as a serial exam, a single field GCS is of limited utility (Winkler et al., 1984). A decreasing GCS is more predictive of poorer outcome than an initially low GCS (Marshall et al., 1991; Servadei et al., 1998), while a GCS trending up is predictive of improved outcomes (Winkler et al., 1984). In children as well, an improving GCS increases rates of survival (White et al., 2001). Patients with an initially high GCS that remains high have the best outcomes and some authors have suggested that in isolated head trauma, patients with a serial GCS of 14 or 15 may not need transport to a designated trauma center (Horowitz and Earle, 2001; Ellis et al., 2007).
However, the GCS has several limitations. Most importantly, it requires an interactive patient. The GCS is best determined by the prehospital provider only after correction of other sources of blunted neurologic response, including opiate overdose, hypoglycemia, hypoxemia, or hypoperfusion. Further, it must be performed prior to sedation or paralysis. If the patient has evidence of airway compromise, a cursory GCS evaluation should be performed in tandem with preparation for airway maneuvers requiring sedatives or paralytics. It also must be stressed that the GCS is intended for serial examinations and outcomes cannot be adequately determined until a trend in GCS has been established.
A further limitation of the GCS is its questionable interrater reliability, especially when performed by prehospital providers (Green, 2011). Recently, a simplified scoring system has been developed and validated in the prehospital setting (Thompson et al., 2011). The simplified motor score (SMS) is a three-point scoring system (Table 23.2) and has been found to predict outcomes in TBI with similar accuracy to the GCS (Thompson et al., 2011). In the future, the SMS may well replace the GCS in prehospital and hospital assessments of patients with TBI. In the meantime, if the GCS is to be used in the field, it is recommended that providers receive ongoing training and access to standardized scorecards in order to maximize interrater reliability (Lane et al., 2002).
Table 23.2. The Simplified Motor Score
| Points | Patient response |
--- |
| 0 | Withdraws to pain (or worse) |
| 1 | Localizes pain |
| 2 | Obeys commands |
(Adapted from Thompson et al., 2011.)
View chapterExplore book
Read full chapter
URL:
Handbook2015, Handbook of Clinical NeurologyScott A. Goldberg, ... Andrew Jagoda
Chapter
Disorders of Consciousness in Children
2017, Swaiman's Pediatric Neurology (Sixth Edition)Stephen Ashwal
Consciousness Rating Scales
The best known and most widely used scale is the Glasgow Coma Scale, which yields a score of 3 to 15, based on best response to stimuli in the these three categories: eye opening, verbal response, and motor response (Table 101-2). In its original form, the Glasgow Coma Scale was not developmentally suitable for assessment of newborns, infants, and younger children, and a variety of alternate scales have been proposed. The Pediatric Coma Scale makes minor changes in the verbal scale of the Glasgow Coma Scale and redefines the “best” score based on developmental and age-appropriate norms (Table 101-3). The Children's Coma Scale included pupillary reflexes, extraocular movements, and apnea in its categories, and is therefore substantially different from the Glasgow Coma (Table 101-4). The Glasgow Coma Scale—Modified for Children is also a valuable tool (Table 101-2) as is the Full Outline of UnResponsiveness (FOUR) Score Coma Scale (Table 101-5). The FOUR Score is divided into four categories (eye response, motor response, brainstem reflexes, and respiration) and provides a measure of consciousness. The Coma Recovery Scale-revised (CRS-R) is considered the best tool to use to assess adult patients with a DOC, but it has not been validated in children.
View chapterExplore book
Read full chapter
URL:
Book2017, Swaiman's Pediatric Neurology (Sixth Edition)Stephen Ashwal
Chapter
Traumatic Brain Injury
2007, Clinical Neurology for Psychiatrists (Sixth Edition)
1
: Physicians use the Glasgow Coma Scale (GCS) to assess three parameters of a patient's level of consciousness following head trauma. Which of the following functions is not assessed by the GCS?
a
: Vital signs
b
: Eye opening
c
: Speaking (best verbal response)
d
: Moving (best motor response)
View chapterExplore book
Read full chapter
URL:
Book2007, Clinical Neurology for Psychiatrists (Sixth Edition)
Chapter
Coma Scales
2014, Encyclopedia of the Neurological Sciences (Second Edition)T.E. Gofton, G.B. Young
Specific Scales
Glasgow Coma Scale
The Glasgow Coma Scale (GCS) (Table 1) is almost universally used in emergency rooms and intensive care units (ICUs), and is by far the most common coma scale cited in the neurosurgical literature. However, it has not been used consistently in different hospitals, and the later versions of the GCS have not been adequately tested for reliability. The GCS was designed for the initial assessment of patients with head injury. Problems with the use of the GCS arise when patients are intubated and cannot respond verbally or if the eyes are swollen shut, preventing ocular assessment. A theoretical disadvantage is the three-dimensional assessment: The total score is obtained by adding the values for three motor activities – eye opening, best motor response, and best verbal response. These are assumed to be independent variables but they are not. Because they covary, their addition may not be valid. Furthermore, to achieve a score of 6–12, there are more than 10 simple combinations of variables, each with very different clinical profiles. It seems unlikely that all patients with specific scores ranging from 6 to 10 will be equivalent in disease severity. Additionally, there is little difference in outcome over several different score values (e.g., between 10 and 15). The GCS is often insufficiently sensitive for the detection of changes in the level of consciousness in patients following head injury or with masses and risk of herniation when they are in lighter stages of impaired consciousness. Furthermore, in the application of the GCS to patients who have been in the ICU for an extended period of time, eye opening does not equal conscious awareness because patients with persistent vegetative state (VS) may show this and patients with seizures show spontaneous eye opening.
Table 1. Glasgow Coma Scale
| Item | Factor | Score |
---
| Best motor response | Obeys | 6 |
| Localizes | 5 |
| Withdraws (flexion) | 4 |
| Abnormal flexion | 3 |
| Extensor response | 2 |
| Nil | 1 |
| Verbal response | Oriented | 5 |
| Confused conversation | 4 |
| Inappropriate words | 3 |
| Incomprehensible sounds | 2 |
| Nil | 1 |
| Eye opening | Spontaneous | 4 |
| To speech | 3 |
| To pain | 2 |
| Nil | 1 |
However, the GCS has been the standard scoring assessment throughout the world for 20 years. It seems unlikely that it will be easily replaced, even by potentially superior scoring systems. The Innsbruck and the Edinburgh-2 Coma Scales have some of the same problems as the GCS but the Reaction Level Scale (RLS 85) has a number of advantages over the others.
Reaction Level Scale
The RLS 85 (Table 2), developed in Sweden in 1985, is the eight-grade single-line ordinal scale for assessment of patients in the ICU. It can be applied to patients who are intubated or whose eyes have swollen shut. There is no addition of covarying values. Although the numerical values are not necessarily separated by steps of equal value, the order appears to be valid. The test compares favorably with the GCS and outcome is inversely related to the achieved score. It is reliable and reproducible, and has also shown good interrater reliability regardless of the etiology for coma. Furthermore, any change in the RLS 85 is related to a significant change in patient status and it is superior to the GCS in this respect.
Table 2. Reaction Level Scale 85
| Score | Clinical descriptor | Qualifying factors |
---
| 1 – Normal | Alert | Alert; not drowsy; oriented |
| No delay in response | Intubated patient: No delay in reaction |
| 2 – Mildly drowsy or confused | Drowsy or confused | Drowsy: Drowsy or shows delay in reaction |
| Responsiveness to light stimulation | Confused: Wrong answer to: What is your name? Where are you? What year and month is it? |
| 3 – Very drowsy or confused; response to strong stimulation | Strong stimulation: Loud verbal or painful stimulation | Mental responsiveness: Arousable |
| | Performs at least one of the following: Obeys commands (including nonverbal response, e.g., “Lift up your arms;” orienting eye movements; wards off painful stimulus) |
| 4 – Unconscious: localizes but does not ward off pain | Unconscious: No mental response | Localizes |
| Cannot perform any activity defined under mental responsiveness | Retromandibular pain: Patient moves one hand above chin level; with pain to nail beds, the other arm crosses the midline |
| 5 – Unconscious: Withdraws | Withdrawing movement: With retromandibular pressure, patient turns head away | With nail bed pressure, patient makes withdrawal movement with abduction at shoulder |
| 6 – Unconscious: Stereotypic flexion movement | Stereotypic movement: With retromandibular or nail bed pressure, patient slowly assumes decorticate posture | The movement is ‘mechanical’ and clearly different from the withdrawal response described previously |
| 7 – Unconscious: Stereotyped extension movements | Stereotyped extension movements: With retromandibular or nail bed pressure, patient extends the limbs | If there is a mixture of extension and flexion, the flexion (best response) is recorded |
| 8 – Unconscious: No response | With retromandibular or nail bed pressure, there is no response of limbs or face | |
The scale cannot be applied to cases in which the patient is clinically or pharmacologically paralyzed but alert (e.g., polyneuropathy, spinal cord lesion, locked-in syndrome, and use of neuromuscular blocking agents) or to those with psychogenic unresponsiveness. These are usually not difficult to identify and exclude.
Full Outline of UnResponsiveness Scale
The Full Outline of UnResponsiveness (FOUR) scale (Table 3) was developed to provide a more comprehensive and clinically relevant scoring system for use in comatose patients. The FOUR score assesses eye reflexes, motor reflexes, brainstem reflexes, and respiratory pattern. There is no assessment of verbal responses, which is appropriate, because comatose patients are generally intubated. The FOUR score remains pertinent in intubated patients and has the ability to identify the locked-in and VSs. It has been validated for use (in multiple languages) in medical ICU and emergency departments by physicians and nurses, for adult and pediatric populations, patients with traumatic brain injury, stroke, and comatose survivors of cardiac arrest. Interrater reliability is equivalent to the GCS among the neurologists, other physicians, residents, and neuroscience nurses. Evidence shows that in serial assessments, the FOUR score is useful in predicting outcome in a similar manner to the GCS. The FOUR score accurately predicts outcome following cardiac arrest, with patients scoring >8 on days 3–5 after cardiac arrest being more likely to survive to hospital discharge.
Table 3. Full Outline of UnResponsiveness Score
| Category | Score |
--- |
| Eye response | E0 – eyelids remain closed with pain |
| E1 – eyelids closed, opens to pain, not tracking |
| E2 – eyelids closed, opens to loud voice, not tracking |
| E3 – eyelids open but not tracking |
| E4 – eyelids open or opened, tracking or blinking to command |
| Motor response | M0 – no response to pain or generalized myoclonus status epilepticus |
| M1 – extensor posturing |
| M2 – flexion response to pain |
| M3 – localizing to pain |
| M4 – thumbs up, fist, or peace sign to command |
| Brainstem reflexes | B0 – absent pupil, corneal, and cough reflex |
| B1 – pupil and corneal reflexes absent |
| B2 – pupil or corneal reflexes absent |
| B3 – one pupil wide and fixed |
| B4 – pupil and corneal reflexes present |
| Respiration | R0 – breathes at ventilator rate or apnea |
| R1 – breathes above ventilator rate |
| R2 – not intubated, irregular breathing pattern |
| R3 – not intubated, Cheyne–Stokes breathing pattern |
| R4 – not intubated, regular breathing pattern |
In the FOUR score, there are four categories for assessment (eye response, motor response, brainstem reflexes, and respiration), each with a maximum of four points (0=worst score, 4=best score). Assessment of eye response involves assessment of tracking or blinking to command in addition to eye opening. For a maximal score in the motor responses category, demonstration of specific hand positions (thumbs up, fist, and peace sign) is required. The brainstem reflex category requires testing of the cough, pupillary, and corneal reflexes, thereby eliciting more information with regard to brainstem involvement. Assessment of respiration involves accounting for intubation and for breathing pattern.
Although, the FOUR score was designed with the intention of assessing comatose patients in a more detailed manner, criticisms of the scale suggest that it would benefit from greater standardization of administration and scoring to improve its administration and use in daily practice.
Coma Recovery Scale-Revised
The Coma Recovery Scale-Revised (CRS-R) is a 23-item scale, which has been validated in multiple languages. It was developed and implemented in a rehabilitation medicine setting. The validation studies were performed in groups of patients under comprehensive acute rehabilitation care following severe brain injury or various etiologies. The original CRS was revised to improve differentiation and monitoring of patients in a VS, a minimally conscious state (MCS), and emerging from MCS. The scale demonstrates good interrater and test–retest reliability, and has been shown to be a reliable measure of neurobehavioral response. The CRS-R has not been tested in patients with severe brain injury during the acute hospitalization and ICU period; thus, its role in an acute care hospital setting is unknown (Table 4).
Table 4. Coma Recovery Scale-Revised
| Function scale | Score |
--- |
| Auditory function scale | 0 – none |
| 1 – auditory startle |
| 2 – localization to sound |
| 3 – reproducible movement to command |
| 4 – consistent movement to command |
| Visual function scale | 0 – none |
| 1 – visual startle |
| 2 – fixation |
| 3 – visual pursuit |
| 4 – object localization: reaching |
| 5 – object recognition |
| Motor function scale | 0 – none/flaccid |
| 1 – abnormal posturing |
| 2 – flexion withdrawal |
| 3 – localization to noxious stimulation |
| 4 – object manipulation |
| 5 – automatic motor response |
| 6 – functional object use |
| Oromotor/verbal function scale | 0 – none |
| 1 – oral reflexive movement |
| 2 – vocalization/oral movement |
| 3 – intelligible verbalization |
| Communication scale | 0 – none |
| 1 – nonfunctional: intentional |
| 2 – functional: accurate |
| Arousal scale | 0 – unarousable |
| 1 – eye opening with stimulation |
| 2 – eye opening without stimulation |
| 3 – attention |
View chapterExplore book
Read full chapter
URL:
Reference work2014, Encyclopedia of the Neurological Sciences (Second Edition)T.E. Gofton, G.B. Young
Chapter
Neuromonitoring and translational research
2022, Perioperative NeuroscienceAnkur Luthra, ... Sameer Sethi
Glasgow coma score
Jennet and Teasdale developed the Glasgow coma scale (GCS) for standardization of the patients with traumatic brain injury (TBI). GCS has been used to classify the TBI as per severity, identify changes in status of patients, and guide in their prognostication (Table 5.1).1,2 It has also been found effective in prognostication of patients after trauma.3 Using GCS scores to classify injuries as mild TBI (GCS scores 13–15), moderate TBI (GCS scores 9–12), or severe TBI (GCS scores ≤8) is an accepted norm. However, the scale has its limitations—the scale is less useful in intubated patients as verbal responses cannot be assessed in them, GCS does not allow assessment of brainstem function and a GCS score of 3 may cover a spectrum of brain injury severity.
Table 5.1. Glasgow Coma Scale and Full Outline of Unresponsiveness score.
| |
| Glasgow Coma Scale |
| Eye response (E) |
| There are four grades: |
| E1—No opening of the eye |
| E2—Eye opening in response to pain stimulus |
| E3—Eye opening to speech |
| E4—Eyes opening spontaneously |
| Verbal response (V) |
| There are five grades: |
| V1—No verbal response |
| V2—Incomprehensible sounds |
| V3—Inappropriate words |
| V4—Confused |
| V5—Oriented |
| Motor response (M) |
| There are six grades: |
| M1—No motor response |
| M2—Abnormal extension in response to pain |
| M3—Abnormal flexion in response to pain |
| M4—Normal flexion |
| M5—Localizes to pain |
| M6—Obeys commands |
| Full Outline of Unresponsiveness score |
| Eye response |
| Eyelids open or opened, tracking, or blinking to command—4 |
| Eyelids open but not tracking—3 |
| Eyelids closed but open to loud voice—2 |
| Eyelids closed but open to pain—1 |
| Eyelids remain closed with pain—0 |
| Motor response |
| Thumbs-up, fist, or peace sign—4 |
| Localizing to pain—3 |
| Flexion response to pain—2 |
| Extension response to pain—1 |
| No response to pain or generalized myoclonus status—0 |
| Brainstem reflexes |
| Pupil and corneal reflexes present—4 |
| One pupil wide and fixed—3 |
| Pupil or corneal reflexes absent—2 |
| Pupil and corneal reflexes absent—1 |
| Absent pupil, corneal, and cough reflex—0 |
| Respiration |
| Not intubated, regular breathing pattern—4 |
| Not intubated, Cheyne–Stokes breathing pattern—3 |
| Not intubated, irregular breathing—2 |
| Breathes above ventilator rate—1 |
| Breathes at ventilator rate or apnea—0 |
View chapterExplore book
Read full chapter
URL:
Book2022, Perioperative NeuroscienceAnkur Luthra, ... Sameer Sethi
Chapter
The Neurologic Examination
2023, Neurologic Localization and DiagnosisKarl E. Misulis, Martin A. Samuels
Glasgow Coma Scale
The GCS is an assessment of LOC routinely used for patients, especially in the acute care setting. An aggregate score is reached with assessment of eye opening, verbal response, and motor response. A score of 3 indicates no function; a score of 15 indicates full function. The elements of the GCS are as follows:
■
: Eye opening is tested by initially observing the patient for spontaneous eye opening. If there is none, then a verbal command is given to the patient to open their eyes. If there is no response to that command, then a noxious stimulation is delivered to the body (e.g., shoulder, extremity). Responses are graded according to level of response:
■
: Spontaneous eye opening = 4 points
■
: Opening to verbal command or speech = 3 points
■
: Opening to pain (but not on the face) = 2 points
■
: No response = 1 point
■
: Verbal response is tested by observing for spontaneous speech. If there is none, then the examiner speaks to the patient, asks questions, and gives commands to determine whether language of any kind can be elicited. Responses are graded on the basis of the speech output:
■
: Speaking and oriented = 5 points
■
: Responds to questions and commands but is confused = 4 points
■
: Responds only with inappropriate words = 3 points
■
: Speech is incomprehensible = 2 points
■
: No verbal response = 1 point
■
: Motor response is observed for spontaneous movement, movement to command, and reflexive movement. Responses are graded as follows:
■
: Moves to command appropriately = 6 points
■
: Makes purposeful movement to noxious stimuli = 5 points
■
: Withdraws to pain = 4 points
■
: Flexion to noxious stimuli (decorticate posturing) = 3 points
■
: Extension to noxious stimuli (decerebrate posturing) = 2 points
■
: No response to command or noxious stimuli = 1 point
■
: Interpretation of the responses is generally as follows:
■
: Mild deficit = GCS 13 or better
■
: Moderate deficit = GCS 9-12
■
: Severe deficit = GCS 8 or less
View chapterExplore book
Read full chapter
URL:
Book2023, Neurologic Localization and DiagnosisKarl E. Misulis, Martin A. Samuels
Chapter
Initial Assessment
2018, Abernathy's Surgical Secrets (Seventh Edition)David J. Skarupa MD, FACS, Marie Crandall MD, MPH, FACS
19 What is the Glasgow Coma Scale and what does it measure?
The Glasgow Coma Scale (GCS) is a common way to score the level of consciousness of a person after a traumatic brain injury (TBI). It is composed of the following variables:
•
: Eyes: 1–4
•
: Motor: 1–6
•
: Verbal: 1–5
The lowest score is 3. The highest score is 15. The highest score if intubated is 11T (T = intubated). If the patient is paralyzed add P (e.g., GCS 3TP).
The GCS helps classify the severity of brain injury into the following:
•
: Mild TBI: 13–15
•
: Moderate TBI: 9–12
•
: Severe TBI: 3–8
View chapterExplore book
Read full chapter
URL:
Book2018, Abernathy's Surgical Secrets (Seventh Edition)David J. Skarupa MD, FACS, Marie Crandall MD, MPH, FACS
Related terms:
Neurotrauma
Coma
Hematoma
Subarachnoid Hemorrhage
Closed Head Injury
Cerebral Hemorrhage
Computed Tomography
Traumatic Brain Injury
Skull
Headache
View all Topics |
189049 | https://artofproblemsolving.com/wiki/index.php/Sequence?srsltid=AfmBOop99mxDBIt_mT7VfasHEtBDnhlsfS7k8unAecdwSxCAmKYWB8tr | Art of Problem Solving
Sequence - 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 Sequence
Page
ArticleDiscussionView sourceHistory
Toolbox
Recent changesRandom pageHelpWhat links hereSpecial pages
Search
Sequence
A sequence is an ordered list of terms. Sequences may be either finite or infinite.
Contents
1 Definition
2 Convergence
3 Monotone Sequences
4 Resources
5 See Also
Definition
A sequence of real numbers is simply a function . For instance, the function defined on corresponds to the sequence .
Convergence
Intuitively, a sequence converges if its terms approach a particular number.
Formally, a sequence of reals converges to if and only if for all positive reals , there exists a positive integer such that for all integers , we have . If converges to , is called the limit of and is written . The statement that converges to can be written as .
A classic example of convergence would be to show that as .
Claim: .
Proof: Let be arbitrary and choose . Then for we see that
which proves that , so as
Monotone Sequences
Many significant sequences have their terms continually increasing, such as , or continually decreasing, such as . This motivates the following definitions:
A sequence of reals is said to be
increasing if for all and strictly increasing if for all ,
decreasing if for all and strictly decreasing if for all ,
monotone if it is either decreasing or increasing.
Resources
Online Encyclopedia of Integer Sequences
See Also
Arithmetic sequence
Geometric sequence
Bolzano-Weierstrass Theorem
This article is a stub. Help us out by expanding it.
Retrieved from "
Category:
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. |
189050 | https://en.wikipedia.org/wiki/Radon_(disambiguation) | Radon (disambiguation) - Wikipedia
Jump to content
[x] Main menu
Main menu
move to sidebar hide
Navigation
Main page
Contents
Current events
Random article
About Wikipedia
Contact us
Contribute
Help
Learn to edit
Community portal
Recent changes
Upload file
Special pages
Search
Search
[x] Appearance
Appearance
move to sidebar hide
Text
Small Standard Large
This page always uses small font size
Width
Standard Wide
The content is as wide as possible for your browser window.
Color (beta)
Automatic Light Dark
This page is always in light mode.
Donate
Create account
Log in
[x] Personal tools
Donate
Create account
Log in
Pages for logged out editors learn more
Contributions
Talk
[x] Toggle the table of contents
Contents
move to sidebar hide
(Top)
1 See also
Radon (disambiguation)
[x] 12 languages
Deutsch
Français
한국어
Bahasa Indonesia
Italiano
עברית
Magyar
Nederlands
日本語
Plattdüütsch
Português
Русский
Edit links
Article
Talk
[x] English
Read
Edit
View history
[x] Tools
Tools
move to sidebar hide
Actions
Read
Edit
View history
General
What links here
Related changes
Upload file
Permanent link
Page information
Cite this page
Get shortened URL
Download QR code
Edit interlanguage links
Print/export
Download as PDF
Printable version
In other projects
Wikidata item
From Wikipedia, the free encyclopedia
Look up Radon or radon in Wiktionary, the free dictionary.
Radon is a chemical element with symbol Rn and atomic number 86.
Radon mitigation, against dangers of gas in buildings
Radon may also refer to:
Radon, Orne, a town in France
Johann Radon, Austrian mathematician
Radon transform, in mathematics
Radon measure, in mathematics
Radon space, a metric space in mathematics
Rodan, (Radon in Japanese), a fictional monster
MSBS Radon, Polish assault rifle
Jaroslav Radoň (born 1986), Czech canoeist
Radon Labs, video game developer in Germany
See also
[edit]
Wikimedia Commons has media related to radon.
Rn (disambiguation)
Isotopes of radon
All pages with titles beginning with Radon
All pages with titles containing Radon
Raydon, a village and civil parish in Suffolk, England
Radeon, a line of GPUs by AMD
Topics referred to by the same term
This disambiguation page lists articles associated with the title Radon.
If an internal link led you here, you may wish to change the link to point directly to the intended article.
Retrieved from "
Category:
Disambiguation pages
Hidden categories:
Commons link is locally defined
Short description is different from Wikidata
All article disambiguation pages
All disambiguation pages
This page was last edited on 12 April 2025, at 15:09(UTC).
Text is available under the Creative Commons Attribution-ShareAlike 4.0 License; additional terms may apply. By using this site, you agree to the Terms of Use and Privacy Policy. Wikipedia® is a registered trademark of the Wikimedia Foundation, Inc., a non-profit organization.
Privacy policy
About Wikipedia
Disclaimers
Contact Wikipedia
Code of Conduct
Developers
Statistics
Cookie statement
Mobile view
Edit preview settings
Search
Search
[x] Toggle the table of contents
Radon (disambiguation)
12 languagesAdd topic |
189051 | https://www.youtube.com/watch?v=Kwn8Kd8LfFg | Subbasis for Some Topology on R^2 that is NOT a Basis or ANY Topology on R^2
Andrew McCrady
5170 subscribers
15 likes
Description
612 views
Posted: 4 Jan 2021
We give an example of a collection of subsets of the plane that forms a subbasis for some topology on the plane, but does not form a basis for any topology on the plane. The definition of subbasis we refer to requires the plane merely be covered by the collection of subsets. But recall a basis further requires that the intersection of any two neighborhoods of a point should contain a (perhaps smaller) neighborhood of the point. In particular, our subbasis violates this property.
2 comments
Transcript:
all right in this video we're going to look at a collection uh fancy s it's my best attempt at a cursive S so some collection of subsets of the plane r two such that fancy s is a Subspace a subasis for some topology on the plane but fancy s is not an actual basis for any topology on the plane R2 so let's look at the proof I guess we'll call it a proof even though we're going to kind of show an example here of such a set s so what do we mean by subbases first of all there's a couple different definitions depending on what book you're you're looking at so the one I'm thinking of here is by subbases we mean that uh just the fancy S covers R2 and what do we mean by covers here we mean that the union of elements and S so in other words if I took the union of all these subsets that live in s then you ought to cover R2 you ought to get R2 back so the unit of all of them gives you R2 so that's what we're going to again take as our definition of a subasis so the first thing we need to do is what's a good candidate for what s is and maybe there's lots of examples of of an S that'll solve this problem so what we're going to do first of all is fix some notation so let's let parentheses a comma B let's let that just be a typical open interval on the real line so just like from college algebra so now let's look at our sub bases so fancy s part of it is going to be the following I'm going to look at a think of these as your X Valu so X is allowed to go from A to B and then this is the cartisian product in the middle here think of these is your y values I'm saying your y-v value can be whatever real number you want and so again where A and B are just some real numbers maybe I ought to be careful and say that a is less than b here so this is going to be part of my subbases here and so just to give me an idea what are those I'm saying I've got kind of these infinite vertical strips so that's how you should think of a typical element of s so far what else I'm going to do I'm going to Union with I'm going to throw some more stuff into s here the other things I'm going to throw is now I'm going to let my X values go from minus infinity to Infinity again cartisian prodct here I'm going to think about my y values as going from say A to B where A and B are any real numbers where again a is less than b and so think of that as again infinite strips one more time but in this case in the horizontal Direction so again uh in the horizontal Direction that's how you can see that the X values again get to go as far to the left and to the right as you like so s again s altogether is the union of all such infinite strips all such infinite horizontal strips and all such infinite vertical strips so to see that this s is a sub basis for some topology on the plane what we need to do again is show that R2 is actually equal to the union of all set strips which is kind of I don't know maybe that's intuitive right like if I had the whole plane it's definitely covered by a bunch of strips that look like this and so how do you do that uh maybe by definition here well it's just kind of a a typical subset proof so let's let X comma y be a point in R2 what we're going to do is we're going to build a set in here that contains X comma y and uh there's lots of ways to do that the way that I thought about doing it was why don't I just take B to be the following set let's go from x -1 to x + one so that X is definitely in this interval here and then y that real number is certainly in the interval minus infinity to Infinity so X comma Y is an element of B got a little picture for you any point you pick at any point in the plane that you pick you could always make some again infinite strip in this case I just picked the vertical one that contains X comma y so there we go so what do we have what have we just done like for any point x comma Y in R2 we can definitely find one of these strips that contains our point and so what does that tell us that tells me that the plane is definitely contained in the union of all such strips so recall that that means that s is a Subspace I'm sorry a sub basis for some topology on R2 now I don't know what the topology is that s is a subbases for I don't know what the topology is that s generates is another way to think about that but the next part of what this problem ask us to do fortunately it doesn't ask us to think about that uh the next thing that it asks us to do though is to uh show that this set that we've constructed s is not a basis for any topology on R2 and this again is going to uh hint at why the notion of subbases versus bases is important in topology that's uh it's a little bit harder to be a basis so again this s that we've talked about here all these different vertical and horizontal infinite strips is not a basis for any topology and the way that we'll do this is we're going to find a specific point in the plane and we're going to find two specific elements in s call them B1 and B2 such that when my point is in the intersection of B1 and B2 but B1 intersect B2 does not contain anybody else from s so remember there's kind of that intersection property that the basis elements have to have if this was a basis uh one of the defining features of a basis is again if you've got a point in the intersection of two basis elements then there's in fact a smaller basis element that contains the point inside of that intersection so we're going to show that s fails that important property of a basis so let's take the origin to be our special point x comma y remember I said we're going to find a specific point so the origin sounds cool let's do that and uh so what else do I need to do I need to choose uh two specific sets that are again in s so that means two specific infinite strips and the ones I'm going to choose I'll have the plane for you here first one I'm going to choose is from B1 goes from minus one to one on the x axis and then y can be anything you like so that's what I've colored for you down there and B2 is going to be similar let's let X be anything we like and Y go from minus1 to one and so those are definitely two elements of s again two infinite strips here but think about the intersection of those two things so the intersection would just be -1 to1 cross -1 to1 and then that picture that's this little unit Square that's in here and so there is a picture of the intersection it definitely contains the origin the origin smack dab in the middle but let's take a look at what happens there it's pretty clear that the elements of s they're all strips right and I've been calling them strips this whole video but all those strips that I have in s have infinite area what about the intersection of B1 and B2 B1 intersect B2 it's got area four and so B1 intersect B2 is well not a strip of infinite area so it's not an element of s so the intersection B1 intersect B2 is definitely not an S and moreover what you could say right is that because B1 intersect B2 does not have an infinite area what I should write here is that it's not possible for B1 intersect B2 to contain an element of s because all the elements of s are strips of infinite area so I can't stuff an infinite area strip into a strip of area four right that's the impossibility here so that's what I mean by uh we're showing that B1 intersect B2 doesn't contain any element of s and so by that logic again uh we've done it so what you conclude at the end of this now is you know we've shown that this set s it fails that important basis property where uh again the intersection of two uh quote quote unquote basis elements should contain a smaller basis element we've shown that that fails here so again conclude that this s is not a basis for any topology on the plane |
189052 | https://en.wikipedia.org/wiki/Skew_binary_number_system | Jump to content
Skew binary number system
Polski
Edit links
From Wikipedia, the free encyclopedia
Non-standard positional numeral system
The skew binary number system is a non-standard positional numeral system in which the nth digit contributes a value of times the digit (digits are indexed from 0) instead of times as they do in binary. Each digit has a value of 0, 1, or 2. A number can have many skew binary representations. For example, a decimal number 15 can be written as 1000, 201 and 122. Each number can be written uniquely in skew binary canonical form where there is only at most one instance of the digit 2, which must be the least significant nonzero digit. In this case 15 is written canonically as 1000.
Examples
[edit]
Canonical skew binary representations of the numbers from 0 to 15 are shown in following table:
| Decimal | Binary | Skew Binary | Ternary |
--- --- |
| 0 | 0 | 0 | 0 |
| 1 | 1 | 1 | 1 |
| 2 | 10 | 2 | 2 |
| 3 | 11 | 10 | 10 |
| 4 | 100 | 11 | 11 |
| 5 | 101 | 12 | 12 |
| 6 | 110 | 20 | 20 |
| 7 | 111 | 100 | 21 |
| 8 | 1000 | 101 | 22 |
| 9 | 1001 | 102 | 100 |
| 10 | 1010 | 110 | 101 |
| 11 | 1011 | 111 | 102 |
| 12 | 1100 | 112 | 110 |
| 13 | 1101 | 120 | 111 |
| 14 | 1110 | 200 | 112 |
| 15 | 1111 | 1000 | 120 |
Arithmetical operations
[edit]
The advantage of skew binary is that each increment operation can be done with at most one carry operation. This exploits the fact that . Incrementing a skew binary number is done by setting the only two to a zero and incrementing the next digit from zero to one or one to two. When numbers are represented using a form of run-length encoding as linked lists of the non-zero digits, incrementation and decrementation can be performed in constant time.
Other arithmetic operations may be performed by switching between the skew binary representation and the binary representation.
Conversion between decimal and skew binary number
[edit]
To convert from decimal to skew binary number, one can use the following formula:
Base case:
Induction case:
Boundaries:
To convert from skew binary number to decimal, one can use the definition of a skew binary number:
, where , st. only least significant bit (lsb) is 2.
C++ code to convert decimal number to skew binary number
[edit]
```
include
include
include
include
using namespace std;
long dp;
//Using formula a(0) = 0; for n >= 1, a(2^n-1+i) = a(i) + 10^(n-1) for 0 <= i <= 2^n-1,
//taken from The On-Line Encyclopedia of Integer Sequences (
long convertToSkewbinary(long decimal){
int maksIndex = 0;
long maks = 1;
while(maks < decimal){
maks = 2;
maksIndex++;
}
for(int j = 1; j <= maksIndex; j++){
long power = pow(2,j);
for(int i = 0; i <= power-1; i++)
dp[i + power-1] = pow(10,j-1) + dp[i];
}
return dp[decimal];
}
int main() {
std::fill(std::begin(dp), std::end(dp), -1);
dp = 0;
//One can compare with numbers given in
//
for(int i = 50; i < 125; i++){
long current = convertToSkewbinary(i);
cout << current << endl;
}
return 0;
}
```
C++ code to convert skew binary number to decimal number
[edit]
```
include
include
using namespace std;
// Decimal = (0|1|2)(2^N+1 -1) + (0|1|2)(2^(N-1)+1 -1) + ...
// + (0|1|2)(2^(1+1) -1) + (0|1|2)(2^(0+1) -1)
//
// Expected input: A positive integer/long where digits are 0,1 or 2, s.t only least significant bit/digit is 2.
//
long convertToDecimal(long skewBinary){
int k = 0;
long decimal = 0;
while(skewBinary > 0){
int digit = skewBinary % 10;
skewBinary = ceil(skewBinary/10);
decimal += (pow(2,k+1)-1)digit;
k++;
}
return decimal;
}
int main() {
int test[] = {0,1,2,10,11,12,20,100,101,102,110,111,112,120,200,1000};
for(int i = 0; i < sizeof(test)/sizeof(int); i++)
cout << convertToDecimal(test[i]) << endl;;
return 0;
}
```
From skew binary representation to binary representation
[edit]
Given a skew binary number, its value can be computed by a loop, computing the successive values of and adding it once or twice for each such that the th digit is 1 or 2 respectively. A more efficient method is now given, with only bit representation and one subtraction.
The skew binary number of the form without 2 and with 1s is equal to the binary number minus . Let represents the digit repeated times. The skew binary number of the form with 1s is equal to the binary number minus .
From binary representation to skew binary representation
[edit]
Similarly to the preceding section, the binary number of the form with 1s equals the skew binary number plus . Note that since addition is not defined, adding corresponds to incrementing the number times. However, is bounded by the logarithm of and incrementation takes constant time. Hence transforming a binary number into a skew binary number runs in time linear in the length of the number.
Applications
[edit]
The skew binary numbers were developed by Eugene Myers in 1983 for a purely functional data structure that allows the operations of the stack abstract data type and also allows efficient indexing into the sequence of stack elements. They were later applied to skew binomial heaps, a variant of binomial heaps that support constant-time worst-case insertion operations.
See also
[edit]
Three-valued logic
Redundant binary representation
n-ary Gray code
Notes
[edit]
^ Sloane, N. J. A. (ed.). "Sequence A169683". The On-Line Encyclopedia of Integer Sequences. OEIS Foundation.
^ Elmasry, Amr; Jensen, Claus; Katajainen, Jyrki (2012). "Two Skew-Binary Numeral Systems and One Application" (PDF). Theory of Computing Systems. 50: 185–211. doi:10.1007/s00224-011-9357-0. S2CID 253736860.
^ The Online Encyclopedia of Integer Sequences. "The canonical skew-binary numbers".
^ Myers, Eugene W. (1983). "An applicative random-access stack". Information Processing Letters. 17 (5): 241–248. doi:10.1016/0020-0190(83)90106-0. MR 0741239.
^ Brodal, Gerth Stølting; Okasaki, Chris (November 1996). "Optimal purely functional priority queues". Journal of Functional Programming. 6 (6): 839–857. doi:10.1017/s095679680000201x.
Retrieved from "
Categories:
Number theory
Functional programming
Computer arithmetic
Non-standard positional numeral systems
Hidden categories:
Articles with short description
Short description matches Wikidata |
189053 | https://clinical-laboratory-diagnostics.com/k45.html | Fungal infections
45 Fungal infections
45.1 Introduction
45.2 Candidiasis
45.2.1 Epidemiology and clinical symptoms
45.2.2 Laboratory findings
45.2.2.1 Antigen detection
45.2.2.2 Antibody detection
45.2.3 Clinical significance
45.3 Cryptococcus neoformans (cryptococcosis)
45.3.1 Epidemiology and clinical symptoms
45.3.2 Laboratory findings
45.3.2.1 Antigen detection
45.3.3 Clinical significance
45.4 Aspergillosis
45.4.1 Epidemiology and clinical symptoms
45.4.2 Laboratory findings
45.4.2.1 Antigen detection
45.4.2.2 Antibody detection
45.4.3 Clinical significance
45.5 Other molds
45.6 Pneumocystis jirovecii
45.6.1 Epidemiology and clinical symptoms
45.6.2 Laboratory findings
45.6.3 Clinical significance
45.7 Histoplasma capsulatum (histoplasmosis)
45.7.1 Epidemiology and clinical symptoms
45.7.2 Laboratory findings
45.7.2.1 Antigen detection
45.7.2.2 Antibody detection
45.7.3 Clinical significance
45.8 Coccidioides immitis, Coccidioides posadasii (coccidioidomycosis, Valley fever, San Joaquin fever, desert rheumatism)
45.8.1 Epidemiology and clinical symptoms
45.8.2 Laboratory findings
45.8.2.1 Antibody detection
45.8.2.2 Antigen detection
45.8.3 Clinical significance
45.9 Blastomyces dermatitidis (Blastomycosis Gilchrist’s disease)
45.9.1 Epidemiology and clinical symptoms
45.9.2 Laboratory findings
45.10 Paracoccidioides brasiliensis (blastomycosis, South-American blastomycosis)
45.10.1 Epidemiology and clinical symptoms
45.10.2 Laboratory findings
45.10.2.1 Antibody detection
45.10.2.2 Skin test
45.11 Sporothrix schenckii (sporotrichosis, Schenck’s disease
45.11.1 Epidemiology and clinical symptoms
45.11.2 Laboratory findings
45.11.2.1 Antibody detection
45.11.2.2 Skin test
– Section 45.2: Candidiasis
– Section 45.3: Cryptococcus neoformans (cryptococcosis)
– Section 45.4: Aspergillosis
– Section 45.5: Other molds
– Section 45.6: Pneumocystis jirovecii
– Section 45.7: Histoplasma capsulatum (histoplasmosis)
– Section 45.8: Coccidioides immitis, Coccidioides posadasii (coccidioidomycosis, Valley fever, San Joaquin fever, Desert rheumatism)
– Section 45.9: Blastomyces dermatitidis (blastomycosis, Gilchrist’s disease)
– Section 45.10: Paracoccidioides brasiliensis (blastomycosis, South-American blastomycosis)
– Section 45.11: Sporothrix schenckii (sporotrichosis, Schenck’s disease)
45 Fungal infections
Evelyn Heintschel von Heinegg, Jörg Steinmann, Jan Buer, Peter-Michael Rath
45.1 Introduction
Fungi are carbon heterotrophic eukaryotic organisms. They decompose dead organic matter and, therefore, play an essential role in the ecological system. Only a fraction of the approximately 1.5 million fungal species estimated to exist have been recorded by taxonomic classification. About 200 species are pathogenic to humans /1, 2, 3, 4, 5, 6, 7, 8, 9/.
The taxonomic nomenclature of fungi is difficult to comprehend for non mycologists. Classical identification of fungal species is based on the macro and micromorphology of their reproductive structures or, in yeast species, their biochemical performance. Fungi can reproduce asexually by mycelial fragmentation or spores as well as sexually.
While the sexual stage is the leading taxonomic criterion for classification of an isolate, historically established species names based on the morphology of asexual reproductive structures continue being used. Thus, a species may happen to have two names. Molecular biological methods have been increasingly used for identification and taxonomic classification, causing the nomenclature of fungi to be reconsidered.
The simple DYMD (dermatophytes, yeasts, molds and dimorphic fungi) scheme, which is considered incorrect from the biological perspective, has become established based on relevance of the fungi in human medicine. The description in this chapter also adheres to this scheme, but only discusses species capable of causing life threatening, invasive infections (synonyms: deep mycoses, systemic mycoses, endomycoses), with the exception of Pneumocystis jirovecii (carinii), a pathogen just recently classified under fungi. Typically, replication of this pathogen is non invasive in alveoli (i.e., invasive), systemic infections are rare. For information on dermatophytes, which only invade dead, keratin containing cells, and for injury mycoses (exogenous infections), please refer to the relevant literature.
Most fungi relevant in human medicine become infectious due to interaction with the biotic or abiotic environment. Person-to-person transmission is rare, except regarding dermatophytes and Pneumocystis. Endogenous infections are only caused by Candida species, which are components of the transient or permanent, mucosal flora in the human body.
The relevant, human pathogenic fungi occurring in Central Europe only cause systemic (invasive) infections in the presence of local or systemic immune system malfunctions. Essential risk factors include congenital or acquired malfunctions of the cellular defense mechanism. Thus, such fungal infections must be classified as opportunistic infections. In Germany, the only occurrences of mycoses caused by dimorphic fungi are imported. They are considered to be obligate pathogenic. However, in immunocompetent patients, most of these infections are mild or even subclinical. None of the discussed pathogens or diseases are subject to mandatory reporting in compliance with the German Infection Protection Act (IfSG).
The diagnosis of invasive mycosis is based on the histological and/or cultural detection in primarily sterile specimens. Cultural detection should always be sought for species identification, on one hand, and for determination of the sensitivity to antimycotic agents, on the other. Microscopic or cultural detection in primarily non sterile specimens (e.g., from the respiratory tract) does not allow to distinguish between contamination, colonization and infection. Culture independent methods such as antigen or antibody detection usually have a lower diagnostic sensitivity and specificity and therefore can only serve as an indication of the presence of infection, except for the highly sensitive and specific antigen detection methods in cryptococcosis. Molecular biological detection methods in mycology are still in the development and validation phase. In most cases, such methods have only been established in specialized laboratories.
Every microbiological laboratory should be capable of detecting fungi relevant to human medicine. The basic spectrum of methods constantly available for diagnostic investigation should always include microscopy and culturing on universal and selective media. It should be possible to identify the most frequently occurring Candida and Aspergillus species to species level based on morphology and/or biochemical performance. For application to cerebrospinal fluid specimens, an antigen assay and culturing media should always be available to assist the diagnosis of cryptococcosis. At least one staining technique should be established for the diagnosis of Pneumocystis. Cooperation with a consultant or reference laboratory is recommended because each fungal isolate from primarily sterile specimens should be identified to species level and also be subjected to a sensitivity test.
Additional techniques can be established in the laboratory depending on the rate of infection and patient specimens received: antigen assays for Candida and Aspergillus, antibody assays, antimycotic agent sensitivity tests, molecular biological methods for pathogen detection and species identification. Diagnostic investigation of extra European systemic mycoses caused by dimorphic fungi should be left to specialized laboratories, the more so since the pathogens of these infections are classified under laboratory biosafety level L3 according to the Ordinance on Biological Agents. Participation in corresponding inter laboratory surveys is mandatory for all laboratories involved in mycological diagnostics.
45.2 Candidiasis
45.2.1 Epidemiology and clinical symptoms
Candida species can cause local (e.g., thrush, diaper dermatitis) and systemic (invasive) infections in humans. The medically relevant species C. albicans is the most frequently encountered cause of mucosal and organ mycoses. However, infections with C. glabrata, C. tropicalis, C. krusei, C. guilliermondii and other species are observed at an increasing rate. C. albicans is the only species to use the human orogastrointestinal tract as a natural habitat /10, 11, 12, 13, 14, 15/.
Routes of infection
The majority of Candida infections arise endogenously (i.e., the normally commensally occurring yeast like organisms gain pathogenic relevance in case of disturbances of the defense mechanisms). Entry ports for the development of Candida mycoses mainly include the orobronchial tract, intravascular catheters or the gastrointestinal tract such as in epithelial lesions due to cytotoxic chemotherapy. Exogenous infections, for example due to contaminated infusion solutions or transmission by medical personnel, have been described.
Prevalence and incidence
Oral and esophageal candidiasis are very common in HIV infected patients, with up to 90% patients estimated to be affected in the pre-HAART (highly active anti-retroviral therapy) era. The rate of infection in tumor patients or patients who received bone marrow transplants is 25–35%. The infection is caused by C. albicans in most cases.
The prevalence of candidemia varies between individual countries and individual hospitals. Relative prevalences of 5–15% in positive blood cultures have been described. In risk groups such as patients with sepsis, a prevalence of 17% has been shown by a German study, and a prevalence between 7% and 30% has been reported in patients with leukemia.
The majority of systemic infections are caused by C. albicans (45–60%); other species are not as common (C. glabrata 15–30%, C. tropicalis 10–30%). However, the percentage distribution of pathogens depends on the patient’s predisposition: For instance, C. parapsilosis is more frequently detected in premature infants and C. glabrata is more common in old, multi morbid patients. An increase in non Candida albicans species has generally been reported in all studies regarding pathogen distribution. The mortality rate in systemic Candida infections is between 15% and 30%.
Clinical symptoms
Candida infections manifest themselves as local mucocutaneous infections in most cases. Clinical symptoms include oral thrush, atrophic oral candidiasis, thrush esophagitis, vaginal infections and balanitis, as well as skin surface (diaper thrush) and nail infections. All of these manifestations have local or systemic disturbances of the defense mechanisms in common. Systemic factors include age extremes, metabolic diseases such as diabetes mellitus and congenital or acquired T cell defects (AIDS, malignant tumors and antineoplastic therapy, systemic steroid therapy, graft-versus-host disease). Other causes are malnutrition, long term administration of antibiotics (> 2 weeks), intravascular catheters, mandatory ventilation, hemodialysis, parenteral nutrition, gastrointestinal perforation and pancreatitis. Local factors include radiotherapy, steroid sprays or artificial dentures. In many cases, long term antibiosis, especially in patients in intensive care, causes colonization of the lower respiratory tract; however, invasive infections exclusively involving the lungs are rare.
Invasive candidiasis /14/
Invasive candidiasis is the most common disease among hospitalized patients in the developed world and is usually preceded by mucocutaneous colonization. Deep seated candidiasis arises from either hematogenous dissemination or direct inoculation of candida species to a sterile site, such as the peritoneal cavity. Mortality among patients with invasive candidiasis is as high as 40%, even when patients receive antifungal therapy. Any inner organ can be affected due to hematogenous or lymphatic dissemination. Hepatosplenic candidiasis and infections in low-perfusion tissue such as the eye (enophthalmitis) or the skeleton (e.g., spondylodiscitis) represent diagnostic and therapeutic challenges.
Invasive candidiasis is characterized by nonspecific clinical symptoms: fever and elevated inflammatory markers are common and multiple organ failure and septic shock are also encountered frequently. Ophthalmologic detection of “cotton-wool” lesions in the retina is typical, but not always possible. Radiological diagnostic examinations usually show a nonspecific infection; abscesses in internal organs or typical lesions in the liver and spleen have been detected in hepatosplenic candidiasis in individual cases.
45.2.2 Laboratory findings
Microscopic and cultural detection of organisms
Specimens suitable for the diagnosis of Candida endomycosis are primarily samples from normally sterile areas such as blood, cerebrospinal fluid as well as aspirated fluid and biopsy material. Respiratory secretions, urine, feces and smears from mucosal surfaces do not allow to differentiate between colonization and clinically overt infection even in the case of quantitative yeast count determinations. Blastospores and possibly pseudo hyphae are recognizable in unstained wet mounts as well as in Gram stained and histological slides /8, 13/.
Candida species will grow at 25–37 °C within 2–7 days on the usual bacteriologic and mycologic culture media. The diagnostic sensitivity of blood cultures is 21% to 71% /14/. Whereas blood cultures may establish a diagnosis when Candida resides in the blood stream, cultures of blood obtained from patients with hematogenous, deep seated infections often yield negative results because Candida has been cleared from the blood stream at the time the blood sample is collected /15/. Blood cultures are further limited by slow turn around times and by the fact that a positive result may revealed only late in the course of disease /14/. Identification is achieved based on morphologic criteria and biochemically by sugar fermentation as well as sugar assimilation. MALDI-TOF mass spectroscopy is also a reliable method for species identification. Molecular biological methods have not been established in every laboratory.
Molecular biological detection
A number of in-house assays with a diagnostic sensitivity of 88–98% and a specificity of 88–95% have been described for the molecular biological detection (PCR) of Candida DNA in the blood or serum. PCR systems are also commercially available (multiplex PCRs for the detection of sepsis pathogens with the inclusion of most common Candida species) /8, 16/.
45.2.2.1 Antigen detection
References /11, 17, 18, 19, 20, 21, 22, 23, 24/.
Latex agglutination for detecting the heat labile antigen
Antibodies obtained by immunization of rabbits with heat destroyed blastospores of Candida albicans are adsorbed to latex particles. Heat labile antigen in the sample binds to the fixed antibodies and thus causes agglutination of the test mixture within 10 minutes.
It is recommended to repeat the test after heating the sample at 56 °C for 15 min. in order to verify that it was indeed the heat labile antigen and not an undiscovered thermoresistant serum factor which reacted in the test. The test can be performed qualitatively using a 1 : 2 sample dilution or quantitatively using serial sample dilutions. Threshold titer is 1 : 4. Diagnostic sensitivity is 30–77% at a specificity of 70–88%.
Latex agglutination for detecting the mannan antigen
The test mixture is composed of latex particles to which monoclonal mannan antibodies (EB-CA1) are adsorbed. The patient serum is heated to 100 °C and centrifuged in order to dissociate immune complexes and destroy rheumatoid factors. If mannan antigen is present in the sample thus prepared, the test mixture will agglutinate within 10 minutes.
A commercially available sandwich ELISA uses the same antibody. The two test systems have similar diagnostic specificity (70–80%), but the sensitivity of ELISA is higher (42–98%).
(1,3)-β-D-glucan detection
Various assays are commercially available for the detection of (1,3)-β-D-glucan, a cell wall component of many fungi. The assays differ in their setup and threshold values. The principle of the assay is the activation of co agglutination in the amebocytes of horseshoe crab hemolymph, similarly to the Limulus assays for endotoxin detection, but following a different metabolic pathway (e.g., using factor G instead of factors B and C). The serine protease inhibitors occurring in serum must be deactivated before the test mixture is prepared. Depending on the assay manufacturer, this is achieved by alkali pretreatment or by adding triton X and heating to 70 °C. Any endotoxin present must be removed by adding factor C or polymyxin.
45.2.2.2 Antibody detection
A multitude of different antigen preparations and probably every serological method have been used for the detection of serum Candida antibodies /13, 21, 25/. Three commonly used test procedures are described in the following. Since neither the production of antigens nor the test techniques are standardized, the specified threshold titers vary from laboratory to laboratory. It has been reported that both diagnostic sensitivity and specificity of antibody detection tests for the diagnosis of invasive Candida infection are low.
Indirect hemagglutination test
Poly saccharides obtained by phenol water extraction are adsorbed to sheep erythrocytes. Interaction between these sensitized erythrocytes and antibodies, mainly immunoglobulin class M (IgM), results in macroscopically visible agglutination. The test can also be used as a micro-method.
Indirect immunofluorescence test
Blastospores fixed on slides are used as antigen. On the fluorescent microscopic view, the presence of antibodies is revealed by fluorescence of the entire cells or cell contours. The immunoglobulin classes (IgG, IgM, IgA) of the antibodies may be differentiated by using heavy chain specific antisera against human immunoglobulin as indicator reagents.
Enzyme immunoassay
Candida antigens are used which are fixed to polystyrene micro titration plates either directly (protein antigens) or by means of specific rabbit antibodies (polysaccharide antigens). Antibodies bound to antigens are detected using enzyme labeled anti-human serum. The immunoglobulin classes of the antibodies may be differentiated by using antisera against human immunoglobulin heavy chains.
45.2.3 Clinical significance
References /18, 19, 26, 27/.
Cultural detection based on primarily sterile specimens should always be sought in suspected invasive Candida infection. The diagnosis is considered to be confirmed if at least one of the following criteria are met:
Repeated detection in blood cultures obtained on different days
Cultural and/or histological detection in tissue biopsy samples
Microscopic and/or cultural detection in normally sterile fluids (e.g., cerebrospinal fluid, joint fluid).
In many cases, however, such confirmation of diagnosis by laboratory findings is not attainable under clinical conditions. Instead, usually a probable diagnosis can only be made based on a review of predisposing factors, the course of the disease, the response to therapy as well as the laboratory diagnostic findings. A conclusive validation of molecular biological tests has not been available to date. Therefore, such tests should be left to specialized laboratories.
Assessments of the diagnostic value of the heat labile antigen differ tremendously. To measure diagnostically conclusive, high antigen titers, repeat serum examinations are recommended. High antigen titers may suggest invasive infection although they may also be observed in conjunction with superficial Candida colonization. False positive reactions can be caused by rheumatoid factors and occur in the presence of high creatinine concentrations. Some authors do not recommend heat labile antigen tests on account of their low diagnostic sensitivity and specificity.
In some studies, Candida mannan detection has been ascribed a higher clinical specificity for the exclusion of invasive candidiasis. Diagnostic sensitivity, however, is only moderate. Regular monitoring of patients at risk may, however, provide crucial clues.
Only a few and inconsistent validation results are available regarding test methods for the detection of (1,3)-β-D-glucans. Positive results are to be expected in invasive Candida and mold infections (not in zygomycoses and cryptococcosis). The antigen is also detectable in serum in pneumocystosis. Diagnostic sensitivities of 58–100% and specificities of 88–98% have been described regarding the detection of invasive mycosis. This test has also been reported to yield a number of false positive results (e.g., due to intravenously administered antibiotics) use of gauze, in hemodialysis, intravenous administration of albumin or β-globulin and in bacteremia. Whether this test is helpful for diagnosis has not been sufficiently documented to date.
The detection of serum Candida antibodies has not produced decisive, clinically useful progress in the diagnosis of Candida endomycoses. No procedure is suited for differentiating with adequate probability between invasive infection, superficial mucocutaneous mycosis or Candida colonization. Despite these limitations, the determination of Candida antibodies has become established for the purposes of monitoring patients at risk of infection. Monitoring must be performed regularly, however (at least once a week). In patients with intact antibody synthesis, a significant increase in the Candida antibody titer (at least fourfold i.e., two titer steps) suggests the presence of enhanced Candida exposure whose clinical relevance must be evaluated by additional tests.
In patients with impaired antibody synthesis, a significant increase in antibodies may not be found despite clinically overt Candida infection. Depending on changes in immunity status, antibody titers may vary considerably.
Monitoring of the antibody levels thus is thought to be of some value in immunocompetent patients (e.g., during the course of intensive care) as an additional diagnostic criterion. In immunocompromised patients, who in particular bear the highest risk for invasive Candida infection, antibody monitoring is the least conclusive. It has been found in several studies that a combination of various test methods will increase diagnostic conclusiveness. This aspect may be elucidated by larger scale studies yet to be performed.
45.3 Cryptococcus neoformans (cryptococcosis)
The genus Cryptococcus comprises 19 species, most of which typically occur as encapsulated yeasts. Besides infections with C. neoformans (teleomorph stage: Filobasidiella neoformans), infections with C. gattii and, more rarely, with C. albidus, C. laurentii and C. adeliensis have also been described /12, 28, 29/.
The gelatinous capsule of C. neoformans and its ability to grow at 37 °C and produce melanin are classical pathogenicity factors.
45.3.1 Epidemiology and clinical symptoms
References /10, 30, 32/.
C. neoformans occurs worldwide. It is frequently found in the feces of birds, in particular of pigeons and parrots, and in bird nesting areas but also on tropical and subtropical plants and fruit. C. gattii, by contrast, has never been found in birds, but primarily on eucalyptus trees and in sewage. The infection affects not only humans, but may also occur in pets, especially cats.
Routes of infection
The infection is spread by inhalation of pathogen contaminated dust. The incubation period may be up to several weeks. Transmission from animals (including infected pets) to humans or from person to person has not been proven.
Prevalence and incidence
Cryptococcosis acquired in Europe is usually caused by C. neoformans. Exposure to C. neoformans apparently occurs often and usually results in latent (asymptomatic) infection. Clinically overt infection is primarily seen in immunocompromised patients with T cell defects. AIDS is the underlying disease in 2.9–13.3% of cryptococcosis cases; however, this rate has markedly decreased since the introduction of the HAART (highly active antiretroviral therapy) in HIV infected patients. Other risk groups include patients following organ transplantation and patients receiving chemotherapy in malignant tumors or on long term steroid medication. An increasing number of infections in non predisposed patients has been reported.
C. gattii infections are restricted to tropical/subtropical regions. However, C. gattii has also been detected in eucalyptus plantations in Mediterranean countries. Due to its higher virulence, it affects immunocompetent individuals more frequently than C. neoformans does.
Clinical symptoms
Pulmonary C. neoformans infections occurring after inhalation of the pathogen show an asymptomatic course or uncharacteristic symptoms. Practically any organ can be affected by hematogenous or lymphatic dissemination. Because of the pathogen’s neurotropism, meningitis or meningoencephalitis are the major manifestations of cryptococcosis. Nonspecific neurological symptoms such as headache, paresthesia, psychiatric abnormalities or focal deficits can indicate the need for adequate diagnostic investigation. The course is acute or chronic, and primary cutaneous infections are also possible.
45.3.2 Laboratory findings
Laboratory diagnostic investigation of cryptococcosis should include cultural detection methods performed in parallel with microscopic and serological antigen diagnostics. Such investigation should always be approached as emergency diagnostics /8/.
Microscopic and cultural detection of organisms
Cerebrospinal fluid (CSF) is primarily used for the microscopic detection of the pathogen. CSF should be centrifuged for 10 min. at 3,500 × g to use the sediment for Cryptococcus antigen detection. The sediment is resuspended with 0.5 mL of CSF for use in microscopy and culture analysis. Respiratory specimens can be used for pathogen detection in the case of pulmonary manifestation /6, 25, 30/.
CSF Gram stains and, in particular, Gram stained slides of respiratory secretions can show artifacts which can easily be mistaken for Cryptococcus cells. Therefore, ink preparations should be used in suspected cryptococcosis. Using ink mounts, the typical budding cells with double refractile cell walls, granulated cytoplasm and thick capsules are detected in 25–50% of patients with cryptococcal meningitis. Every microscopic finding, therefore, must be confirmed by culturing the pathogen.
Culture is also indicated in the case of a negative microscopic result. CSF, respiratory secretions such as broncho alveolar lavage fluid, blood, urine and tissue samples should be used as specimens.
Molecular biological detection
PCR detection of C. neoformans DNA from tissue has been established, but is not commercially available /33/.
45.3.2.1 Antigen detection
References /25, 29, 30, 34/.
C. neoformans capsular poly saccharides (glucurono-xylomannan) appear mostly in serum and CSF, and more rarely in urine and other body fluids. Polyclonal and monoclonal antibodies are used for the detection of these antigens. Rheumatoid factors are eliminated and immune complexes are split by pretreatment of the samples using pronase. Thus, false positive and false negative reactions and the occurrence of a prozone phenomenon are reduced. Infections with C. gattii are also reliably detected with the antigen test. Enzyme immunoassay and latex agglutination test show equivalent diagnostic sensitivity and specificity in antigen detection.
Latex agglutination test
The test mixture consists of latex particles to which Cryptococcus antibodies are bound. Agglutination occurs in the presence of antigens in the sample. The lower detection limit in serum is 50 ng/mL. In the case of CSF, no pretreatment is necessary.
Enzyme immunoassay
Polyclonal Cryptococcus antibodies which are adsorbed to wells of micro titer plates bind Cryptococcus antigens in the sample. Labeled monoclonal Cryptococcus antibodies are used to detect the antigen-antibody binding. The test can be performed quantitatively. The enzyme immunoassay is also suited for the reliable detection of C. gattii antigen.
The use of monoclonal C. neoformans antibodies, however, may yield false negative test results in C. gattii infections.
The two test methods have a diagnostic sensitivity of 98–100% and a clinical specificity of 98–99%.
Immunochromatography
Cryptococcus antigen can be detected in serum, plasma, cerebrospinal fluid (CSF) or urine at high diagnostic sensitivity and specificity using rapid immunochromatographic tests (lateral flow assay) /34/. The test strips are coated with 4 different Cryptococcus antigens. These tests have the advantage of being easy to handle.
45.3.3 Clinical significance
The culture method should always be sought for pathogen detection. CSF ink preparations are suited for the rapid detection of the pathogen in suspected Cryptococcus meningitis. However, false negative results have to be anticipated in up to 2/3 of patients with this method. Molecular biological methods have only been established as in-house tests in specialized laboratories /8, 30/.
The detection of C. neoformans antigens in the CSF is considered to be a sensitive and specific sign of Cryptococcus meningitis; in half of the cases, the antigens are also found in serum and urine. Antigen detection in serum alone indicates extra meningeal infection with a high level of probability even though the detection limit of the test is lower in these cases. Repeat examinations increase the degree of clinical certainty and are valuable for the purpose of monitoring. Rising antigen titers are interpreted as an indicator for progression of the infection and thus are associated with poor prognosis. If antifungal therapy is successful, a decrease in the antigen titer will ensue.
False positive serum reactions have been described in disseminated Trichosporon asahii (beigelii) infection and false positive CSF reactions have sporadically been seen in patients with bacteremia and/or malignancy. Nonspecific agglutination of the latex test mixture was observed in samples of CSF which were contaminated with agar and/or powder from latex gloves. Nonspecific agglutination can also occur, if the loop was already used to spread the sample on culture media and is again immersed into the sample. Moreover, the storage of the specimens and enzyme mixture for sample pretreatment in siliconized tubes can cause nonspecific agglutination. Cryptococcus antigen detection is not widespread in diagnostic laboratories and no test systems are commercially available.
45.4 Aspergillosis
References /1, 10, 12, 28, 35, 36/.
Molds of the genus Aspergillus occur worldwide. They are ubiquitous in the soil, air and water and, in particular, in dead and rotting organic material. Most species reproduce asexually (conidiophores and conidia); for some species, however, a teleomorph stage (sexual reproduction) has been described justifying their taxonomic classification under the phylum ascomycota. Thirty-three of the roughly 180 species described are human pathogens. A. fumigatus, less frequently A. flavus, A. terreus or A. nidulans cause life threatening infections with a high mortality rate (30–87%) in immunocompromised patients. Other clinically relevant conditions include aspergilloma and allergy symptoms.
45.4.1 Epidemiology and clinical symptoms
Routes of infection
Transmission is primarily airborne and the lungs are the primary organ affected by invasive infection. Aspergillus spores (conidia) are omnipresent in the ambient air. They are present in especially high concentrations near compost heaps. Construction work in hospitals and contaminated air conditioning systems may also cause increased air pollution with Aspergillus spores. Therefore, plant cultures in potting soil should not be placed near immunocompromised risk patients. HEPA (high-efficiency particulate air) filtration helps to provide spore free indoor air. The sporadic detection of Aspergillus sp. in respiratory specimens is insignificant because of its ubiquitous occurrence, unless there is a risk constellation for infection or clinical symptoms are present. Contaminated food, particularly nuts and cereals, may also be a source of infection, although gastrointestinal infection is very rare. Person-to-person transmission is very rarely encountered.
Prevalence and incidence
Allergic bronchopulmonary aspergillosis occurs in approximately 1–2% of asthma patients and a mean of 7% mucoviscidosis patients. Preformed cavities that may, for example, occur in patients with chronic pulmonary disease (e.g., bullous emphysema, sarcoidosis, tuberculosis, bacterial lung abscess) predispose to the development of aspergilloma.
Exact data on the prevalence and incidence of Aspergillus infections are not available. An incidence of 1–23% has been reported depending on the investigating center and the constellation risk. In cases of aspergillosis occurring during prolonged neutropenia of more than 14 days duration (below 1.0 × 109/L), one of the main risk factors for the disease, the incidence of the invasive form of infection in patients with hemato-oncological diseases may be 5–20%. Other risk groups include patients following organ transplantation (1–20%), AIDS patients, tumor patients receiving immunosuppressants, patients with chronic obstructive pulmonary disease and patients with autoimmune disease receiving high doses of corticosteroids.
According to a German postmortem study covering the years 2001 to 2005, approximately 20% of patients with hematologic neoplasia were found to have invasive mycosis. Two thirds of the invasive mycoses were only diagnosed by autopsy. Histologically, two thirds of the cases were aspergilloses. The mortality rate is 30–80% depending on the underlying disease.
Clinical symptoms
Allergic bronchopulmonary aspergillosis is characterized by pulmonary deterioration in the presence of an underlying disease (asthma, mucoviscidosis), transient pulmonary infiltrates and positive IgE antibodies. Aspergillomas are usually asymptomatic. In the presence of vascular erosion, however, life threatening hemoptysis may occur Surgical intervention is the treatment of choice. In case the para nasal sinuses are involved, the clinical symptoms correspond to those of chronic sinusitis.
Due to the predominantly airborne route of infection, invasive types usually manifest as pulmonary aspergilloses with nonspecific symptoms: cough, dyspnea, antibiotic resistant fever, hemoptyses. Risk patients with therapy refractory fever should be examined by CT imaging of the lungs at an early stage. The pathogens can disseminate via the blood stream or continuously. Cerebral aspergillosis is associated with extremely high mortality.
45.4.2 Laboratory findings
References /35, 37, 38/.
Microscopic and cultural detection of organisms
Invasive Aspergillus infection is confirmed by pathogen detection in tissue samples by microscopy (characteristic septate hyphae branching at angles of 45°). The detection procedure is based on Grocott-Gomori silver or PAS staining, but does not allow species identification. Microscopic detection using primarily non sterile specimens such as respiratory secretions is not as conclusive. However, the detection of Aspergillus sp. in a risk patient may indicate the presence of infection. Staining with optical brightening agents such as calcofluor white is recommended as a more sensitive detection method. The fluorescent stains bind to poly saccharides of the fungal cell wall. The hyphae are difficult to detect by the common staining method according to Gram.
Cultural detection for species identification and resistance testing are the procedures of choice. Aspergillus species grow on universal and selective culture media within a few days at incubation temperatures generally used in bacteriology. Species identification is based on macroscopic and microscopic colony morphology. Specialized laboratories use molecular biology methods (DNA chip technology, ITS region sequencing). Although rarely used to date, isolates with triazole resistance have been described. Resistance testing is only helpful in isolates from invasive infections.
Molecular biological detection
A number of in-house assays with a diagnostic sensitivity of 40–100% and a specificity of 65–100% have been described for the molecular biological detection of pathogen DNA in the blood, serum, respiratory secretions or tissue samples. Commercial systems are also available (multiplex PCR for the detection of sepsis pathogens with the inclusion of A. fumigatus and/or Aspergillus specific PCR tests). However, no conclusive validation of these systems is available /39, 40, 41, 42/.
Molecular biology methods for pathogen DNA detection should, therefore, be left to specialized laboratories. Result interpretation is only meaningful in the context of risk constellation and other laboratory and radiological findings.
45.4.2.1 Antigen detection
References /40, 43, 44, 45, 46/.
ELISA is most frequently used to detect circulating serum galactomannan, a cell wall component of many hyphomycetes. The assay is not Aspergillus specific and does not detect mucoromycotina (zygomycetes). It is based on a monoclonal antibody directed against the β-D-(1,5)-galactofuranoside side chain. The antigen is detected by a second, peroxidase labeled antibody. The specimen must be incubated at 100 °C prior to analysis for immunocomplex dissociation. A non dimensional index is calculated based on the absorption values of the specimen. The detection limit is 1 ng/mL of galactomannan.
Another method available includes various tests for the detection of (1,3)-β-D-glucan, a cell wall component of many fungi. The underlying principle of this method is described in Section 45.2.
45.4.2.2 Antibody detection
Indirect hemagglutination tests or ELISA are commercially available for the detection of antibodies. Other serological methods include tests based on immunoblot, immunodiffusion, immunoelectrophoresis or immunofluorescence /35, 47/.
A positive antibody test can be helpful for diagnosing immunocompetent patients and in cases of suspected aspergilloma or allergic bronchopulmonary aspergillosis. However, the specificity of these test methods is limited due to the high number of cross-reacting antigens. Test results in immunocompromised patients are not very conclusive. Clinical sensitivities of 14–36% and specificities of 72–99% have been described in hemato-oncological patients with suspected invasive aspergillosis. Because of their poor clinical sensitivity, antibody diagnostics are not helpful in such patients.
45.4.3 Clinical significance
References /40, 43, 44, 46, 47/.
The cultural detection of Aspergillus species in primarily sterile specimens confirms the diagnosis of invasive infection and allows species identification and resistance testing.
Microscopic or cultural detection in primarily non sterile respiratory specimens (sputum, respiratory secretion, lavage fluids) does not allow to differentiate between contamination, colonization or infection. However, the detection in patients with a risk constellation may indicate the need for further clinical and radiological investigation. Microscopic or cultural detection is practically never successful in blood and CSF samples.
Only a small number of commercially available, direct molecular genetic detection tests have been validated. Such tests should be left to specialized laboratories. Result interpretation is only meaningful in the context of risk constellation and other laboratory and radiological findings.
The galactomannan ELISA is the most frequently used for antigen detection. It has only been adequately validated for hemato-oncological patients. In these patients, the diagnostic sensitivity for the detection of invasive infection is 70–92% and the specificity is approximately 90%. The diagnostic sensitivity in patients under antimycotic therapy has been improved by reducing the index from 1.5 to 0.5. Little information is available on other groups of risk patients (e.g., patients following organ transplantation or intensive care patients). However, it indicates a markedly lower diagnostic sensitivity, probably based on the rapid degradation of circulating galactomannan in undisturbed cell poiesis.
False positive results can be caused by β-lactam antibiotics, nutritional solutions or translocation of fungal antigens from the gastrointestinal tract. Therefore, a positive result only indicates the presence of an infection and should be confirmed by a second sample in the near term. One or two screening tests at weekly intervals appear to be justified in patients with prolonged neutropenia.
Investigations using broncho alveolar lavage specimens have a markedly higher diagnostic sensitivity compared to serum specimens. A study showed a diagnostic sensitivity of 96% and a specificity of 88% for an ELISA index threshold value of 1.0 in patients with hemato-oncological disease. CSF specimens can also be analyzed. However, the test is currently only approved for serum. The antigen concentration decreases rapidly under sufficient treatment. Thus, the test can be used for treatment monitoring.
Only a few and inconsistent validation results are available regarding test methods for the detection of (1,3)-β-D-glucans. Positive results are to be expected in invasive Candida and mold infections (not in mucoromycotina infections and cryptococcosis). The antigen is also detectable in serum in pneumocystosis. Diagnostic sensitivities of 58–100% and specificities of 88–98% have been described regarding the detection of invasive mycosis. False positive results are obtained due to intravenously administered antibiotics, use of OR gauze and bacteremia.
Clinical manifestations and risk factors of Aspergillus infection are shown in Tab. 45.4-1 – Risk factors for and clinical manifestation of Aspergillus infections.
45.5 Other molds
Other molds (Mucoromycotina sp., Fusarium sp., Scedosporium/Pseudallescheria sp.) rarely act as pathogens of invasive infection, but have been observed increasingly often /10, 48/. Besides immunosuppression (neutropenia, steroid therapy), risk factors also include diabetes mellitus (Mucoromycotina sp.), hemochromatosis (Mucoromycotina sp.) or almost-drowning (Scedosporium sp., Pseudallescheria sp.). The mortality rate is 40–90% depending on the underlying disease. The diagnosis is based on histologic and cultural detection. Commercial tests for antigen or antibody detection are not available. Molecular biological detection systems have been described.
45.6 Pneumocystis jirovecii
Pneumocystis jirovecii (formerly carinii ssp. humanis) is an important pathogen of pulmonary infection (pneumocystosis) in immunocompromised patients /2/. This pathogen cannot be cultured in vitro and was considered to be a parasite for a long time. Based on molecular biological research data, however, it is now classified as a fungus (ascomycete). Pneumocystis sp. can be detected in practically all mammals, but its host specificity is high. It occurs in two basic structural forms: an ameboid trophozoite and a cyst with eight nuclei. The trophozoite binds to type 1 alveolar epithelial cells in a mucous extracellular matrix which impairs the gas exchange. The cysts are thought to additionally induce inflammatory response.
45.6.1 Epidemiology and clinical symptoms
Routes of infection
Infection is presumably acquired by inhalation. The human is the only reservoir for the human-pathogenic Pneumocystis species P. jirovecii detectable to date /2, 49/. The structure of the infectious pathogen has not been elucidated to date. Outbreaks in risk patients have been reported.
Prevalence and incidence
According to serological analysis, the clinically inapparent primary infection in immunocompetent individuals occurs in early childhood. Information obtained by molecular genetic investigation indicates that pneumocysts can in many cases be detected in immunocompetent individuals at obviously low microorganism concentrations without overt clinical symptoms. The main risk factors for a clinically apparent infection include a decreased CD4+T cell count (below 0.2 × 109/L) and immunosuppressive therapy. AIDS patients, hemato-oncological patients, transplantation patients and patients with autoimmune disease belong, therefore, to the risk groups. Recent investigations indicate a correlation between clinical deterioration in COPD and Pneumocysts.
Clinical symptoms
Partial to global respiratory insufficiency with low fever and unproductive cough is the predominant clinical symptom. Elevated LD values are typical, but not specific to pneumocystosis. Radiological examination reveals atypical pneumonia. From a differential diagnostic point of view, Mycoplasma, Chlamydophila, Legionella infection or mycobacteriosis must be taken into consideration. Systemic infections are rare.
45.6.2 Laboratory findings
References /50, 51, 52/.
Detection of organisms by microscopy
A number of staining methods are available for the detection of the pathogen in respiratory specimens such as broncho alveolar lavage (BAL) and induced sputum. The Giemsa staining method will detect both trophozoites and cysts, whereas methods using Grocott-Gomori silver staining, toluidine blue O staining and staining with optical brightening agents (e.g., calcofluor white) will only detect cysts. The diagnostic specificity of these methods is higher than 99% and the sensitivity in BAL is 70–80%. The sensitivity of immunofluorescence techniques is higher (90%). Since the pathogens tend to form clusters in many cases, all respiratory specimens should be liquefied and several preparations be examined.
Molecular biological detection
Molecular biological methods are also commercially available. Real-time PCR allows to detect pneumocysts from respiratory tract specimens even if microscopy yielded a negative result. BAL specimens have the highest sensitivity for detection; PCR can, however, also be used to analyze tracheal secretion and induced sputum.
Antigen detection
Test methods for the detection of (1,3)-β-D-glucans seem to be helpful for laboratory diagnostics in patients with clinically relevant P. jirovecii infection. A recent study showed a diagnostic sensitivity of 92% and a specificity of 65%. Refer to Section 45.2.2 – Laboratory findings.
45.6.3 Clinical significance
Microscopy is the gold standard for the laboratory detection of Pneumocystis pneumonia. Positive results correlate well with the clinical picture. The choice of staining method depends on the experience of the laboratory. Commercially available molecular biological procedures have not yet been adequately validated. Since the level of infection is obviously high, also in the healthy population, a positive test result in the presence of a negative microscopy result should be interpreted with reservation regarding the need for treatment because microscopy does not allow to differentiate between colonization and an infection requiring treatment.
The serum detection of (1,3)-β-D-glucans has a diagnostic sensitivity > 90%. However, the specificity is limited because the test also covers many other fungi (e.g., Candida, Aspergillus) and yields false positive findings in many cases. This test method is commendable for therapy monitoring.
Test procedures for antibody detection are not commercially available and are unnecessary for diagnosis because of the high seroprevalence, on the one hand, and immunosuppression of affected patients, on the other.
45.7 Histoplasma capsulatum (histoplasmosis)
References /12, 53, 54/
Histoplasma capsulatum var. capsulatum (teleomorph stage: Ajellomyces capsulatus) is an obligate pathogenic, dimorphic fungus. The distinction of the H. capsulatum varieties (H. capsulatum var. duboisii and H. capsulatum var. farcinimosum) is no longer valid because of recent phylogenetic findings. Certain genotypes of H. capsulatum occur at different frequencies. In natural environments and in vitro at 25–30 °C, the fungus grows in the saprophytic mycelial phase, forming infectious microconidia (2–5 μm) and characteristic tuberculate macroconidia (8–15 μm). In vitro at 37 °C, the parasitic yeast phase develops which is composed of globose and oval budding cells with a diameter of 2–4 μm.
45.7.1 Epidemiology and clinical symptoms
References /53, 55, 56, 57, 58, 59/
Geographic distribution
The American histoplasmosis is endemic to the Midwest regions of the USA, in particular to the valleys of the Mississippi, Ohio and Missouri, and to parts of Central and South America. Sporadic cases have been reported worldwide. The large cell variant of H. capsulatum (formerly H. capsulatum var. duboisii) is endemic to Central and West Africa. In a study on histoplasmoses in Europe, 46 cases were reported in Germany between 1995 and 1999. The majority of patients came from endemic regions.
Routes of infection
The natural reservoir of H. capsulatum consists of soil with high nitrate contents (including bird and bat feces which are a particularly potent growth enhancer) in warm and humid regions. The infection is acquired by inhalation of dust containing hyphal fragments and conidia. Direct person-to-person or bird-to-person transmission does not occur. Birds and bats can be intestinally colonized by Histoplasma, but only bats become diseased because of their lower body temperature.
Prevalence and incidence
According to studies in endemic regions (Tennessee, Kentucky) using the histoplasmin skin test, the prevalence of infection can amount up to 90%. Men and women are exposed equally frequently. However, apparent infection is more frequently encountered in men. In the USA, an estimated number of 500,000 individuals are infected each year.
Clinical symptoms
Depending on the location, differentiation is made between acute pulmonary, chronic pulmonary and disseminated types of histoplasmosis as well as between primary and reactivation induced types. The mean incubation period is 10 days. In 90% of cases, the primary type of the disease is asymptomatic or produces influenza like symptoms.
The course and severity of the disease depend on the infection dose and the immune status of the infected individual. Complex courses can lead to lobar pneumonia or pericarditis.
After infection, calcifications of the hilar and/or mediastinal lymph nodes or pulmonary lesions (so-called coin lesions) can remain.
Chronic pulmonary histoplasmosis usually occurs in older patients (above 50 years) with preexisting pulmonary emphysema, manifests as fibrosing or cavitary pulmonary disease and takes a clinical course resembling tuberculosis.
The disseminated type (1 in 2,000–5,000 clinically manifest infections) occurs primarily in immunocompromised patients and especially in those with AIDS. Besides the lung, there is involvement of the lymph nodes, liver, spleen, bone marrow, meninges, gastrointestinal tract, skin and mucosal surfaces. Chronic types of the disease are characterized by nonspecific symptoms over the course of many years. Disseminated histoplasmosis is an AIDS defining disease. From the differential diagnosis point of view, pulmonary histoplasmosis must be distinguished from sarcoidosis, tuberculosis, atypical mycobacteriosis, blastomycosis as well as paracoccidioidosis and coccidioidosis.
In cases of African histoplasmosis, pulmonary involvement is the exception. Instead, subcutaneous foci of infection, granulomatous alterations within the oral cavity and osseous lesions are predominant.
45.7.2 Laboratory findings
Microscopic and cultural detection of organisms
The microscopic detection of the organism is most likely accomplished by examining histological slides, biopsy samples and bone marrow smears using the Giemsa staining or the Grocott-Gomori silver staining technique. The intracellular location (macrophages) of the budding cells is characteristic /53, 54/.
Blood, bone marrow, respiratory secretions and tissue samples are suited to detect the pathogen by culture. Standard culture media such as Sabouraud glucose agar can be used. An incubation period of 4–6 weeks at 30 °C should be allowed due to the slow growth of the organism. The yeast stages are cultured at 37 °C on agars containing blood and cystine.
Successful cultural detection can primarily be expected in disseminated and chronic pulmonary histoplasmosis. Even in cases of disseminated histoplasmosis in patients with AIDS, positive cultures of the organism are found in approximately 90% of cases. The organism can be identified based on the above mentioned micromorphological characteristics of the macro and microconidia.
The time consuming cultural detection of dual phase growth has been replaced by highly specific, commercially available gene probes. Specific handling of the pathogen is only permitted in laboratories under laboratory biosafety level L3 (Ordinance on Biological Agents) because of the risk of laboratory infection.
45.7.2.1 Antigen detection
Enzyme immunoassay
H. capsulatum antigen (heat-stable polysaccharide) can be detected in urine, serum, cerebrospinal fluid and broncho alveolar lavage fluid. The commercially available test is suited to diagnose disseminated histoplasmosis. Diagnostic sensitivities are /30, 60/:
Approximately 90% in disseminated histoplasmosis
Approximately 40% in chronic pulmonary histoplasmosis
Approximately 20% in acute pulmonary histoplasmosis.
A diagnostic specificity of 98% has been reported for urine samples.
False positive reactions can occur in coccidioidomycosis, para coccidioidomycosis or blastomycosis. Rheumatoid factors and treatment with rabbit thymoglobulin can cause false positive reactions in serum. Antigen detection is suited as a marker for disease and therapy monitoring in AIDS patients.
45.7.2.2 Antibody detection
References /54, 59, 61/.
Antibody tests play an important role in the diagnosis of the various types of histoplasmosis. Filtrates of H. capsulatum mycelial cultures (histoplasmin) and extracts of the yeast phase are used as antigen preparations.
Complement fixation test
The test is performed using serum and the customary complement fixation technique. A four-fold increase in the titer confirms the suspicion of an active infection process. Cross reactions with coccidioidomycosis, blastomycosis and para coccidioidomycosis are possible.
Immunodiffusion test
The test is performed in agarose according to the Ouchterlony technique. Differentiation is made between two diagnostically relevant precipitate bands. The M-band may occur during the early stage of the disease and shows prolonged persistence even in inactive processes. An additional H-band is considered to be an indicator for active infection. Immunodiffusion tests are less sensitive (80%) than the complement fixation tests but superior as far as clinical specificity is concerned.
Latex agglutination test
Latex particles coated with histoplasmin are agglutinated by antibodies in the serum. IgM antibodies are primarily detected. Therefore, the test is positive mostly in acute infections.
Enzyme immunoassay
This test uses yeast phase antigen. The diagnostic sensitivity is 97% and the specificity is 84%. False positive reactions are seen in the presence of invasive aspergillosis, blastomycosis and nonspecific pulmonary diseases.
45.7.3 Clinical significance
Antibodies occur 2–4 weeks after exposure and may persist for a long time. Depending on the type of the disease, 60–80% of the cases can be seroreactive. Seroconversion may not occur in immunocompromised patients (particularly AIDS patients). Cross reactions with other fungi have been observed. Compared to the standard methods (complement fixation and immunodiffusion tests), the latex agglutination test and the enzyme immunoassay do not achieve any improvement in serodiagnostics.
The histoplasmin skin test plays a significant role in epidemiological investigations in endemic regions. It is, however, not recommended for diagnostic purposes.
PCR detection of histoplasma DNA has been described, but is not yet commercially available.
45.8 Coccidioides immitis, Coccidioides posadasii (coccidioidomycosis, Valley fever, San Joaquin fever, Desert rheumatism)
C. immitis and C. posadasii are obligate pathogenic, dimorphic fungi causing coccidioidomycosis /12/. In the saprophytic mycelial phase, they form infectious arthroconidia (2–5 μm) which are propelled by the slightest movements of the air. Arthroconidia are defined as a genetically programmed disarticulation of septate hyphae to produce a chain of conidia. Arthroconidia are produced by dermatophytes, both in culture and in lesions, under certain predisposing conditions. In their host, the arthroconidia transform into round structures with thick walls, so-called spherules (10–80 μm), containing up to 600 endospores (2–5 μm). Following the rupture of spherules, each endospore may give rise to the development of a new spherule.
45.8.1 Epidemiology and clinical symptoms
References /62, 63, 64/.
Geographic distribution
Coccidioidomycosis is endemic to deserts, semi deserts, and plains (arid regions) of the American continent, especially in the Southwest of the USA (California, Arizona, Nevada, New Mexico, Utah, Texas) and the adjacent regions of northern Mexico.
Routes of infection
The natural reservoir of C. immitis and C. posadasii is dry, sandy and dusty soil containing the dwellings of rodents which provide especially suitable conditions for the development of the organism. Infection is acquired by inhalation of arthroconidia which are spread, in particular, by sand storms and construction work.
It is not necessary to isolate individuals afflicted by this disease since the tissue forms of the organism are not infectious. Excreted endospores may, however, grow into mycelia and infectious arthroconidia, thus necessitating disinfection of body secretions or specimens contaminated with them.
Prevalence and incidence
In the USA, the estimated numbers are approximately 100,000 infected individuals/year and approximately 500–5000 individuals afflicted by disease/year. Most infections occur in the late summer or early fall.
Clinical symptoms
About 60% of infected individuals have an asymptomatic course of the infection which manifests by transient, influenza like symptoms. In approximately 25% of infected individuals, acute infection of the lower respiratory tract develops 1–3 weeks after exposure and may be associated with arthralgias (desert rheumatism), erythema multiforme or erythema nodosum. This acute, pulmonary (primary) form of the disease usually heals spontaneously in immunocompetent individuals after a few weeks. In rare cases, there is a transition from the acute to a chronic pulmonary form of the disease with cavity formation, pleural empyema and bronchopleural fistulae. Extrapulmonary dissemination, which may occur both in the acute and chronic forms of the disease, is frequently encountered in immunocompromised individuals (approximately 0.5% of infected patients). The granulomatous or abscess like lesions mostly affect the skin, soft tissues, bones and meninges. Coccidioidomycosis closely resembles tuberculosis but other nonspecific pulmonary infections, sporotrichosis, blastomycosis, leishmaniasis, paracoccidioidomycosis and malignant tumors also need to be considered.
45.8.2 Laboratory findings
References /53, 56, 63, 65, 66/.
Microscopic and cultural detection of organisms
Sputum, purulent secretions, aspirated fluid, cerebrospinal fluid and tissue samples are primarily used as specimens. Silver staining according to Grocott-Gomori or labeling with optical brightening agents allow the detection of the characteristic spherules (20 to 70 μm) with endospores (2–5 μm). Cultural detection is achieved on customary nutrient culture media within an incubation period of 1 week at 20–30 °C (mycelial phase). Identification of the species necessitates transformation into the yeast phase using special media or animal tests.
A commercially available test for the specific detection of C. immitis DNA by culture yields results within one day. Specific handling of C. immitis and C. posadasii is only permitted in laboratories under laboratory biosafety level L3 (Ordinance on Biological Agents) because of the risk of laboratory infection.
45.8.2.1 Antibody detection
Autolysates and cultural filtrates of the mycelial phase (i.e., so-called coccidioidin) are used as antigen preparations. Serum and cerebrospinal fluid (CSF) are primarily suited as specimens.
Complement fixation test
Diluted, inactivated (56 °C, 30 min.) serum or CSF is incubated with coccidioidin solution and a defined amount of complement for 2 h at 37 °C (alternatively for 18 h at 4 °C) followed by another incubation period of 1 h at 37 °C after the addition of hemolysin sensitized erythrocytes. A lack of hemolysis indicates the presence of antibodies. The test detects IgG antibodies. Threshold titer is 1 : 8 in serum and 1 : 2 in CSF. Titers lower than this require confirmation by means of immunodiffusion testing. Data concerning diagnostic sensitivity and specificity are not available.
Immunodiffusion test
Using the Ouchterlony technique, application sites punched out in agarose gel are filled with serum or CSF on the one side and with coccidioidin solution on the opposite side. Pre diffusion of the test fluid prior to the application of antigen is beneficial. After incubation, an antigen-antibody reaction is revealed by the occurrence of precipitate bands. If heated (60 °C, 30 min.) coccidioidin is used, IgM antibodies will react as they do in the tube immunoprecipitation test. This modification of immunodiffusion (ID) is therefore referred to as IDTP. If non heated antigens are used, IgG antibodies, in particular, will react. This ID modification is called IDCF. Diagnostic sensitivity is higher than in the complement fixation reaction.
Latex agglutination test
Latex particles coated with heated coccidioidin serve as the test mixture. The test primarily detects IgM antibodies. The test is considered to be a sensitive screening method but it has the disadvantage of false positive reactions at a rate of approximately 15%. Positive reactions, therefore, must be confirmed by the more specific immunodiffusion tests.
Enzyme immunoassay
Wells of micro titer plates are coated with coccidioidin. Bound antibodies in diluted serum samples are detected by monovalent IgM or IgG specific conjugates. Diagnostic sensitivity is 100% in the case of joint evaluation of IgM and IgG reactions. Diagnostic specificity is 96%. False positive reactions are observed in the presence of blastomycosis and pulmonary infections of other causes.
45.8.2.2 Antigen detection
The Coccidioides galactomannan is detectable in the serum and urine of patients with a severe course of the disease. The test can also be helpful for diagnosing recently acquired infections when antibody production cannot yet be detected. Antigen detection is also a surrogate marker of fungal load in disseminated infection. In a study /65/, it was shown that EDTA heat pretreatment of samples improves the diagnostic sensitivity to 73% in serum samples and to 50% in urine samples. The test achieved a high specificity in healthy individuals, but cross reactions were seen in 22% of patients with histoplasmosis or blastomycosis /56, 65/.
Skin test
Coccidioidin and spherulin are used as test antigens. Performance and interpretation are analogous to the tuberculin skin test. The skin test does not allow to differentiate between latent and overt infection. Anergy can occur in massive dissemination and progressive course of the disease. The test is used to determine the level of infection in endemic regions, but is not commercially available /62, 63, 69/.
45.8.2.3 Molecular biological detection
PCR detection of Coccidioides DNA has been described, but is not commercially available /53, 63, 66/.
45.8.3 Clinical significance
IgM antibodies in the serum (tube immunoprecipitation, IDTP, latex agglutination) are detectable within 2–3 weeks after the occurrence of symptoms. They are usually no longer present after 6 months but may appear again and/or persist in the case of relapse and dissemination. IgG antibodies (complement fixation, IDCF) usually do not occur until 2–3 months after the onset of the disease. A complement fixation titer above 16 is considered to be a sign of florid infection and raises the suspicion of extra pulmonary dissemination. Complement fixation examinations at 3–4 week intervals are recommended for monitoring. A significant titer decrease under therapy is interpreted as a favorable prognostic sign. Detection of IgM antibodies in cerebrospinal fluid, in the case of meningitis, is the exception rather than the rule whereas IgG antibodies are usually detectable although sometimes not until several weeks after the onset of the disease /59, 62, 67, 68, 69/.
45.9 Blastomyces dermatitidis (Blastomycosis Gilchrist’s disease)
References /12, 70, 71/.
Blastomyces dermatitidis (teleomorph stage: Ajellomyces dermatitidis) is a temperature dependent, dimorphic fungus. B. dermatitidis grows in its natural environment and in vitro at 25–30 °C in the mycelial phase and at 37 °C in the yeast phase. The yeast cells can vary in size (8–15 μm), are thick walled and characteristically bear one daughter cell with a wide bud base.
45.9.1 Epidemiology and clinical symptoms
References /9, 53, 70, 71/.
Geographic distribution
No precise information is available about the incidence and epidemiology of blastomycosis because the pathogen is difficult to detect in the environment and no sensitive skin test is available. Case reports have suggested that the distribution of blastomycosis is similar to that of histoplasmosis. Blastomycosis is endemic to the USA, especially to the regions of the Mississippi and Ohio Rivers and the Midwest, well as the Canadian regions bordering on the Great Lakes and along the St. Lawrence River. Sporadic cases have been reported from Hawaii, Central and South America, Africa and the Middle East. No case of blastomycosis acquired in Germany has been reported to date.
Routes of infection
Moist soil containing decomposing plants and decaying wood (river banks, beaver dams) is the natural reservoir of the fungus. The infection is most commonly acquired by conidia inhalation. Direct infections due to skin lesions during outdoor activity or after a dog bite have also been described. Person-to-person transmission is a rarity.
Prevalence and incidence
Blastomycosis is a rare disease. In a study from Wisconsin covering the years 2000 and 2004, the rate of clinically overt infections in urban US areas such as Milwaukee was estimated to be 1–3/100,000 inhabitants. The incidence in individuals living near rivers was 74/100,000 inhabitants.
Clinical symptoms
Blastomycosis can be asymptomatic, manifest as acute or chronic pneumonia or take a disseminated form of the disease. Acute infection of the lung manifests after a 30–45 day incubation period with uncharacteristic, influenza like symptoms. If untreated, blastomycosis in many cases assumes a chronic form manifesting in the lungs as granulomatous, abscess like lesions and resembling the clinical picture of tuberculosis. Extra pulmonary manifestations involve the skin (verrucous or ulcerative lesions) in 40–80% of cases, bone (osteomyelitis) in 10–15% of cases, urogenital tract (primarily prostatitis, epididymitis in men) and, more rarely, the central nervous system.
45.9.2 Laboratory findings
Direct microscopic and cultural detection of organisms
Respiratory secretions, exudates, tissue samples and urine are suitable specimens for examination /70/. The budding cells with a wide bud based daughter cell are characteristic findings in the direct microscopic preparation (native or with optical brightening agent). Cultural detection is performed at 25–30 °C and at 37 °C on Sabouraud glucose agar and blood enriched brain-heart infusion agar for a time period of at least 14 days. Culturing from primarily non sterile specimens should be performed on culture medium containing inhibitors such as chloramphenicol or cycloheximide agar.
Antibody detection
The detection of antibodies to Blastomyces dermatitidis antigens is possible by immunodiffusion (Ouchterlouny) as well as by enzyme immunoassay and complement fixation test. Immunodiffusion is the only commercially available test. Extracts from the yeast phase are used as antigens in immunodiffusion testing. In the agarose gel technique according to Ouchterlony, antibodies in the serum react by forming precipitin lines (A and B bands). The diagnostic sensitivity in pulmonary blastomycosis is 65–80% at a specificity of approximately 30% /59, 70, 72, 73/
Antigen detection, molecular biological detection
A test kit for urine Blastomyces antigen detection is commercially available. Clinical sensitivity is 93% at low specificity. The test yields a positive result even in the presence of blastomycosis, histoplasmosis, para coccidioidomycosis and penicillinosis. A test kit for pathogen detection by PCR is not commercially available. Broncho alveolar lavage, urine, blood and other body fluid samples can be sent to the consultant laboratory for pathogens of extra European systemic mycoses at the Robert-Koch Institute for immunological and molecular biological analyses /56, 70, 74/.
45.10 Para coccidioides brasiliensis (Blastomycosis, South-American blastomycosis)
References /12, 53, 75, 76, 77/.
The pathogen P. brasiliensis is a temperature dependent, dimorphic fungus. It grows in cultures at 37 °C and in vivo in the yeast form with characteristic round or oval cells (up to 60 μm) surrounded by multiple peripheral buds (2–10 μm) also referred to as pilot wheel. At low temperatures, a mycelium is formed from thin septate hyphae with differently shaped conidia (3.5–5 μm).
45.10.1 Epidemiology and clinical symptoms
Routes of infection
The natural reservoir of P. brasiliensis is not exactly known. To date, the fungus has only been identified twice in soil samples. Epidemiological and experimental data suggest that the ecological niches are located within humid, warm and forested areas. Fish, amphibians and armadillos possibly play a role in the life cycle of the fungus. The infection is probably acquired by inhaling conidia present in the environment. Person-to-person transmission does not occur /53, 76, 78/.
Prevalence and incidence
Para coccidioidomycosis is endemic to regions from Mexico to Argentina. Approximately 80% of recorded infections occurred in Brazil, followed by Colombia and Venezuela. All clinically overt infections diagnosed outside these regions, in many cases after years of latency, can be traced back to stays in endemic regions. Infections are usually inapparent in immunocompetent individuals. In many cases, patients aged 30–50 years are affected. The disease occurs 15 times more often in men than in women. In endemic regions, approximately 5 clinically overt cases of the disease per 1 million people occur each year. The infection rate in the population is estimated to be 10–20%.
Clinical symptoms
The clinical picture of this chronic progressive disease resembles that of tuberculosis. Differentiation is made between a juvenile form of the disease with an acute or subacute course (3–5% of the cases) and an adult form with a chronic course (> 90% of cases). In the juvenile, rapidly generalizing form, pulmonary dissemination into the spleen, liver, lymph nodes and bone marrow is the main characteristic. The adult form of the disease, which develops over the course of months and years, only affects the lung in approximately 25% of the cases (unifocal form). It leads to emphysema and fibrous abnormalities in the lung. Dissemination (multi focal form) occurs mostly to the oral and nasal mucosa, skin, lymph nodes and adrenal glands.
45.10.2 Laboratory findings
Microscopic and cultural detection
References /75, 77, 79/.
Sputum, broncho alveolar lavage fluid, exudates, cerebrospinal fluid and tissue samples can be used as specimens. Microscopic examination is performed using native preparations (potassium hydroxide solution) and histologically prepared slides using various staining techniques (e.g., Grocott-Gomorri staining). These methods allow to detect typical yeast cells with multiple budding.
Sabouraud agar is suitable to culture the organism. An incubation period of 10–15 days at 37 °C (yeast form) and of 20–30 days at 25 °C (mycelial form) is necessary for colonies to occur. The diagnostic sensitivity of microscopic and of cultural detection are reported as 85–100% and 86–100%, respectively.
45.10.2.1 Antibody detection
Intact cells of the yeast form, extracts of the cell wall and the cytoplasm as well as culture filtrates are used as test antigens in numerous serological test methods /77, 79, 80/.
However, these methods often show cross reactions. More recently, a glycoprotein (GP43) was identified, whose peptide component is a carrier of epitopes specific for P. brasiliensis.
Immunodiffusion test
According to the Ouchterlony technique, application wells are punched out in agarose gel and filled with antigen solution (e.g., 1 μl of GP43, and/or serum). The diagnostic sensitivity is approximately 90% and the specificity is 100%.
Indirect hemagglutination test
Antigen GP43 adsorbed to sheep erythrocytes is incubated with diluted serum for 1 h at room temperature. Antibodies cause agglutination of the loaded erythrocytes. The diagnostic sensitivity and specificity are 100%.
Immunoblot
The available information allows the conclusion that the detection of antibodies to the GP43 antigen clearly indicates the presence of manifest para coccidioidomycosis. The level of antibody titers or titer changes might be helpful to assess disease activity and the success of therapy.
45.10.2.2 Skin test
Numerous different antigen preparations (para coccidioidins) have been used for skin tests /53/. By using a polysaccharide antigen, positive reactions were obtained in 67% of confirmed cases of the disease but also in 87% of the references, indicating the presence of subclinical or cross-reacting infections. Skin tests are generally not helpful for diagnosis due to their unspecificity. Cross reactions with histoplasmin have been reported.
45.11 Sporothrix schenckii (Sporotrichosis)
The pathogen Sporothrix schenckii is a temperature-dependent, dimorphic fungus /12/. In rich and liquid cultures and in vivo at 37 °C, it grows in the yeast phase, forming characteristic oval or tuberculate daughter cells. At low temperatures, it forms mycelium from septate hyphae with conidia of various shapes that develop a brown/black pigment (melanin) when incubated for an extended period of time.
45.11.1 Epidemiology and clinical symptoms
References /81, 82/.
Routes of infection
Sporotrichosis occurs worldwide. However, most cases have been reported from the tropical and subtropical regions in America. S. schenkii can be found in its natural environment on plants and in the soil. For this reason, sporotrichosis affects farmers, florists, nursery workers, landscapers, greenhouse workers, leisure gardeners (i.e., individuals working with plants, peat or soil). S. schenckii invades the subcutaneous tissue through puncture injuries caused by splinters or thorns. Transmissions from animals to humans have been reported.
Prevalence and incidence
Systematic information has not been available to date because of the low specificity of the reactivity of skin tests using sporotrichin or antibody detection tests. The disease occurs in humans of any age, preferably in young individuals aged between 10 and 40 years.
Infections are usually sporadic. They have often been described in connection with forest or peat work. Epidemic outbreaks following feline contact, infection after armadillo hunting and person-to-person transmission have been reported. Epidemic outbreaks in pets such as dogs and cats have also been described.
Clinical symptoms
An ulcerative, verrucous or reddish lesion, in some cases associated with lymphangitis, develops at the site of invasion.
After inhalation of the spores, the fungus can cause granulomatous, often bullous, pneumonia with symptoms resembling tuberculosis. Hematogenous dissemination via central nervous, articular and ocular infiltrate has been reported. Disseminated sporotrichosis has been described to occur in immunocompromised individuals.
45.11.2 Laboratory findings
Microscopic and cultural detection of organisms
Cultural detection is the gold standard for the detection of sporotrichosis. Biopsy samples or tissue scrapings from skin lesions, respiratory secretions and cerebrospinal fluid are suited as specimens for culturing and analysis. Culturing should at first be performed at room temperature and subsequently at 37 °C to confirm the identity of the biphasic fungus /83, 84, 85/.
45.11.2.1 Antibody detection
The detection of antibodies by immunodiffusion and/or immunoelectrophoresis has already been used successfully for the disseminated form of sporotrichosis. However, because of its low sensitivity to the most frequently encountered cutaneous form of the disease, antibody detection has not been established in routine diagnostics. Reliable diagnosis of the cutaneous form of sporotrichosis has been achieved by using an antigen (SsCBF) derived from a peptido-rhamnomannan component of the cell wall. Based on this antigen, an ELISA was developed showing a diagnostic sensitivity of 90% and a specificity of 86% in various forms of sporotrichosis. However, the usability of this ELISA is limited because of the difficult antigen preparation and the low robustness of the test. At the same time, an antibody assay was developed which is based on the easily obtained exoantigens of the mycelial phase. These antigens do not show cross reactivity in the presence of coccidioidomycosis, histoplasmosis or paracoccidioidomycosis and/or chromoblastomycosis and leishmaniosis. However, cross reactions with numerous molds have been observed in antigen preparations.
45.11.2.2 Skin test
Sporotrichin is the yeast form of a cell wall glycopeptide already used since the early days of sporotrichosis diagnostics. In many cases, however, it showed false positive as well as false negative reactions. Therefore, the skin test should not be applied for routine serodiagnosis /81/.
References
Lass-Flörl C. The changing face of epidemiology of invasive fungal disease in Europe. Mycoses 2009; 52: 197–205.
Carmona EM, Limper AH. Update on the diagnosis and treatment of Pneumocystis pneumonia. Ther Adv Respir Dis 2011; 5: 41–59.
Kauffman CA. Fungal infections in older adults. Clin Infect Dis 2001; 33: 550–5.
Marr KA. Fungal infections in oncology patients: update on epidemiology, prevention, and treatment. Curr Opin Oncol 2010; 22: 138–142.
Shoham S, Levitz SM. The immune response to fungal infections. Brit J Haematol 2005; 129: 569–82.
Stevens DA. Diagnosis of fungal infections: current status. J Antimicrob Chemother 2002; 49 (S1): 11–19.
Yeo SF, Wong B. Current status of nonculture methods for diagnosis of invasive fungal infections. Clin Microbiol Rev 2002; 15: 465–84.
Haase G, Borg-von Zepelin M, Bernhardt H, Fegeler W, Harmsen D, Kappe R, et al. Qualitätsstandards in der mikrobiologisch-infektiologischen Diagnostik. Pilzinfektionen Teil I (MiQ 14), Teil II (MiQ 15), 2001. München Jena Urban & Fischer.
Lilienfeld-Toal M, Wagener J, Einsele H, Cornely OA, Kurzai O. Invasive fungal infection: new treatments to meet new challenges Dtsch Arztebl Int 2019; 116: 271–8.
Cornely OA. Aspergillus to Zygomycetes: causes, risk factors, prevention, and treatment of invasive fungal infections. Infection 2008; 36: 296–313.
Groll AH, Buchheidt D, Cornely O, Glöckner A, Heinz W, Höhl R, et al. Diagnose und Therapie von Candida-Infektionen. Chemotherapie Journal 2011; 20: 67–93.
de Hoog GS, Guarro J, Gené J, Figueras MJ. Atlas of Clinical Fungi. 2nd ed. Utrecht; Centralbureau voor Schimmelcultures 2001.
Jones JM. Laboratory diagnosis of invasive candidiasis. Clin Microbiol Rev 1990; 3: 32–45.
Kullberg BJ, Arendrup MX. Invasive candidiasis. N Engl J Med 2015; 373: 1445–56.
Clancy CJ, Nguyen MH. Finding the missing 50% of invasive candidiasis: how nonculture diagnostics will improve understanding of disease spectrum and transform patient care. Clin Infect Dis 2013; 56: 1284–92.
Avni T, Leibovici L, Paul M. PCR Diagnosis of invasive candidiasis: Systematic review and meta analysis. J Clin Microbiol 2011; 49: 665–70.
Ellepola AN, Morrison CJ. Laboratory diagnosis of invasive candidiasis. J Microbiol 2005; 43: 65–84.
Pemán J, Zaragoza R. Current approaches to invasive candidiasis in critical care settings. Mycoses 2009; 53: 424–33.
Prella M, Bille J, Pugnale M, Duvoisin B, Cavassini M, Calandra T, et al. Early diagnosis of invasive candidiasis with mannan antigenemia and antimannan antibodies. Diag Microbiol Infect Dis 2005; 51: 95–101.
Sendid B, Poirot JL, Tabouret M, Bonnin A, Caillot D, Camus D, et al. Combined detection of mannanaemia and anti-mannan antibodies as a strategy for the diagnosis of systemic infection caused by pathogenic Candida species. J Med Microbiol 2002; 51: 433–42.
Warnock DW. Fungal complications of transplantation: diagnosis, treatment and prevention. J Antimicrob Chemother 1995;36(suppl B): 73–90.
Miyazaki T, Kohno S, Mitsutake K, Maesaki S, Tanak K-I, Ishikawa N, et al. Plasma (1–3)-β-D-glucan and fungal antigenemia in patients with candidemia, aspergillosis, and cryptococcosis. J Clin Microbiol 1995; 33: 3115–8.
Ostrowsky-Zeichner L, Alexander BD, Kett DH, Vazquez J, Pappas PG, Saeki F, et al. Multicenter clinical evaluation of the (1–3) β-D-Glucan assay as an aid to diagnosis of fungal infections in humans. Clin Infect Dis 2005; 41: 654–9.
Pickering JW, Sant HW, Bowles CAP, Roberts WL, Woods GL. Evaluation of a (1–3)-β-D-Glucan assay for diagnosis of invasive fungal infections. J Clin Microbiol 2005; 43: 5957–62.
de Repentigny L. Serodiagnosis of candidiasis, aspergillosis, and cryptococcosis. Clin Infect Dis 1992; 14 (Suppl 1): S11–S22.
Sendid B, Caillot D, Baccouch-Humbert B, Klingspor L, Grandjean M, Bonnin A, et al. Contribution of the Platelia Candida-Specific Antibody and Antigen Tests to early diagnosis of systemic Candida tropicalis infection in neutropenic patients. J Clin Microbiol 2003; 41: 4551–8.
Clancy CJ, Nguyen M-L, Cheng S, Huang H, Fan G, et al. Immunoglobulin G responses to a panel of Candida albicans antigens as accurate and early markers for the presence of systemic candidiasis. J Clin Microbiol 2008; 46: 1647–54.
Donhuijsen K, Petersen P, Schmid KW. Trendwende in der Mykosefrequenz bei hämatologischen Neoplasien. Dt Ärztebl 2008; 105: 501–6.
Tintelnot K, Adler S, Bergmann F, Schönherr K, Seibold M. Case reports. Disseminated cryptococcoses without cryptococcal antigen detection. Mycoses 2000; 43: 203–7.
Perfect JR. Cryptococcus neoformans. In: Mandell, Douglas and Bennett’s Principles and Practice of Infectious Disease. Mandell GL, Bennett JE, Dolin R eds. Philadelphia; Churchill Livingstone 2010. p. 3287–3301.
Tintelnot K, Losert H. Isolation of Cryptococcus adeliensis from clinical samples and the environment in Germany. J Clin Microbiol 2005; 43: 1007.
Tintelnot K. Kryptokokkose: Cryptococcus gattii Infektionen können importiert werden und schwer verlaufen. Epidemiol Bulletin 2008; 20: 159.
Bialek R, Weiss M, Bekure-Nemariam K, Najvar LK, Alberdi MB, Graybill JR, et al. Detection of Cryptococcus neoformans DNA in tissue samples by nested and real-time PCR assays. Clin Diag Lab Immunol 2002; 9: 461–9.
Lindsay MD, Mekha N, Baggett HC, Surinthong Y, Autthateinchai R, Sawatwong P, et al. Evaluation of a newly developed lateral flow immunoassay for the diagnosis of cryptococcosis. Clin Infect Dis 2011; 53: 321–5.
Hope WW, Walsh TJ, Denning DW. Laboratory diagnosis of invasive aspergillosis. Lancet Infect Dis 2005; 5: 609–22.
Patterson TF. Aspergillus species. In: Mandell, Douglas and Bennett’s Principles and Practice of Infectious Disease. Mandell GL, Bennett JE, Dolin R eds. Philadelphia; Churchill Livingstone 2010. p. 3241–55.
Erjavec Z, Kluin-Nelemans H, Verweij PE. Trends in invasive fungal infections, with emphasis on invasive aspergillosis. Clin Microbiol Infect 2009; 15: 625–33.
Thompson III GR, Young Jo-Anne H. Aspergillus infections. N Engl J Med 2021; 385: 1496–509.
Buchheidt D, Baust C, Skladny H, Ritter J, Suedhoff T, Baldus M, et al. Detection of Aspergillus species in blood and bronchoalveolar lavage samples from immunocompromised patients by means of 2-step polymerase chain reaction: Clinical results. Clin Infect Dis 2001; 33: 428–35.
Chen SC-A, Kontoyiannis DP. New molecular and surrogate biomarker-based tests in the diagnosis of bacterial and fungal infection in febrile neutropenic patients. Curr Opin Infect Dis 2010; 23: 567–77.
Tuon FF. A systematic literature review on the diagnosis of invasive aspergillosis using polymerase chain reaction (PCR) from bronchoalveolar lavage clinical samples. Rev Iberoam Micolog 2007; 24: 89–94.
Spies B, Seifarth W, Hummel M, Frank O, Fabarius A, Zheng C, et al. DNA Microarray-based detection and identification of fungal pathogens in clinical samples from neutropenic patients. J Clin Microbiol 2007; 45: 3743–53.
Maertens J, Maertens V, Theunissen K, Meersseman W, Meersseman P, Meers S, et al. Bronchoalveolar lavage fluid galactomannan for the diagnosis of invasive pulmonary aspergillosis in patients with hematologic diseases. Clin Infect Dis 2009; 49: 1688–93.
Marr KA, Balajee SA, McLaughlin L, Tabouret M, Bentsen C, Walsh TJ. Detection of galactomannan antigenemia by enzyme immunoassay for the diagnosis of invasive aspergillosis: variables that affect performance. J Infect Dis 2004; 190: 641–9.
Mennink-Kersten MASH, Ruegebrink D, Klont RR, Warris A, Blijlevens NAM et al. Improved detection of circulating Aspergillus antigen by use of a modified pretreatment procedure. J Clin Microbiol 2008; 46: 1391–97.
Mennink-Kersten MASH, Donnelly JP, Verweij PE. Detection of circulating galactomannan for the diagnosis and management of invasive aspergillosis. Lancet Infect Dis 2004; 4: 349–57.
Kappe R, Schulze-Berge A, Sonntag H-G. Evaluation of eight antibody tests and one antigen test for the diagnosis of invasive aspergillosis. Mycoses 1996; 39: 13–23.
Chamilos G, Luna M, Lewis RE, Bodey GP, Chemaly R, Tarrand JJ, et al. Invasive fungal infections in patients with hematologic malignancies in a tertiary care cancer center: an autopsy study over a 15-year period (1989–2003). Haematologica 2006; 91: 986–9.
Morris A, Wei K, Afshar K, Huang L. Epidemiology and clinical significance of Pneumocystis colonization. J Infect Dis 2008; 197: 10–17.
Krajicek BJ, Limper AH, Thomas Jr CF. Advances in the biology, pathogenesis and identification of Pneumocystis pneumonia. Curr Opin Pul Med 2008; 14: 228–34.
Rath PM, Schmid EN, Ansorg R. Vergleich zwischen mikroskopischen Methoden und Polymerasekettenreaktion zum Nachweis von Pneumocystis carinii in bronchoalveolären Lavage-Flüssigkeiten. J Lab Med 1996; 20: 553–9.
Sax PE, Komarow L, Finkelman MA, Grant PM, Andersen J, Scully E, et al. Blood (1–3)-β-D-Glucan as a diagnostic test for HIV-related Pneumocystis jirovecii pneumonia. Clin Infect Dis 2011; 53: 197–202.
Bonifaz A, Vázquez González D, Perusquia-Ortiz AM. Endemic systemic mycoses: coccidioidomycosis, histoplasmosis, paracoccidioidomycosis and blastomycosis. J Dtsch Dermatol Ges 2011; 9: 705–15.
Kauffman CA. Histoplasmosis: a clinical and laboratory update. Clin Microbiol Rev 2007; 20: 115–32.
Ashbee HR, Evans EGV, Ashbee HR, Evans EGV, Viviani MA, Dupont B, et al. Histoplasmosis in Europe: Report on an epidemiological survey from the European Confederation of Medical Mycology Working Group. Med Mycol 2008; 46: 57–65.
Durkin MM, Connolly PA, Karimi K, Wheat E, Schnizlein-Bick C, Allen SD, et al. Pathogenic differences between North American and Latin American strains of Histoplasma capsulatum var. capsulatum in experimentally infected mice. J Clin Microbiol 2004; 42: 4370–3.
Karimi K, Wheat LJ, Connolly P, Cloud G, Hajjeh R, Wheat E, et al. Differences in histoplasmosis in patients with acquired immunodeficiency syndrome in the United States and Brazil. J Infect Dis 2002; 186: 1655–60.
Spitzer ED, Keath EJ, Travis SJ, Painter AA, Kobayashi GS, Medoff G. Temperature-sensitive variants of Histoplasma capsulatum isolated from patients with acquired immunodeficiency syndrome. J Infect Dis 1990; 162: 258–61.
Wheat J. Endemic mycoses in AIDS: a clinical review. Clin Microbiol Rev 1995; 8: 146–59.
Hage CA, Davis TE, Fuller D, Egan L, Witt JR, Wheat LJ, et al. Diagnosis of histoplasmosis by antigen detection in BAL. Chest 2010; 137: 62–8.
Sekhon AS, Kaufman L, Kobayashi GS, Moledina N, Jalbert M, Notenboom RH. Comparative evaluation of the Premier enzyme immunoassay, micro-immunodiffusion and complement fixation tests for the detection of Histoplasma capsulatum var. capsulatum antibodies. Mycoses 1994; 37: 313–6.
Stevens DA. Coccidioidomycosis. N Engl J Med 1995; 332: 1077–82.
Galgiani JN. Coccidioides species. In: Mandell, Douglas and Bennett’s Principles and Practice of Infectious Disease. Mandell GL, Bennett JE,Dolin R eds. Philadelphia; Churchill Livingstone 2010. p. 3333–44.
Parish JM, Blair JE. Coccidioidomycosis. Mayo Clin Proc 2008; 83: 343–9.
Durkin M, Estok L, Hospenthal D, Crum-Cianflone N, Swartzentruber S, Hackett E, et al. Detection of Coccidioides antigenemia following dissociation of immune complexes. Clin Vaccine Immunol 2009; 16: 1453–6.
Binnicker MJ, Popa AS, Catania J, Alexov M, Tsaras G, Lloyd F, et al. Meningeal coccidioidomycosis diagnosed by real-time polymerase chain reaction analysis of cerebrospinal fluid. Mycopathologia 2011; 171: 285–9.
Kaufman L, Sekhon AS, Moledina N, Jalbert M, Pappagianis D. Comparative evaluation of commercial premier EIA and microimmunodiffusion and complement fixation tests for Coccidioides immitis antibodies. J Clin Microbiol 1995; 33: 618–9.
Pappagianis D, Zimmer BL. Serology of coccidioidomycosis. Clin Microbiol Rev 1990; 3: 247–68.
Snyder LS, Galgiani JN. Coccidioidomycosis: the initial pulmonary infection and beyond.Sem Resp Crit Care Med 1997; 18: 235–48.
Saccente M, Woods GL. Clinical and laboratory update on blastomycosis. Clin Microbiol Rev 2010; 23: 367–81.
Baumgardner DJ, Knavel EM, Steber D, Swain GR. Geographic distribution of human blastomycosis cases in Milwaukee, Wisconsin, USA: association with urban watersheds. Mycopathologia 2006; 161: 275–82.
Kwong-Chung KJ, Bennett JE. Medical Mycology. Philadelphia: Philadelphia; Lea and Febiger 1992.
Pappas PG. Blastomycosis in the immunocompromised patient. Semin Respir Infect. 1997; 12: 243–51.
Bariola JR, Hage CA, Durkin M, Bensadoun E, Gubbins PO, Wheat LJ, et al. Detection of Blastomyces dermatitidis antigen in patients with newly diagnosed blastomycosis. Diag Microbiol Infect Dis 2011; 69: 187–91.
Brummer E, Castaneda E, Restrepo A. Paracoccidioidomycosis: an update. Clin Microbiol Rev 1993; 6: 89–117.
Restrepo A, Tobon AM. Paracoccidioides brasiliensis. In: Mandell, Douglas and Bennett’s Principles and Practice of Infectious Disease. Mandell GL, Bennett JE,Dolin R editors. 7th ed. Philadelphia; Churchill Livingstone 2010. p. 3357–3366.
San-Blas G. Paracoccidioidomycosis and its etiologic agent Paracoccidioides brasiliensis. Med Mycol 1993; 31: 99–113.
Bagagli E, Theodoro RC, Bosco SMG, McEwen JG. Paracoccidioides brasiliensis: phylogenetic and ecological aspects. Mycopathologia 2008; 165: 197–207.
de Camargo ZP. Serology of paracoccidioidomycosis. Mycopathologia 2008; 165: 289–302.
Taborda CP, Camargo ZP. Diagnosis of paracoccidioidomycosis by passive haemagglutination assay of antibody using a purified and specific antigen-gp43. Med Mycol 1993; 31: 155–60.
Lopes-Bezerra LM, Schubach A, Costa RO. Sporothrix schenckii and sporotrichosis. An Acad Bras Ciénc 2006; 78: 293–308.
Rex JH, Okhuysen PC. Sporothrix schenckii. In: Mandell, Douglas and Bennett’s Principles and Practice of Infectious Disease. Mandell GL, Bennett JE,Dolin R editors. 7th ed. Philadelphia; Churchill Livingstone 2010. p. 3357–3366.
Almeida-Paes R, Pimenta MA, Pizzini CV, Monteiro PCF, Peralta JM, Nosanchuk JD, et al. Use of mycelial-phase Sporothrix schenckii exoantigens in an enzyme-linked immunosorbent assay for diagnosis of sporotrichosis by antibody detection. Clin Vaccine Immunol 2007; 14: 244–9.
Bernardes-Engemann AR, Orofino Costa RC, Miguens BP, Penha CVL, Neves E, Pereira BAS, et al. Development of an enzyme-linked immunosorbent assay for the serodiagnosis of several clinical forms of sporotrichosis. Med Mycol 2005; 43: 487–93.
Fernandes GF, Lopes-Bezerra LM, Bernardes-Engemann AR, Schubach TMP, Dias MAG, Pereira SA, et al. Serodiagnosis of sporotrichosis infection in cats by enzyme-linked immunosorbent assay using a specific antigen, SsCBF, and crude exoantigens. Vet Microbiol 2011; 147: 445–9.
Tabelle 45.4-1 Clinical manifestations and risk factors of Aspergillus infections. Courtesy of Lit. /38/
| | Risk factors | Clinical elements of infection |
| Allergic bronchopulmonary aspergillosis | Reactive airway condition | Hypersensitivity to inhaled airborne candida causes allergic bronchopulmonary aspergillosis |
| | Structural lung disease | Fungus balls develop inside lung cavities (e.g. from tuberculosis) |
| Colonization of the pulmonary tree | Inhaled glucocorticoids, bronchiektasis, cystic fibrosis | Recurrent recovery in culture without compatible symptoms or radiographic findings |
| Cutaneous disease | Tissue damage, traumatic, foreign body | Iatrogenic infections of wounds through contaminations |
| Eye disease | Tissue damage, surgery, foreign body | Topical keratitis, endophthalmitis or extension of sinus disease |
| | Most commonly in lung-transplant recipients | Cough, hemoptysis, wheezing, dyspnoe |
| Chronic pulmonary aspergillosis | Structural lung disease, underlying cancer, malnutrition, impaired mucociliary clearance after recent pulmonary infection | Fever, cough, shortness of breath, with or without chest pain |
| Sinus disease | Neutropenia, diabetes, excessive alcohol use,immuno-compromized state immunocompromised | Difficult to differentiate from other infections |
| Disease of the central nervous system | Develops with hematogenous dissemination | Neurologic deficits consistent with the region of involvement |
| Invasive infection | Immunosuppression often related to chemotherapy | Vegetative hyphae invade through tissue planes into blood vessels | |
189054 | https://www.youtube.com/watch?v=9UjW6XYiyZA | Pre-Calculus: Review Exponents and Radicals
Michael Porinchak - AP Statistics & AP Precalculus
28800 subscribers
567 likes
Description
59567 views
Posted: 23 Jul 2012
A very brief review of exponent rules, working with radicals, and rationalizing the denominator.
35 comments
Transcript:
all right we're going to fly through a quick lesson here on exponents and radicals it's going to go pretty fast um but again hopefully all this is nothing more of a review from what you've learned in algebra and even in Algebra 2 um real quick let's make sure we understand the basic rules of what an exponent is so if we have a to the m that basically means we have tons and tons and tons of A's being multiplied M times so we would have M of these A's whatever that is let's also make sure we understand negative exponents remember negative exponents have nothing to do with negative numbers so a to the netive m simply means that this repeating a like much like up here would be on the denominator of a fraction so it' be 1 over a to the positive M which would be 1 over a a a again M and having negative exponents is um bad math grammar um so we try to avoid it as much as we can so final answers we usually try to make sure there's no negative exponents so let's fly through the um rules real quick the first rule when dealing with exponents is that if you are multiplying with the same base base a base a you can add the exponents it' be a to the m + n so real quick example here if we had x to the 7th X to 3 we have a total of 10 X's 7 here and 3 here 7 + 3 is 10 x 10th also can work with negatives if we have x to the 7th and x^ the -3 again you could treat this as 1/ X 3r or you can simply add them 7 + -3 would be X to 4th and since that's a positive exponent we would leave it alone second major rule here is with division if you have the same base on top same base on the bottom you could subtract the exponents so it' be a to the m minus n so do a quick example if we had uh let's see here let's use y y to the 10th over y to the 3r we would just subtract 10 - 3 is 7 so we have 10 y's on top three on the bottom there's only one term on top here one term on the bottom no plus signs no minus sign so three of the Y's on top would cancel which is where we get the Y to the 7th from be careful though because if we did it like this with the y 3 on top y the 10 on the bottom again you could just use the idea that three of these y's on the bottom are going to cancel leaving seven y's on the bottom or 3 - 10 is y to the ne7 and then we say wait a minute we don't like that negative exponent so we'd put 1 over y to the 7th as our final answer all right next rule number three here um would be that any value raised to the zero power is always one um anything to the zero is always one no matter what easy easy easy number four here we have a B in parentheses with an exponent outside the parentheses basically that exponent goes to both of those that means we have M A's and mb's just like that um be careful I want to just quickly show you there's no parentheses like 3x^2 that means there's just three and then XX two x's 3 x X the two does not go to three because there's parentheses whereas if there were parentheses around both of them then all of a sudden we have 3^ squared 2 3es and we also have x^2 which would shorten to 9 x^2 obviously so those parentheses do make a huge difference um we also have the next rule which is a power to a power so if we have a to the m raised to the N then that is when you multiply you have a to the m n so quick example of that if we have x to the 3 raised to the 6 think about that we have x x x on the insid that's 3 X's 6 times that's obviously a total of 18 X's x to the 18 could also do it with negative exponents if we got y^ -2 rais to the 3r again we have multiplying we get y to the -6 but wait a minute we don't like leaving negative exponents so that's 1 over y to the 6 should make complete sense um we also have the same rule right here I'll throw it up here rule number six the same rule with division I A over B to the m again because it's parentheses the a gets the m and so does the B so a to the M over um B to the m now be careful again this is all implying notice through none of these problems has there been a plus sign it's all been one term one term on the top one term on the bottom or one term inside the parentheses so I quickly want to um kind of go over here what I'm talking about so let me kind of stop with those rules real quick this is pretty important here if I have x + 3 squar now all of a sudden I have two terms in here that does not mean x^2 + 3^ 2 that's not what the rule says the rule said if it's x 3 then it's X2 3^ s but this is a plus that means this is a quantity this is two terms so we have no idea what this number is until somebody tells us what x is so this would literally mean x + 3 x + 3 and then we can go ahead and foil that out and get X SAR on the inside we get a 3X on the outside we would get a 3x on the final last terms we get a 9 so we'd get x^2 + 6x + 9 so please do not confuse this and think that that rues what we stated earlier there's a plus sign there two terms makes it completely different when there's only one term in here then the two can go to everything but there's two different terms here it does not work that way so that's an important rule we want to make sure everybody knows please all right moving on um we can do some quick uh reduction rules here um for example we want to reduce a quantity here so here's a quick example so we have uh -3 a b to the 4th on parentheses Time 4 a b to the -3 all right so these are multiplying no you know it's a term times a term I don't no plus signs in here anything like that there's nothing outside the parentheses that I need to handle first so basically I can start multiplying I got a -3 a 4 is -12 I got one a here one B here so it's a total of a squ a a is a squ now I got these B's same base so what do I do with the exponents I add them so 4 + -3 is 1 so I have B to the 1st or you don't necessarily have to write that one right there but that would be your final answer uh we could do a second example here where we have something like 2 x sorry my handwriting is getting a little sloppy I'll try to be neater we got 2 x^2 y all raised to the 3 4 x y to the 7th now we have to take care of this three first we cannot multiply these threes because we cannot multiply these terms together because we got this three right here there's nothing right here so I don't have to worry about it but this guy do have to worry about there's only one term in here one term 2x^2 Y is one term remember terms or anything set right by plus sign so there's no plus sign in there so now that means I have to do 2 the 3r 2 3r is 2 2 2 which is 8 x that's already X2 three times that's multiplying 2 3 is 6 and then there's y to the first right here so 3 1 is 3 now I can bring in the 4xy to 7th now I can start multiplying 8 4 is 32 x 6 x is 1 X is here that's x to the 7th 3 and 7 for the y y the 3r Y the 7th is a total of Y to the 10th so that is um some quick combining rules right there so that um you guys know how to do that we can do one more here with um some division here so we got 5 x to the 3 / y all to the second power so remember one term in here no plus sign no minus sign one term so everybody gets this two so 5^ SAR is 25 3 X's 2quared would be x to the 6 CU we got 3 X's taken twice X the 6 and then it even goes the bottom with the y^2 um the last thing we want to go over is just some quick ones with uh subtraction or some negative exponents so we got 12 8 to the 3r B to -4 / 4 a to -2 uh B so first thing we going to do is these are just whole numbers divide them 12 divided 4 3 no big deal next we got the same base what we do here is remember just subtract include the negative don't ignore it so 3 minus -2 3 - -2 is actually going to turn into 3 + 2 which is 5 so we have a to the 5ifth -4 -1 -4 -1 there really one right there -4 -1 is5 so we have B to theg5 now we don't want to leave our answer like that because this B has a negative expon so he is going to move to the bottom the 3A to the 5th will stay on top but the B will move to the bottom with the Positive five so watch out for problems like that with negative exponents all right next we want to quickly very quickly look at um square roots and radical rules kind of a idea here um just remember that um uh square roots and any power Roots could also be looked at as fractions so the square root of x could be looked at as x to the one2 power uh we could do the cubed root of x that would be x to the 1/3 power or if we had like the cubed root of x^2 that would be X to 2 2/3 power again two is a power three is a root roots on the bottom power's on top so that's why the power I'm sorry the power of two was on top the root of three would be on the bottom so don't forget about using those rules right there as well hopefully you guys all know how to reduce radicals for example reducing something like 48 the square root of 48 is not a perfect number but 48 is 16 3 so now the 16 can come out as a four so I get four radical Al 3 as a final answer we need to reduce all square roots in this class I'm not going to spend a lot more time on reducing square roots hopefully that's something very easy for you guys to do the only other thing I want to talk about real quick is what we call rationalizing the denominator anytime we have a radical in the denominator we want to get rid of it so for example if I have 3 over the square < TK of 5 I do not want to leave that as a final answer so I want to rationalize rationalizing denominator means getting rid of square roots so something like this I just multiply the top and bottom by the < TK of 5 now the < TK of 5 divid theun of 5 is equal to one so I'm really multiplying by one which according to our rules is not changing anything so on top I get 3 radical 5 on the bottom you could look at this in several different ways you could look at radical 5 radical 5 is radical 25 which is a perfect five that's the idea is the the rule is that the radical x radical X as long as these values are the same you just get X um if you did something like radical 3 radical 2 you'd get radical 6 but nothing you could reduce with that but obviously that's the reason why we chose radical 5 let's do a couple more here if I have 2 over 5 radical 7 now again I don't this five is fine it's the radical 7 I don't like so I'm going to multiply top and bottom by radical 7 on top I get 2 radical 7 on the bottom leave the five alone for a second radical 7 radical 7 is radical 49 the square otk of 49 or radical 49 is 7 times the 5 makes 35 so that's where that comes from again the radical 7s cancel and just get seven the 5 that was there makes the 35 we also need to talk about reducing when you have um two terms on the bottom so something like two over 3+ radical 7 again there's two terms down here one of course is a radical and one's not this is a little bit trickier we need to multiply by What's called the conjugate which would be 3 minus radical 7 and you're going to see why in 1 second here now again minus radical 7 this is still one so I'm not really changing anything but on top I got 2 3 - radical 7 on the bottom I have a two terms time two terms so I got to get end up with four terms here 3 3 is 9 outside I get min -3 radical 7 inside I get plus 3 radical 7 on the ends I get a Min - 7 once again radical 7 and R 7 is radical 49 which is 7 notice these terms cancel which is exactly why I chose to multiply by the conjugate which if this is a plus I use the minus so on the bottom I get 9 - 7 which is just two so I get 2 over 3 - radical 7 / 2 and this doesn't happen too often but this works out really nice here where the twos ended up cancelling doesn't always work that way but it did there so my final answer to be 3 minus radical 7 all right and that is it for exponents and radicals sh |
189055 | https://www.engineersedge.com/materials/densities_of_metals_and_elements_table_13976.htm | Engineers Edge utilizes cookies to enable essential site functionality, and targeted advertising. To learn more, see our Privacy Policy.
Related Resources: materials
Densities of Metals and Elements Table
Engineering Materials
Densities of Metals and Elements Table
Density is defined as the mass per unit volume
For density in lb/ft3, multiply lb/in.3 by 1728; for g/cm3, multiply density in lb/in.3 by 27.68; for kg/m3, multiply density in lb/in.3 by 27679.9
| | | |
---
| Metal / Element or Alloy Density | Density g/cm3 | Density kg/m3 |
| Actinium | | |
| Admiralty Brass | | |
| Aluminum | | |
| Aluminum - 1100 | | |
| Aluminum - 6061 | 2.7 | 2720 |
| Aluminum - 7050 | | |
| Aluminum - 7178 | 2.8 | |
| Aluminum bronze (3-10% Al) | 7.8 - 8.6 | 7800 - 8650 |
| Aluminum foil | 2.7 | |
| Antimony | | |
| Babbitt | | |
| Barium | 3.62 | 3595 |
| Beryllium | 1.85 | 1850 |
| Beryllium Copper | 8.5 | 8500 |
| Bismuth | 9.79 | 9790 |
| Brass - casting | 8.5 | 8500 |
| Brass - rolled and drawn | 8.5 | 8500 |
| Brass 60/40 | 8.52 | 8520 |
| Bronze - lead | 7.7 - 8.7 | 7700 - 8700 |
| Bronze - phosphorous | 8.7 - 8.9 | 8700 - 8900 |
| Bronze (8-14% Sn) | 7.4 - 8.9 | 7400 - 8900 |
| Cadmium | 8.69 | 8690 |
| Caesium | 1.87 | 1870 |
| Calcium | 1.54 | 1540 |
| Cast Iron | 6.85 - 7.75 | 6850 - 7750 |
| Cerium | 6.77 | 6770 |
| Cesium | 1.93 | 1930 |
| Chromium | 7.15 | 7150 |
| Cobalt | 8.86 | 8860 |
| Constantan | 8.9 | 8900 |
| Columbium | 8.55 | 8550 |
| Constantan | 8.8 | 8800 |
| Copper | 8.96 | 8960 |
| Cupronickel | 8.9 | 8900 |
| Duralumin | 2.78 | 2780 |
| Dysprosium | 8.55 | 8550 |
| Electrum | 8.5 - 8.8 | 8500 - 8800 |
| Erbium | 9.07 | 9070 |
| Europium | 5.24 | 5240 |
| Gadolinium | 7.90 | 7900 |
| Gallium | 5.91 | 5910 |
| Germanium | 5.3 | 5300 |
| Gold | 19.3 | 19300 |
| Hafnium | 13.3 | 13300 |
| Hastelloy | 9.25 | 9250 |
| Holmium | 8.80 | 8800 |
| Indium | 7.31 | 7310 |
| Inconel | 8.5 | 8500 |
| Incoloy | 8.03 | 8003 |
| Iridium | 22.5 | 22500 |
| Iron | 7.87 | 7870 |
| Lanthanum | 6.15 | 6150 |
| Lead | 11.3 | 11,300 |
| Lithium | 0.53 | 530 |
| Lutetium | 9.84 | 9840 |
| Magnesium | 1.74 | 1740 |
| Manganese | 7.3 | 7300 |
| Manganese Bronze | 8.37 | 8730 |
| Manganin | 8.55 | 8550 |
| Mercury | 13.53 | 13530 |
| Molybdenum | 10.2 | 10200 |
| Monel | 8.37 - 8.82 | 8370 - 8820 |
| Neodymium | 7.01 | 7010 |
| Neptunium | 20.2 | 20200 |
| Nichrome | 8.45 | 8450 |
| Nickel | 8.90 | 8900 |
| Nickeline | 8.7 | 8700 |
| Nimonic | 8.1 | 8100 |
| Niobium | 8.57 | 8570 |
| Osmium | 22.59 | 22590 |
| Palladium | 12.0 | 12000 |
| Phosphor Bronze | 8.9 | 8900 |
| Platinum | 21.5 | 21500 |
| Plutonium | 19.7 | 19700 |
| Polonium | 9.20 | 9200 |
| Potassium | 0.89 | 890 |
| Praseodymium | 6.77 | 6770 |
| Promethium | 7.26 | 7260 |
| Protactinium | 15.4 | 1540 |
| Radium | 5 | 500 |
| Red Brass | 8.75 | 8720 |
| Rhenium | 20.8 | 20800 |
| Rhodium | 12.4 | 12400 |
| Rubidium | 1.53 | 1530 |
| Ruthenium | 12.1 | 12100 |
| Samarium | 7.52 | 7520 |
| Scandium | 2.99 | 2990 |
| Silver | 10.5 | 10500 |
| Sodium | 0.97 | 970 |
| Solder 50/50 Pb Sn | 8.88 | 8880 |
| Stainless Steel | 7.48 - 7.950 | 7480 - 7950 |
| Steel | 7.860 | 7860 |
| Strontium | 2.64 | 2640 |
| Tantalum | 16.4 | 16400 |
| Technetium | 11 | 11000 |
| Terbium | 8.23 | 8230 |
| Thallium | 11.8 | 11800 |
| Thorium | 11.7 | 11700 |
| Thulium | | |
| Tin | 7.26 | 7260 |
| Titanium | | |
| Tungsten | 19.3 | 19300 |
| Uranium | | |
| Vanadium | | |
| White Metal | | |
| Wrought Iron | | |
| Yellow Brass | | |
| Ytterbium | | |
| Yttrium | | |
| Zinc | | |
| Zirconium | | |
Air Density and Specific Weight Table, Equations and Calculator
Density Equation
| |
| |
| |
| Home Engineering Book Store Engineering Forum Applications and Design Beam Deflections and Stress Bearing Apps, Specs & Data Belt Design Data Calcs Civil Engineering Design & Manufacturability Electric Motor Alternators Engineering Calculators Engineering Terms Excel App. Downloads Flat Plate Stress Calcs Fluids Flow Engineering Friction Engineering Gears Design Engineering General Design Engineering Hardware, Imperial, Inch Hardware, Metric, ISO Heat Transfer Hydraulics Pneumatics HVAC Systems Calcs Economics Engineering Electronics Instrumentation Engineering Mathematics Engineering Standards Finishing and Plating Friction Formulas Apps Lubrication Data Apps Machine Design Apps Manufacturing Processes Materials and Specifications Mechanical Tolerances Specs Plastics Synthetics Power Transmission Tech. Pressure Vessel Pumps Applications Re-Bar Shapes Apps Section Properties Apps Strength of Materials Spring Design Apps Structural Shapes Threads & Torque Calcs Thermodynamics Physics Vibration Engineering Videos Design Manufacture Volume of Solids Calculators Welding Stress Calculations Training Online Engineering --- Copyright Notice | |
189056 | https://www.fishbase.se/summary/speciessummary.php?id=14417 | This website uses cookies to enhance your browsing experience and ensure the functionality of our site. For more detailed information about the types of cookies we use and how we protect your privacy, please visit our Privacy Information page..
Accept All Accept Necessary Cookies Only Cookie Settings
×
Cookie Settings
This website uses different types of cookies to enhance your experience. Please select your preferences below:
Strictly Necessary- [x]
These cookies are essential for the website to function properly. They include session cookies, which help maintain your session while you navigate the site, as well as cookies that remember your language preferences and other essential functionalities. Without these cookies, certain features of the website cannot be provided.
Performance- [x]
These cookies help us understand how visitors interact with our website by collecting and reporting information anonymously. For example, we use Google Analytics to generate web statistics, which helps us improve our website's performance and user experience. These cookies may track information such as the pages visited, time spent on the site, and any errors encountered.
Save and Close
Latris lineata, Striped trumpeter : fisheries, aquaculture
You can sponsor this page
About this page
More Info
Plus d'info
Mais info
Languages
Arabic
Bahasa/Malay
Bangla
Chinese(Si)
Chinese(Tr)
Deutsch
English
Español
Farsi
Français
Greek
Hindi
Italiano
Japanese
Lao
Nederlands
Português(Br)
Português(Pt)
Russian
Swedish
Thai
Vietnamese
User feedbacks
Comments & Corrections
Fish Forum
Guest Book
Facebook
Citation
Uploads
Attach website
Upload photo
Upload video
Upload references
Fish Watcher
Related species
Species in Latris
Species in Latridae
Classification -
Latridae
Centrarchiformes
Teleostei
Chordata
Animalia
Latrislineata (Forster, 1801)
Striped trumpeter
Add your observation in Fish Watcher
Native range | All suitable habitat | Point map | Year 2050
This map was computer-generated and has not yet been reviewed.
Latris lineataAquaMaps Data sources: GBIFOBIS
Upload your photosandvideos
Pictures | Google image
Latris lineata
Picture by CSIRO
Classification / Names Common names | Synonyms | Catalog of Fishes(genus, species) | ITIS | CoL | WoRMS | Cloffa
Teleostei (teleosts) >Centrarchiformes (Basses) >Latridae (Trumpeters)
Etymology: Latris:Greek, latris = slave (Ref. 45335);lineata:Named for the on the distinctive dark lines running longitudinally across the body.
More on author: Forster.
Environment: milieu / climate zone / depth range / distribution range Ecology
Marine; reef-associated; depth range 5 - 400 m (Ref. 76767). Temperate; 17°S - 56°S, 13°W - 175°W
Distribution Countries | FAO areas | Ecosystems | Occurrences | Point map | Introductions | Faunafri
Southern Hemisphere: Atlantic Ocean (from Tristan da Cunha and Gough I.); Indian Ocean (Walter Shoals and Saint Paul's I. and Amsterdam Is., Western Australia, across South Australia to southern New South Wales and Tasmania; Pacific Ocean (New Zealand, Bay of Plenty, east to the Chatham Is, and south to the Snares Is. and the subantarctic Auckland Is.). Latris hecateia has been listed in Chile, but records are questionable (Ref. 76767).
Length at first maturity / Size / Weight / Age
Maturity: L m54.3, range 51 - ? cm
Max length : 120 cm TL male/unsexed; (Ref. 9563); max. published weight: 25.0 kg (Ref. 26433); max. reported age: 43 years (Ref. 53584)
Short description Identification keys | Morphology | Morphometrics
Dorsalspines (total): 17 - 19; Dorsalsoft rays (total): 33 - 38; Analspines: 3; Analsoft rays: 26 - 30; Vertebrae: 37. This species can be distinguished from all other latrids and cheilodactylids (and most percoids) by having anal fin soft rays 26-30; vertebrae 16+21 = 37; a striped body colour pattern comprising three dark bands running longitudinally on the upper body and onto the opercular series and head, with a broader faint band on the flank and a narrow dorsal median band running along the nape and head in front of the dorsal fin (Ref. 76767).
Body shape(shape guide): fusiform / normal.
Biology Glossary (e.g. epibenthic)
Adults apparently prefer broken rocky reefs, rich in 'coral' and other invertebrate life. Reported to be frequently found over rocky places, especially narrow channels between two cliffs in fjords of New Zealand. May spawn closer to shore (Ref. 9563) but probably in areas of deeper waters, otherwise they seem to migrate offshore to spawn (Ref. 85801). Collected by bottom trawls, dropline, and rod and line (Ref. 76767). In Australia, adults are found over rocky reefs at depths from 5 to 300 m (Ref. 53584).
Life cycle and mating behavior Maturity | Reproduction | Spawning | Eggs | Fecundity | Larvae
Females are multiple spawners with group synchronous oocyte development (Ref. 95206).
Main reference Upload your references | References | Coordinator | Collaborators
Roberts, C.D., 2003. A new species of trumpeter (Teleostei; Percomorpha; Latridae) from the central South Pacific Ocean, with a taxonomic review of the striped trumpeter Latris lineata. J. R. Soc. N. Z. 33(4):731-754. (Ref. 76767)
IUCN Red List Status (Ref. 130435: Version 2025-1)
Not Evaluated
CITES
Not Evaluated
CMS (Ref. 116361)
Not Evaluated
Threat to humans
Harmless
Human uses
Fisheries: commercial; aquaculture: experimental
FAO - Publication: search | FishSource |
More information
Trophic ecology
Food items (preys)
Diet composition
Food consumption
Food rations
Predators
Ecology
Ecology
Home ranges
Population dynamics
Growth parameters
Max. ages / sizes
Length-weight rel.
Length-length rel.
Length-frequencies
Mass conversion
Recruitment
Abundance
Life cycle
Reproduction
Maturity
Maturity/Gills rel.
Fecundity
Spawning
Spawning aggregations
Eggs
Egg development
Larvae
Larval dynamics
Distribution
Countries
FAO areas
Ecosystems
Occurrences
Introductions
BRUVS - Videos
Anatomy
Gill area
Brain
Otolith
Physiology
Body composition
Nutrients
Oxygen consumption
Swimming type
Swimming speed
Visual pigments
Fish sound
Diseases & Parasites
Toxicity (LC50s)
Genetics
Genome
Genetics
Heterozygosity
Heritability
Human related
Aquaculture systems
Aquaculture profiles
Strains
Ciguatera cases
Stamps, coins, misc.
Outreach
Collaborators
Taxonomy
Common names
Synonyms
Morphology
Morphometrics
Pictures
References
References
Tools
E-book | Field guide | Length-frequency wizard | Life-history tool | Point map | Catch-MSY |
Special reports
Check for Aquarium maintenance | Check for Species Fact Sheets | Check for Aquaculture Fact Sheets
Download XML
Summary page | Point data | Common names | Photos
Internet sources
AFORO (otoliths) | Aquatic Commons | BHL | Cloffa | Websites from users | Check FishWatcher | CISTI | Catalog of Fishes: genus, species | DiscoverLife | ECOTOX | FAO - Publication: search | Faunafri | Fishipedia | Fishtrace | GloBI | Google Books | Google Scholar | Google | IGFA World Record | OneZoom | Open Tree of Life | Otolith Atlas of Taiwan Fishes | PubMed | Reef Life Survey | Socotra Atlas | TreeBase | Tree of Life | Wikipedia: Go, Search | Zoological Record
Estimates based on models
Preferred temperature (Ref. 123201): 10.6 - 21.2, mean 14.9 °C (based on 181 cells).
Phylogenetic diversity index (Ref. 82804):PD 50 = 0.7500 [Uniqueness, from 0.5 = low to 2.0 = high].
Bayesian length-weight: a=0.01349 (0.00825 - 0.02206), b=3.05 (2.90 - 3.20), in cm total length, based on LWR estimates for this species & (Sub)family-body (Ref. 93245).
Trophic level (Ref. 69278):3.5 ±0.2 se; based on size and trophs of closest relatives
Generation time: 15.8 ( na - na) years. Estimated as median ln(3)/K based on 2growth studies.
Resilience (Ref. 120179):Very Low, minimum population doubling time more than 14 years (K=0.06-0.4; tmax=43; tm=5-7; Fec=205,054 (batch)).
Fishing Vulnerability (Ref. 59153):High to very high vulnerability (67 of 100). 🛈
Price category (Ref. 80766):Very high.
Random Species
Back to Search
Comments & Corrections
Back to Top
Accessed through: 14417
FishBase mirror site : localhost
Page last modified by : mrius-barile - 20 July 2016 Total processing time for the page : 0.0313 seconds
Search FishBaseclose
Common name (e.g. trout)
Genus + Species (e.g. Gadus morhua)
close |
189057 | https://www.pa.uky.edu/~crawford/phy416_fa14/notes_01_1a.pdf | Section 1.1 - Vector Algebra Linear (vector) space ~ linear combination: ( ) is the basic operation ~ basis: ( or ) # basis elements = dimension independence: not collapsed into lower dimension closure: vectors span the entire space ~ components: in matrix form: Metric (inner, dot product) - distance and angle ~ orthonormality and completeness - two fundamental identities help to calculate components, implicitly in above formulas ~ properties: 1) scalar valued - what is outer product? 2) bilinear form 3) symmetric ~ orthogonal projection: a vector divides the space into geometric view: dot product is length of along Projection operator: acts on x: ~ generalized metric: for basis vectors which are not orthonormal, collect all nxn dot products into a symmetric matrix (metric tensor) in the case of a non-orthonormal basis, it is more difficult to find components of a vector, but it can be accomplished using the reciprocal basis (see HW1) ~ all other structure is added on as multilinear (tensor) extensions ~ Einstein notation: implicit summation over repeated indices ~ direct sum: add one vector from each independent space to get vector in the product space (not simply union) ~ projection: the vector has a unique decomposition (’coordinates’ in ) - relation to basis/components?
=0 =0 1 1 -1 -1 -2 -2 2 2 where etc (usually one upper, one lower index) whole space Kroneker delta: components of the identity matrix Exterior Products - higher-dimensional objects cross product (area) ~ properties: 1) vector-valued 2) bilinear 3) antisymmetric ~ components: ~ orthogonal projection: projects to and rotates by 90 ~ where is the metric in x? vector x vector = pseudovector symmetries act more like a ’bivector ‘ can be defined without metric triple product (volume of parallelpiped) - base times height ~ completely antisymmetric - definition of determinant ~ why is the scalar product symmetric / vector product antisymmetric? ~ vector vector x vector = pseudoscalar (transformation properties) ~ acts more like a ’trivector ‘ (volume element) ~ again, where is the metric? (not needed!) exterior algebra (Grassman, Hamilton, Clifford) ~ extended vector space with basis elements from objects of each dimension ~ pseudo-vectors, scalar separated from normal vectors, scalar magnitude, length, area, volume scalar, vectors, bivectors, trivector ~ what about higher-dimensional spaces (like space-time)? can’t form a vector ‘cross-product’ like in 3-d, but still have exterior product ~ all other products can be broken down into these 8 elements most important example: BAC-CAB rule (HW1: relation to projectors) Levi-Civita tensor - completely antisymmetric: where (RH-rule) where (parallel) (oriented) ijk even permutation ijk odd permutation repeated index (ijk cyclic) |
189058 | https://www.albert.io/blog/first-order-differential-equations-what-to-know-for-studying-calculus/ | Skip to content
First Order Differential Equations: What to Know for Studying Calculus
First-order differential equations are equations involving some unknown function and its first derivative. The main purpose of this Calculus III review article is to discuss the properties of solutions of first-order differential equations and to describe some effective methods for finding solutions.
Standard Form
The standard form for a first-order differential equation for the unknown function y(t) is:
\dfrac{dy}{dt} = f(t,y)
Here, f is some function of two variables. Many, but not all, first-order differential equations can be written in standard form by algebraically solving for \dfrac{dy}{dt} and then setting f(t,y) equal to the right side of the resulting equation.
Any differentiable function y = y(t) that satisfies this equation for all t in some interval is called a solution. Some differential equations have no solutions, whereas other differential equations have infinitely many solutions. It is also possible that a differential equation has exactly one solution. The general solution of a differential equation is the set of all solutions. A differential equation along with a subsidiary condition y(t_0) = y_0 , given at some value of the independent variable t = t_0 , constitutes an initial value problem. A solution to an initial value problem is a function y(t) that both solves the differential equation and satisfies the given subsidiary condition y(t_0) = y_0 .
Equations with Separable Variables
There is no universally applicable procedure for solving first-order differential equations in standard form with an arbitrary f(t,y) . Here, we consider a subset of first-order equations that can be directly integrated.
This is possible if the function
can be presented in the form
f(t,y) = g(t) h(y)
Here, g is a function of only t , and h is a function of only y . The differential equation
\dfrac{dy}{dt} = g(t) h(y)
is said to be separable. We can write it in the differential form
\dfrac{dy}{h(y)} - g(t) dt = 0
The general solution of this equation is given by the following integral:
\int \dfrac{dy}{h(y)} - \int g(t) dt = C
Here, C represents the arbitrary constant of integration. The integrals obtained in this expression may be impossible to evaluate. In such a case, numerical methods can be used to obtain an approximate solution. Even if the indicated integrations can be performed, it may not be algebraically possible to solve for y explicitly in terms of t . In that case, the solution is left in implicit form.
Example 1
Let’s solve the equation:
\dfrac{dy}{dt} = \dfrac{t^2+2}{y}
This first-order differential equation is separable, with
g(t) = t^2+2 and h(y) = \dfrac{1}{y} .
Its solution is
\int y dy - \int (t^2+2) dt = C
\dfrac{y^2}{2} - \dfrac{t^3}{3} - 2 t = C
Solving for the function y explicitly, we obtain two solutions:
y = \sqrt{\dfrac{2t^3}{3} + 4t + K} \qquad \text{and} \qquad y = -, \sqrt{\dfrac{2t^3}{3} + 4t + K}
The constant K is related to the constant
C via K=2C.
Example 2
Consider the following initial value problem:
\dfrac{dy}{dt} = \dfrac{y cos t}{1+2y^2} ,, \qquad y(0) = 1
To find the solutions, assume that y neq 0 and write the differential equation in the form:
\int \dfrac{1+2y^2}{y} ,dy - \int cos t, dt = 0
Then, integrating the left side with respect to y and the right side with respect to t , we obtain:
ln|y| + y^2 = sin t + C
To satisfy the initial condition, we substitute t=0 and y=1 ; this gives C=1 . Hence, we can present a solution in the form:
ln|y| + y^2 = sin t + 1
Homogeneous Equations
A differential equation in standard form
\dfrac{dy}{dt} = f(t,y)
is homogeneous if
f(\alpha t,\alpha y) = f(t,y)
for any real number \alpha .
Such an equation can always be transformed into a separable equation by the change of an independent variable
y = t,z
along with its corresponding derivative:
\dfrac{dy}{dt} = t ,\dfrac{dz}{dt} + z
The resulting equation with the variables z and t
can be solved as a separable differential equation, because the function f
after such a substitution appears to be a function with a single variable, z .
Let’s illustrate this method through examples.
Example 1
Consider the equation:
\dfrac{dy}{dt} = \dfrac{y+t}{t}
First, we verify the condition for homogeneity:
f(\alpha t,\alpha y) = \dfrac{\alpha y+\alpha t}{\alpha t} = \dfrac{y+t}{t} = f(t, y)
Second, we introduce a new dependent variable,
z , so that z=y/t :
\dfrac{dy}{dt} = t ,\dfrac{dz}{dt} + z = z+1
The equation
t,\dfrac{dz}{dt} = 1
is a first-order differential equation with separable variables, which can be directly integrated:
z = ln|Kt|
Here, we have set the constant of integration
C=-,ln|K| ,
and have noted that
ln|t| + ln|K| = ln|Kt| .
Finally, substituting
z=y/t
back into this solution, we obtain the solution to the original differential equation as
y=tln|Kt| .
Example 2
The following equation is also homogeneous:
\dfrac{dy}{dt} = \dfrac{t^2+3y^2}{2ty}
For the variable
z=y/t ,
we have:
t ,\dfrac{dz}{dt} = \dfrac{1+z^2}{2z}
Integrating, we get the following solution:
z^2 + 1 = C t,,\qquad C = \text{constant}
Now, replacing z by y/t ,
we obtain the solution for the original equation:
y = C t^3 - t^2
Linear Equations with Variable Coefficients
Consider a first-order differential equation in standard form:
\dfrac{dy}{dt} = f(t,y) .
A differential equation is linear if f(t,y)
can be written as a function of t times y,
plus another function of
t : f(t,y) = - ,p(t) y + q(t) .
Consequently, a linear differential equation can always be expressed as:
\dfrac{dy}{dt} + p(t) y = q(t)
Here, p and q
are given functions of the independent variable t .
First-order linear differential equations cannot be solved by straightforward integration methods,because the variables are not separable.As a result, we need to use a different method of solution. The first step is to multiply the linear differential equation by an undetermined function, \mu(t) :
\mu(t) ,\dfrac{dy}{dt} + \mu(t) p(t) y = \mu(t) q(t)
The question now is whether we can choose \mu(t) so that the left side of this equation is recognizable as the derivative of some particular expression. Let’s note the following equalities:
\dfrac{d}{dt} [\mu(t)y] = \mu(t) \dfrac{dy}{dt} + \dfrac{d\mu(t)}{dt},y = \mu(t) ,\dfrac{dy}{dt} + \mu(t) p(t) y
Here, the second equality is valid under the condition that \mu(t) satisfies the equation:
\dfrac{d\mu(t)}{dt} = p(t) \mu(t)
This is a first-order differential equation with separable variables that can be directly integrated:
\int \dfrac{d\mu}{\mu} - \int p(t) dt = 0
If we assume temporarily that \mu(t) is positive, then we have:
ln\mu(t) = \int p(t) dt + K
By choosing the arbitrary constant K to be zero, we obtain the simplest possible function for \mu .Namely:
\mu(t) = exp\left(\int p(t) dt\right)
Note that the integrating factor \mu(t)
is positive for all t ,
as we had assumed. Returning to the linear differential equation, we have:
\dfrac{d}{dt} [\mu(t)y] = \mu(t) q(t)
Hence, the general solution is:
y = \dfrac{\int \mu(s) q(s) ds + C}{\mu(t)}
Observe that two integrations are required to find the solution of a linear differential equation: one to obtain the integrating factor \mu(t) , and the other to obtain y .
We will now consider some examples.
Example 1
Let’s solve the differential equation:
\dfrac{dy}{dt} + \dfrac{4}{t},y = t^4
This differential equation is a first-order linear differential equation, with p(t) = 4/t and q(t) = t^4 . We can determine
\int p(t) dt = \int \dfrac{4}{t} ,dt = 4 ln|t| = ln t^4
So, the integrating factor in this case is:
\mu(t) = exp\left(\int p(t) dt\right) = e^{ln t^4} = t^4
Multiplying the differential equation by the integrating factor, we get:
t^4 \dfrac{dy}{dt} + 4 t^3 ,y = t^8 \qquad \text{or} \qquad \dfrac{d}{dt} (t^4 y) = t^8
Integrating both sides of this last equation with respect to t , we obtain the solution:
y = \dfrac{t^5}{9} + \dfrac{C}{t^4}
Example 2
Consider the following initial value problem:
t, \dfrac{dy}{dt} + 2y = 4 t^2,, \qquad y(1) = 2
Dividing both sides by t , we can write this equation in standard form (for linear equations):
\dfrac{dy}{dt} + \dfrac{2}{t},y = 4 t,, \qquad y(1) = 2
Thus, p(t) = 2/t and q(t) = 4 t . To solve this differential equation, we first compute the integrating factor:
\mu(t) = exp\left(\int \dfrac{2}{t} dt\right) = e^{2ln |t|} = t^2
Uponmultiplying the equation by \mu(t) = t^2 , we obtain
t^2, \dfrac{dy}{dt} + 2ty = \dfrac{d}{dt} (t^2 y) = 4 t^3
Therefore:
y = t^2 + \dfrac{C}{t^2} ,,\qquad C=\text{constant}
is the general solution of the initial differential equation. To satisfy the initial condition y(1) = 2 , it is necessary to choose C=1 .
Thus, y = t^2 + \dfrac{1}{t^2}
Bernoulli Equations
In some cases, we can solve a nonlinear equation by changing the dependent variable so that the equation becomes linear. Here, we will investigate the equation that has the form:
\dfrac{dy}{dt} + p(t) y = q(t) y^n
Here, n is supposed to be a real number, restricted by the condition n\neq 0, 1 . Such an equation is called a Bernoulli equation, after Jacob Bernoulli. The substitution z=y^{1-n} reduces a Bernoulli equation to a linear equation. The practical use of this method will be demonstrated in the examples.
Example 1
Let’s solve the differential equation:
\dfrac{dy}{dt} + t y = t y^2
This equation is not linear. It is, however, a Bernoulli differential equation, with p(t)=q(t)=t and n=2 . The required substitution in this case is:
z = y^{1-2} = \dfrac{1}{y},, \qquad \dfrac{dy}{dt} = -,\dfrac{1}{z^2} ,\dfrac{dz}{dt}
Substituting these relations into the initial equation, we obtain:
\dfrac{dz}{dt} - t z = - t
This last equation is linear in the unknown function z . The integrating factor is:
\mu(t) = exp\left(\int(-t)dt \right) = e^{-t^2/2}
Multiplying the differential equation by the integrating factor, we obtain:
\dfrac{d}{dt}\left(z e^{-t^2/2}\right) = -te^{-t^2/2}
The solution of the differential equation is then:
z(t) = C e^{t^2/2} + 1,, \qquad y = \dfrac{1}{z} = \left( C e^{t^2/2} + 1 \right)^{-1}
Example 2
Consider the equation:
\dfrac{dy}{dt} + \dfrac{2y}{t} = \dfrac{y^3}{t^2}
This equation is a Bernoulli differential equation, with
p=2/t , q=1/t^2 and n=3 .
So, we make the substitution:
z = y^{1-3} = \dfrac{1}{y^2},, \qquad \dfrac{dy}{dt} = -,\dfrac{1}{2z^{3/2}} ,\dfrac{dz}{dt}
The differential equation for the variable z is:
\dfrac{dz}{dt} - \dfrac{4}{t} z = - ,\dfrac{2}{t^2}
For this linear differential equation, we can define the following integrating factor:
\mu(t) = exp\left(-,\int\dfrac{4}{t}dt \right) = \dfrac{1}{t^4}
Then, the solution for z(t) is:
\dfrac{d}{dt}\left(\dfrac{z}{t^4}\right) = -,\dfrac{2}{t^6},, \qquad z(t) = \dfrac{2}{5t} + Ct^4,,\qquad C=\text{const}
Finally, we can present the solution for the original differential equation in the form:
y = pm\sqrt{\dfrac{5t}{2+5Ct^5}}
Exact Equations and Integrating Factors
Consider the following first-order differential equation:
m(t,y) + n(t,y) ,\dfrac{dy}{dt} = 0
We suppose it is neither linear nor separable, so the methods suitable for those types of equations are not applicable here. But suppose that we can identify a function Psi(t,y) such that:
\dfrac{\partial Psi}{\partial t} = m(t,y) ,,\qquad \dfrac{\partial Psi}{\partial y} = n(t,y)
Then, the differential equation becomes:
m(t,y) + n(t,y) ,\dfrac{dy}{dt} = \dfrac{\partial Psi}{\partial t} + \dfrac{\partial Psi}{\partial y},\dfrac{dy}{dt} = \dfrac{d}{dt} ,Psi[t, y(t)] = 0
Such a differential equation is called an exact equation. The solution for an exact equation is given implicitly by
Psi(t,y) = C ,
where, as usual, C
represents an arbitrary constant.
A systematic way of determining whether a given differential equation is exact is provided by the following test. If m(t,y) and n(t,y) are continuous functions, then a first-order differential equation of the form:
m(t,y) + n(t,y) ,\dfrac{dy}{dt} = 0
is exact if and only if:
\dfrac{\partial m(t,y)}{\partial y} = \dfrac{\partial n(t,y)}{\partial t}
In some cases, a differential equation that is not exact can be transformed into an exact equation. Such a transformation is possible if we multiply the equation by a suitable integrating factor. To investigate the possibility of implementing this idea more generally, let us multiply the equation by a function \mu , then try to choose \mu so that the resulting equation:
\mu(t,y) m(t,y) + \mu(t,y) n(t,y) ,\dfrac{dy}{dt} = 0
passes the test of exactness:
\dfrac{\partial}{\partial y}[\mu m] = \dfrac{\partial}{\partial t}[\mu n]
While in principle, integrating factors are powerful tools for solving differential equations, in practice they can be found only in special cases. The most important situation in which integrating factors can be found occurs when \mu is a function of only one of the variables t or y , instead of both. Assuming that \mu is a function of only t , we have:
\dfrac{d\mu}{dt} = g \mu ,, \qquad g = \dfrac{1}{n}\left( \dfrac{\partial m}{\partial y} - \dfrac{\partial n}{\partial t} \right)
If g is a function of only t , then the integrating factor \mu can be defined as:
\mu(t) = exp\left(\int g(t) dt\right)
A similar procedure can be used to determine the condition under which the differential equation has an integrating factor dependent only on y . In this case, we find:
\mu(t) = exp\left(-,\int h(t) dt\right),, \qquad \text{where} \qquad \dfrac{1}{m}\left( \dfrac{\partial m}{\partial y} - \dfrac{\partial n}{\partial t} \right) = h(y)
Example 1
Solve the differential equation:
2ty + (1+t^2) \dfrac{dy}{dt} = 0
This equation has the standard form with
m(t,y) = 2ty and n(t,y) = 1+t^2 .
We can easily verify that:
\dfrac{\partial m(t,y)}{\partial y} = \dfrac{\partial n(t,y)}{\partial t} = 2t
So, the differential equation is exact. Thus, there is a Psi(t,y) such that:
\dfrac{\partial Psi}{\partial t} = m (t,y) = 2ty ,, \qquad \dfrac{\partial Psi}{\partial y} = n (t,y) = 1+t^2
Integrating the first of these equations, we obtain:
Psi(t,y) = t^2 y + \varphi(y)
Note that when integrating with respect to t , the constant (with respect to t) of integration can depend on y . We can determine \varphi(y) by differentiating Psi(t,y) with respect to y :
\dfrac{\partial Psi}{\partial y} = t^2 + \dfrac{d\varphi}{dy} = 1+t^2,, \qquad \dfrac{d\varphi}{dy} = 1
Integrating this last equation with respect to
y , we obtain \varphi(y) = y + C_1 ( C_1 = \text{constant} ).
So, we have
Psi(t,y) = (1+t^2) y + C_1
Finally, the solution to the differential equation, which is given implicitly as
Psi(t,y)=C , is:
y = \dfrac{C_2}{1+t^2},,\qquad C_2 = C - C_1
Example 2
Consider the first-order differential equation:
(3ty + y^2) + (t^2+ty) \dfrac{dy}{dt} = 0
This equation has the standard form with
m = 3ty + y^2 and n = t^2+ty .
We can easily calculate:
\dfrac{\partial m(t,y)}{\partial y} - \dfrac{\partial n(t,y)}{\partial t} = (3t+2y) - (2t+y) = t+y
The equation is not exact, but:
g = \dfrac{1}{n}( \dfrac{\partial m}{\partial y} - \dfrac{\partial n}{\partial t} ) = \dfrac{t+y}{t^2+ty} = \dfrac{1}{t}
is a function of t only. Thus, there is an integrating factor \mu that also depends only on t :
\mu(t) = exp\left(\int g(t) dt\right) = exp\left({\int \dfrac{dt}{t}}\right) = t
Multiplying the differential equation by this integrating factor, we obtain:
(3t^2y + ty^2) + (t^3+t^2y) \dfrac{dy}{dt} = 0
We can easily verify that this differential equation is exact. Thus, there is a Psi(t,y) such that:
\dfrac{\partial Psi}{\partial t} = 3t^2y + ty^2 ,, \qquad \dfrac{\partial Psi}{\partial y} = t^3+t^2y
Integrating this equation, we obtain:
Psi(t,y) = t^3 y + \dfrac{1}{2},t^2 y^2 + \varphi(t)
Differentiating Psi(t,y) with respect to t , we have
\dfrac{\partial Psi}{\partial y} = 3t^2y + ty^2 + \dfrac{d\varphi(t)}{dt} = 3t^2y + ty^2 \quad \Rightarrow \quad \dfrac{d\varphi}{dt} = 0
Finally, we can write the solution:
t^3 y + \dfrac{1}{2},t^2 y^2 = C
Wrapping Everything Up
In this calculus III review article, we have investigated different types of first order differential equations that can be taken. Now, you will be able to determine the type of differential equation, then apply the correct method for solving it. We hope this post gives you greater confidence in your knowledge of first order differential equations and facilitates your study of Calculus III.
Let’s put everything into practice. Try this Differential Equations practice question:
Looking for more Differential Equations practice?
You can find thousands of practice questions on Albert.io. Albert.io lets you customize your learning experience to target practice where you need the most help. We’ll give you challenging practice questions to help you achieve mastery in Differential Equations.
Start practicing here.
Are you a teacher or administrator interested in boosting Differential Equations student outcomes?
Learn more about our school licenses here. |
189059 | https://openoregon.pressbooks.pub/biochemistry/chapter/2-2-structure-function-amino-acids-biology-libretexts/ | Skip to content
2.2: Structure and Function – Amino Acids
All of the proteins on earth are made up of the same 20 amino acids. Linked together in long chains called polypeptides, amino acids are the building blocks for the vast assortment of proteins found in all living cells.
“It is one of the more striking generalizations of biochemistry …that the twenty amino acids and the four bases, are, with minor reservations, the same throughout Nature.” – Francis Crick
All amino acids have the same basic structure, shown in Figure 2.1. At the center of each amino acid is a carbon called the α carbon and attached to it are four groups – a hydrogen, a carboxylic acid group, an amine group, and an R-group, sometimes referred to as a variable group or side chain. The α carbon, carboxylic acid, and amino groups are common to all amino acids, so the R-group is the only variable feature. With the exception of glycine, which has an R-group consisting of a hydrogen atom, all of the amino acids in proteins have four different groups attached to them and consequently can exist in two mirror isomeric forms.
The designations used in organic chemistry are not generally applied to amino acid nomenclature, but a similar system uses L and D to describe these enantiomers. Nature has not distributed the stereoisomers of amino acids equally. Instead, with only very minor exceptions, every amino acid found in cells and in proteins is in the L configuration.
Figure 2.1 – General amino acid structure
There are 22 amino acids that are found in proteins and of these, only 20 are specified by the universal genetic code. The others, selenocysteine and pyrrolysine use tRNAs that are able to base pair with stop codons in the mRNA during translation. When this happens, these unusual amino acids can be incorporated into proteins. Enzymes containing selenocysteine, for example, include glutathione peroxidases, tetraiodothyronine 5′ deiodinases, thioredoxin reductases, formate dehydrogenases, glycine reductases, and selenophosphate synthetase. Pyrrolysine-containing proteins are much rarer and are mostly confined to archaea.
Essential and non-essential
Nutritionists divide amino acids into two groups – essential amino acids and non-essential amino acids. Essential amino acids must be included in our diet because our cells can’t synthesize them. What is essential varies considerably from one organism to another and even differ in humans, depending on whether they are adults or children. Table 2.1 shows essential and non-essential amino acids in humans.
Some amino acids that are normally nonessential, may need to be obtained from the diet in certain cases. Individuals who do not synthesize sufficient amounts of arginine, cysteine, glutamine, proline, selenocysteine, serine, and tyrosine, due to illness, for example, may need dietary supplements containing these amino acids.
Table 2.1 – Essential and non-essential amino acids
Non-protein amino acids
There are also amino acids found in cells that are not incorporated into proteins. Common examples include ornithine and citrulline. Both of these compounds are intermediates in the urea cycle, an important metabolic pathway.
R-group chemistry
Table 2.2 – Amino acid categories (based on R-group properties)
Amino acids can be classified based on the chemistry of their R-groups. It is useful to classify amino acids in this way because it is these side chains that give each amino acid its characteristic properties. Thus, amino acids with (chemically) similar side groups can be expected to function in similar ways, for example, during protein folding. The specific category divisions may vary, but all systems are attempts to organize and understand the relationship between an amino acid’s structure and its properties or behavior as part of a larger system.
Non-polar amino acids
Amino acids in this group include:
Alanine (Ala/A)
Glycine (Gly/G)
Isoleucine (Ile/I)
Leucine (Leu/L)
Methionine (Met/M)
Valine (Val/V)
The amino acids in this group have nonpolar, hydrophobic R groups. When incorporated into globular proteins they tend to pack inward among other hydrophobic groups. In proteins that embed themselves into or through membranes, these amino acids can orient themselves toward hydrophobic portions of the inside of the membrane.
The small R groups here are more readily packed into tight formations. Proline is exceptional in that it has an R group that folds back and covalently bonds to the backbone of the amino acid, creating a more rigid element in a protein chain that reduces free movement of the polypeptide chain. Additionally, proline can undergo hydroxylation reactions, stabilizing the protein structure. This occurs in collagen with the aid of ascorbic acid (Vitamin C). One symptom of the vitamin C deficiency syndrome ‘scurvy’ is the reduced quality of collagen in tissues, including the skin and gums. This can lead to the deterioration and loss of teeth.
Figure 2.2 – Amino acid side chain properties Wikipedia
Figure 2.3 – Non-polar amino acids
Acidic Amino Acids (Carboxylic acid side chains)
Amino acids in this group include:
Aspartic acid (Asp/D)
Glutamic acid (Glu/E)
Figure 2.4 – Carboxyl amino acids
These amino acids each contain a carboxylic acid group as part of the variable group. At physiological pH, these groups exist primarily in their deprotonated state. It is easy to be confused if they are drawn in this state, because their names include “acid” while the structure shows no ionizable proton and the charge on the R group is negative.
In addition to its role as a building block in proteins, glutamic acid (with the deprotonated form named “glutamate”) is a neurotransmitter. It also is recognized by a receptor in our mouths, contributing to a taste sensation described as “umami.” Many foods contain appreciable amounts of glutamate that are recognized by our taste receptors, and encourage us to eat these substances. Those foods frequently contain protein that has broken down to some degree: cooked meats, fermented sauces like Worcestershire or soy, tahini, broths, and yeast extracts.
Basic amino acids (Nitrogen-containing side chains)
Included in this group of amino acids are:
Arginine (Arg/R)
Histidine (His/H)
Lysine (Lys/K)
Figure 2.5 – Amine amino acids
The variable group in each of these amino acids contains nitrogen, which imparts to the group the ability to exist in protonated and deprotonated states. They are frequently called basic, but also are often drawn in their protonated state which is more prevalent at physiological pH.
Arginine (Arg/R) is interesting due to the fact it is an essential dietary amino acid for premature infants, who cannot synthesize it. In addition, surgical trauma, sepsis, and burns increase demand for arginine and proper healing can require dietary intake.
Histidine contains a nitrogen-containing imidazole functional group that has a pKa of 6. This means it can pick up or donate hydrogen ions in response to small changes in pH. In proteins, histidine frequently has important roles participating directly in reactions involving hydrogen ion transfer.
The R group on lysine is frequently chemically modified in order for it to make unusual linkages to other chemical groups or to take part in specific chemical reactions. Lysine is often added to animal feed because it is a limiting amino acid and is necessary for optimizing growth of animals raised for consumption.
Aromatic amino acids
Figure 2.6 – Aromatic amino acids
Amino acids with aromatic side chains include:
Phenylalanine (Phe/ F)
Tryptophan (Trp/W)
Tyrosine (Tyr/Y)
These amino acids are included in protein structures but also serve as precursors in some important biochemical pathways, leading to the production of hormones such as L-Dopa and serotonin.
Hydroxyl amino acids
This group includes
Threonine (Thr/T)
Serine (Ser/S)
Tyrosine (already discussed as an aromatic amino acid)
The amino acids in this group contain alcohol groups, which can engage in hydrogen-bonding interactions. As part of protein molecules they are hydrophilic and can be oriented outward in watery environments. The alcohol group is subject to chemical reactions or modifications, for instance when carbohydrate groups are covalently linked to proteins.
Figure 2.7 – Hydroxyl amino acids
Figure 2.8 – Amino acid properties Wikipedia
Other amino acids
Asparagine (Asn/N) is a polar amino acid. The amide on the functional group is not basic.
Cysteine (Cys/C)
Glutamine (Gln/Q)
Figure 2.9 – Other amino acids
Cysteine, which contains a thiol. Thiols can react with one another via oxidation, forming disulfide links containing two covalently-linked sulfur atoms. Variable groups on methionine in protein chains can undergo such reactions, covalently tying the chains to one another with a short tether. Such disulfide links or bridges restrict the mobility of protein chains and contribute to more defined structures.
Selenocysteine (Sec/U) is a component of selenoproteins found in all kingdoms of life. Twenty five human proteins contain selenocysteine. It is a component in several enzymes, including glutathione peroxidases and thioredoxin reductases. It is not coded for by the standard genetic code.
Pyrrolysine (Pyl/O) is a twenty second amino acid, but is rarely found in proteins. Like selenocysteine, it is not coded for in the genetic code and must be incorporated by unusual means.
Ionizing groups
Some, but not all amino acids have R-groups that can ionize. The charge of a protein then arises from the charges of the amine group, the carboxyl group, and the sum of the charges of the ionized R-groups. Titration/ionization of aspartic acid is depicted in Figure 2.10. Ionization (or deionization) within a protein’s structure can have significant effect on the overall conformation of the protein and, since structure is related to function, a major impact on the activity of a protein.
Figure 2.10 – Titration curve for aspartic acid Image by Penelope Irving
Building Polypeptides
Although amino acids serve other functions in cells, their most important role is as constituents of proteins. Proteins, as we noted earlier, are polymers of amino acids.
Amino acids are linked to each other by peptide bonds, in which the carboxyl group of one amino acid is joined to the amino group of the next, with the loss of a molecule of water. Additional amino acids are added in the same way, by formation of peptide bonds between the free carboxyl on the end of the growing chain and the amino group of the next amino acid in the sequence. A chain made up of just a few amino acids linked together is called an oligopeptide (oligo=few) while a typical protein, which is made up of many amino acids is called a polypeptide (poly=many). The end of the peptide that has a free amino group is called the N-terminus (for NH2), while the end with the free carboxyl is termed the C-terminus (for carboxyl).
Figure 2.16 Formation of a peptide bond
As we’ve noted before, function is dependent on structure, and the string of amino acids must fold into a specific 3-D shape, or conformation, in order to make a functional protein. The folding of polypeptides into their functional forms is the topic of the next section. |
189060 | https://hsmpgrid.files.wordpress.com/2019/03/piano_inclinato_simple.pdf | EQUILIBRIO SU UN PIANO INCLINATO Marco Monaci 1Liceo Scientifico G. Marconi (1F) In questi appunti valuteremo brevemente le condi-zioni di equilibrio di un corpo appoggiato su un piano inclinato.
Figura 1: Situazione dell’esercizio proposto.
Consideriamo la situazione riportata in Figura 1. Ab-biamo una massa m appoggiata sulla superficie di un piano inclinato, con angolo pari a θ. Il coefficiente di attrito statico fra il piano e la massa lo indichiamo con µ. Innanzitutto possiamo qualitativamente considerare le forze che agiranno sulla massa: • La forza peso tenderà a far muovere verso il basso il corpo; • La forza di attrito tenderà a frenare il corpo, quindi si opporrà al moto.
Conviene innanzitutto scomporre la forza peso in due componenti, una parallela al piano inclinato e una per-pendicolare al piano inclinato. La scomposizione e la rappresentazione di tutte le forze è riportata invece in Figura 2.
Figura 2: La schematizzazione delle forze nel caso del nostro problema.
Il piano inclinato è rappresentato dalla linea sfumata grigia, mentre i due assi perpendicolari su cui conviene scomporre tutte le forze sono indicate con delle linee tratteggiate. Descriviamo brevemente le forze che sono in gioco.
La forza peso l’abbiamo indicata con FP , ed è sempre rivolta verso il basso. Come si vede è però sconveniente utilizzare questa forza, perché la sua direzione non è per-pendicolare al piano, quindi non possiamo direttamente calcolare la forza di attrito. Possiamo però scomporla in due componenti perpendicolari tra loro, proiettando il vettore forza peso sui due assi che abbiamo considerato prima, uno parallelo al piano inclinato e uno perpen-dicolare ad esso. Iniziamo scomponendo la forza peso nella componente perpendicolare al piano, che abbia-mo indicato con F⊥. Consideriamo proprio il triangolo formato dalla forza peso FP e dalla forza perpendicola-re F⊥: per un bellissimo colpo di fortuna geometrica, questo triangolo è simile al piano inclinato, ovvero gli angoli sono identici. Ruotando la testa si può quindi riconoscere fra la forza perpendicolare e la forza peso lo stesso angolo θ del piano inclinato.
Poiché vogliamo trovare il cateto vicino all’angolo (in-fatti notiamo come il vettore F⊥sia vicino all’angolo θ), dobbiamo calcolare il coseno di θ. Quindi possiamo scrivere la forza perpendicolare come: F⊥= FP cos θ Ovvero: F⊥= mg cos θ E la forza perpendicolare l’abbiamo sistemata. Pas-siamo alla forza parallela, che però è facile da trovare, in quanto è pari al cateto opposto all’angolo, quindi sarà pari a: F∥= mg sin θ Ribadiamo che questa scomposizione è utilissima, in quanto ora le forze in gioco sono o perpendicolari o parallele al piano inclinato, e questo è molto comodo.
Adesso che abbiamo la forza perpendicolare e la forza parallela possiamo calcolare le altre due forze in gioco.
Partiamo dalla forza di attrito, che nel disegno abbiamo indicato con FA. Sappiamo che in generale la forza di attrito è pari a: FA = µN Dove µ è il coefficiente di attrito del piano inclinato, mentre la forza N è la forza normale al piano, ovvero la forza perpendicolare al piano; qui capiamo bene il perché della scomposizione della forza peso, infatti così facendo abbiamo trovato la forza normale che ci serve per calcolare la forza d’attrito! Se avessimo usato diret-tamente la forza peso avremmo sbagliato, in quanto tale forza non è perpendicolare al piano.
A questo punto possiamo calcolare la forza di attrito come: FA = µN = µF⊥ = µmg cos θ Abbiamo sistemato tutte le forze lungo il piano in-clinato.
Vediamo ora le forze che agiscono lungo la direzione perpendicolare al piano inclinato: abbiamo la F⊥che però viene "contrastata" da una forza gene-rata dal piano, chiamata reazione vincolare, indicata nel disegno come R. La reazione vincolare di un piano possiamo interpretarla come la forza che si oppone allo "sprofondamento" del corpo all’interno del piano incli-nato. Prendiamo una situazione più semplice, ovvero una penna appoggiata sopra un tavolo. Tale penna sarà sicuramente sottoposta alla sua forza peso, eppure la penna resta in equilibrio. Affinché però un oggetto sia in equilibrio è necessario che la somma di tutte le forze applicate a tale oggetto sia pari a 0; ciò significa che sulla penna agisce un’altra forza, chiamata per l’appun-to reazione vincolare esercitata dal tavolo che è opposta alla forza peso.
Morale della favola: le forze che agiscono perpendico-larmente al piano hanno somma totale pari a 0, e questo ci piace: infatti il corpo non decolla e non sprofonda dentro il piano inclinato.
Facciamo ora un piccolo backup: sono rimaste in gioco solo due forze, ovvero la forza di attrito e la forza parallela al piano: FA = µmg cos θ F∥= mg sin θ Affinché il corpo stia in equilibrio queste due for-ze devono essere uguali ed opposte. Questo significa scrivere: FA = F∥ E sostituendo le espressioni che abbiamo trovato prima: µmg cos θ = mg sin θ Ed ora ecco la magia della fisica: le masse si semplificano, esattamente come l’accelerazione di gravità!
µ cos θ = sin θ In altre parole il sistema considerato è indipendente dalla massa e dall’accelerazione di gravità! Questo si-stema funziona sia utilizzando un blocchetto di plastica che un TIR; così come funziona se siamo sulla Luna o su Marte: otterremmo sempre lo stesso risultato.
Scriviamo quindi in definitiva il coefficiente di attrito statico come: µ = sin θ cos θ = tan θ Questo risultato fra le altre cose ci permette di cal-colare facilmente il coefficiente di attrito di qualunque materiale: basta misurare l’angolo per il quale il corpo inizia a muoversi, effettuando la tangente troveremo il coefficiente di attrito cercato.
Viceversa se conosciamo il coefficiente di attrito sap-piamo calcolare al volo quale è l’angolo massimo oltre il quale il corpo inizia a scivolare.
NOTA. Se il coefficiente di attrito è pari a 0, la con-dizione di equilibrio non è mai rispettata: il corpo si muove per qualsiasi angolo (tranne nel caso estremo di angolo nullo).
NOTA. La forza di attrito dipende dall’angolo del piano inclinato. E’ massima quando l’angolo θ = 0, infatti tutta la forza peso viene per così dire "convertita" in forza perpendicolare, che quindi va a "creare" la forza di attrito; aumentando l’angolo la forza peso viene in parte anche convertita in forza parallela, a scapito della forza perpendicolare. Aumentando l’angolo si riduce quindi la componente perpendicolare, e di conseguenza la forza di attrito: questo ci torna, infatti per θ = 90o abbiamo un corpo in caduta libera, dove l’attrito con il piano inclinato si annulla completamente.
Esercizio. Consideriamo un piano inclinato con an-golo θ pari a 27o. Su di esso viene posizionato un corpo di massa m = 1kg. Il coefficiente di attrito sia µ = 0.4.
Il corpo scivola oppure no? Se invece il coefficiente di attrito è pari a µ = 0.6? Quale è il coefficiente di attri-to minimo affinché il corpo stia in equilibrio sul piano inclinato? |
189061 | https://www.ixl.com/math/grade-1/add-three-numbers-word-problems | IXL | Add three numbers - word problems | 1st grade math
SKIP TO CONTENT
[x] - [x]
IXL Learning
Sign in- [x] Remember
Sign in nowJoin now
IXL Learning
Learning
Math
Skills
Lessons
Videos
Games
Fluency Zone New!
Language arts
Skills
Videos
Games
Science
Social studies
Spanish
Recommendations
Recommendations wall
Skill plans
IXL plans
South Carolina state standards
Textbooks
Test prep
Awards
Student awards
Assessment
Analytics
Takeoff
Inspiration
Learning All
Learning
Math
Language arts
Science
Social studies
Spanish
Recommendations
Skill plans Learning
Skill plans
IXL plans
South Carolina state standards
Textbooks
Test prep
Awards
Assessment
Analytics
Takeoff
Inspiration
Membership
Sign in
Math Math
Language arts Language arts
Science Science
Social studies Social studies
Spanish Spanish
Recommendations Recommendations
Skill plans Skill plans
IXL plans
South Carolina state standards
Textbooks
Test prep
Awards Awards
First grade
R.4
Add three numbers - word problems
Z7S
Share skill
Copy the link to this skill
share to facebook share to twitter
Time to get in the zone!
Your teacher would like you to focus on skills in .
Let's pick a skill from these categories.
Let's go!
R.4 Add three numbers - word problems Z7S
Share video
Copy the link to this video
share to facebook share to twitter
You are watching a video preview. Become a member to get full access!
You've reached the end of this video preview, but the learning doesn't have to stop!
Join IXL today!
Become a memberSign in
Incomplete answer
You did not finish the question. Do you want to go back to the question?
Go back
Submit
Learn with an example
Lee donated money to various charities last year. First, he donated $1 to an elementary school and $6 to repair local playgrounds. Then he donated $2 to a state park. How much money did Lee donate in all?
$
Submit
Back to practice
ref_doc_title.
Back to practice
Learn with an example
Learn with an example
question
Abby had to practice for a math test. She did 1 math problem on Monday, 6 math problems on Tuesday, and 2 more problems on Wednesday. How many problems did Abby do in all?
problems
solution
Add the numbers of problems.
1 + 6 + 2
First, add 1 and 6:
1 + 6 + 2 = ?
7 + 2 = ?
Now add 7 and 2:
7 + 2 = 9
The sum is 9.
Abby did 9 problems in all.
Learn with an example
Excellent! You got that right!
Continue
Learn with an example
Jumping to level 1 of 1
Excellent!
Now entering the Challenge Zone—are you ready?
Questions
answered Questions
0
Time
elapsed Time
00 00 02
hr min sec
SmartScore
out of 100
IXL's SmartScore is a dynamic measure of progress towards mastery, rather than a percentage grade. It tracks your skill level as you tackle progressively more difficult questions. Consistently answer questions correctly to reach excellence (90), or conquer the Challenge Zone to achieve mastery (100)! Learn more
0
You met your goal!
Teacher tools
Group Jam
Live Classroom
Leaderboards
Work it out Not feeling ready yet? These can help:Q.15 Add three numbersQ.15 Add three numbers - First grade RL2R.2 Addition word problems - sums up to 20R.2 Addition word problems - sums up to 20 - First grade KY5
Company | Membership | Blog | Help center | User guides | Tell us what you think | Testimonials | Careers | Contact us | Terms of service | Privacy policy
© 2025 IXL Learning. All rights reserved. Follow us
First time here?
1 in 4 students uses IXL
for academic help and enrichment.
Pre-K through 12th grade
Sign up nowKeep exploring |
189062 | https://www.americaspg.com/article/download/2334 | Prospects for Applied Mathematics and data Analysis ( PAMDA) Vol. 02, No. 02, PP. 41-44, 2023
41
Doi: 5
Received: April 17, 2023 Accepted: October 2 5, 2023
A New Proof of Feuerbach’s Theorem
Ion Patrascu 1, Florentin Smarandache 2
1
“Frații Buzești” National College , Bld. Stirbei Voda, nr. 5, Craiova, Dolj, Romania
2
University of New Mexico , Mathematics, Physics and Natural Sciences Division,
705 Gurley Ave., Gallup, NM 87301, USA
Email: patrascu_ion@yahoo.com ; smarand@unm.edu
Abstract
Feuerbach’s theorem on the tangent of the circle of the nine points and the inscribed and exinscribed circle is
considered one of the most beautiful theorems in geometry. In this paper, we offer a basic proof of this theorem
starting from one of Gh. Buicli u’s ideas .
Keywords: Feuerbach’s theorem : circle of the nine points ; inscribed circle s; exinscribed circle s
The Theorem : The circle of the nine points of a triangle is tangent to the inscribed circle and to the exinscribed
circles of the triangle (K. Feuerbach – 1822).
Proof: May O9 be the center of the circle of the 9 points of the ABC triangle and I the center of its inscribed circle.
We denote by D the contact of the inscribed circle with BC and with D’ its diameter within the inscribed circle.
May AH a be the height from A (see Figure 1 ); we will denote by K and M the intersections of the parallel going
through I to AD’ with AH a, respectively with BC and L the intersection of the parallel at AD ’, drawn through D,
with AH a
We also note with P the intersection of the IL and BC lines and with Q the intersection of the half -line (D’K with
the inscribed circle).
Figure 1
We will prove that Q is the tangency point of the inscribed circle with the circle of the 9 points. ( Q is one of
Feuerbach’s points.) From the fact that the inscribed circle ( I) and the exinscribed circle A (Ia) are homothetical
through the homothety of pole A, we know that the point E is the exinscribed (I a) circle’s contact with BC, and that Prospects for Applied Mathematics and data Analysis ( PAMDA) Vol. 02, No. 02, PP. 41-44, 2023
42
Doi: 5
Received: April 17, 2023 Accepted: October 2 5, 2023
points D and E are isotomical (symmetrical to M, the middle of BC ). We note that α = m (𝐼𝐷 ’𝑄 ̂ ), because ID’ =
IQ , thus m (𝐼𝑄𝐷 ’̂ ) = α. The ID’AK and IDLK quadrilaterals are parallelograms.
ID’ ∥AK și ID’ = KL = r (the ray of the inscribed circle), thus the ID’KL quadrilater is a parallelogram, so m (𝐼𝑄𝐷 ’̂ )
= m (𝐿 𝐼𝑄 ̂ ) = α. Likewise, m (𝐷𝐼𝐿 ̂ ) = m (𝐿 𝐼𝑄 ̂ ) = α, which, along with ID = IQ = r , shows that the points D and Q
are symmetrical to IP . Moreover, because PD is tangent in D to the (I) circle, we have that PQ is tangent in Q to
the (I) inscribed circle, so PD = PQ .
The fact that the (MI; DL) and (DD’; AH a) lines are parallel leads to: 𝑃𝑀
𝑃𝐷
= 𝑃𝐼
𝑃𝐿
= 𝑃𝐷
𝑃 𝐻 𝑎
.
We remember that PD 2 = PM ∙PH a, and seeing as how PD = PQ , we have that PQ 2 = PM ∙PH a, which shows PQ is
tangent to the circumscribed circle of the QMH a triangle .
We prove this circle is the circle of the 9 points of the ABC triangle.
It is sufficient to prove that Q belongs to the circle of the 9 points, so we show that O9Q = 𝑅
2
(R being the ray of
the circumscribed circle and O9 the center of the circle of the 9 points). We denote by X the orthogonal projection
of O9 on BC . We have that O9X2 + PX 2 = PO 92.
On another note, O9M = 𝑅
2
and O9X2 = 𝑅 2
4
− 𝑀 𝑋 2 . We know that X este is the middle of the MH a segment
(because O9 is the middle of OH – where O and H are , respectively, the center of the circumscribed circle and the
orthocenter of ABC).
We obtain PO 92 = PX 2−MX 2 + 𝑅 2
4
.
But PX 2−MX 2 = (PX+MX)(PX -MX)= PM ∙(PH a+H aX-MX) .
However, MX=MH a, thus PX 2−MX 2 = PM ∙PH a = PD 2 = PQ 2.
We obtain PO 92 = PQ 2 + 𝑅 2
4
.
We interpret this relationship as follows: the triangle with the length sides PO 9, PQ and 𝑅
2
is rectangular. Because
PQ is the tangent of the circumscribed circle to the MH aQ triangle, we obtain the 𝑅
2
length segment with an end in
Q and the other on the IQ line . But the circle circumscribed to the MH aQ triangle has its center at the intersection
of the bisection of the MH a segment with perpendiculars in Q on PQ , just as O9 is the bisection of MH a and is at
the distance of O9M = 𝑅
2
of M and Ha, thus O9Q = 𝑅
2
, so O9 is the center of the circumscribed circle of the MH aQ
triangle.
We continue to prove that the circle of the 9 points is tangent to the A -exinscribed circle in a Feuerbach which we
note by Qa. We build on the E point the orthogonal projection of Ia on BC and E’ its diameter in the (I a) circle.
We draw the parallels to AE’ through Ia and E and we denote by Ka respectively La their intersections with the
height AH a of the ABC triangle (see Figure 2 ). The intersection of the IaKa lines and BC is M – the middle of BC ,
but the intersection of the IaLa line with BC we denote by Pa. We affirm that the Qa Feuerbach point is the
intersection of the exinscribed A -circle with the E’K a segment. Prospects for Applied Mathematics and data Analysis ( PAMDA) Vol. 02, No. 02, PP. 41-44, 2023
43
Doi: 5
Received: April 17, 2023 Accepted: October 2 5, 2023
Figure 2
We prove that the Qa point is symmetrical to Pa on the IaKa line. From what we have so far, the E’I aLaKa
quadrilateral is a parallelogram, so ∢𝐸 𝐼 𝑎 𝑃 𝑎 = ∢𝐼 𝑎 𝐸 ′𝑄 𝑎 (alt. int.). because the IaF’Q a triangle is isosceles we have
that: ∢𝐼 𝑎 𝐸 ′𝑄 𝑎 = ∢𝐼 𝑎 𝑄 𝑎 𝐸 ′.
The ∢𝐸 𝐼 𝑎 𝑄 𝑎 angle is exterior to the E’I aQa triangle, so ∢𝐸 𝐼 𝑎 𝑄 𝑎 = 2 ∙ ∢𝐼 𝑎 𝑄 𝑎 𝐸 ′ and seeing as ∢𝐸 𝐼 𝑎 𝑃 𝑎 = ∢𝐼 𝑎 𝑄 𝑎 𝐸 ′,
we have that ∢𝐸 𝐼 𝑎 𝑃 𝑎 = ∢𝐼 𝑎 𝑄 𝑎 𝐸 ′, relationship which, together with IaE = IaQa implies ∆𝑃 𝑎 𝐼 𝑎 𝐸 ≡ ∆𝑄 𝑎 𝐼 𝑎 𝑃 𝑎 , and Prospects for Applied Mathematics and data Analysis ( PAMDA) Vol. 02, No. 02, PP. 41-44, 2023
44
Doi: 5
Received: April 17, 2023 Accepted: October 2 5, 2023
from here we obtain that PaQa = P aE and PaQa⊥IaKa. From the parallelism of the EL a line with the IaKa line and
EE’ with AH a we have that: 𝑃 𝑎 𝑀
𝑃 𝑎 𝐸
= 𝐼 𝑎 𝑃 𝑎
𝑃 𝑎 𝐿 𝑎
= 𝑃 𝑎 𝐸
𝑃 𝑎 𝐻 𝑎
.
We remember that PaE2=PaM∙PaHa, thus PaQa2=PaM∙PaHa.
This relationship shows that the A -exinscribed circle and the circumscribed circle of the MH aQa triangle are
exterior tangents in the Qa point.
We continue to prove that the circumscribed circle of the MH aQa triangle is the circle of the 9 points (O 9).
We denote by T the orthogonal projection of O9 on BC , we have PaO92 = PaT2+O9T2.
But O9T2 = O9M2−MT 2 = 𝑅
4
− 𝑀 𝑇 2. We obtain PaT2−MT 2+𝑅
4
= PaO92. But MT = HaT = 1
2
MH a, we have PaT2−MT 2
= (P aT+TH a)(P aT−MT) = PaHa ∙ PaM. We obtained PaO92 = PaQa2 + 𝑅 2
4
.
We interpret this relationship as follows: we can build a rectangular triangle with the hypothenuses PaO9 with a
PaQa cathetus and a 𝑅
2
length cathetus. Because m (𝑃 𝑎 𝑄 𝑎 𝐼 𝑎 ̂ ) = 90 o, the rectangular triangle with the PaO9
hypothenuses will have a 90 o angle in Qa and the third corner will be on the prolongment of the IaQa segment. That
corner will be on the bisection of the MH a segment at the distance of OM = OH a = 𝑅
2
. Because O9M = O9Ha = 𝑅
2
,
we have that O9 is the circumscribed center of the MH aQa triangle, meaning this last circle is the circle of the 9
points of the ABC triangle. The theorem is proved the same way for the B -exinscribed and C -exinscribed circles.
References
. Gh. Buicliu: Probleme de construcții geometrice cu rigla și compasul . Tehnica (publishing house), Bucharest,
1967 .
. Florentin Smarandache, Ion Patrascu: The Geometry of Homological Triangles . Education Publisher,
Colombus, Ohio, 2012 . |
189063 | https://dannyreviews.com/h/Ants.html | The Ants (Bert Hölldobler, Edward Wilson) - book review
Danny Yee's Book Reviews
Subjects | Titles | Authors | Best Books | Search | Latest
The Ants
Bert Hölldobler + Edward O. Wilson
Harvard University Press 1990
A book review by Danny Yee © 2004
The only professional science work to have won a Pulitzer Prize — for general nonfiction in 1991 — The Ants is a sweeping survey of the ethology, ecology and evolutionary biology of ants. About a third of it is taken up by reference material that is probably only for the myrmecologists and entomologists, but the remainder should keep anyone with a serious interest in biology engrossed.
Apart from a brief overview of ant phylogeny in chapter one, there is no separate discussion of evolution — rather, evolutionary ideas pervade the entire work. Similarly, there is no separate chapter on ant anatomy, which is dealt with where relevant: the various glands, for example, are covered in the chapter on communication. And ants are connected throughout to more general biological theory, most notably sociobiology.
"Ant castes are exceptionally well suited for optimization studies in sociobiology. ... individual worker ants are full organisms with ordinary, whole patterns of social behavior, yet they are also clearly specialized for particular well-defined tasks. ... Thus well-defined anatomical structures and behavioral acts can be more readily assayed with reference to the four elements of optimization models; we really are able to specify a restricted and relatively easily defined state space, a set of strategies, testable fitness functions, and measurable constraints."
It is not a major focus, but aspects of the human and economic significance of ants are touched on. The dominant animals in many habitats, they are sometimes a major constraint on agriculture — and not always a negative one.
"Records from southern China show that weaver ant nests have been gathered, sold, and placed in selected citrus trees to combat insect pests for approximately 1,700 years. ... the oldest known instance of the biological control of insects in the history of agriculture."
And extracts from travel narratives, or the more personal stories of myrmecologists, are used in a few places to illustrate aspects of ant behaviour.
A few opening pages on the importance of ants are followed by the least digestible chunks of the work: nearly 80 pages of taxonomic key and 60 pages of line drawings of the major genera. The other specialist material is scattered through the remainder of the book, in the form of tables showing the taxonomic distribution of particular traits or features.
The bulk of the book falls roughly into two parts, with the chapters in the first covering general features of ant life cycles, ecology, and so forth and those in the second different ant specialisations. The following summary does no justice to this at all.
A typical ant colony life cycle consists of the nuptial flight and mating, followed by colony foundation, growth, and movement; there are, however, many species that don't fit this pattern.
"By almost any conceivable standard, the single most important feature of insect social behavior is the existence of the nonreproductive worker caste." The debate over the origin and persistence of this and associated altruism, of eusociality, has seen kin selection first downplayed as an explanation and then revived.
A critical factor in ant social behaviour is the extent to which ants can recognise kin, other ants of the same colony, and different brood stages. But "the pattern of diversity in recognition systems of ants is still mostly unknown".
Many ant species exhibit polygyny, which can arise through founding queens remaining together, later adoption of new queens, or fusion of colonies. The number of queens is a critical feature of a colony, among other things influencing conflicts between queens and workers over the proportion of investment in new queens and males.
"The typical ant is a walking battery of exocrine glands", with more than ten organs varying in form and use among different ant groups — the six most important are "Dufour's gland, the poison gland, the pygidial gland, the sternal glands, the mandibular glands, and the metapleural glands". The pheromones from these are central to ant communication, lending themselves to ritualization and modulatory communication, but ants also use auditory and visual modes. Communication underpins alarm signalling, recruitment, adult transport, trails and border marking, food-sharing, and corpse removal, among other behaviors.
Ergonomic analysis of ant colonies reveals a range of tasks and roles, allocated between different castes. As well as males, workers, and queens (or an ergatogyne reproductive caste), there are often subdivisions of workers into minors, medias, and majors, with the latter often highly specialised.
"The greatest size variation of nestmates ever recorded in ants occurs in the Asian marauder ant Pheidologeton diversus. The minor worker depicted in this scanning electron micrograph has a head width exactly 1/10 that of the major on which it sits, and a dry weight only about 1/500 that of the larger ant."
And there are often temporal castes, with roles changing over individual lifetimes. Caste determination is complex, but colony demographics are clearly adaptive.
While ants are capable of some learning — but don't play — it is often helpful to consider the ant colony as a superorganism, with its own homeostatic responses and flexibility in behaviour. Notable examples are in thermoregulation of nests and control of humidity.
A "temperature-humidity" envelope constrains the daily cycle of ants: with very few exceptions "they function poorly below 20 degrees Celsius and not at all below 10". Foraging strategies also face a tradeoff between predation and energy optimisation. Competition is another factor in population regulation: it produces overdispersion of colonies and a range of different territorial strategies.
And this applies between species as well: "Competition is the hallmark of ant ecology. Many, perhaps most, ant species employ aggressive techniques up to and including organized warfare." Coexistence can involve differences in niche, density, and size, with ants displaying "a startling variety of mechanisms that appear to adjust species to one another and hence to organize local communities".
Two hundred pages cover some of the specialisations of ants. There are three chapters on ant "symbioses", covering the full range from mutualisms to parasitic relationships, and five on trophic specialisations.
Ants exhibit many kinds of social parasitism, of which slave-taking is the best known, though "no verified examples are yet known of mutualism, in which two species cooperate to the benefit of both." Hölldobler and Wilson analyse the different stages of inquilinism (where the parasite spends its entire life cycle in the host nest) and their possible evolution.
When it comes to symbioses with other arthropods, individual ants have predators and parasites, but in some cases it is useful to see the ant colony as an ecosystem for symbionts. Myrmecophilous insects include beetles and the larvae of many lycaenid butterflies, with relationships ranging from mutualistic to parasitic; perhaps the most startling are the trophobiont homopterans (aphids) "herded" by some ant species.
Relationships with plants are equally varied: ants protect plants, plants shelter or feed ants, ants feed plants, ants disperse plants, and so forth. There are also fungi and microorganisms that are parasitic on ants.
There are ants specialised for a vast range of prey — arthropod eggs, termites, cockroaches, collembolans, and more. The evolution of feeding specialisations can be correlated with social behavior: the trend within the Dacetine ants has been a "shift from open foraging on the ground and low vegetation to an increasingly cryptic, subterranean existence. ... associated with a reduction in the variety of arthropods taken as prey".
Army ants are among the best known — if not infamous — ants. As well as the well-studied swarm raider Eciton burchelli and other swarm and column raiders in the same genus, there are African driver ants Dorylus and many others that combine migration and group predation. The origin of legionary behaviors remains unclear, however.
Among the most complex colonies are those of the leaf-cutter fungus growers Atta, with up to seven castes performing 29 distinct tasks.
"The ultimate size reached by the _Atta_nests is enormous. ... [one] _A. sexdens_nest, 77 months old, contained 1,920 chambers, of which 390 were occupied by fungus gardens and ants. The loose soil that had been brought out and piled on the ground by the ants during the excavation of their nest ... occupied 22.72 cubic meters and weighed approximately 40,000 kilograms."
The harvesting ants are the source of classical tropes about the industry and diligence of ants. And the weaver ants Oecophylla use silk produced by their larvae to glue leaves together to make nests, an adaptation that has helped them to become one of the most abundant social insects.
Hölldobler and Wilson end with some brief tips on collecting and farming ants. There's a sixty-five page bibliography and — more importantly for the non-specialists — a ten page glossary.
RepleteAn individual ant whose crop is greatly distended with liquid food, to the extent that the abdominal segments are pulled apart and the intrasegmental membranes are stretched tight. Repletes usually serve as living reservoirs, regurgitating food on demand to their nestmates. Also used as an adjective, e.g., replete worker.
The Ants has an excellent range of black and white photographs, backed up by sixteen separate pages of dramatic colour photographs. Effective use is also made of line diagrams, graphs, maps, plans, and so forth. It is a large format work, nearly 30cm square and weighing 3.5 kilograms, and on the expensive side, so it was never going to be a bestseller. Any university library should have a copy, however.
Hölldobler and Wilson have also written a much shorter and more accessible popular book, Journey to the Ants. This omits most of the fascinating biological detail as well as the reference material and technical apparatus, focusing on some of the more dramatic and unusual ants. It also includes some autobiographical and travel details not in the larger work.
May 2004
External links: - buy from Amazon.com or Amazon.co.uk
buy Journey to the Ants from Amazon.com or Amazon.co.uk
Related reviews: - more insects + entomology
books about evolution
books published by Harvard University Press
%T The Ants
%A Hölldobler, Bert
%A Wilson, Edward O.
%I Harvard University Press
%D 1990
%O hardcover, large format, photos, references, index
%G ISBN 0674040759
%P xii,732pp
Subjects | Titles | Authors | Best Books | Search | Latest
Book Reviews by Danny Yee |
189064 | https://brainly.in/question/55807578 | Prove that the tangent and normal at any point of an ellipse bisect - Brainly.in
Skip to main content
Ask Question
Log in
Join for free
For parents
For teachers
Honor code
Textbook Solutions
Brainly App
Sruthypotter5129
15.03.2023
Physics
Secondary School
answered
Prove that the tangent and normal at any point of an ellipse bisect
1
See answer
See what the community says and unlock a badge.
Add answer+8 pts
0:00
/
0:15
Read More
Sruthypotter5129 is waiting for your help.
Add your answer and earn points.
Add answer +8 pts
Answer
1 person found it helpful
aadhysanthosh12
aadhysanthosh12
Answer:
Explanation:
To prove that the tangent and normal at any point of an ellipse bisect the angle between the two focal radii passing through that point, we can use the following steps:
Let P be any point on the ellipse, and let F1 and F2 be the two foci of the ellipse. Draw the focal radii FP1 and FP2 passing through P.
Let T be the point of intersection between the tangent to the ellipse at P and the line segment joining F1 and F2. Let N be the point of intersection between the normal to the ellipse at P and the line segment joining F1 and F2.
Draw the line segments TP and NP, and let X be the point of intersection between TP and FP1, and Y be the point of intersection between NP and FP2.
To prove that TP and NP bisect the angle between FP1 and FP2, we need to show that angles XPF1 and YPF2 are equal.
Note that angles F1PX and F2PY are right angles, since TP and NP are perpendicular to the tangent and normal to the ellipse, respectively.
Since F1P = F2P (by definition of the ellipse), triangles F1PX and F2PY are congruent by the hypotenuse-leg (HL) criterion.
Therefore, angles XPF1 and YPF2 are congruent (since they are corresponding angles in congruent triangles), which implies that TP and NP bisect the angle between FP1 and FP2.
Therefore, we have proved that the tangent and normal at any point of an ellipse bisect the angle between the two focal radii passing through that point.
Explore all similar answers
Thanks 1
rating answer section
Answer rating 0.0
(0 votes)
Find Physics textbook solutions?
See all
Class 12
Class 11
Class 10
Class 9
Class 8
Class 7
Class 6
500 Selected Problems In Physics for JEE Main & Advanced
877 solutions
Selina - Concise Physics - Class 9
1224 solutions
Lakhmir Singh, Manjit Kaur - Physics 10
1964 solutions
Selina - Concise Physics - Class 8
715 solutions
NEET Exam - Physics
359 solutions
Lakhmir Singh, Manjit Kaur - Physics 9
1131 solutions
NCERT Class 11 Physics Part I
378 solutions
Physics
520 solutions
Selina - Physics - Class 7
519 solutions
Physical Science
670 solutions
SEE ALL
Advertisement
Still have questions?
Find more answers
Ask your question
New questions in Physics
explain"fluids"with one example
Find current flowing through loop analysis.(using KVL)
Q.I Suppose potential energy between electron and proton at separation 'r' is given by UK In (r), where K is a constant. For such a hypothetical
Why we put a constant when we proportionality is removed
01. A uniform rod of mass 6M and length 61 is bent to make an equilateral hexagon. Its M.I. about an axis passing through the centre of mass and
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) |
189065 | https://stackoverflow.com/questions/10397574/efficiently-compute-mean-and-standard-deviation-from-a-frequency-table | r - Efficiently compute mean and standard deviation from a frequency table - Stack Overflow
Join Stack Overflow
By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy.
Sign up with Google
Sign up with GitHub
OR
Email
Password
Sign up
Already have an account? Log in
Skip to main content
Stack Overflow
1. About
2. Products
3. For Teams
Stack Overflow for Teams Where developers & technologists share private knowledge with coworkers
Advertising Reach devs & technologists worldwide about your product, service or employer brand
Knowledge Solutions Data licensing offering for businesses to build and improve AI tools and models
Labs The future of collective knowledge sharing
About the companyVisit the blog
Loading…
current community
Stack Overflow helpchat
Meta Stack Overflow
your communities
Sign up or log in to customize your list.
more stack exchange communities
company blog
Log in
Sign up
Let's set up your homepage Select a few topics you're interested in:
python javascript c#reactjs java android html flutter c++node.js typescript css r php angular next.js spring-boot machine-learning sql excel ios azure docker
Or search from our full list:
javascript
python
java
c#
php
android
html
jquery
c++
css
ios
sql
mysql
r
reactjs
node.js
arrays
c
asp.net
json
python-3.x
.net
ruby-on-rails
sql-server
swift
django
angular
objective-c
excel
pandas
angularjs
regex
typescript
ruby
linux
ajax
iphone
vba
xml
laravel
spring
asp.net-mvc
database
wordpress
string
flutter
postgresql
mongodb
wpf
windows
xcode
amazon-web-services
bash
git
oracle-database
spring-boot
dataframe
azure
firebase
list
multithreading
docker
vb.net
react-native
eclipse
algorithm
powershell
macos
visual-studio
numpy
image
forms
scala
function
vue.js
performance
twitter-bootstrap
selenium
winforms
kotlin
loops
express
dart
hibernate
sqlite
matlab
python-2.7
shell
rest
apache
entity-framework
android-studio
csv
maven
linq
qt
dictionary
unit-testing
asp.net-core
facebook
apache-spark
tensorflow
file
swing
class
unity-game-engine
sorting
date
authentication
go
symfony
t-sql
opencv
matplotlib
.htaccess
google-chrome
for-loop
datetime
codeigniter
perl
http
validation
sockets
google-maps
object
uitableview
xaml
oop
visual-studio-code
if-statement
cordova
ubuntu
web-services
email
android-layout
github
spring-mvc
elasticsearch
kubernetes
selenium-webdriver
ms-access
ggplot2
user-interface
parsing
pointers
c++11
google-sheets
security
machine-learning
google-apps-script
ruby-on-rails-3
templates
flask
nginx
variables
exception
sql-server-2008
gradle
debugging
tkinter
delphi
listview
jpa
asynchronous
web-scraping
haskell
pdf
jsp
ssl
amazon-s3
google-cloud-platform
jenkins
testing
xamarin
wcf
batch-file
generics
npm
ionic-framework
network-programming
unix
recursion
google-app-engine
mongoose
visual-studio-2010
.net-core
android-fragments
assembly
animation
math
svg
session
intellij-idea
hadoop
rust
next.js
curl
join
winapi
django-models
laravel-5
url
heroku
http-redirect
tomcat
google-cloud-firestore
inheritance
webpack
image-processing
gcc
keras
swiftui
asp.net-mvc-4
logging
dom
matrix
pyspark
actionscript-3
button
post
optimization
firebase-realtime-database
web
jquery-ui
cocoa
xpath
iis
d3.js
javafx
firefox
xslt
internet-explorer
caching
select
asp.net-mvc-3
opengl
events
asp.net-web-api
plot
dplyr
encryption
magento
stored-procedures
search
amazon-ec2
ruby-on-rails-4
memory
canvas
audio
multidimensional-array
random
jsf
vector
redux
cookies
input
facebook-graph-api
flash
indexing
xamarin.forms
arraylist
ipad
cocoa-touch
data-structures
video
azure-devops
model-view-controller
apache-kafka
serialization
jdbc
woocommerce
razor
routes
awk
servlets
mod-rewrite
excel-formula
beautifulsoup
filter
docker-compose
iframe
aws-lambda
design-patterns
text
visual-c++
django-rest-framework
cakephp
mobile
android-intent
struct
react-hooks
methods
groovy
mvvm
ssh
lambda
checkbox
time
ecmascript-6
grails
google-chrome-extension
installation
cmake
sharepoint
shiny
spring-security
jakarta-ee
plsql
android-recyclerview
core-data
types
sed
meteor
android-activity
activerecord
bootstrap-4
websocket
graph
replace
scikit-learn
group-by
vim
file-upload
junit
boost
memory-management
sass
import
async-await
deep-learning
error-handling
eloquent
dynamic
soap
dependency-injection
silverlight
layout
apache-spark-sql
charts
deployment
browser
gridview
svn
while-loop
google-bigquery
vuejs2
dll
highcharts
ffmpeg
view
foreach
makefile
plugins
redis
c#-4.0
reporting-services
jupyter-notebook
unicode
merge
reflection
https
server
google-maps-api-3
twitter
oauth-2.0
extjs
terminal
axios
pip
split
cmd
pytorch
encoding
django-views
collections
database-design
hash
netbeans
automation
data-binding
ember.js
build
tcp
pdo
sqlalchemy
apache-flex
mysqli
entity-framework-core
concurrency
command-line
spring-data-jpa
printing
react-redux
java-8
lua
html-table
ansible
jestjs
neo4j
service
parameters
enums
material-ui
flexbox
module
promise
visual-studio-2012
outlook
firebase-authentication
web-applications
webview
uwp
jquery-mobile
utf-8
datatable
python-requests
parallel-processing
colors
drop-down-menu
scipy
scroll
tfs
hive
count
syntax
ms-word
twitter-bootstrap-3
ssis
fonts
rxjs
constructor
google-analytics
file-io
three.js
paypal
powerbi
graphql
cassandra
discord
graphics
compiler-errors
gwt
socket.io
react-router
solr
backbone.js
memory-leaks
url-rewriting
datatables
nlp
oauth
terraform
datagridview
drupal
oracle11g
zend-framework
knockout.js
triggers
neural-network
interface
django-forms
angular-material
casting
jmeter
google-api
linked-list
path
timer
django-templates
arduino
proxy
orm
directory
windows-phone-7
parse-platform
visual-studio-2015
cron
conditional-statements
push-notification
functional-programming
primefaces
pagination
model
jar
xamarin.android
hyperlink
uiview
visual-studio-2013
vbscript
google-cloud-functions
gitlab
azure-active-directory
jwt
download
swift3
sql-server-2005
configuration
process
rspec
pygame
properties
combobox
callback
windows-phone-8
linux-kernel
safari
scrapy
permissions
emacs
scripting
raspberry-pi
clojure
x86
scope
io
expo
azure-functions
compilation
responsive-design
mongodb-query
nhibernate
angularjs-directive
request
bluetooth
reference
binding
dns
architecture
3d
playframework
pyqt
version-control
discord.js
doctrine-orm
package
f#
rubygems
get
sql-server-2012
autocomplete
tree
openssl
datepicker
kendo-ui
jackson
yii
controller
grep
nested
xamarin.ios
static
null
statistics
transactions
active-directory
datagrid
dockerfile
uiviewcontroller
webforms
discord.py
phpmyadmin
sas
computer-vision
notifications
duplicates
mocking
youtube
pycharm
nullpointerexception
yaml
menu
blazor
sum
plotly
bitmap
asp.net-mvc-5
visual-studio-2008
yii2
electron
floating-point
css-selectors
stl
jsf-2
android-listview
time-series
cryptography
ant
hashmap
character-encoding
stream
msbuild
asp.net-core-mvc
sdk
google-drive-api
jboss
selenium-chromedriver
joomla
devise
cors
navigation
anaconda
cuda
background
frontend
binary
multiprocessing
pyqt5
camera
iterator
linq-to-sql
mariadb
onclick
android-jetpack-compose
ios7
microsoft-graph-api
rabbitmq
android-asynctask
tabs
laravel-4
environment-variables
amazon-dynamodb
insert
uicollectionview
linker
xsd
coldfusion
console
continuous-integration
upload
textview
ftp
opengl-es
macros
operating-system
mockito
localization
formatting
xml-parsing
vuejs3
json.net
type-conversion
data.table
kivy
timestamp
integer
calendar
segmentation-fault
android-ndk
prolog
drag-and-drop
char
crash
jasmine
dependencies
automated-tests
geometry
azure-pipelines
android-gradle-plugin
itext
fortran
sprite-kit
header
mfc
firebase-cloud-messaging
attributes
nosql
format
nuxt.js
odoo
db2
jquery-plugins
event-handling
jenkins-pipeline
nestjs
leaflet
julia
annotations
flutter-layout
keyboard
postman
textbox
arm
visual-studio-2017
gulp
stripe-payments
libgdx
synchronization
timezone
uikit
azure-web-app-service
dom-events
xampp
wso2
crystal-reports
namespaces
swagger
android-emulator
aggregation-framework
uiscrollview
jvm
google-sheets-formula
sequelize.js
com
chart.js
snowflake-cloud-data-platform
subprocess
geolocation
webdriver
html5-canvas
centos
garbage-collection
dialog
sql-update
widget
numbers
concatenation
qml
tuples
set
java-stream
smtp
mapreduce
ionic2
windows-10
rotation
android-edittext
modal-dialog
spring-data
nuget
doctrine
radio-button
http-headers
grid
sonarqube
lucene
xmlhttprequest
listbox
switch-statement
initialization
internationalization
components
apache-camel
boolean
google-play
serial-port
gdb
ios5
ldap
youtube-api
return
eclipse-plugin
pivot
latex
frameworks
tags
containers
github-actions
c++17
subquery
dataset
asp-classic
foreign-keys
label
embedded
uinavigationcontroller
copy
delegates
struts2
google-cloud-storage
migration
protractor
base64
queue
find
uibutton
sql-server-2008-r2
arguments
composer-php
append
jaxb
zip
stack
tailwind-css
cucumber
autolayout
ide
entity-framework-6
iteration
popup
r-markdown
windows-7
airflow
vb6
g++
ssl-certificate
hover
clang
jqgrid
range
gmail
Next You’ll be prompted to create an account to view your personalized homepage.
Home
Questions
AI Assist Labs
Tags
Challenges
Chat
Articles
Users
Jobs
Companies
Collectives
Communities for your favorite technologies. Explore all Collectives
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
Collectives™ on Stack Overflow
Find centralized, trusted content and collaborate around the technologies you use most.
Learn more about Collectives
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
Efficiently compute mean and standard deviation from a frequency table
Ask Question
Asked 13 years, 5 months ago
Modified13 years, 5 months ago
Viewed 10k times
Part of R Language Collective
This question shows research effort; it is useful and clear
9
Save this question.
Show activity on this post.
Suppose I have the following frequency table.
```r
print(dat)
V1 V2
1 1 11613
2 2 6517
3 3 2442
4 4 687
5 5 159
6 6 29
V1 = Score
V2 = Frequency
```
How can I efficiently compute the Mean and standard deviation? Yielding: SD=0.87 MEAN=1.66. Replicating the score by frequency takes too long to compute.
R Language Collective
r
statistics
mean
Share
Share a link to this question
Copy linkCC BY-SA 3.0
Improve this question
Follow
Follow this question to receive notifications
edited May 1, 2012 at 12:39
Paul Hiemstra
61.2k 12 12 gold badges 145 145 silver badges 151 151 bronze badges
asked May 1, 2012 at 12:22
neversaintneversaint
64.5k 145 145 gold badges 324 324 silver badges 494 494 bronze badges
Add a comment|
4 Answers 4
Sorted by: Reset to default
This answer is useful
7
Save this answer.
Show activity on this post.
Mean is easy. SD is a little trickier (can't just use fastmean() again because there's an n-1 in the denominator.
```r
dat <- data.frame(freq=seq(6),value=runif(6)100)
fastmean <- function(dat) {
+ with(dat, sum(freqvalue)/sum(freq) )
+ }
fastmean(dat)
55.78302
fastRMSE <- function(dat) {
+ mu <- fastmean(dat)
+ with(dat, sqrt(sum(freq(value-mu)^2)/(sum(freq)-1) ) )
+ }
fastRMSE(dat)
34.9316
To test
expanded <- with(dat, rep(value,freq) )
mean(expanded)
55.78302
sd(expanded)
34.9316
```
Note that fastRMSE calculates sum(freq) twice. Eliminating this would probably result in another minor speed boost.
Benchmarking
```r
microbenchmark(
+ fastmean(dat),
+ mean( with(dat, rep(value,freq) ) )
+ )
Unit: microseconds
expr min lq median uq max
1 fastmean(dat) 12.433 13.5335 14.776 15.398 23.921
2 mean(with(dat, rep(value, freq))) 21.225 22.3990 22.714 23.406 86.434
dat <- data.frame(freq=seq(60),value=runif(60)100)
dat <- data.frame(freq=seq(60),value=runif(60)100)
microbenchmark(
+ fastmean(dat),
+ mean( with(dat, rep(value,freq) ) )
+ )
Unit: microseconds
expr min lq median uq max
1 fastmean(dat) 13.177 14.544 15.8860 17.2905 54.983
2 mean(with(dat, rep(value, freq))) 42.610 48.659 49.8615 50.6385 151.053
dat <- data.frame(freq=seq(600),value=runif(600)100)
microbenchmark(
+ fastmean(dat),
+ mean( with(dat, rep(value,freq) ) )
+ )
Unit: microseconds
expr min lq median uq max
1 fastmean(dat) 15.706 17.489 25.8825 29.615 79.113
2 mean(with(dat, rep(value, freq))) 1827.146 2283.551 2534.7210 2884.933 26196.923
```
The replicating solution appears to be O( N^2 ) in the number of entries.
The fastmean solution appears to have a 12ms or so fixed cost after which it scales beautifully.
More benchmarking
```r
Comparison with dot product.
dat <- data.frame(freq=seq(600),value=runif(600)100)
dbaupp <- function(dat) {
total.count <- sum(dat$freq)
as.vector(dat$freq %% dat$value) / total.count
}
microbenchmark(
fastmean(dat),
mean( with(dat, rep(value,freq) ) ),
dbaupp(dat)
)
Unit: microseconds
expr min lq median uq max
1 dbaupp(dat) 20.162 21.6875 25.6010 31.3475 104.054
2 fastmean(dat) 14.680 16.7885 20.7490 25.1765 94.423
3 mean(with(dat, rep(value, freq))) 489.434 503.6310 514.3525 583.2790 30130.302
```
Share
Share a link to this answer
Copy linkCC BY-SA 3.0
Improve this answer
Follow
Follow this answer to receive notifications
edited May 1, 2012 at 13:20
answered May 1, 2012 at 12:52
Ari B. FriedmanAri B. Friedman
73k 35 35 gold badges 183 183 silver badges 238 238 bronze badges
Comments
Add a comment
This answer is useful
7
Save this answer.
Show activity on this post.
I might be missing something, but this seems to work very quickly, even substituting millions in the frequency column:
```r
dset <- data.frame(V1=1:6,V2=c(11613,6517,2442,687,159,29))
mean(rep(dset$V1,dset$V2))
1.664102
sd(rep(dset$V1,dset$V2))
0.8712242
```
Share
Share a link to this answer
Copy linkCC BY-SA 3.0
Improve this answer
Follow
Follow this answer to receive notifications
answered May 1, 2012 at 12:50
thelatemailthelatemail
94.3k 12 12 gold badges 139 139 silver badges 197 197 bronze badges
Comments
Add a comment
This answer is useful
6
Save this answer.
Show activity on this post.
How about:
```r
m = sum(dat$V1 dat$V2) / sum(dat$V2)
m
1.664102
sigma = sqrt(sum((dat$V1 - m)2 dat$V2) / (sum(dat$V2)-1))
sigma
0.8712242
```
No replication here.
Share
Share a link to this answer
Copy linkCC BY-SA 3.0
Improve this answer
Follow
Follow this answer to receive notifications
edited May 1, 2012 at 13:04
answered May 1, 2012 at 12:55
OliverOliver
150 4 4 bronze badges
Comments
Add a comment
This answer is useful
5
Save this answer.
Show activity on this post.
The following code doesn't use replication, and it uses R builtins (for the dot product especially) as much as possible so it is probably more efficient that solutions that use sum(V1 V2). (Edit: this is possibly false: @gsk3's solution seems to be about 1.5 - 2 times faster from my testing.)
Mean
The definition of mean (or expectation) is sum(n freq(n)) / total.count where n is the "score" and freq(n) is the frequency of n (total.count is just sum(freq(n))). The sum in the numerator is precisely the dot product of the scores with the frequencies.
The dot product in R is %% (it returns a matrix, but this can basically be treated at a vector for most purposes):
```r
total.count <- sum(dat$V2)
mean <- dat$V1 %% dat$V2 / total.count
mean
[,1]
[1,] 1.664102
```
SD
There is a formula at the end of this section of the Wikipedia article, which translates to the following code
```r
sqrt(dat$V1^2 %% dat$V2 / total.count - mean^2)
[,1]
[1,] 0.8712039
```
Share
Share a link to this answer
Copy linkCC BY-SA 3.0
Improve this answer
Follow
Follow this answer to receive notifications
edited May 1, 2012 at 13:21
answered May 1, 2012 at 13:04
huonhuon
103k 24 24 gold badges 238 238 silver badges 229 229 bronze badges
10 Comments
Add a comment
Ari B. Friedman
Ari B. FriedmanOver a year ago
Also, I'm not sure it's any faster (see my benchmarking). I rather disagree with the implication that built-ins are always faster (see mean()!). But it's a lovely solution and it scales nicely.
2012-05-01T13:12:29.657Z+00:00
0
Reply
Copy link
huon
huonOver a year ago
@gsk3, nope, total.count is the total number of things/samples so dat$V2. And yep, I had literally just done some benchmarking myself and had come to the same conclusion (I was just editing my answer to that affect). It's a little peculiar that the operation that gives R more information (i.e. %%) is slower than the naive one!
2012-05-01T13:16:22.277Z+00:00
0
Reply
Copy link
Ari B. Friedman
Ari B. FriedmanOver a year ago
Oh that's funny. I'd been working on the opposite premise when I made up my example, but the initial question clearly states V2 is frequency. Didn't matter since I renamed them anyway :-). And it's unfortunate that R built-ins are not always (often?) optimized for speed. I wish someone would spend some time optimizing things for Summer of Code or something.
2012-05-01T13:18:53.013Z+00:00
0
Reply
Copy link
huon
huonOver a year ago
It looks like Julia might be an appropriate substitute sometime soon (and they are focusing on performance from the start!). Also, taking out the as.vector calls makes them a little faster (but not as fast as I thought, if anyone saw my edit, haha).
2012-05-01T13:29:01.007Z+00:00
0
Reply
Copy link
Ari B. Friedman
Ari B. FriedmanOver a year ago
@Tommy , no, didn't you know that all Julia code is at least 60x faster than R code that the Julia creators wrote to compare it against? Here's their version: Rmean <- function(dat) { Sys.sleep(10); fastmean(dat) }
2012-05-01T16:51:42.403Z+00:00
2
Reply
Copy link
Add a comment|Show 5 more comments
Your Answer
Thanks for contributing an answer to Stack Overflow!
Please be sure to answer the question. Provide details and share your research!
But avoid …
Asking for help, clarification, or responding to other answers.
Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Draft saved
Draft discarded
Sign up or log in
Sign up using Google
Sign up using Email and Password
Submit
Post as a guest
Name
Email
Required, but never shown
Post Your Answer Discard
By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.
Start asking to get answers
Find the answer to your question by asking.
Ask question
Explore related questions
r
statistics
mean
See similar questions with these tags.
R Language Collective See more
This question is in a collective: a subcommunity defined by tags with relevant content and experts.
The Overflow Blog
The history and future of software development (part 1)
Getting Backstage in front of a shifting dev experience
Featured on Meta
Spevacus has joined us as a Community Manager
Introducing a new proactive anti-spam measure
New and improved coding challenges
New comment UI experiment graduation
Policy: Generative AI (e.g., ChatGPT) is banned
Report this ad
Report this ad
Linked
1dealing with data table with redundant rows
Related
1Find the mean and standard deviation of all values in a data frame
5R computing mean, median, variance from file with frequency distribution
0Calculating means with list data in R
0How to calculate standard deviation of a table in R?
2Calculate the mean and sd of a column within a list
2A simpler way to achieve a frequency count with mean, sum, length and sd in R
6How to get Mean and Standard deviation from a Frequency Distribution table
0Compute mean across variable using R
0Frequency Table with Column of Average in R
0Calculating the mean from frequency table
Hot Network Questions
Checking model assumptions at cluster level vs global level?
Interpret G-code
Do we need the author's permission for reference
Weird utility function
Making sense of perturbation theory in many-body physics
If Israel is explicitly called God’s firstborn, how should Christians understand the place of the Church?
For every second-order formula, is there a first-order formula equivalent to it by reification?
How different is Roman Latin?
An odd question
How to start explorer with C: drive selected and shown in folder list?
Is it safe to route top layer traces under header pins, SMD IC?
Drawing the structure of a matrix
Overfilled my oil
How do you create a no-attack area?
Can peaty/boggy/wet/soggy/marshy ground be solid enough to support several tonnes of foot traffic per minute but NOT support a road?
Bypassing C64's PETSCII to screen code mapping
Are there any legal strategies to modify the nature and rules of the UN Security Council in the face of one, or more SC vetos?
Can a cleric gain the intended benefit from the Extra Spell feat?
What can be said?
Proof of every Highly Abundant Number greater than 3 is Even
Numbers Interpreted in Smallest Valid Base
The geologic realities of a massive well out at Sea
Is existence always locational?
In Dwarf Fortress, why can't I farm any crops?
Question feed
Subscribe to RSS
Question feed
To subscribe to this RSS feed, copy and paste this URL into your RSS reader.
lang-r
Why are you flagging this comment?
Probable spam.
This comment promotes a product, service or website while failing to disclose the author's affiliation.
Unfriendly or contains harassment/bigotry/abuse.
This comment is unkind, insulting or attacks another person or group. Learn more in our Code of Conduct.
Not needed.
This comment is not relevant to the post.
Enter at least 6 characters
Something else.
A problem not listed above. Try to be as specific as possible.
Enter at least 6 characters
Flag comment Cancel
You have 0 flags left today
Stack Overflow
Questions
Help
Chat
Products
Teams
Advertising
Talent
Company
About
Press
Work Here
Legal
Privacy Policy
Terms of Service
Contact Us
Your Privacy Choices
Cookie Policy
Stack Exchange Network
Technology
Culture & recreation
Life & arts
Science
Professional
Business
API
Data
Blog
Facebook
Twitter
LinkedIn
Instagram
Site design / logo © 2025 Stack Exchange Inc; user contributions licensed under CC BY-SA. rev 2025.9.29.34589
By clicking “Accept all cookies”, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy.
Accept all cookies Necessary cookies only
Customize settings |
189066 | https://www.tckpublishing.com/irony/ | Select Page
Irony in Literature: Types and Examples Explained
by Kaelyn Barron
While living abroad, I found myself in a pretty frustrating situation.
For exactly one year following my graduation, I was unable to seek employment while I awaited the approval of my residency permit. As the months passed and all my friends back home were finding stable work, I grew increasingly anxious and began to feel like a total failure.
Then, on almost the exact one-year anniversary of my graduation, I was offered a full-time remote position with a U.S. company—my status abroad wouldn’t matter.
Two days later, my work visa was approved.
Oh, the irony.
Irony in Literature
Perhaps one of the reasons we love a good case of irony in film or literature is that our own lives are so often filled with it.
The effect can be frustrating, comical, or even tragic—but we can’t escape it in real life, so why not embrace it in literature, too?
When used skillfully, irony can add both depth and substance to your writing.
What Is Irony?
Irony is a figure of speech in which the intended meaning of words is different from their actual meaning.
If you’re fluent in sarcasm, this might sound like the same thing. Sarcasm is in fact a type of verbal irony, but whereas sarcasm only characterizes someone’s speech, irony can be found in words, situations, or circumstances.
Simply put, irony is the difference between appearance (or expectations) and reality.
(Did you just think of that scene from 500 Days of Summer? Me too.)
Irony can add an exciting twist, strike at our emotions, and remind us of our own lives, making a work of fiction more relatable and reflective of the human experience.
Types of Irony
There are three main types of irony that frequently appear in literature and art. They are verbal, situational, and dramatic.
Verbal Irony
Verbal irony is the form that probably reminds you most of sarcasm. When you say the opposite of what you mean, you are using verbal irony.
For example:
After your sister says something foolish, you respond, “What a great idea, genius!”
Verbal irony usually produces a comic effect, although often at the expense of the speaker or someone else.
Lemony Snicket frequently uses verbal irony, which is used to paint an image while simultaneously producing a comedic effect. Take this example from his Unauthorized Autobiography:
“Today was a very cold and bitter day, as cold and bitter as a cup of hot chocolate; if the cup of hot chocolate had vinegar added to it and were placed in a refrigerator for several hours.”
Verbal irony forces readers to think a bit harder about the writer’s intentions, and usually offers them a snicker or smirk in return for their efforts.
Situational Irony
Situational irony involves sharp contrasts or contradictions between the audience’s expectations and the actual outcome of a situation.
A pilot with a fear of heights is an example of situational irony, since we would expect someone who works at an altitude of 37,000 feet to be a bit more accustomed to heights.
This kind of irony is commonly employed in sitcoms or comedic films. However, situational irony can also take a tragic direction, like if a fire station burns to the ground, or a man in need of help is run over by an ambulance.
Dramatic Irony
When dramatic irony is used in films or literature, the audience knows more about what is really going on (or what is to come) than the oblivious characters.
More often than not, dramatic irony produces a tragic effect. For example, in William Shakespeare’s Romeo and Juliet, we the readers know that Juliet has taken a sleeping potion to fake her death, but this is unbeknownst to Romeo, who believes she is really dead and proceeds to actually kill himself.
Dramatic irony fills readers with anticipation and heightens their interest. They know what is to come, but are powerless in the situation and cannot intervene or advise the characters.
Examples of Irony in Literature and Film
Can you guess what type of irony is being used in the following examples?
Example 1: The Wonderful Wizard of Oz by L. Frank Baum
In The Wonderful Wizard of Oz, three of the supporting characters, the Scarecrow, Tin Man, and Lion, all wish for traits that they already possess.
Scarecrow thinks he’s unintelligent, but by the end discovers he is actually a genius; Tin Man wishes for a heart, but comes to realize he is very much capable of love; Lion seems cowardly, but turns out to be extremely courageous.
This is an example of:
A. Verbal Irony
B. Situational Irony
C. Dramatic Irony
Example 2: Othello by William Shakespeare
Othello trusts Iago, who has become his main advisor. Unbeknownst to Othello, however, Iago has long despised him, claiming he was unfairly passed over for a promotion to the rank of Othello’s lieutenant.
The audience knows that Iago is manipulating Othello and plotting against him, but Othello does not see this.
This is an example of:
A. Verbal Irony
B. Situational Irony
C. Dramatic Irony
Example 3: Julius Caesar by William Shakespeare
In his speech, Marc Antony refers to Brutus several times as an “honorable man,” knowing that Brutus aided in the murder of Caesar.
This is an example of:
A. Verbal Irony
B. Situational Irony
C. Dramatic Irony
Example 4: Beauty and the Beast
In the Disney animated film Beauty and the Beast, viewers know that the Beast is actually a prince who was once very handsome, but Belle has no idea that this is the case.
This is an example of:
A. Verbal Irony
B. Situational Irony
C. Dramatic Irony
Example 5: Fahrenheit 451 by Ray Bradbury
In the dystopian novel Fahrenheit 451, firefighters burn books rather than extinguishing fires.
(Also ironic? This book, which makes a statement about the perils of censorship, was banned for a time in the United States.)
This is an example of:
A. Verbal Irony
B. Situational Irony
C. Dramatic Irony
Quiz Answers:1.B; 2.C; 3.A; 4.C; 5.B.
Writing with Irony
For readers, irony in literature can be extremely appealing because it reflects the unexplainable ironies and coincidences that we encounter in our own lives. It also builds suspense, peaking our interest and at at times making it impossible to put the book down.
By using irony in your own creative writing, you can add meaning to your text and make your story more intriguing for audiences.
What’s your favorite example of irony in film or literature? Share it with us in the comments below!
If you enjoyed this post, then you might also like:
3 Killer Plot Twists in Fiction: And How They Blow Our Minds
What is Foreshadowing? The Opposite of a Flashback
How to Write Fiction from Multiple Viewpoints: Picking the Right Point of View
Writing Quiz: Are You a Plotter or a Pantser?
Kaelyn Barron
As a blog writer for TCK Publishing, Kaelyn loves crafting fun and helpful content for writers, readers, and creative minds alike. She has a degree in International Affairs with a minor in Italian Studies, but her true passion has always been writing. Working remotely allows her to do even more of the things she loves, like traveling, cooking, and spending time with her family.
Learn More About
Best Books Lists (412)
Fiction (242)
Nonfiction (78)
Blogging (50)
Book Marketing (181)
Book Promotion (30)
How to Get Reviews (9)
Book Publishing (213)
Audiobooks (17)
Book Design (11)
Ebook Publishing (13)
Hybrid Publishing (8)
Print Publishing (9)
Self Publishing (71)
Traditional Publishing (54)
Editing (36)
How to Find an Editor (12)
Health (52)
Fitness (4)
Mindfulness and Meditation (8)
Miscellaneous (125)
New Releases (17)
Personal Development (364)
Career Development (88)
Online Courses (46)
Productivity (56)
Personal Finance (21)
Podcast (179)
Poetry (38)
Poetry Awards Contest (3)
Publishing News (9)
Readers Choice Awards (5)
Reading Tips (146)
Tools and Resources (409)
Software (18)
Technology (18)
Writing (890)
Contests (4)
Grammar (69)
Word Choice (78)
Writing a Book (137)
Writing Fiction (210)
Writing Nonfiction (87) |
189067 | https://www.youtube.com/watch?v=IICR-w1jYcA | Newton's Law of Cooling | First order differential equations | Khan Academy
Khan Academy
9070000 subscribers
2754 likes
Description
542598 views
Posted: 24 Sep 2014
Another separable differential equation example.
Watch the next lesson:
Missed the previous lesson?
Differential Equations on Khan Academy: Differential equations, separable equations, exact equations, integrating factors, homogeneous equations.
About Khan Academy: Khan Academy offers practice exercises, instructional videos, and a personalized learning dashboard that empower learners to study at their own pace in and outside of the classroom. We tackle math, science, computer programming, history, art history, economics, and more. Our math missions guide learners from kindergarten to calculus using state-of-the-art, adaptive technology that identifies strengths and learning gaps. We've also partnered with institutions like NASA, The Museum of Modern Art, The California Academy of Sciences, and MIT to offer specialized content.
For free. For everyone. Forever. #YouCanLearnAnything
Subscribe to KhanAcademy’s Differential Equations channel::
Subscribe to KhanAcademy:
100 comments
Transcript:
[Voiceover] Let's think
about another scenario that we can model with the
differential equations. This is a scenario where we take an object that is hotter or cooler than
the ambient room temperature, and we want to model how
fast it cools or heats up. And the way that we'll think about it is the way that Newton thought about it. And it is described as
Newton's Law of Cooling. And in a lot of ways, it's common sense. It states that the rate
of change of temperature should be proportional
to the difference between the temperature of the object
and the ambient temperature. So let me write that
in mathematical terms. So Newton's Law of Cooling tells us, that the rate of change of temperature, I'll use that with a capital T, with respect to time, lower case t, should be proportional to the difference between the temperature of the object and the ambient temperature. So that is a mathematical
description of it. And once again, it's common sense. If something is much, much hotter than the ambient temperature,
the rate of change should be pretty steep, it should be declining in temperature quickly. If something is much, much cooler, it should be increasing
in temperature quickly. And if something is
close, if these two things are pretty close, well
maybe this rate of change shouldn't be so big. Now I know one thing that you're thinking. You're like, okay, if
the temperature is hotter than the ambient temperature,
then I should be cooling. My temperature should be decreasing. And a decreasing temperature would imply a negative instantaneous change. So how will this be a negative value in the case where our
temperature of our object is greater than our ambient temperature? And the way that that would happen is, you would have to have a negative k. If you don't like thinking
in terms of a negative k, you can just put a
negative right over here and now you would have a positive k. Now it makes sense. If our thing is hotter, if
it has a higher temperature than the ambient temperature,
so this is a positive, then our rate of change will be negative, will be getting cooler. If it was the other way around, if our temperature of our
object is cooler than our ambient temperature, then this thing is going to be a negative,
and then the negative of that is going to be a positive,
we're assuming a positive k, and our temperature will be increasing. So hopefully, this makes
some intuitive sense. And our constant k could depend on the specific heat of the
object, how much surface area is exposed to it, or whatever else. But now I'm given this,
let's see if we can solve this differential equation
for a general solution. And I encourage you to pause
this video and do that, and I will give you a clue. This is a separable differential equation. So I assume you've had a go at it, so let's now work through it together. So, we just have to
algebraically manipulate this so all my Ts and dTs are on one side. I should say, so all my capital
Ts and dTs are on one side, this is going to be a
little bit more confusing because I have a capital
T and a lower case t. Capital T for temperature,
lower case t for time. But hopefully we'll be
able to work through it. And then I'm going to have
all my time differentials and time variables on the other side. So one thing I could is
I could divide both sides by T minus ambient
temperature, minus T sub a. Remember this is just
going to be a constant based on what our ambient temperature is. We're going to assume
our ambient temperature doesn't change as a function of time, it's just such a big
room that our cup of tea is not going to actually warm up the room. So that's just one of these assumptions that we're going to make. So if we do that, if we
divide both sides by this, we are going to have... So I'm going to divide both sides, I'm going to do this in a new color. If I divide both sides
by that, I get one over T minus T sub a, and let
me multiply both sides times the time differential. So I'm going to have, that dT,
our temperature differential. Times our temperature differential,
is going to be equal to negative k times our time differential. So once again, to separate the variables, all I did was divide both sides by this, and multiply both sides by that. Now I can integrate both sides, we've seen this show before. So I can integrate both sides. And the integral of this is going to be the natural log of the absolute value of what we have in the denominator. And you can do u substitution if you want. If we said u is equal to T minus T sub a, then du is just going to be one dT, and so this is essentially,
you could say the integral of one over u du, and so
it would be the natural log of the absolute value of u, and this right over here is u. So this is the natural
log of the absolute value of T minus T sub a, is equal to, and once again I could
put a constant here, but I'm going to end up with a constant on the right hand side too so
I'm just going to merge them into the constant on the right hand side. So that is going to be equal to, now here, this is going to be negative kt, and once again we have plus C. And now we can raise e
to both of these powers, or another way of interpreting this is if e to this thing is
going to be the same as that. So we can write this
as, the absolute value, let me do that in that same blue color. We can write this as the
absolute value of T minus T sub a is equal to e, something about e I always think of the color green. e to the negative kt plus C. This of course is the same thing as, this is equal to e to the negative kt, we've done this multiple times before. Negative kt times e to the C power. And we could just call this
another arbitrary constant. If we called this C1,
then we could just call this whole thing C. So this we could say is
Ce to the negative kt. So at least it's starting
to resemble what we did when we were modelling population. We'll see it's a little bit different. Instead of just temperature
on this left hand side, we have temperature minus
our ambient temperature. And so, we can do a couple of things. If, in a world, say we were
dealing with a hot cup of tea, something that's hotter than
the ambient temperature. So we could imagine a world where T is greater than or equal to
our ambient temperature. So that means this is hot, or it's hotter, I guess we could say. So if we're dealing with something hotter than the ambient temperature, then this absolute value
is going to be positive or the thing inside the absolute value is going to be positive. So we don't need the absolute value. Or the absolute value of it is going to be the same thing as it. And then we can just add
T sub a to both sides, and then we would have our temperature, and I can even write this
as a function of time, is going to be equal to this business, is going to be equal to Ce, let me do that in that same color. Ce to the negative kt plus T sub a. All I did is I'm assuming that this inside the absolute value
is going to be positive, so the absolute value is not
going to change the value. And I added T sub a to
both sides to get this. So this right over here is going
to be our general solution, in the case where we start
with something that is hotter than the ambient room temperature. And if we want to look at the
case where something is cooler than the ambient room temperature, so that's the situation,
let's say T is less than our ambient room temperature. Then the absolute value of
T, then this thing over here is going to be negative, and
so the absolute value of it's going to be the negative of that. So then this up here
results in T sub a minus T, that's going to be the same
thing as the absolute value, it's going to be the
negative of the negative. So then that is going to be
equal to e to the negative k plus, actually let me just do it... T sub a minus T is going to be equal to Ce to the negative kt, so
this is equal to that. I'm just assuming that
T is less than T sub a. And so then, to solve for T,
you could add T to both sides and subtract this from both sides. You would have T as a function
of t is going to be equal to, let's see, if this went onto that side and this goes over here,
you would have T sub a minus Ce to the negative kt. Did I do that right? So yep, that looks right. So this is the situation where
you have something that is cooler than the ambient temperature. So this right over
here, based on the logic of Newton's Law of Cooling,
these are the general solutions to that differential equation. In the next video we can actually apply it to model how quickly something
might cool or heat up. |
189068 | https://www.youtube.com/watch?v=3o5WmeIPZ8k | Rewriting Quadratic Functions To Vertex Form y = a(x - h)² + k: 9th Grade Math
MATH TEACHER GON
655000 subscribers
1744 likes
Description
124363 views
Posted: 18 Oct 2023
Rewriting Quadratic Functions To Vertex Form y = a(x - h)² + k: 9th Grade Math
Follow me on my social media accounts:
Facebook:
Facebook:
Tiktok:
YouTube:
quadraticfunctions
grade9math
mathteachergon
87 comments
Transcript:
hi guys it's me teacheron in our today's video we will talk about the vertex form of quadratic function in the entire video we'll you will be encountering quadratic function and we will convert that into the form y = to a x - h^ 2 + K or the vertex form of any quadratic function so without further Ado let's do this topic so what we have here is a given example wherein we are asked to rewrite y = x^2 - 4x - 10 to the form Y is = a x - h^ 2 + K and this is the vertex form currently this given quadratic function is in general form and we will convert this into this now what are the things that you need to remember in this kind of situation you need to remember or you need to be to master the completing the square process okay so what is the first step the first step is first I will copy this one y = to x^2 - 4x - 10 now first step is to regroup the terms we have have x^2 - 4x - 10 gr terms with the X variable separated with a constant so it goes like this you have your Y is equal to the quantity of x² - 4x - 10 as you can see I separated this term and this term because they are the terms with the variable X compared to the constant now Target see in this x^2 - 4x to is to make it a perfect square trinomial remember square is another term that will make this binomial a perfect square trinomial so we have the Formula b/ 2 squared here in x^2 - 4x are B is -4 so that is -4 / 2 2ar -4 divided by 2 is -2 and then Square it as you can see the answer is four this is now the constant that we will put here okay so what we have here is y is equal to the quantity of x^2 minus 4X and then plus + 4 this plus 4 it came here perfect square trinomial x - h 2 and then minus 10 and is we will put here minus 4 so let me explain 4 remember Square Four for example four S right side plus four on the left side but it will become that's and another reason to negate this plus4 for example from this equation into this equation this equation or this function is different from this function plusus 4 here Z 4us 4 is z and this function is the same as this function okay so continue since we we already have this trinomial or perfect square trinomial we can express this like this one so you have y equal to square binomial that is x - 2^ 2 get the square root of this that is x the square root of four is two copy the middle sign put your exponent here then simplify this -10 - 4 is4 and as you can see they look like the same now import when you know how to transform a quadratic function into vertex form values a H and K and this letters a H and K are essential in graphing quadratic function soy identify here a visible that is one H and K H and K uh those serves as the vertex or the coordinates of the vertex paraba h and k h k inverse if this is -2 for H that would be positive2 and as for the K number as is which is netive 14 so I hope guys way on how to write a quadratic function into its vertex form in the next part of our video I will give you another example for you to master this kind of topic now let's move on with item number two we are asked here to convert y = 2x^2 - 4 x + 4 to Y = to a x - h^ 2 + K now the difference here the difference of this example from our previous example is that D the coefficient of the quadratic term is two we don't have which just one process so what we need to do here is to regroup first so we have y is equal to the quantity of 2 x^2 - 4x + 4 after grouping it terms with X variable factor out value a okay factor out nothing for these two terms so it will become Y is equal to 2 2 2x^2 / 2 is x^2 the -4x / 2 that is - 2x and then + 4 so let me repeat the process value a to get this 2x^2 / 2 x^2 -4x / 2 is -2X so what's next here is perfect square trinomial again using the formula negative sorry b b/ 2 squar B is -2 so that is -2 over 2 and then squared so what we have here is -2 / 2 is -1 then squared and the answer is positive1 so we have Y is = 2 x^2 - 2x + 1 to make it a perfect square trinomial then + 4 first example remember four we subtracted four here in this case since Plus one we need to subtract here a onee that is greater than one this term multiply to this coefficient so 1 2 that is equal to 2 per minus 2 how did we get two instead of one one one 2 that is 2 per minus to make it zero and then Express square of binomial you have your Y is equal to 2 Square x s 1 is 1us minus to then squar 4 - 2 is + 2 and this is now the vertex form if you want to identify your a your a is equal to 2 the vertex h k is 1 and two so that's it guys I hope you learn something from this video and as part of our routine guys I want you to convert this function y is equal to x^2 - 4x + 1 into the vertex form comment video what is your answer here so guys if you're new to my channel don't forget to like And subscribe but hit Bell button for you to be updated sating dat uploads again it's meeron bye-bye |
189069 | https://falmatasaba.files.wordpress.com/2013/04/hydraulic-st-chapter-three-assie-kemal.pdf | Chapter 3. Concrete Gravity Dams oromiyaan ni bilisoomti 1 4/9/2013 3.1 Definition A concrete gravity dam is a massive concrete structure, roughly triangular in shape, and designed so that its weight ensures structural stability against the hydrostatic pressure of the impounded water and other forces that may act on the dam. Gravity dams may be classified by plan as straight gravity dams and curved gravity dams, depending up on the axis alignment. In the earlier periods, gravity dams were constructed from masonry. In recent years, however, gravity dams are constructed from concrete. Gravity dams are permanent structures that require little maintenance constructed to greater heights oromiyaan ni bilisoomti 2 4/9/2013 oromiyaan ni bilisoomti 3 Reservoir Capacity 400,000,000 m3 Catchment area 46 km2 Surface area 4 km2 Max. water depth 284 m Power station Installed capacity 2,069 MW Grande Dixence Dam, Switzerland The largest concrete gravity Dam Height 285 m Length 700 m Base width 200 m Volume 6,000,000 m3 4/9/2013 oromiyaan ni bilisoomti 4 Headwater Crest Heel Gallery Outlet Tailwater Toe Foundation Crest: Top of the dam Heel: Dam contact with foundation on the u/s side Toe: Dam contact with foundation on the d/s side Abutment: sides of the valley which the structure of the dam meets Gallery: opening or passage left in the body of the dam for inspection and drainage purposes Outlets: opening to discharge water Headwater: Impounded water Axis 0f Dam 4/9/2013 AbutmentAbutmentNon overflow partHeadwater oromiyaan ni bilisoomti 5 Overflow part /Spillway 4/9/2013 3.2 Types of Concrete Gravity Dams 3.2.1 Conventional Concrete Dams (CC Dams) These are dams constructed with Mass concrete Mass Concrete is any volume of concrete with dimensions large enough to require that measures be taken to cope with generation of heat from hydration of the cement and attendant volume change to minimizing cracking. (American Concrete Institute ACI) Cement Hydration is a very exothermic process, leading to a rise in temperature at the core of very large pours. Expected to reach the maximum with in 1 to 3 days after placement. oromiyaan ni bilisoomti 6 High core temperature during hydration Cooled surface and thremal cracking if the temperature rises to 70 oc Delayed ettringite formation (DEF) or Ii the surface temperature is allowed to deviate greatly from that of the core, i.e., temperature difference between the interior and exterior reaches to 19 oc , thermal cracking will develop. 4/9/2013 Cracks affects water tightness, durability, internal stresses. Several methods for controlling cracks due to thermal stresses exist Block construction: 15m x 15m x 1.5m Mix design that limits heat of hydration reduced cement content using special low heat cement use of pozollana and other admixtures Embedded pipe cooling system The main disadvantages of CC dams include They are expensive They require long construction time oromiyaan ni bilisoomti 7 4/9/2013 3.2.2 Roller Compacted Concrete Dams(RCC Dams) Roller-compacted concrete is simply concrete constructed with the use of earthfill methods. It was introduced in late 1970s The traditional method of placing, compacting, and consolidating mass concrete is a slow process. Improvements in earth moving equipments made the construction of earth and rock fill dams speedier and more cost efficient. According to ACI, RCC is a concrete compacted by roller compaction. The concrete, in its unhardened state, will support a roller while being compacted RCC thus differs from conventional concrete in its consistency requirement (zero slump): dry enough to support roller wet enough to permit adequate distribution of the binder mortar during mixing and vibration oromiyaan ni bilisoomti 8 4/9/2013 oromiyaan ni bilisoomti 9 Advantages of RCC dams include Reduced cost (25 % - 50% less than CC) oLesser cement consumption, less thermal stresses oLess formwork o Transportation, placement, and compaction is easier. Reduced construction time (1-2yrs) o transportation, placement, and compaction is done in highly mechanized way Characteristic RCC Conventional Cement (kg/m3) <150 150-230 Water to Cement ratio 0.5-0.6 0.5-0.7 90 days strength (MN/m2) 20-40 18-40 Layer (m) 0.3 1.5-2.5 4/9/2013 oromiyaan ni bilisoomti 10 4/9/2013 oromiyaan ni bilisoomti 11 4/9/2013 oromiyaan ni bilisoomti 12 4/9/2013 oromiyaan ni bilisoomti 13 4/9/2013 2.3. Loads on Gravity Dams The structural integrity of a dam must be maintained across the range of circumstances or events likely to arises in service. The design is therefore determined through considerations of corresponding spectrum of loading conditions. Gravity dams are subjected to the following main loads / forces 1.
Water pressure (water load) 2.
Weight of the dam 3.
Uplift pressure 4.
Silt pressure 5.
Wave pressure 6.
Ice pressure 7.
Wind load 8.
Earthquake load Headwater Tailwater silt P1 P1 P2 P3 P8 P6 P5 P4 oromiyaan ni bilisoomti 14 P8 4/9/2013 Loads can be classified in terms of applicability or relative importance as primary, secondary, and exceptional loads Primary loads: are identified as those of major importance to all dams irrespective of type. e.g., Water load, self weight, and uplift Secondary loads: are universally applicable although of a lesser magnitude or alternatively are of major importance to certain types of dam. e.g. Silt load, wave pressure, thermal pressure Exceptional loads: are so designed on the basis of limited general applicability of occurrence. e.g. Earthquake loads oromiyaan ni bilisoomti 15 1m For convenience in analysis loads are expressed per metre length of dam, i.e. they are determined for a two dimensional transverse section with unit width parallel to the dam axis. It is similarly convenient to account for some loads in terms of resolved horizontal and vertical components, identified by the use of appropriate subscripts, Ph and Pv respectively. 4/9/2013 2.3.1 Primary Loads 2.3.1.1 Water loads A. Non-overflow section PH PH’ H Z PV’ H’ γw H γw H’ i) U/s vertical face a) Upstream face: Horizontal force The force acts horizontally at from the basis of the dam b) Downstream face: Horizontal and vertical forces Horizontal component The force acts horizontally at from the basis of the dam Vertical component d/s face The force acts vertically at from the toe of the dam H – Headwater depth H’ – Tailwater depth PH – Horizontal Headwater Pressure Force PH ‘– Horizontal Tailwater Pressure Force PV ‘– Vertical Tailwater Pressure Force C D E oromiyaan ni bilisoomti 16 4/9/2013 Vertical force It acts vertically at from the toe of the dam It acts vertically at from the toe of the dam ii) U/s inclined face a) Upstream Face: Horizontal force It acts horizontally at from the basis of the dam PH PH’ H Z PV’ H’ γw H γw H’ c) Downstream face Horizontal force It acts vertically at from the base of the dam Vertical Force it acts vertically at from the toe of the dam C D E PV1 PV2 E F G H J I G’ oromiyaan ni bilisoomti 17 4/9/2013 oromiyaan ni bilisoomti 18 H2 H1 h B. Overflow section Exercise: calculate the water load on an overflow section of a gravity dam 4/9/2013 2.3.1.2 Self weight H H’ The weight of the dam is given by where γc is unit weight of concrete A is the x-sectional area of the dam The force acts through the centroid of the x-sectional area. It will include weight of ancillary structures. oromiyaan ni bilisoomti 19 Pm 4/9/2013 2.3.1.3 Uplift Force Both the dam body and foundation material are permeable. Water in the reservoir percolates through the dam body (lift and construction joints) and foundation material Headwater Tailwater Concrete permeability (cm/s) Rock permeability (cm/s) Soil permeability (cm/s) Cement Granite 5 x 10-9 Gravel 0.01-1 Fresh 2 x 10-4 Sandstone 1.2 x 10-8 Sand 10-3 to 0.1 Ultimate 6 x 10-11 Silt 10-5 to 10-3 Concrete 10-6 to 10-8 Clay < 10-6 oromiyaan ni bilisoomti 20 4/9/2013 T he percolating water exerts an uplift pressure with in the dam body and at the base of the dam. The uplift in the dam body is small compared to at the base. The uplift force due to the uplift pressure at the base depends on two factors A) Area factor : the fraction of the actual area of the base over which the uplift pressure is supposed to act. (Link) B) Intensity factor: the intensity of pressure acting on any point of the base expressed as a fraction of the total head Uplift at the heel = hydrostatic pressure at the u/s headwater Uplift at the heel = hydrostatic pressure at the tailwater Uplift pressure at any intermediate point = linear interpolation Uplift force = Average pressure intensity x Area x Area factor (η) It acts at from the toe Tailwater Pu H H’ B oromiyaan ni bilisoomti 21 4/9/2013 Effects of drains on uplift pressure To reduce uplift pressure, drains are formed through the body of the dam and also drainage holes are drilled in the foundation rock Uplift at the heel Uplift at the toe Uplift pressure at the line of drain Hd – mean effective head at the line of drain Kd is a function of drain geometry The uplift force and acts at Tailwater Pu H H’ B a line of drain oromiyaan ni bilisoomti 22 4/9/2013 2.3.2 Secondary Loads 2.3.2.1 Earth and Silt Load Gravity dams are sometimes subjected to earth pressures on either u/s or d/s face, where the foundation trench is backfilled. Such pressures usually have minor effect on the stability of the structure, and may be ignored in design. Practically all streams transport silts or fine sediments, particularly during floods. The gradual accumulation of fine sediments against the face of the dam generates a resultant horizontal force Psh. The magnitudes of Psh, which is in addition to the water load, is a function of the sediment depth, hs, the submerged unit weight, γ’s , and the active lateral pressure coefficient Ks. When the u/s face have flaring, the sediments will generate vertical force Psv. Psh Psv Psh hs oromiyaan ni bilisoomti 23 4/9/2013 The horizontal force is given by where is the submerged unit weight of the silt sediment is active lateral earth pressure coefficient and ϕs is the angle of shearing resistance hs is the height of the silt sediment The horizontal force act at above the base of the dam The vertical force Just after construction of the dam, the depth (hs) of the silt is zero. It increases gradually with time and finally it becomes equal to the height of the dead storage. It is usual practice to assume the value of hs is equal to the height of the dead storage above the base. oromiyaan ni bilisoomti 24 4/9/2013 oromiyaan ni bilisoomti 25 2.3.2.2 Waver pressure Waves are generated on the surface of the reservoir by the blowing winds, which causes a pressure towards the downstream side Wave pressure depends on the wave height (hw) which depends on fetch (F) and wind velocity (U) a c b hw F Dam Reservoir 4/9/2013 oromiyaan ni bilisoomti 26 a c b hw The wave height hw is given by hw- height of wave in m U – wind velocity km/hr F – Fetch in km The maximum pressure intensity (Smax) due to wave action may be given by The pressure distribution may be assumed to be triangular of base 5/3 hw. The total force due to wave action The force acts a distance of 3/8 hw above the reservoir surface 5/3 hw 1/3 hw 4/3 hw Pwv 3/8 hw Smax 4/9/2013 oromiyaan ni bilisoomti 27 2.3.3 Exceptional loads 2.3.3.1 Earth quake forces Earthquake represents the release of built up stress in the lithosphere. It occurs along fault lines. The released energy propagates in form of seismic waves that causes the ground to shake. Earthquake link Ground motions associated with earthquakes can be characterized in terms of acceleration. The earthquake ground acceleration are expressed as fraction of gravitational acceleration. ground acceleration = αg Although the seismic waves propagate in all direction, for design purpose, the accelerations are resolved in horizontal and vertical components. Horizontal acceleration = αh g Vertical acceleration = αv g Earthquake Seismic waves Ground Motion Equivalent to imparting acceleration to the ground in direction of the wave 4/9/2013 oromiyaan ni bilisoomti 28 During earth quake , as the ground under a dam moves, the dam must also move with it to avoid rupture. This means that the dam has to resist the inertial force caused by the sudden movement of the earth crest. Inertial forces always act opposite to the direction earthquake movement. Two methods for estimating seismic loads exist Pseudostatic (seismic coefficient method) Dynamic method Earthquake forces on foundation-dam-reservoir act in two ways inertial forces / earth quake force on the dam body increase in water pressure 4/9/2013 oromiyaan ni bilisoomti 29 A. Earthquake forces on the body of the dam A.1 Effects of horizontal acceleration The horizontal acceleration can occur in upstream or downstream direction Because a dam is designed for the worst case, the horizontal acceleration is assumed to occur in the direction which would produce the worst combination of the forces. Reservoir Empty Earthquake movement: Upstream Inertial force acts in : Downstream Ground motion Reservoir Full Peh Peh Earthquake movement: Downstream Inertial force acts in : Upstream It acts at the cenetroid of the dam 4/9/2013 oromiyaan ni bilisoomti 30 Pev A.2 Effects of Vertical Acceleration It acts at the cenetroid of the dam Its effect is to modify the weight Dam is separated from the foundation 4/9/2013 oromiyaan ni bilisoomti 31 B. Earthquake forces on the body of water B.1 Effects of horizontal acceleration Dam and foundation move in upstream direction, they push against the reservoir momentarily increasing the water pressure An initial estimate of these forces can be obtained using a parabolic approximation to the theoretical pressure distribution as analysed in Westergaard (1933). Relative to any elevation at depth Z below the water surface, hydrodynamic pressure pewh is determined by Ground motion where zmax is the maximum depth of water at the section of dam considered Ce is a dimensionless pressure factor, and is a function of Z/Zmax and ϴ, the angle of inclination of the upstream face to the vertical. . Zmax Z 4/9/2013 oromiyaan ni bilisoomti 32 The resultant hydrodynamic load is given by And the load acts at 0.4 Z B. Effect of vertical acceleration If the dam has an upstream flare / batter the resultant vertical hydrodynamic load, Pewv, effective above an upstream face batter or flare may be accounted for by application of the appropriate seismic coefficient to vertical water load, Pwv. It is considered to act through the centroid of area thus: ϴ, the angle of inclination of the upstream face to the vertical. 4/9/2013 oromiyaan ni bilisoomti 33 2.3.4 Load Combinations All the forces which are discussed in the preceding sections may not act simultaneously on a dam. A concrete dam should be designed with regard to the most rigorous adverse groupings or combination of load which gave a reasonable probability of simultaneous occurrence. Three nominated load combinations are sufficient for almost all circumstances. In ascending order of severity they may be designated as normal (sometimes usual), unusual and extreme load combinations, here denoted as NLC, ULC and ELC respectively, Load source Qualification Load combination NLC ULC ELC Primary 1.
Headwater At Design Flood Level (MWL) X At Full Reservoir Level (FRL) X X 2. Tail Water At maximum tail water level X X Minimum tail water level X 3. Self weight - X X X 4. Uplift Drains functioning X X Drains inoperative X X X Secondary 5. Silt - X X X Exceptional Seismic - X 4/9/2013 oromiyaan ni bilisoomti 34 The Indian Standard Criteria (IS: 6512-1972) : Load Combination Load A. – Const Condition Dam completed but no water in reservoir and no tail water. B – Normal Operating Condition Full reservoir elevation, normal dry weather tail water, normal uplift, ice and silt (if applicable). C – Flood Discharge Condition Reservoir at maximum flood elevation, all gates open, tail water at flood elevation, normal uplift and silt (if applicable). D Combination A with earthquake. E Combination B with earthquake. F Combination C, but with extreme uplift. G Combination E, but with extreme uplift. 4/9/2013 oromiyaan ni bilisoomti 35 2.4. Modes of failure and criteria for structural stability 2.4.1 Modes of failures A gravity dam fail in the following ways Overturning about the toe Sliding or shear failure Crushing and cracking 2.4.1.1 Overturning When the resultant (R) of all the vertical and horizontal forces acting on a dam at any given section passes outside of the toe, the dam will rotate and overturn at the toe. Generally, a dam is safe against overturning, if the resultant lies with in the middle third of the section R Pwv PH Pm R 4/9/2013 oromiyaan ni bilisoomti 36 2.4.1.2 Sliding or Shear failure When the shear stress developed at any potential path/plane due to the applied horizontal and vertical forces exceed the shearing strength of the material along the path or sliding plane, the dam will fail by sliding Sliding may occur a) At a horizontal lift joint in the dam b) At the base of the dam i.e. dam – foundation interface c) Weak joints and seams at joints and strata in the rock PWV PH PWV PH PWV PH 4/9/2013 oromiyaan ni bilisoomti 37 2.4.1.3 Crushing When the stress that are developed at any point in the dam exceed the strength of concrete, the dam may fail by crushing / cracking a) Compressive stress exceeding compressive strength of concrete b) tensile stress exceeding tensile strength of concrete c) When the stress developed at the foundation exceed the bearing capacity of the foundation 4/9/2013 oromiyaan ni bilisoomti 38 2.4.2 Stability Requirements of Gravity Dam 2.4.2.1 Symbols and sign conventions y z Symbols M – moment ΣH, ΣV – Horizontal & vertical forces C – Cohesion Φ – angle of friction α - angle of sliding plane τ– Shear stress As – Area of shearing surface Convention Restoring Moment Overturning Moment +ve Horizontal force -ve Horizontal force +ve Vertical force -ve Vertical force + - + + - - +y +z +τyz + τ zy Stabilizing Forces : Weight of the dam (Dead Load) The thrust of the tail water. Destabilizing Forces : Head water pressure, Uplift, Wave pressure in the reservoir, Earth and Silt pressure, Seismic forces, +ve Shear 4/9/2013 oromiyaan ni bilisoomti 39 2.4.2.2. Structural Equilibrium Applied Force = Reactive Force ΣM = 0 No rotational movement ΣFx = 0 No translational movement ΣFy = 0 Pwv R PH PU Pm Phs PV’ PH’ R’ R = Resultant of all loads R’ = Foundation Reaction 4/9/2013 oromiyaan ni bilisoomti 40 2.4.2.3 Assumptions in Stability Analysis 1.
Concrete used homogeneous, isotropic and elastic 2.
Dam consist of a number of vertical cantilivers of unit length. The cantilevers act independently 3.
Perfect bond between dam and foundation 4.
All loads are transferred by cantilever action by the foundation. No beam action 5.
The foundation is strong and unyielding. No movement caused in the foundation due to the imposed loads 6.
Small openings, galleries, shafts, do not affect the over all stability. 4/9/2013 oromiyaan ni bilisoomti 41 2.4.2.4 Stability Requirements A gravity dam must be designed such that it is safe against all possible modes of failures, with adequate factor of safety. A dam may fail a. Overturning b. Sliding and shear c. Crushing 2.4.2.5 Overturning Stability (Fo) The moments are about the toe of any horizontal plane ΣM-ve includes the moment generated by uplift. + - Criteria Fo > 1.25 Acceptable Fo > 1.50 Desirable 4/9/2013 oromiyaan ni bilisoomti 42 2.4.2.6 Sliding Stability It is a function of the loading pattern and of the resistant to translational displacement which can be mobilized at any plane. Three methods are available. Sliding Factor (Fss) Shear Friction Factor (FSF) Limit Equilibrium Method (FLE) A. Sliding Factor Used by dam designers in 1900-1930 Resistance to sliding is assumed to be purely frictional with no cohesion a) When the sliding plane is horizontal ΣH ΣV 4/9/2013 oromiyaan ni bilisoomti 43 b) When the sliding plane is at an angle +ve α c) When the sliding plane is at an angle -ve α Criteria Fss ≤ 0.75 for NLC Fss ≤ 0.90 for ELC ΣH ΣH sin αΣH cosα ΣV ΣV cosαΣV sin α α ΣH ΣH sin αΣH cosα ΣV ΣV cosαΣV sin α α 4/9/2013 oromiyaan ni bilisoomti 44 B- Shear Friction Factor (FSF) Introduced by Henny in 1933 Considers both friction and cohesion for shear resistance From Mohr –Coulomb Shear friction factor is the total resistance to shear and sliding to the horizontal load a) When the sliding plane is horizontal b) When the sliding plane is inclined +ve ΣH ΣV α 4/9/2013 oromiyaan ni bilisoomti 45 c) When the sliding plane is inclined –ve Criteria NLC ULC ELC Dam/foundation 3 2 >1 α 4/9/2013 oromiyaan ni bilisoomti 46 C) Limit equilibrium Factor (FLE) It follows the conventional soil mechanics logic τa is the shear strength available τ shear stress developed under the applied loading τa is expressed by mohr-coulomb failure criteria a) When the sliding plane is horizontal b) When the sliding plane is inclined at an angle +ve α c) When the sliding plane is inclined at angle –ve α Criteria FLE = 2 NLC FLE = 1.3 ELC 4/9/2013 oromiyaan ni bilisoomti 47 3.4.2.7 Stress Analysis Gravity method is commonly used 2d elastic dam on rigid foundation stress evaluated in the analysis The various stresses that are determined in gravity method are 1. Vertical normal stress, σz , on horizontal planes; 2. Horizontal normal stress, σy , on vertical planes; 3. Horizontal and vertical shear stresses τZy & τyz; 4. Principal stresses σ1 & σ3; +z σz σz σy σ1 τyz τzy τzy τyz σy +y 4/9/2013 oromiyaan ni bilisoomti 48 A. Vertical Normal Stress Modified beam theory is employed (i.e., combined Axial Load and Bending Moment) T 1 m y Area = T 1 = T ΣM : sum of moment w.r.t cenetroid of the plane y: distance from the cenetroid to point of consideration I: second moment of area of the plane w.r.t the centeroid ΣV : sum of vertical loads excluding uplift For rectangular 2D plane section of unit width The second moment of area and the eccentricity is given by The vertical normal stress is then given by The maximum and minimum normal stresses are at y = T/2 Plan 4/9/2013 A.2 For reservoir empty condition U/s face D/s face oromiyaan ni bilisoomti 49 A.1 For reservoir full condition U/s face D/s face R σzu σzd R σzd σzu For e > T/6; -ve stress develop at u/s face, i.e., Tensile stress 4/9/2013 oromiyaan ni bilisoomti 50 B. Shear Stresses Linearly varying normal stress generate numerically equal and complementary τyz horizontal shear τzy vertical shear Adequate to establish the shear stresses at the boundary Taking a small element on the u/s and d/s faces: link U/s face D/s face 4/9/2013 oromiyaan ni bilisoomti 51 C. Horizontal Normal Stress Horizontal normal stress on vertical planes can be determined from consideration of horizontal shear forces. Differences in horizontal shear forces is balanced by normal stresses on vertical planes U/s D/s 4/9/2013 oromiyaan ni bilisoomti 52 D. Principal Stresses The vertical normal stresses calculated at the boundaries are not the maximum stresses produced anywhere in the dam. The maximum normal stress is the major principal stress that will be developed on the major principal planes Given σz, σy, τ The major and minor principal stresses are σy σz σz σy σ1 τ τ τ τ 4/9/2013 oromiyaan ni bilisoomti 53 The upstream and downstream faces are places of zero shear, and therefore planes of principal stresses. Link. The boundary values of σ1 & σ3 are then determined as follows: For U/S face For D/s face assuming no tail water Criteria Concrete Rock NLC 3 4 ULC 2 2.7 ELC 1 1.3 4/9/2013 |
189070 | https://tutorial.math.lamar.edu/classes/alg/lines.aspx | Paul's Online Notes
Custom Search
| |
| |
Go To
Notes
Practice Problems
Assignment Problems
Show/Hide
Show all Solutions/Steps/etc.
Hide all Solutions/Steps/etc.
Sections
Graphing
Circles
Chapters
Solving Equations and Inequalities
Common Graphs
Classes
Algebra
Calculus I
Calculus II
Calculus III
Differential Equations
Extras
Algebra & Trig Review
Common Math Errors
Complex Number Primer
How To Study Math
Cheat Sheets & Tables
Misc
Contact Me
MathJax Help and Configuration
Notes Downloads
Complete Book
Practice Problems Downloads
Complete Book - Problems Only
Complete Book - Solutions
Assignment Problems Downloads
Complete Book
Other Items
Get URL's for Download Items
Print Page in Current Form (Default)
Show all Solutions/Steps and Print Page
Hide all Solutions/Steps and Print Page
Paul's Online Notes
Home
/
Algebra
/
Graphing and Functions
/ Lines
Prev. Section
Notes
Practice Problems
Assignment Problems
Next Section
Show Mobile Notice
Show All Notes Hide All Notes
Mobile Notice
You appear to be on a device with a "narrow" screen width (i.e. you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best viewed in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (you should be able to scroll/swipe to see them) and some of the menu items will be cut off due to the narrow screen width.
Section 3.2 : Lines
Let’s start this section off with a quick mathematical definition of a line. Any equation that can be written in the form,
Ax+By=CAx+By=C
where we can’t have both AA and BB be zero simultaneously is a line. It is okay if one of them is zero, we just can’t have both be zero. Note that this is sometimes called the standard form of the line.
Before we get too far into this section it would probably be helpful to recall that a line is defined by any two points that are on the line. Given two points that are on the line we can graph the line and/or write down the equation of the line. This fact will be used several times throughout this section.
One of the more important ideas that we’ll be discussing in this section is that of slope. The slope of a line is a measure of the steepness of a line and it can also be used to measure whether a line is increasing or decreasing as we move from left to right. Here is the precise definition of the slope of a line.
Given any two points on the line say, (x1,y1)(x1,y1) and (x2,y2)(x2,y2), the slope of the line is given by,
m=y2−y1x2−x1m=y2−y1x2−x1
In other words, the slope is the difference in the yy values divided by the difference in the xx values. Also, do not get worried about the subscripts on the variables. These are used fairly regularly from this point on and are simply used to denote the fact that the variables are both xx or yy values but are, in all likelihood, different.
When using this definition do not worry about which point should be the first point and which point should be the second point. You can choose either to be the first and/or second and we’ll get exactly the same value for the slope.
There is also a geometric “definition” of the slope of the line as well. You will often hear the slope as being defined as follows,
m=riserunm=riserun
The two definitions are identical as the following diagram illustrates. The numerators and denominators of both definitions are the same.
Note as well that if we have the slope (written as a fraction) and a point on the line, say (x1,y1)(x1,y1), then we can easily find a second point that is also on the line. Before seeing how this can be done let’s take the convention that if the slope is negative we will put the minus sign on the numerator of the slope. In other words, we will assume that the rise is negative if the slope is negative. Note as well that a negative rise is really a fall.
So, we have the slope, written as a fraction, and a point on the line, (x1,y1)(x1,y1). To get the coordinates of the second point, (x2,y2)(x2,y2) all that we need to do is start at (x1,y1)(x1,y1) then move to the right by the run (or denominator of the slope) and then up/down by rise (or the numerator of the slope) depending on the sign of the rise. We can also write down some equations for the coordinates of the second point as follows,
x2=x1+runy2=y1+risex2=x1+runy2=y1+rise
Note that if the slope is negative then the rise will be a negative number.
Let’s compute a couple of slopes.
Example 1 Determine the slope of each of the following lines. Sketch the graph of each line.
The line that contains the two points (−2,−3)(−2,−3) and (3,1)(3,1).
The line that contains the two points (−1,5)(−1,5) and (0,−2)(0,−2).
The line that contains the two points (−3,2)(−3,2) and (5,2)(5,2).
The line that contains the two points (4,3)(4,3) and (4,−2)(4,−2).
Show All Solutions Hide All Solutions
Show Discussion
Okay, for each of these all that we’ll need to do is use the slope formula to find the slope and then plot the two points and connect them with a line to get the graph.
a The line that contains the two points (−2,−3)(−2,−3) and (3,1)(3,1). Show Solution
Do not worry which point gets the subscript of 1 and which gets the subscript of 2. Either way will get the same answer. Typically, we’ll just take them in the order listed. So, here is the slope for this part.
m=1−(−3)3−(−2)=1+33+2=45m=1−(−3)3−(−2)=1+33+2=45
Be careful with minus signs in these computations. It is easy to lose track of them. Also, when the slope is a fraction, as it is here, leave it as a fraction. Do not convert to a decimal unless you absolutely have to.
Here is a sketch of the line.
Notice that this line increases as we move from left to right.
b The line that contains the two points (−1,5)(−1,5) and (0,−2)(0,−2). Show Solution
Here is the slope for this part.
m=−2−50−(−1)=−71=−7m=−2−50−(−1)=−71=−7
Again, watch out for minus signs. Here is a sketch of the graph.
This line decreases as we move from left to right.
c The line that contains the two points (−3,2)(−3,2) and (5,2)(5,2). Show Solution
Here is the slope for this line.
m=2−25−(−3)=08=0m=2−25−(−3)=08=0
We got a slope of zero here. That is okay, it will happen sometimes. Here is the sketch of the line.
In this case we’ve got a horizontal line.
d The line that contains the two points (4,3)(4,3) and (4,−2)(4,−2). Show Solution
The final part. Here is the slope computation.
m=−2−34−4=−50=undefinedm=−2−34−4=−50=undefined
In this case we get division by zero which is undefined. Again, don’t worry too much about this it will happen on occasion. Here is a sketch of this line.
This final line is a vertical line.
We can use this set of examples to see some general facts about lines.
First, we can see from the first two parts that the larger the number (ignoring any minus signs) the steeper the line. So, we can use the slope to tell us something about just how steep a line is.
Next, we can see that if the slope is a positive number then the line will be increasing as we move from left to right. Likewise, if the slope is a negative number then the line will decrease as we move from left to right.
We can use the final two parts to see what the slopes of horizontal and vertical lines will be. A horizontal line will always have a slope of zero and a vertical line will always have an undefined slope.
We now need to take a look at some special forms of the equation of the line.
We will start off with horizontal and vertical lines. A horizontal line with a yy-intercept of bb will have the equation,
y=by=b
Likewise, a vertical line with an xx-intercept of aa will have the equation,
x=ax=a
So, if we go back and look that the last two parts of the previous example we can see that the equation of the line for the horizontal line in the third part is
y=2y=2
while the equation for the vertical line in the fourth part is
x=4x=4
The next special form of the line that we need to look at is the point-slope form of the line. This form is very useful for writing down the equation of a line. If we know that a line passes through the point (x1,y1)(x1,y1) and has a slope of mm then the point-slope form of the equation of the line is,
y−y1=m(x−x1)y−y1=m(x−x1)
Sometimes this is written as,
y=y1+m(x−x1)y=y1+m(x−x1)
The form it’s written in usually depends on the instructor that is teaching the class.
As stated earlier this form is particularly useful for writing down the equation of a line so let’s take a look at an example of this.
Example 2 Write down the equation of the line that passes through the two points (−2,4)(−2,4) and (3,−5)(3,−5).
Show Solution
At first glance it may not appear that we’ll be able to use the point-slope form of the line since this requires a single point (we’ve got two) and the slope (which we don’t have). However, that fact that we’ve got two points isn’t really a problem; in fact, we can use these two points to determine the missing slope of the line since we do know that we can always find that from any two points on the line.
So, let’s start off my finding the slope of the line.
m=−5−43−(−2)=−95m=−5−43−(−2)=−95
Now, which point should we use to write down the equation of the line? We can actually use either point. To show this we will use both.
First, we’ll use(−2,4)(−2,4). Now that we’ve gotten the point all that we need to do is plug into the formula. We will use the second form.
y=4−95(x−(−2))=4−95(x+2)y=4−95(x−(−2))=4−95(x+2)
Now, let’s use (3,−5)(3,−5).
y=−5−95(x−3)y=−5−95(x−3)
Okay, we claimed that it wouldn’t matter which point we used in the formula, but these sure look like different equations. It turns out however, that these really are the same equation. To see this all that we need to do is distribute the slope through the parenthesis and then simplify.
Here is the first equation.
y=4−95(x+2)=4−95x−185=−95x+25y=4−95(x+2)=4−95x−185=−95x+25
Here is the second equation.
y=−5−95(x−3)=−5−95x+275=−95x+25y=−5−95(x−3)=−5−95x+275=−95x+25
So, sure enough they are the same equation.
The final special form of the equation of the line is probably the one that most people are familiar with. It is the slope-intercept form. In this case if we know that a line has slope mm and has a yy-intercept of (0,b)(0,b) then the slope-intercept form of the equation of the line is,
y=mx+by=mx+b
This form is particularly useful for graphing lines. Let’s take a look at a couple of examples.
Example 3 Determine the slope of each of the following equations and sketch the graph of the line.
2y−6x=−22y−6x=−2
3y+4x=63y+4x=6
Show All Solutions Hide All Solutions
Show Discussion
Okay, to get the slope we’ll first put each of these in slope-intercept form and then the slope will simply be the coefficient of the xx (including sign). To graph the line we know the yy-intercept of the line, that’s the number without an xx (including sign) and as discussed above we can use the slope to find a second point on the line. At that point there isn’t anything to do other than sketch the line.
a 2y−6x=−22y−6x=−2 Show Solution
First solve the equation for yy. Remember that we solved equations like this back in the previous chapter.
2y=6x−2y=3x−12y=6x−2y=3x−1
So, the slope for this line is 3 and the yy-intercept is the point (0,−1)(0,−1). Don’t forget to take the sign when determining the yy-intercept. Now, to find the second point we usually like the slope written as a fraction to make it clear what the rise and run are. So,
m=3=31=riserun⇒rise=3,run=1m=3=31=riserun⇒rise=3,run=1
The second point is then,
x2=0+1=1y2=−1+3=2⇒(1,2)x2=0+1=1y2=−1+3=2⇒(1,2)
Here is a sketch of the graph of the line.
b 3y+4x=63y+4x=6 Show Solution
Again, solve for yy.
3y=−4x+6y=−43x+23y=−4x+6y=−43x+2
In this case the slope is −43−43 and the yy-intercept is (0,2)(0,2). As with the previous part let’s first determine the rise and the run.
m=−43=−43=riserun⇒rise=−4,run=3m=−43=−43=riserun⇒rise=−4,run=3
Again, remember that if the slope is negative make sure that the minus sign goes with the numerator. The second point is then,
x2=0+3=3y2=2+(−4)=−2⇒(3,−2)x2=0+3=3y2=2+(−4)=−2⇒(3,−2)
Here is the sketch of the graph for this line.
The final topic that we need to discuss in this section is that of parallel and perpendicular lines. Here is a sketch of parallel and perpendicular lines.
Suppose that the slope of Line 1 is m1m1 and the slope of Line 2 is m2m2. We can relate the slopes of parallel lines and we can relate slopes of perpendicular lines as follows.
parallel : m1=m2perpendicular : m1m2=−1or m2=−1m1parallel : m1=m2perpendicular : m1m2=−1or m2=−1m1
Note that there are two forms of the equation for perpendicular lines. The second is the more common and in this case we usually say that m2m2 is the negative reciprocal of m1m1.
Example 4 Determine if the line that passes through the points (−2,−10)(−2,−10) and (6,−1)(6,−1) is parallel, perpendicular or neither to the line given by 7y−9x=157y−9x=15.
Show Solution
Okay, in order to do answer this we’ll need the slopes of the two lines. Since we have two points for the first line we can use the formula for the slope,
m1=−1−(−10)6−(−2)=98m1=−1−(−10)6−(−2)=98
We don’t actually need the equation of this line and so we won’t bother with it.
Now, to get the slope of the second line all we need to do is put it into slope-intercept form.
7y=9x+15y=97x+157⇒m2=977y=9x+15y=97x+157⇒m2=97
Okay, since the two slopes aren’t the same (they’re close, but still not the same) the two lines are not parallel. Also,
(98)(97)=8156≠−1(98)(97)=8156≠−1
so the two lines aren’t perpendicular either.
Therefore, the two lines are neither parallel nor perpendicular.
Example 5 Determine the equation of the line that passes through the point (8,2)(8,2) and is,
parallel to the line given by 10y+3x=−210y+3x=−2
perpendicular to the line given by 10y+3x=−2.
Show All Solutions Hide All Solutions
Show Discussion
In both parts we are going to need the slope of the line given by 10y+3x=−2 so let’s actually find that before we get into the individual parts.
10y=−3x−2y=−310x−15⇒m1=−310
Now, let’s work the example.
a parallel to the line given by 10y+3x=−2 Show Solution
In this case the new line is to be parallel to the line given by 10y+3x=−2 and so it must have the same slope as this line. Therefore, we know that,
m2=−310
Now, we’ve got a point on the new line, (8,2), and we know the slope of the new line, −310, so we can use the point-slope form of the line to write down the equation of the new line. Here is the equation,
y=2−310(x−8)=2−310x+2410=−310x+4410y=−310x+225
b perpendicular to the line given by 10y+3x=−2. Show Solution
For this part we want the line to be perpendicular to 10y+3x=−2 and so we know we can find the new slope as follows,
m2=−1−310=103
Then, just as we did in the previous part we can use the point-slope form of the line to get the equation of the new line. Here it is,
y=2+103(x−8)=2+103x−803y=103x−743
| | |
--- |
| | | |
189071 | https://flexbooks.ck12.org/cbook/cbse-biology-class-10/section/4.2/primary/lesson/ecosystem-and-its-components/ | 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?
Science
Grade K to 5
Earth Science
Life Science
Physical Science
Biology
Chemistry
Physics
Advanced Biology
FlexLets
Math FlexLets
Science FlexLets
English
Writing
Spelling
Social Studies
Economics
Geography
Government
History
World History
Philosophy
Sociology
More
Astronomy
Engineering
Health
Photography
Technology
College
College Algebra
College Precalculus
Linear Algebra
College Human Biology
The Universe
Adult Education
Basic Education
High School Diploma
High School Equivalency
Career Technical Ed
English as 2nd Language
Country
Bhutan
Brasil
Chile
Georgia
India
Translations
Spanish
Korean
Deutsch
Chinese
Greek
Polski
EXPLORE
Flexi
A FREE Digital Tutor for Every Student
FlexBooks 2.0
Customizable, digital textbooks in a new, interactive platform
FlexBooks
Customizable, digital textbooks
Schools
FlexBooks from schools and districts near you
Study Guides
Quick review with key information for each concept
Adaptive Practice
Building knowledge at each student’s skill level
Simulations
Interactive Physics & Chemistry Simulations
PLIX
Play. 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
4.2
Ecosystem and Its Components
Written by:Deepthi Telikicherla, Ph.D. | Jean Brainard, Ph.D. |
Fact-checked by:The CK-12 Editorial Team
Last Modified: Sep 01, 2025
Lesson
What nonliving things are essential for life?
Living organisms cannot exist without the nonliving aspects of the environment. For example: air, water, and sunlight, which are all nonliving, are all essential to living organisms. Both nonliving and living things make up an ecosystem.
What is an Ecosystem?
An ecosystem consists of all the nonliving factors and living organisms interacting in the same habitat. Recall that living organisms are biotic factors. The biotic factors of an ecosystem include all the populations in a habitat, such as all the species of plants, animals, and fungi, as well as all the micro-organisms. Also recall that the nonliving factors are called abiotic factors. Abiotic factors include temperature, water, soil, and air. Ecology is the study of ecosystems. That is, ecology is the study of how living organisms interact with each other and with the nonliving parts of their environment.
You can find an ecosystem in a large body of fresh water or in a small aquarium. You can find an ecosystem in a large thriving forest or in a small piece of dead wood. Examples of ecosystems are as diverse as the rainforest, the savanna, the tundra, or the desert (Figure below). The differences in the abiotic factors, such as differences in temperature, rainfall, and soil quality, found in these areas greatly contribute to the differences seen in these ecosystems. Ecosystems can include well-known sites, such as the Himalayas in the northern region of India, the Sunderban Mangroves in the east, the Thar desert in the west, and the Eastern and Western Ghats in the south of India.
Ecosystems need energy. Many ecosystems get their energy in the form of sunlight, which enters the ecosystem through photosynthesis. This energy then flows through the ecosystem, passed from producers to consumers. Plants are producers in many ecosystems. Energy flows from plants to the herbivores that eat the plants, and then to carnivores that eat the herbivores. The flow of energy depicts interactions of organisms within an ecosystem.
Matter is also recycled in ecosystems. Biogeochemical cycles recycle nutrients, like carbon and nitrogen, so they are always available. These nutrients are used over and over again by organisms. Water is also continuously recycled. The flow of energy and the recycling of nutrients and water are examples of the interactions between organisms and the interactions between the biotic and abiotic factors of an ecosystem.
Progress
0 / 3
1.
Which among the following are the abiotic factors in a marine ecosystem?
Select all that apply
a
fish larvae
b
dissolved oxygen levels
c
seaweed
d
depth of water
Summary
An ecosystem consists of all the living things and nonliving things interacting in the same area.
Matter is also recycled in ecosystems; recycling of nutrients is important so they can always be available.
Review
Define an ecosystem.
Distinguish between abiotic and biotic factors. Give examples of each.
Where does the energy come from for many ecosystems?
Name two nutrients that are recycled through an ecosystem.
NOTES / HIGHLIGHTS
Please Sign In to create your own Highlights / Notes
Asked by Students
Here are the top questions that students are asking Flexi for this concept:
Overview
An ecosystem consists of all the living things and nonliving things interacting in the same area.
Matter is also recycled in ecosystems; recycling of nutrients is important so they can always be available.
Vocabulary
organism
habitat
population
species
animal
fungus
abiotic factor
soil
ecology
energy
photosynthesis
producer
herbivore
carnivore
biogeochemical cycle
nutrient
Test Your Knowledge
Question 1
What goods do ecosystems provide to humans?
Select all that apply
a
construction materials
b
genetic biodiversity
c
foods
d
medicines
more tangible and less tangible
Question 2
Different species of shark have different types of teeth that can provide clues about what type of food a species of shark commonly eats. What does a shark’s teeth describe about the species of shark in terms of the ecosystem it lives in?
a
habitat
b
niche
c
abiotic factor
d
competitive exclusion principle
The shape of sharks' teeth differ according to their diet. Species that feed on mollusks and crustaceans have dense flattened teeth for crushing, those that feed on fish have needle-like teeth for gripping, and those that feed on larger prey, such as mammals, have pointed lower teeth for gripping and triangular upper teeth with serrated edges for cutting. Sharks continually shed and replace their teeth, with some shedding as much as 35,000 teeth in a lifetime.
Asked by Students
Here are the top questions that students are asking Flexi for this concept:
Related Content
Ecosystems
How Ecosystems Work
Food Pyramid
The Ecosystem - Advanced
Key Ecology Terms
ECOSYSTEM: The Dr. Binocs Show
Ecosystem Ecology: Links in the Chain
| Image | Reference | Attributions |
---
| | | Credit: PavanPrasad_IND Source: License: Pixabay License |
| | | Credit: Photo by form PxHere Source: License: CC0 / Public Domain |
Ask me anything!
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)28/ 100
This lesson has been added to your library.
No Results Found
Your search did not match anything in .
|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 |
189072 | https://ardianumam.wordpress.com/2017/09/27/newtons-method-optimization-derivation-and-how-it-works/ | Newton’s Method Optimization: Derivation and How It Works – Ardian Umam blog
Skip to content
Ardian Umam blog
.Tempat berbagi catatan. Indonesian.
Facebook
IG
YouTube
LinkedIn
GitHub
Soundcloud
Menu
Home
About Me
Science & Engineering
Computer Vision & Image Processing
Machine Learning & Pattern Recognition
Data Mining & Big Data Analitycs
Programming
Electronics
My Thought
Newton’s Method Optimization: Derivation and How It Works
ardianumamMachine Learning, Science & EngineeringSeptember 27, 2017 November 8, 2017 4 Minutes
In machine learning, we usually define a cost function that measures our error between our prediction and the ground truth data. And we will use the cost function to build a good hypothesis function, either for classifier or regressor, by minimizing the cost function. We already discuss here, hereand herehow we can use LSE (Least Square Estimation) to minimize our cost function for regression problem. Those problem are opimization problems, finding the optimal parameter that gives the minimal error of cost function. And now, we will learn another powerful and important method for optimization, which is newton’s method. I will divide this discussion into three parts: (1) newton’s method, (2) taylor series, and (3) newton’s method for optimization.
(1) Newton method – finding root of an equation
Newton method is originally intended to find root of an equation. For example, we have an equation , it is easy to find roots of this equation, by decomposing , we get that the roots of this equation are and . But, what about complex equation to be solved by computer? We can use numerical methodto find the root of an equation, one of them is by using Newton’s method. See picture below.
We have like picture above, we can find root location by using Newton’s method. Suppose we are now in axis , first, we will find tangent line of in axis . Tangent line in this picture is the red-dash line. This first tangent line intersects axis in . Then, we find next tangent line of in . This tangent line will intersect axis in . By iterating this process, we will reach the root location. That is the idea how Newton’s method for finding root.
Now, let’s try to derive the mathematical formula of Newton’s method. Tangent line of in is a linear line that is parallel with in axis . In high school mathematics, we already know that to get such linear function, we can find by:
, where gradient is same with gradient of at
To find the value of , we can plug a pair x-y values of to. In this case, our and . Here we go.
After getting , our tangent line becomes.
We know that our tangent line intersects axis at . Thus, by plugging this coordinate to our tangent line function, we get:
Thus, our Newton’s method is by iterating this equation.
(2) Taylor series – approximating a function
Taylor series is an approximation of a function using series expansion. See picture below.
Source : en.wikipedia.org/wiki/Taylor_series
Picture above illustrates approximation of red line using Taylor series with some order (blue line) in . Higher order, more precise it will be. From this illustration, we can approximate any function using Taylor series. Taylor series of function around axis can be written below.
(3) Newton’s method for optimization
After we know how Newton’s method works for finding root, and Taylor series for approximating a function, we will try to expand our Newton’s method for optimization, finding the minimal value of a function. For how it works, see picture below.
Blue line is our cost function, and we want to minimize it until reaching the optimal point. This will be done very similar with what we already discuss in Newton’s method for finding root before. We can say that Newton’s method for finding root, it uses first order method. In this case, tangent line can be considered as first order approximation of . Whereas, Newton’s method for optimization, it use second order method. In other words, we approximate using Taylor series with second order approximation. This approximation is shown with orange and red-dash lines in picture above.
Suppose we are now in , first, we find second order approximation of in , then, find the minimal value location of it. In picture above, it would be at . This can be done by taking first differential, and make it equal to zero. Then, we take next approximation of in , and find the minimal value location of it. Iterate this process until we reach optimal point.
Now, it’s time to derive the mathematical formula of Newton’s method for optimization. Second approximation of around axis is as follows.
Then, in order to get the minimal value location of approximation above, we take the first differential, and make it equal to zero. Here we go.
Voila! We just derived our Newton’s method for optimization. To find minimal value of our cost function in machine learning, we can iterate using this equation .
Share this:
Click to share on Facebook (Opens in new window)Facebook
Click to share on X (Opens in new window)X
Like Loading...
Related
Published by ardianumam
View all posts by ardianumam
PublishedSeptember 27, 2017 November 8, 2017
Post navigation
Previous Post Some Important Information Theory Concepts in Machine Learning
Next Post Some Important Probability Distributions to Understand Online Learning in Bayesian Inference
3 thoughts on “Newton’s Method Optimization: Derivation and How It Works”
rslzsays: October 9, 2019 at 1:09 pm 2 0 i Rate This
Unfortunately the graphic for Newton’s method is really misleading. The shown parabolas are angled, which is wrong. A parabola cannot be angled, it is always symmetric around it’s minimum. The thing that can change is it’s curvature, controlled by the leading coefficient. Reply
ProVen NutraVestasays: January 29, 2021 at 3:06 pm 0 0 i Rate This
If some one wishes expert view regarding blogging afterward i suggest him/her to pay a visit this web site, Keep up the pleasant job.| Reply
carbofix scamsays: February 9, 2021 at 5:49 am 0 0 i Rate This
I know this site gives quality based articles or reviews and other data, is there any other web page which provides these things in quality?| Reply
Leave a comment Cancel reply
Δ
Search for:
Blog hits
98,423 hits
New book (in Indonesian language)
Recent posts
Image Processing & Computer Vision Video Series
Apache Spark Tutorial for Big Data Analytics
Machine Learning from The Scratch using Python
Understanding How Mask RCNN Works for Semactic Segmentation
Understanding Faster R-CNN for Object Detection
Categories
Categories
Tags
Apache Sparkbayesianbayesian regressionBig Data Analyticsc++classificationclusteringCMakeComputer Visionconfusion matrixData MiningDeep LearningdoolittleElectronicsEM algorithmestimatorFaster RCNNGaussian distributiongaussian mixture modelgradient descentImage ProcessingJacobian matrixkernel methodkNNleast square errorlinear regressionlogistic regressionMachine LearningMask RCNNmathMaximum Likelihood Estimationmutivariate gaussianObject Detectiononline learningOpenCVoptimizationpolar coordinatePolynomial Regressionpower supplypredictive distributionprobabilityProgrammingRegressionregularizationSegmentationTutorialvideo seriesVisual Studio
Website Powered by WordPress.com.
Comment
Reblog
SubscribeSubscribed
Ardian Umam blog
Sign me up
Already have a WordPress.com account? Log in now.
Ardian Umam blog
SubscribeSubscribed
Sign up
Log in
Copy shortlink
Report this content
View post in Reader
Manage subscriptions
Collapse this bar
Loading Comments...
Write a Comment...
Email (Required) Name (Required) Website
%d
Design a site like this with WordPress.com
Get started |
189073 | https://math.stackexchange.com/questions/34931/what-does-the-notation-n-mean | Stack Exchange Network
Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.
Visit Stack Exchange
Teams
Q&A for work
Connect and share knowledge within a single location that is structured and easy to search.
Learn more about Teams
What does the notation n mean?
Ask Question
Asked
Modified 4 years, 8 months ago
Viewed 13k times
3
$\begingroup$
Are there any conventions about the use of $n^$ as notation of a variable? I have seen it for the first time here.
notation
Share
edited Apr 25, 2011 at 1:38
Kevin Reid
1,00677 silver badges1919 bronze badges
asked Apr 24, 2011 at 23:14
Eli FreyEli Frey
13311 gold badge11 silver badge55 bronze badges
$\endgroup$
3
1
$\begingroup$ typically it is used to show that something is new or different. Yet maintains a lot of the aspects of the old. $\endgroup$
picakhu
– picakhu
2011-04-24 23:19:52 +00:00
Commented Apr 24, 2011 at 23:19
2
$\begingroup$ In the post you linked to it's already been defined as the function described in the post. In general it means a lot of things depending on context. $\endgroup$
Qiaochu Yuan
– Qiaochu Yuan
2011-04-24 23:41:27 +00:00
Commented Apr 24, 2011 at 23:41
$\begingroup$ Huh, the original link broke. Oh well, that's life. $\endgroup$
Willie Wong
– Willie Wong
2013-04-29 15:51:59 +00:00
Commented Apr 29, 2013 at 15:51
Add a comment |
3 Answers 3
Reset to default
15
$\begingroup$
There's no general convention. The use in the webpage you cited certainly isn't widely known or used.
Generally, mathematicians like to use adornments to variable names to indicate a particular operation. Mathematicians especially like the use of adornments when the operation is a duality relation. (Perhaps the most prominent counterexample to this rule is the prime-notation for derivatives, where $f'$ is the first derivative of the function $f$ and $f''$ is the second derivative.)
Some examples of adornments that signifies a duality relation include:
$p$ and $\bar{p}$ for complex conjugation
$A$ and $A^t$ for matrix transposition
$A$ and $A^$ for operator adjoints
$f$ and $\hat{f}$ for the Fourier transformation (technically the double Fourier transform is not the identity, so this is not a real symmetric transformation between an object and its dual, but it is close enough)
$\omega$ and ${}^\omega$ for the Hodge dual of forms
A given symbol can however take on different meanings in different contexts. The bar-over-a-symbol and the prime are two very commonly overloaded symbols. The meaning of $$ for the adjoint operation on operations (or matrices) is sometimes applied to the case of a linear operator on the one-dimensional complex vector space; in this case the linear operators can be identified with the complex numbers and we sometimes see the notation $z$ and $z^$ for the complex conjugate pairs.
Aside from some of the uses of the star adornment which are already mentioned in other answers, there is one other use that is common in mathematical analysis, especially in conjunction with the study of partial differential equations (which is as close to a convention as I know for $p$ and $p^$ when $p$ is a real number):
For a fixed integer $n$ (representing the dimension of the space $\mathbb{R}^n$ one is working on), the number $p^$ is defined to be $$ p^ = \frac{np}{n-p} $$ for any $1 \leq p < n$. This number comes up from scaling considerations on function spaces, and is often called the Sobolev conjugate of $p$ due to its appearance in Sobolev inequalities. Personally I think this notation is lousy, but it is somewhat established in the literature.
Share
answered Apr 24, 2011 at 23:37
Willie WongWillie Wong
75.5k1313 gold badges167167 silver badges263263 bronze badges
$\endgroup$
3
6
$\begingroup$ Ohh, I couldn't possibly agree more with "I think this notation is lousy" :) $\endgroup$
t.b.
– t.b.
2011-04-24 23:41:32 +00:00
Commented Apr 24, 2011 at 23:41
1
$\begingroup$ "I think this notation is lousy" - yes, ohh yessss... $\endgroup$
J. M. ain't a mathematician
– J. M. ain't a mathematician
2011-04-24 23:48:17 +00:00
Commented Apr 24, 2011 at 23:48
1
$\begingroup$ As an alternative notation for $\frac{np}{n-p}$, I would suggest letting $n^$ be defined to mean the function $[1,n) \rightarrow \mathbb{R}$ given by $n^(p) = np/(n-p)$. So instead of $p^$, we can write $n^(p)$. $\endgroup$
goblin GONE
– goblin GONE
2015-08-06 02:23:30 +00:00
Commented Aug 6, 2015 at 2:23
Add a comment |
5
$\begingroup$
It can mean:
the adjoint matrix, if $n$ is a matrix
the set of all powers of $n$ is $n$ is a word or letter in a regular expression
the dual space if $n$ is a vector space
the unit elements if $n$ is a ring
...
or it can just be used to denote the image of $n$ under a function called as in your example.
Share
answered Apr 24, 2011 at 23:20
PhiraPhira
21.3k22 gold badges6161 silver badges109109 bronze badges
$\endgroup$
Add a comment |
2
$\begingroup$
Without context it's difficult to say. One common use of the star is for the optimal value or point in an optimization problem. But in the link you provided it's just something related to n.
Share
edited Apr 24, 2011 at 23:33
answered Apr 24, 2011 at 23:26
lhflhf
222k2020 gold badges254254 silver badges585585 bronze badges
$\endgroup$
Add a comment |
You must log in to answer this question.
Start asking to get answers
Find the answer to your question by asking.
Ask question
Explore related questions
notation
See similar questions with these tags.
Featured on Meta
Introducing a new proactive anti-spam measure
Spevacus has joined us as a Community Manager
stackoverflow.ai - rebuilt for attribution
Community Asks Sprint Announcement - September 2025
Related
What does the notation $P[X\in dx]$ mean?
2 What does the following notation mean? (ODEs)
1 What does $<\cdot>$ notation mean in gradient?
1 What does the notation: function[a,b] mean?
1 What does the notation $h \ll 1$ mean?
1 What does the apostrophe here mean in the set?(it's not "complement")
Hot Network Questions
What were "milk bars" in 1920s Japan?
Best solution to prevent loop between tables for granular relations
Traversing a curve by portions of its arclength
How to design a circuit that outputs the binary position of the 3rd set bit from the right in an 8-bit input?
Determine which are P-cores/E-cores (Intel CPU)
Another way to draw RegionDifference of a cylinder and Cuboid
How can I get Remote Desktop (RD) to scale properly AND set maximum windowed size?
Are there any alternatives to electricity that work/behave in a similar way?
An odd question
How do I disable shadow visibility in the EEVEE material settings in Blender versions 4.2 and above?
RTC battery and VCC switching circuit
ICC in Hague not prosecuting an individual brought before them in a questionable manner?
how do I remove a item from the applications menu
Is direct sum of finite spectra cancellative?
"Unexpected"-type comic story. Aboard a space ark/colony ship. Everyone's a vampire/werewolf
Why include unadjusted estimates in a study when reporting adjusted estimates?
How to locate a leak in an irrigation system?
How to home-make rubber feet stoppers for table legs?
Bypassing C64's PETSCII to screen code mapping
What meal can come next?
What "real mistakes" exist in the Messier catalog?
What is this chess h4 sac known as?
How exactly are random assignments of cases to US Federal Judges implemented? Who ensures randomness? Are there laws regulating how it should be done?
Who is the target audience of Netanyahu's speech at the United Nations?
more hot questions
Question feed |
189074 | https://www.mdpi.com/2077-0383/11/6/1583 | Next Article in Journal
Quality of Life in Patients with End-Stage Renal Disease Undergoing Hemodialysis
Previous Article in Journal
Brain Sex in Transgender Women Is Shifted towards Gender Identity
Active Journals Find a Journal Journal Proposal Proceedings Series
## Topics
For Authors For Reviewers For Editors For Librarians For Publishers For Societies For Conference Organizers
Open Access Policy Institutional Open Access Program Special Issues Guidelines Editorial Process Research and Publication Ethics Article Processing Charges Awards Testimonials
## Author Services
Sciforum MDPI Books Preprints.org Scilit SciProfiles Encyclopedia JAMS Proceedings Series
Overview Contact Careers News Press Blog
Sign In / Sign Up
Notice
clear
Notice
You are accessing a machine-readable page. In order to be human-readable, please install an RSS reader.
Continue Cancel
clear
All articles published by MDPI are made immediately available worldwide under an open access license. No special permission is required to reuse all or part of the article published by MDPI, including figures and tables. For articles published under an open access Creative Common CC BY license, any part of the article may be reused without permission provided that the original article is clearly cited. For more information, please refer to
Feature papers represent the most advanced research with significant potential for high impact in the field. A Feature Paper should be a substantial original Article that involves several techniques or approaches, provides an outlook for future research directions and describes possible research applications.
Feature papers are submitted upon individual invitation or recommendation by the scientific editors and must receive positive feedback from the reviewers.
Editor’s Choice articles are based on recommendations by the scientific editors of MDPI journals from around the world. Editors select a small number of articles recently published in the journal that they believe will be particularly interesting to readers, or important in the respective research area. The aim is to provide a snapshot of some of the most exciting work published in the various research areas of the journal.
Original Submission Date Received: .
Journals
Active Journals
Find a Journal
Journal Proposal
Proceedings Series
Topics
Information
For Authors
For Reviewers
For Editors
For Librarians
For Publishers
For Societies
For Conference Organizers
Open Access Policy
Institutional Open Access Program
Special Issues Guidelines
Editorial Process
Research and Publication Ethics
Article Processing Charges
Awards
Testimonials
Author Services
Initiatives
Sciforum
MDPI Books
Preprints.org
Scilit
SciProfiles
Encyclopedia
JAMS
Proceedings Series
About
Overview
Contact
Careers
News
Press
Blog
Sign In / Sign Up Submit
Journals
JCM
Volume 11
Issue 6
10.3390/jcm11061583
Submit to this Journal Review for this Journal Propose a Special Issue
► ▼ Article Menu
Article Menu
Academic Editors
Romain Coriat
Matthias Biebl
Subscribe SciFeed
Recommended Articles
Related Info Links
PubMed/Medline
Google Scholar
More by Authors Links
on DOAJ
Yeh, P.
Wu, R.
Chen, C.
Chiu, C.
Lai, M.
Chen, C.
Kuo, C.
Hsu, J.
Su, M.
Le, P.
on Google Scholar
Yeh, P.
Wu, R.
Chen, C.
Chiu, C.
Lai, M.
Chen, C.
Kuo, C.
Hsu, J.
Su, M.
Le, P.
on PubMed
Yeh, P.
Wu, R.
Chen, C.
Chiu, C.
Lai, M.
Chen, C.
Kuo, C.
Hsu, J.
Su, M.
Le, P.
/ajax/scifeed/subscribe
Article Views
Citations -
Table of Contents
Altmetric share Share announcement Help format_quote Cite question_answer Discuss in SciProfiles
Need Help?
Support
Find support for a specific problem in the support section of our website.
Get Support
Feedback
Please let us know what you think of our products and services.
Give Feedback
Information
Visit our dedicated information section to learn more about MDPI.
Get Information
clear
JSmol Viewer
clear
first_page
Download PDF
settings
Order Article Reprints
Font Type:
Arial Georgia Verdana
Font Size:
Aa Aa Aa
Line Spacing:
Column Width:
Background:
Open AccessArticle
Risk Factors, Clinical and Endoscopic Features, and Clinical Outcomes in Patients with Cytomegalovirus Esophagitis
by
Pai-Jui Yeh
Pai-Jui Yeh
SciProfiles Scilit Preprints.org Google Scholar
1,
Ren-Chin Wu
Ren-Chin Wu
SciProfiles Scilit Preprints.org Google Scholar
2,
Chien-Ming Chen
Chien-Ming Chen
SciProfiles Scilit Preprints.org Google Scholar
3,
Cheng-Tang Chiu
Cheng-Tang Chiu
SciProfiles Scilit Preprints.org Google Scholar
4,5,
Ming-Wei Lai
Ming-Wei Lai
SciProfiles Scilit Preprints.org Google Scholar
1,5,6,
Chien-Chang Chen
Chien-Chang Chen
SciProfiles Scilit Preprints.org Google Scholar
1,
Chia-Jung Kuo
Chia-Jung Kuo
SciProfiles Scilit Preprints.org Google Scholar
4,5,
Jun-Te Hsu
Jun-Te Hsu
SciProfiles Scilit Preprints.org Google Scholar
7,
Ming-Yao Su
Ming-Yao Su
SciProfiles Scilit Preprints.org Google Scholar
5,8 and
Puo-Hsien Le
Puo-Hsien Le
SciProfiles Scilit Preprints.org Google Scholar
4,5,6,
1
Department of Pediatric Gastroenterology, Chang Gung Memorial Hospital, Linkou Branch, Taoyuan 333, Taiwan
2
Department of Pathology, Chang Gung Memorial Hospital, Linkou Branch, Taoyuan 333, Taiwan
3
Department of Medical Imaging and Interventions, Chang Gung Memorial Hospital, Linkou Branch, Taoyuan 333, Taiwan
4
Department of Gastroenterology and Hepatology, Chang Gung Memorial Hospital, Linkou Branch, Taoyuan 333, Taiwan
5
Taiwan Association of the Study of Small Intestinal Disease, Taoyuan 333, Taiwan
6
Liver Research Center, Chang Gung Memorial Hospital, Linkou Branch, Taoyuan 333, Taiwan
7
Department of General Surgery, Chang Gung Memorial Hospital, Linkou Branch, Taoyuan 333, Taiwan
8
Department of Gastroenterology and Hepatology, New Taipei City Municipal Tucheng Hospital, New Taipei City 236, Taiwan
Author to whom correspondence should be addressed.
J. Clin. Med. 2022, 11(6), 1583;
Submission received: 31 January 2022 / Revised: 1 March 2022 / Accepted: 9 March 2022 / Published: 13 March 2022
(This article belongs to the Section Gastroenterology & Hepatopancreatobiliary Medicine)
Download keyboard_arrow_down
Download PDF Download PDF with Cover Download XML Download Epub
Browse Figures
p = 0.044). (B) Patients who required ICU care had a worse survival rate, although it was not statistically significant (log-rank p = 0.101). AKI—acute kidney injury; CMV—cytomegalovirus; ICU—intensive care unit.
" href="
Versions Notes
Abstract
Cytomegalovirus (CMV) esophagitis is the second most common CMV disease of the gastrointestinal tract. This study aims to comprehensively analyze risk factors, clinical characteristics, endoscopic features, outcomes, and prognostic factors of CMV esophagitis. We retrospectively collected data of patients who underwent esophageal CMV immunohistochemistry (IHC) staining between January 2003 and April 2021 from the pathology database at the Chang Gung Memorial Hospital. Patients were divided into the CMV and non-CMV groups according to the IHC staining results. We enrolled 148 patients (44 CMV and 104 non-CMV patients). The risk factors for CMV esophagitis were male sex, immunocompromised status, and critical illness. The major clinical presentations of CMV esophagitis included epigastric pain (40.9%), fever (36.4%), odynophagia (31.8%), dysphagia (29.5%), and gastrointestinal bleeding (29.5%). Multiple diffuse variable esophageal ulcers were the most common endoscopic feature. The CMV group had a significantly higher in-hospital mortality rate (18.2% vs. 0%; p < 0.001), higher overall mortality rate (52.3% vs. 14.4%; p < 0.001), and longer admission duration (median, 24 days (interquartile range (IQR), 11–47 days) vs. 14 days (IQR, 7–24 days); p = 0.015) than the non-CMV group. Acute kidney injury (odds ratio (OR), 174.15; 95% confidence interval (CI), 1.27–23,836.21; p = 0.040) and intensive care unit admission (OR, 26.53; 95% CI 1.06–665.08; p = 0.046) were predictors of in-hospital mortality. In conclusion, the mortality rate of patients with CMV esophagitis was high. Physicians should be aware of the clinical and endoscopic characteristics of CMV esophagitis in high-risk patients for early diagnosis and treatment.
Keywords:
acute kidney injury; cytomegalovirus; endoscopy; esophagitis; prognostic factor
1. Introduction
Cytomegalovirus (CMV) esophagitis is the second most common CMV disease of the gastrointestinal tract and the third leading cause of infectious esophagitis [1,2]. It is usually diagnosed in immunocompromised patients but is also seen in immunocompetent patients [3,4]. Because of the limited number of cases, most studies discussed CMV disease of the upper gastrointestinal tract or infectious esophagitis instead of focusing on CMV esophagitis [3,5,6]. The largest case series of CMV esophagitis enrolled 25 patients and narrated the clinical characteristics and endoscopic features of CMV esophagitis . No study has investigated the risk factors and prognostic factors of CMV esophagitis. This is the first retrospective cohort study to provide comprehensive information on CMV esophagitis, including risk factors, clinical characteristics, endoscopic features, treatments, outcomes, and prognostic factors.
2. Materials and Methods
2.1. Compliance with Ethical Standards
The study protocol was approved by the Institutional Review Board (IRB) of the Chang Gung Medical Foundation (approval document No. 202101234B0. “Clinical presentations and outcome of cytomegalovirus, herpes simplex virus, Epstein-Barr virus, and clostridioides infection”) for the period from 28 July 2021 to 27 July 2022. Due to the retrospective nature of the study, the IRB waived the requirement of signed informed consent from individual patients to review medical records from the electronic medical record system. The study protocol conformed to the ethical guidelines of the 1975 Declaration of Helsinki, as reflected in the prior approval by the institution’s human research committee.
2.2. Patients
In this retrospective cohort study, we enrolled all patients with esophageal CMV immunohistochemistry (IHC) staining results from the pathology database at the Linkou Chang Gung Memorial Hospital who underwent esophageal CMV immunohistochemistry (IHC) staining between January 2003 and April 2021. Patients were divided into the CMV and non-CMV groups according to the IHC staining results. CMV esophagitis was diagnosed based on positive CMV IHC staining of the esophageal tissue, with or without viral inclusion bodies, using hematoxylin and eosin staining (Figure 1). CMV IHC was performed using monoclonal antibodies directed against the CMV pp65 antigen (Novocastra™ lyophilized mouse monoclonal antibody; Leica Microsystems, Wetzlar, Germany).
2.3. Data Collection
Medical records of eligible patients were reviewed for data on age; sex; patient source (outpatient or inpatient); admission duration; date of diagnosis (the date of final pathological confirmation); acquisition time (interval between the date of admission or outpatient clinic visit to the date of diagnosis); recurrence; death or last follow-up; presence of critical conditions, such as shock, pneumonia, and respiratory distress requiring intubation within 1 week before diagnosis; requirement of intensive care unit (ICU) admission; underlying disease; medication history; major clinical presentation; endoscopic findings including lesion characteristics, location, number, and concomitant mucosal findings; histopathological results including presence of malignancy or findings indicating other etiology of esophagitis; results of other imaging tests such as upper gastrointestinal series study and computed tomography; treatment and therapeutic duration; complications; outcomes including admission duration, in-hospital mortality rate, and overall mortality rate; hematological parameters including total white blood cell count, absolute neutrophil count, absolute lymphocyte count, platelet count, and hemoglobin level; biochemical parameters including creatinine, aspartate aminotransferase, alanine aminotransferase, bilirubin, albumin, and C-reactive protein (CRP) levels; CMV pp65 antigenemia; CMV DNA ( Light-Mix® Kit human cytomegalovirus (TIB Molbiol, Berlin, Germany, cut-off: Cp 35, 226 bp segment on glycoprotein B gene), COBAS® AmpliPrep/COBAS® TaqMan® CMV Test (Roche Diagnostics, Branchburg, NJ, USA, cut-off: 150 copies/mL)); and CMV serology.
2.4. Definitions
The locations of esophageal lesions (upper, middle, and lower third) were categorized according to the cancer staging manual of the American Joint Committee on Cancer . Barrett’s esophagus was diagnosed according to the American College of Gastroenterology guideline . Patients were defined as “immunocompromised” if they were documented to have primary immunodeficiency, human immunodeficiency virus (HIV) infection, underlying malignancy with exposure to radiotherapy or chemotherapeutic agents within 6 months, use of immunosuppressants including corticosteroids (oral or intravenous administration ≥20 mg/day of prednisolone or any equivalent for >2 weeks), or were recipients of solid organ or bone marrow transplantation [9,10].
2.5. Statistical Analyses
Numerical data are presented as mean ± standard deviation or median (interquartile range [IQR]), while categorical data are expressed as absolute numbers and percentages. An independent t-test or Mann–Whitney U test was used to compare continuous variables, while the χ2 and Fisher’s exact tests were used for categorical variables. Logistic regression models were used to identify independent risk factors for in-hospital mortality. Statistical significance was set at p < 0.05. The results are presented as odds ratios (ORs), 95% confidence intervals (CIs), and p-values. Survival outcomes were evaluated using the Kaplan–Meier survival curve analysis and log-rank test. All statistical calculations were performed using SPSS statistical software (version 21.0; IBM Corp., Armonk, NY, USA).
3. Results
3.1. Baseline Characteristics of Patients with CMV Esophagitis
A total of 148 patients were enrolled, including 44 and 104 patients in the CMV and non-CMV groups, respectively. The CMV group was predominated by men (77.3%), with a mean age of 59.5 ± 18.5 years. Three-quarters were hospitalized, and 15.9% required ICU care. Up to 77.3% of this group was considered immunocompromised, and the major underlying diseases were malignancy, gastroesophageal reflux disease (GERD), and hypertension. Most malignancies in the CMV group were located in the neck and chest areas, including seven esophageal cancers, six lung cancers, two breast cancers, and one orbital melanoma with thoracic spine metastasis. All 16 patients in this group underwent radiation therapy and had radiation exposure in the region of the esophagus. HIV infection, solid organ transplant (kidney), and autoimmune disease (systemic lupus erythematosus) were noted in eight, three, and two patients, respectively. No patient with Crohn’s disease or prior esophageal surgery was noted in this study. Seven patients had other CMV diseases, including CMV gastritis (three patients), extra-alimentary diseases (three patients; two patients with hepatitis and one patient with retinitis), and one patient with both retinitis and gastritis. The most common baseline medications were antibiotics (75%), proton pump inhibitors (PPIs) (63.6%), and steroids (52.3%). Other details are listed in Table 1.
3.2. Clinical Manifestations of CMV Esophagitis
The most common symptoms were epigastric pain (40.9%), fever (36.4%), odynophagia (31.8%), dysphagia (29.5%), and gastrointestinal bleeding (29.5%). No patient developed esophageal perforation, fistula, stricture, or mediastinitis.
3.3. Results of Laboratory Examinations in Patients with CMV Esophagitis
Anemia, hypoalbuminemia, and elevated CRP levels indicated a poor general condition with inflammation/infection in patients with CMV esophagitis. Although CMV serology tests were valuable, the data were incomplete for a real-world study. According to available data, the positivity rates of CMV IgM, IgG, antigenemia, and viremia were 21.4% (3/14), 92.9% (13/14), 66.7% (8/12), and 57.1% (4/7), respectively.
3.4. Endoscopic Findings in Patients with CMV Esophagitis
All specimens in the study were obtained from endoscopic biopsies without surgical resection. With regard to endoscopic features, diffuse and multiple (≥2) ulcers with variable morphologies and sizes were the most common finding (Figure 2). The lower (86.4%) and middle (61.4%) esophagus were commonly involved areas. Reflux esophagitis, Barrett’s esophagus, and esophageal candidiasis were noted in 38.6%, 9.1%, and 18.2% of patients, respectively. Moreover, 29.5% of the patients had concurrent gastric ulcers. However, no malignant cells were observed in any specimen.
3.5. Radiological Examination of Patients with CMV Esophagitis
In the CMV group, an upper gastrointestinal series was performed for two patients, and nine patients underwent computed tomography for underlying diseases but not for CMV esophagitis. These findings were not related to CMV esophagitis.
3.6. Treatments and Outcomes in Patients with CMV Esophagitis
The median acquisition time was 8 days (IQR, 5–14 days). After diagnosis, 26 (59.1%) patients were administered antiviral treatment (valganciclovir, oral form; ganciclovir, intravenous or oral form), and 31 (70.5%) patients were prescribed PPIs. None of the patients required surgical intervention for CMV-related complications. The median admission duration was 24 days (IQR, 11–47 days). The in-hospital and overall mortality rates were 18.2% and 52.3%, respectively. The median follow-up duration was 276 days (IQR, 64.8–738.8 days) days, and no disease recurrence was observed.
3.7. Differences between the CMV and Non-CMV Groups
With regard to risk factors, there was a significantly higher male to female ratio; more inpatients; more immunocompromised patients, including patients with HIV infection, malignancy, organ transplant, steroid usage, chemotherapy, and radiotherapy; and more critically ill patients, including patients with shock, pneumonia, respiratory failure, and ICU requirement in the CMV group. Antibiotics, PPIs, other antacids, and mucosal protectants were more commonly used in the CMV group before the diagnosis of CMV disease. Clinically, CMV esophagitis was more likely to present with fever, hematemesis, gastrointestinal bleeding, odynophagia, and abdominal fullness. Laboratory examination revealed that patients with CMV esophagitis had lower blood cellular counts, lower albumin levels, and higher ALT levels. These findings suggested poor nutritional status and more severe infection. Regarding endoscopic features, upper and lower esophageal involvement, diffuse/multiple esophageal ulcers, concurrent esophageal candidiasis, and gastric ulcers were more common in the CMV group. Regarding clinical outcomes, patients with CMV esophagitis had a longer hospital stay and higher in-hospital and overall mortality rates.
3.8. Prognostic Factors for In-Hospital Mortality in Patients with CMV Esophagitis
Acute kidney injury (AKI) (OR, 174.148; 95% CI, 1.272–23,836.208; p = 0.04) and ICU requirement (OR, 26.526; 95% CI, 1.058–665.083; p = 0.046) were independent prognostic factors for in-hospital mortality (Table 2). In the Kaplan–Meier survival curve analysis, patients with AKI showed significantly worse survival rates than those without AKI (log-rank p = 0.044) (Figure 3).
4. Discussion
CMV esophagitis affects not only immunocompromised patients but also immunocompetent hosts, although it is predominant and leads to worse outcomes in the former. It is important to understand the risk factors and other information for early diagnosis and treatment, but related studies are limited. Most studies on CMV esophagitis have focused on patients with organ transplantation or HIV infection, [4,11,12,13] and provided clinical presentations and endoscopic findings in these groups of patients. Regarding the general population, the largest and latest case series enrolled 25 patients with CMV esophagitis, of which 4% were immunocompetent . This report presented the clinical features, endoscopic findings, and clinical outcomes of patients with CMV esophagitis. However, it was difficult to identify specific risk factors, clinical characteristics, and endoscopic features without a control group. Therefore, we conducted the largest retrospective cohort study with 148 patients to provide comprehensive information on CMV esophagitis, from risk factors to clinical outcomes.
In this cohort study, we found that male sex, immunocompromised status, and critical illness were risk factors for CMV esophagitis. Many esophageal diseases, including GERD, Barrett’s esophagus, and esophageal adenocarcinoma, also exhibit male predominance [14,15]. This might be attributed to lifestyle, dietary habits, genes, or estrogen. Estrogen contributes to esophageal epithelial resistance to causative insults and modulates cytokine-induced barrier dysfunction, thereby decreasing the risk of esophageal epithelial infection [15,16,17]. As mentioned in previous reports, immunocompromised status, including concurrent chemoradiotherapy, increases the risk of CMV esophagitis because of leukopenia and mucosal defects [1,18]. Critical illness (shock and respiratory failure) leads to multiple organ dysfunction, hypoxia, and hypoperfusion; it is also regarded as an immunodeficient condition and is, therefore, a risk factor for CMV esophagitis. Regarding symptoms and endoscopic findings, when inpatients have fever, hematemesis, odynophagia, diffuse or multiple esophageal ulcers, and esophageal candidiasis, physicians should perform biopsy and CMV IHC staining to rule out CMV esophagitis.
There were 20 cases (45.5%) with malignant diseases in the CMV group, which was higher than in the non-CMV group. Among these patients, 80% of them had received radiation therapy, and all of them had radiation exposure in the area of the esophagus. Additionally, 75% of patients had received chemotherapy. Therefore, there were three possible explanations for the association between malignancies and CMV esophagitis: (1) chemotherapy suppresses immunity and substantially increases the risk of opportunistic infections; CMV reactivation in patients with esophageal cancer has been reported [18,19,20]; (2) local radiotherapy leads to mucosal injury and increases the risk of invasive CMV diseases [1,21]; (3) some studies noted the role of CMV in the pathogenesis of esophageal and other cancers, including immune escapes, tumor microenvironment, and oncomodulation [22,23].
ICU requirement and AKI were independent prognostic factors for in-hospital mortality. Patients with AKI also had poorer Kaplan–Meier survival curves. ICU admission suggests that the patients had severe disease activity and critical conditions, leading to a higher mortality rate. In our previous study of CMV gastritis, AKI was also a negative prognostic factor . In critically ill patients, AKI is associated with higher ICU admission and mortality rates . Furthermore, renal injury leads to T-cell exhaustion and dysfunction, resulting in imbalanced immune regulation and impaired clearance of pathogens [26,27]. Both perspectives were possible explanations for AKI being a prognostic factor for in-hospital mortality. Therefore, we should be alert to the risk factors, symptoms, and endoscopic features of CMV esophagitis and prevent AKI by avoiding agents causing renal toxicity and administering adequate hydration to improve the in-hospital mortality rate. Although some studies mentioned that antiviral therapies improved the in-hospital survival in cytomegalovirus diseases of the whole gastrointestinal tract, it showed no significant benefit for in-hospital mortality in this study [10,28]. This could be caused by a small case number, the heterogeneity of treatment course, immunity, variable disease severity, and the side effects of antiviral agents. A prospective study with a larger case number will help to clarify this issue.
In this largest cohort study of CMV esophagitis, we applied strict diagnostic criteria and comprehensively presented risk factors, clinical presentations, laboratory findings, endoscopic features, treatments, prognostic factors, and outcomes. Because we only enrolled the patients with IHC staining results and most biopsies were performed due to endoscopic findings, selection bias might have existed. Other limitations of this study included a single-center, retrospective study design and a relatively small number of cases.
5. Conclusions
Physicians should watch for the clinical and endoscopic characteristics of CMV esophagitis in high-risk patients for early diagnosis and treatment. In the therapeutic course, the prevention of AKI is important for improving the in-hospital mortality rate.
Author Contributions
Planning and conducting the study: P.-H.L. and P.-J.Y.; collecting data and statistical analysis: P.-J.Y., C.-T.C. and J.-T.H.; literature review: P.-H.L., P.-J.Y. and C.-C.C.; data interpretation: P.-H.L., P.-J.Y. and M.-Y.S.; drafting the manuscript: P.-J.Y.; radiological consultation and review: C.-M.C.; pathological consultation and review: R.-C.W.; critical revision of manuscript: P.-H.L., M.-W.L. and C.-J.K. All authors have read and agreed to the published version of the manuscript.
Funding
This work was supported by Chang Gung Memorial Hospital (grant CMRPG3L0471).
Institutional Review Board Statement
The study was conducted according to the guidelines of the Declaration of Helsinki and approved by the Institutional Review Board of the Chang Gung Medical Foundation (approval document No. 202101234B0).
Informed Consent Statement
Being a retrospective study, individual informed consent was waived according to the requirement by the IRB of the Chang Gung Medical Foundation.
Data Availability Statement
The data presented in this study are available on request from the corresponding author.
Acknowledgments
The authors are grateful to Yu-Bin Pan (Biostatistical Section, Clinical Trial Center, Chang Gung Memorial Hospital “MOHW110-TDU-B-212-124005”) for the statistical support.
Conflicts of Interest
The authors declare no conflict of interest relevant to this manuscript.
References
Wang, H.W.; Kuo, C.J.; Lin, W.R.; Hsu, C.M.; Ho, Y.P.; Lin, C.J.; Su, M.Y.; Chiu, C.T.; Wang, C.L.; Chen, K.H. The clinical characteristics and manifestations of cytomegalovirus esophagitis. Dis. Esophagus 2016, 29, 392–399. [Google Scholar] [CrossRef] [PubMed]
Hoversten, P.; Kamboj, A.K.; Katzka, D.A. Infections of the esophagus: An update on risk factors, diagnosis, and management. Dis. Esophagus 2018, 31, doy094. [Google Scholar] [CrossRef] [PubMed]
Hoversten, P.; Kamboj, A.K.; Wu, T.T.; Katzka, D.A. Risk Factors, Endoscopic Features, and Clinical Outcomes of Cytomegalovirus Esophagitis Based on a 10-year Analysis at a Single Center. Clin. Gastroenterol. Hepatol. 2020, 18, 736–738. [Google Scholar] [CrossRef] [PubMed]
Li, L.; Chakinala, R.C. Cytomegalovirus Esophagitis. In StatPearls; StatPearls Publishing LLC.: Treasure Island, FL, USA, 2021. [Google Scholar]
Iwamuro, M.; Kondo, E.; Tanaka, T.; Hagiya, H.; Kawano, S.; Kawahara, Y.; Otsuka, F.; Okada, H. Endoscopic Manifestations and Clinical Characteristics of Cytomegalovirus Infection in the Upper Gastrointestinal Tract. Acta Med. Okayama 2017, 71, 97–104. [Google Scholar] [CrossRef] [PubMed]
Jung, K.H.; Choi, J.; Gong, E.J.; Lee, J.H.; Choi, K.D.; Song, H.J.; Lee, G.H.; Jung, H.Y.; Chong, Y.P.; Lee, S.O.; et al. Can endoscopists differentiate cytomegalovirus esophagitis from herpes simplex virus esophagitis based on gross endoscopic findings? Medicine 2019, 98, e15845. [Google Scholar] [CrossRef] [PubMed]
Esophagus. In AJCC Cancer Staging Manual; Greene, F.L., Page, D.L., Fleming, I.D., Fritz, A.G., Balch, C.M., Haller, D.G., Morrow, M., Eds.; Springer: New York, NY, USA, 2002; pp. 91–98. [Google Scholar]
Shaheen, N.J.; Falk, G.W.; Iyer, P.G.; Gerson, L.B. ACG Clinical Guideline: Diagnosis and Management of Barrett’s Esophagus. Off. J. Am. Coll. Gastroenterol. ACG 2016, 111, 30–50. [Google Scholar] [CrossRef] [PubMed]
Le, P.H.; Lin, W.R.; Kuo, C.J.; Wu, R.C.; Hsu, J.T.; Su, M.Y.; Lin, C.J.; Chiu, C.T. Clinical characteristics of cytomegalovirus colitis: A 15-year experience from a tertiary reference center. Ther. Clin. Risk Manag. 2017, 13, 1585–1593. [Google Scholar] [CrossRef]
Chaemsupaphan, T.; Limsrivilai, J.; Thongdee, C.; Sudcharoen, A.; Pongpaibul, A.; Pausawasdi, N.; Charatcharoenwitthaya, P. Patient characteristics, clinical manifestations, prognosis, and factors associated with gastrointestinal cytomegalovirus infection in immunocompetent patients. BMC Gastroenterol. 2020, 20, 22. [Google Scholar] [CrossRef] [PubMed]
Wilcox, C.M.; Straub, R.F.; Schwartz, D.A. Prospective endoscopic characterization of cytomegalovirus esophagitis in AIDS. Gastrointest. Endosc. 1994, 40, 481–484. [Google Scholar] [CrossRef]
Balthazar, E.J.; Megibow, A.J.; Hulnick, D.; Cho, K.C.; Beranbaum, E. Cytomegalovirus esophagitis in AIDS: Radiographic features in 16 patients. Am. J. Roentgenol. 1987, 149, 919–923. [Google Scholar] [CrossRef]
Chen, Y.-J.; Huang, H.-C.; Chen, T.-C.; Cheng, H.-T. Cytomegalovirus esophagitis with symptoms of gastroesophageal reflux disease in a kidney transplant recipient. Kaohsiung J. Med. Sci. 2020, 36, 859–860. [Google Scholar] [CrossRef] [PubMed]
Litle, V.R.; Rice, T.W. The esophagus: Do sex and gender matter? Semin. Thorac. Cardiovasc. Surg. 2011, 23, 131–136. [Google Scholar] [CrossRef] [PubMed]
Asanuma, K.; Iijima, K.; Shimosegawa, T. Gender difference in gastro-esophageal reflux diseases. World J. Gastroenterol. 2016, 22, 1800–1810. [Google Scholar] [CrossRef] [PubMed]
Wheeler, J.C.; Vanoni, S.; Zeng, C.; Waggoner, L.; Yang, Y.; Wu, D.; Uddin, J.; Karns, R.; Kottyan, L.; Mukkada, V.; et al. 17β-Estradiol protects the esophageal epithelium from IL-13-induced barrier dysfunction and remodeling. J. Allergy Clin. Immunol. 2019, 143, 2131–2146. [Google Scholar] [CrossRef] [PubMed]
Iijima, K. Does Estrogen Contribute to the Esophageal Barrier Function in Women? Gut Liver 2018, 12, 373–374. [Google Scholar] [CrossRef] [PubMed]
Aizawa, R.; Matsumoto, S.; Uneno, Y.; Nishikawa, Y.; Ozaki, Y.; Mori, Y.; Kanai, M.; Ishida, Y.; Sakanaka, K.; Hiraoka, M.; et al. Severe esophagitis associated with cytomegalovirus during concurrent chemoradiotherapy for esophageal cancer. Jpn. J. Clin. Oncol. 2017, 47, 885–888. [Google Scholar] [CrossRef] [PubMed]
Kitagawa, K.; Okada, H.; Miyazaki, S.; Funakoshi, Y.; Sanada, Y.; Chayahara, N.; Mayahara, H.; Fujii, M. Cytomegalovirus reactivation in esophageal cancer patients receiving chemoradiotherapy: A retrospective analysis. Cancer Med. 2021, 10, 7525–7533. [Google Scholar] [CrossRef] [PubMed]
Umemoto, K.; Kojima, Y.; Nagata, N.; Yokoi, C.; Sakurai, T.; Kobayakawa, M.; Iizuka, T.; Igari, T.; Yanase, M.; Akiyama, J. Cytomegalovirus esophagitis developing during chemoradiotherapy for esophageal cancer: Two case reports. J. Med. Case Rep. 2016, 10, 259. [Google Scholar] [CrossRef] [PubMed][Green Version]
Yeh, P.J.; Chiu, C.T.; Lai, M.W.; Wu, R.C.; Chen, C.M.; Kuo, C.J.; Hsu, J.T.; Su, M.Y.; Lin, W.P.; Chen, T.H.; et al. Clinical manifestations, risk factors, and prognostic factors of cytomegalovirus enteritis. Gut Pathog. 2021, 13, 53. [Google Scholar] [CrossRef] [PubMed]
Sarvari, J.; Habibi, A.; Moattari, A.; Eftehkar Sadat, A.T.; Ahangar Oskouee, M. Association of Epstein–Barr Virus and Cytomegalovirus Infections with Esophageal Carcinoma. Int. J. Cancer Manag. 2021, 14, e114566. [Google Scholar] [CrossRef]
El Baba, R.; Herbein, G. Immune Landscape of CMV Infection in Cancer Patients: From “Canonical” Diseases Toward Virus-Elicited Oncomodulation. Front. Immunol. 2021, 12, 730765. [Google Scholar] [CrossRef]
Yeh, P.J.; Chiu, C.T.; Lai, M.W.; Wu, R.C.; Kuo, C.J.; Hsu, J.T.; Su, M.Y.; Le, P.H. Cytomegalovirus gastritis: Clinicopathological profile. Dig. Liver Dis. 2021, 53, 722–728. [Google Scholar] [CrossRef] [PubMed]
Hoste, E.A.; Bagshaw, S.M.; Bellomo, R.; Cely, C.M.; Colman, R.; Cruz, D.N.; Edipidis, K.; Forni, L.G.; Gomersall, C.D.; Govil, D.; et al. Epidemiology of acute kidney injury in critically ill patients: The multinational AKI-EPI study. Intensive Care Med. 2015, 41, 1411–1423. [Google Scholar] [CrossRef] [PubMed]
Betjes, M.G.H. Immune cell dysfunction and inflammation in end-stage renal disease. Nat. Rev. Nephrol. 2013, 9, 255–265. [Google Scholar] [CrossRef]
Hartzell, S.; Bin, S.; Cantarelli, C.; Haverly, M.; Manrique, J.; Angeletti, A.; Manna, G.L.; Murphy, B.; Zhang, W.; Levitsky, J.; et al. Kidney Failure Associates With T Cell Exhaustion and Imbalanced Follicular Helper T Cells. Front. Immunol. 2020, 11, 583702. [Google Scholar] [CrossRef] [PubMed]
Wetwittayakhlang, P.; Rujeerapaiboon, N.; Wetwittayakhlung, P.; Sripongpun, P.; Pruphetkaew, N.; Jandee, S.; Chamroonkul, N.; Piratvisuth, T. Clinical Features, Endoscopic Findings, and Predictive Factors for Mortality in Tissue-Invasive Gastrointestinal Cytomegalovirus Disease between Immunocompetent and Immunocompromised Patients. Gastroenterol. Res. Pract. 2021, 2021, 8886525. [Google Scholar] [CrossRef] [PubMed]
Figure 1. Diagnosis of CMV esophagitis using CMV IHC staining and/or CMV inclusion bodies in H&E staining. (A) H&E staining (40× objective) showing typical intranuclear (owl’s eye) and intracytoplasmic (eosinophilic punctiform) CMV inclusions within the circles. (B) IHC staining (40× objective) with 1:200 diluted Novocastra™ lyophilized mouse monoclonal antibody against CMV pp65 antigen shows strong focal CMV immunoreactivity with brownish areas. CMV—cytomegalovirus; H&E—hematoxylin and eosin; IHC—immunohistochemistry.
Figure 1. Diagnosis of CMV esophagitis using CMV IHC staining and/or CMV inclusion bodies in H&E staining. (A) H&E staining (40× objective) showing typical intranuclear (owl’s eye) and intracytoplasmic (eosinophilic punctiform) CMV inclusions within the circles. (B) IHC staining (40× objective) with 1:200 diluted Novocastra™ lyophilized mouse monoclonal antibody against CMV pp65 antigen shows strong focal CMV immunoreactivity with brownish areas. CMV—cytomegalovirus; H&E—hematoxylin and eosin; IHC—immunohistochemistry.
Figure 2. Endoscopic features of CMV esophagitis. (A) Inflammation; (B) polypoid lesion; (C–F) variable morphologies of esophageal ulcers. CMV—cytomegalovirus.
Figure 2. Endoscopic features of CMV esophagitis. (A) Inflammation; (B) polypoid lesion; (C–F) variable morphologies of esophageal ulcers. CMV—cytomegalovirus.
Figure 3. Kaplan–Meier survival curve analysis of patients with CMV esophagitis. (A) Patients with AKI (solid line) had a significantly worse survival rate than those without AKI (dash line) (log-rank p = 0.044). (B) Patients who required ICU care had a worse survival rate, although it was not statistically significant (log-rank p = 0.101). AKI—acute kidney injury; CMV—cytomegalovirus; ICU—intensive care unit.
Figure 3. Kaplan–Meier survival curve analysis of patients with CMV esophagitis. (A) Patients with AKI (solid line) had a significantly worse survival rate than those without AKI (dash line) (log-rank p = 0.044). (B) Patients who required ICU care had a worse survival rate, although it was not statistically significant (log-rank p = 0.101). AKI—acute kidney injury; CMV—cytomegalovirus; ICU—intensive care unit.
Table 1. Baseline characteristics of patients with and without CMV esophagitis.
Table 1. Baseline characteristics of patients with and without CMV esophagitis.
| Characteristics | Overall (n = 148) | CMV Esophagitis (n = 44) | Non-CMV Esophagitis (n = 104) | p-Value |
--- ---
| Age, years | 56.7 ± 18.9 | 59.5 ± 18.5 | 56.9 ± 19.1 | 0.489 |
| Gender (M/F) | 96 (64.9%)/52 (35.1%) | 34 (77.3%)/10 (22.7%) | 62 (59.6%)/42 (40.4%) | 0.040 |
| OPD patients | 71 (48%) | 11 (25%) | 60 (57.7%) | <0.001 |
| Acquisition time (day) | 8 (5–14) | 11.5 (6.8–26) | 7 (4–12) | 0.005 |
| General conditions | | | | |
| Shock | 10 (6.8%) | 8 (18.2%) | 2 (1.9%) | 0.001 |
| Pneumonia | 25 (16.9%) | 16 (36.4%) | 9 (8.7%) | <0.001 |
| Intubation | 9 (6.1%) | 6 (13.6%) | 3 (2.9%) | 0.02 |
| ICU required | 10 (6.8%) | 7 (15.9%) | 3 (3%) | 0.008 |
| Immunocompromised | 68 (45.9%) | 34 (77.3%) | 34 (32.7%) | <0.001 |
| Underlying diseases | | | | |
| Diabetes mellitus | 29 (19.6%) | 8 (18.2%) | 21 (20.2%) | 0.78 |
| Hypertension | 54 (36.5%) | 19 (43.2%) | 35 (33.7%) | 0.27 |
| Autoimmune disease | 7 (4.7%) | 2 (4.5%) | 5 (4.8%) | 1 |
| Crohn’s disease | 1 (0.6%) | 0 (0%) | 1 (1%) | 1 |
| Ulcerative colitis | 2 (1.4%) | 1 (2.3%) | 1 (1%) | 0.51 |
| Coronary artery disease | 8 (5.4%) | 5 (11.4%) | 3 (2.9%) | 0.05 |
| COPD | 8 (5.4%) | 4 (9.1%) | 4 (3.8%) | 0.24 |
| Renal disease | | | | |
| AKI | 13 (8.8%) | 5 (11.4%) | 8 (7.7%) | 0.47 |
| CKD | 24 (16.2%) | 7 (15.9%) | 17 (16.3%) | 0.95 |
| ESRD | 11 (7.4%) | 4 (9.1%) | 7 (6.7%) | 0.73 |
| HIV infection | 11 (7.4%) | 8 (18.2%) | 3 (2.9%) | 0.003 |
| Malignancies | 44 (29.7%) | 20 (45.5%) | 24 (23.1%) | 0.01 |
| Transplantation | 3 (2%) | 3 (6.8%) | 0 (0%) | 0.03 |
| GERD | 91 (61.5%) | 20 (45.5%) | 71 (68.3%) | 0.01 |
| Immunosuppressive therapies | | | | |
| Steroid | 35 (23.6%) | 23 (52.3%) | 12 (11.5%) | <0.001 |
| Immunosuppressant | 11 (7.4%) | 6 (13.6%) | 5 (4.8%) | 0.06 |
| Chemotherapy | 27 (18.2%) | 15 (34.1%) | 12 (11.5%) | 0.001 |
| Radiotherapy | 29 (19.6%) | 16 (36.4%) | 13 (12.5%) | 0.001 |
| Other medications | | | | |
| Antibiotics | 67 (45.3%) | 33 (75%) | 34 (32.7%) | <0.001 |
| Proton pump inhibitor | 75 (50.7%) | 28 (63.6%) | 47 (45.2%) | 0.04 |
| Other antacids | 24 (16.2%) | 14 (31.8%) | 10 (9.6%) | 0.001 |
| Mucosal protectant | 31 (20.9%) | 14 (31.8%) | 17 (16.3%) | 0.03 |
| Clinical presentation | | | | |
| Fever | 23 (15.5%) | 16 (36.4%) | 7 (6.7%) | <0.001 |
| Epigastric pain | 48 (32.4%) | 18 (40.9%) | 30 (28.8%) | 0.15 |
| Vomiting | 34 (23%) | 11 (25%) | 23 (22.1%) | 0.70 |
| Hematemesis | 13 (8.8%) | 9 (20.5%) | 4 (3.8%) | 0.002 |
| GI bleeding † | 29 (19.6%) | 13 (29.5%) | 16 (15.4%) | 0.05 |
| Dysphagia | 37 (25%) | 13 (29.5%) | 24 (23.1%) | 0.41 |
| Odynophagia | 27 (18.2%) | 14 (31.8%) | 13 (12.5%) | 0.005 |
| Abdominal fullness | 13 (8.8%) | 9 (20.5%) | 4 (3.8%) | 0.002 |
| Laboratory data | | | | |
| WBC count (/μL) | 6900 (4250–9750) | 4900 (2700–8575) | 7600 (5200–10,300) | 0.004 |
| Hemoglobin (g/dL) | 10.5 (9.2–12.1) | 10.4 (9–11.6) | 10.7 (9.4–13.2) | 0.026 |
| Platelets (×1000/mm3) | 228 (136–284.5) | 173.5 (116.5–250) | 250 (183–295) | 0.005 |
| Creatinine (mg/dL) | 0.9 (0.7–1.3) | 0.73 (0.6–1.11) | 0.9 (0.7–1.3) | 0.071 |
| ALT (IU/L) | 20 (14–33) | 24 (17.3–33) | 17 (12–29) | 0.022 |
| Albumin (g/dL) | 2.9 (2.5–3.7) | 2.8 (2.3–2.9) | 3.3 (2.8–4) | 0.004 |
| C-reactive protein (mg/dL) | 32.4 (6.9–63.9) | 43 (8.1–73.2) | 29.2 (3.4–58.5) | 0.415 |
| Endoscopic features | | | | |
| Main findings | | | | |
| Ulcer | 131 (88.5%) | 39 (88.6%) | 92 (88.5%) | 0.98 |
| Diffuse/multiple ulcers | 90 (60.8%) | 33 (75%) | 57 (54.8%) | 0.02 |
| Inflammation | 5 (3.4%) | 1 (2.3%) | 4 (3.8%) | 1 |
| Polypoid lesion | 10 (6.8%) | 6 (13.6%) | 4 (3.8%) | 0.07 |
| Location of lesion | | | | |
| Upper 3rd | 31 (20.9%) | 14 (31.8%) | 17 (16.3%) | 0.03 |
| Middle 3rd | 85 (57.4%) | 27 (61.4%) | 58 (55.8%) | 0.53 |
| Lower 3rd | 110 (74.3%) | 38 (86.4%) | 72 (69.2%) | 0.03 |
| Concurrent findings | | | | |
| Reflux esophagitis | 75 (50.7%) | 17 (38.6%) | 58 (55.8%) | 0.06 |
| Esophageal candidiasis | 12 (8.1%) | 8 (18.2%) | 4 (3.8%) | 0.01 |
| Barrett esophagus | 9 (6.1%) | 4 (9.1%) | 5 (4.8%) | 0.45 |
| Gastric ulcer | 25 (16.9%) | 13 (29.5%) | 12 (11.5%) | 0.01 |
| CMV gastritis | 4 (2.7%) | 4 (9.1%) | 0 (0%) | 0.01 |
| CMV, others ‡ | 4 (2.7%) | 4 (9.1%) | 0 (0%) | 0.01 |
| Outcomes | | | | |
| Follow-up duration (days) | 351 (112.5–1187.3) | 276 (64.8–738.8) | 451.1 (141.3–1345.5) | 0.198 |
| Hospital stay (days) | 17.5 (9–35.3) | 24 (11–47) | 14 (7–24) | 0.02 |
| In-hospital mortality | 8 (5.4%) | 8 (18.2%) | 0 (0%) | <0.001 |
| Overall mortality | 38 (25.7%) | 23 (52.3%) | 15 (14.4%) | <0.001 |
p < 0.05. Age is presented as mean ± standard deviation. Laboratory data, acquisition time, admission duration, and follow-up duration are presented as median (IQR). The remaining data are presented as numbers (percentages). † defined as hematemesis, tarry stool, or bloody stool. ‡ included two patients with CMV retinitis and two patients with CMV hepatitis. AKI—acute kidney injury; ALT—alanine aminotransferase; CKD—chronic kidney disease; CMV—cytomegalovirus; COPD—chronic obstructive pulmonary disease; ESRD—end-stage renal disease; F—female; GERD—gastroesophageal reflux disease; GI—gastrointestinal; HIV—human immunodeficiency virus; ICU—intensive care unit; IQR—interquartile range; M—male; OPD—outpatient department; SD—standard deviation; WBC—white blood cell.
Table 2. Analysis of clinical factors associated with in-hospital mortality in patients with CMV esophagitis.
Table 2. Analysis of clinical factors associated with in-hospital mortality in patients with CMV esophagitis.
| Characteristics | Univariable Analysis | Multivariable Analysis |
---
| OR | 95% CI | p-Value | OR | 95% CI | p-Value |
| Age | 1.04 | 0.99–1.09 | 0.11 | | | |
| Gender (Male) | 0.86 | 0.14–5.10 | 0.87 | | | |
| Acquisition time | 1.06 | 1.00–1.13 | 0.04 | | | |
| General conditions | | | | | | |
| Shock | 18.33 | 2.87–117.33 | 0.002 | | | |
| Intubation | 17 | 2.33–124.19 | 0.01 | | | |
| ICU required | 28.33 | 3.76–213.70 | 0.001 | 26.53 | 1.06–665.08 | 0.05 |
| Immunocompromised | 0.86 | 0.14–5.10 | 0.87 | | | |
| Underlying diseases | | | | | | |
| Diabetes mellitus | 1.67 | 0.27–10.33 | 0.58 | | | |
| Hypertension | 2.62 | 0.54–12.72 | 0.23 | | | |
| Acute kidney injury | 10.2 | 1.35–76.93 | 0.02 | 174.15 | 1.27–23,836.21 | 0.04 |
| Chronic kidney disease | 2.07 | 0.32–13.25 | 0.44 | | | |
| End-stage renal disease | 0 | 0–0 | 0.1 | | | |
| Malignancy | 1.25 | 0.27–5.80 | 0.78 | | | |
| Chemotherapy | 0.22 | 0.03–2.03 | 0.18 | | | |
| Radiotherapy | 1.06 | 0.22–5.18 | 0.94 | | | |
| Steroid usage | 1.67 | 0.35–8.04 | 0.53 | | | |
| GERD | 0.13 | 0.01–1.15 | 0.07 | | | |
| CMV gastritis | 5.67 | 0.66–48.33 | 0.11 | | | |
| Clinical symptoms | | | | | | |
| Fever | 0.52 | 0.09–2.97 | 0.47 | | | |
| Epigastric pain | 2.95 | 0.61–14.38 | 0.18 | | | |
| Vomiting | 0.37 | 0.04–3.42 | 0.38 | | | |
| Bloody vomiting | 6.2 | 1.16–33.17 | 0.03 | | | |
| GI bleeding | 12.43 | 2.05–75.24 | 0.01 | | | |
| Laboratory data | | | | | | |
| WBC count | 1 | 1–1 | 0.13 | | | |
| Hemoglobin | 0.77 | 0.48–1.21 | 0.25 | | | |
| Platelet | 0.10 | 0.99–1.01 | 0.76 | | | |
| Creatinine | 0.85 | 0.42–1.73 | 0.66 | | | |
| ALT | 1.04 | 0.99–1.09 | 0.09 | | | |
| Albumin | 0.21 | 0.04–1.05 | 0.06 | | | |
| C-reactive protein | 0.99 | 0.97–1.01 | 0.37 | | | |
| Endoscopic features | | | | | | |
| Diffuse/multiple ulcers | 2.69 | 0.29–24.75 | 0.38 | | | |
| Esophageal candidiasis | 0.59 | 0.06–5.63 | 0.65 | | | |
| Barrett’s esophagus | 1.57 | 0.14–17.42 | 0.71 | | | |
| Gastric ulcer | 12.43 | 2.05–75.24 | 0.01 | | | |
| Antiviral therapy | 1.19 | 0.246–5.764 | 0.828 | | | |
p < 0.05, calculated using logistic regression analysis. ALT—alanine aminotransferase; CI—confidence interval; CMV—cytomegalovirus; F—female; GERD—gastroesophageal reflux disease; GI—gastrointestinal; ICU—intensive care unit; M—male; OR—odds ratio; WBC—white blood cell.
| |
| Publisher’s Note: MDPI stays neutral with regard to jurisdictional claims in published maps and institutional affiliations. |
© 2022 by the authors. Licensee MDPI, Basel, Switzerland. This article is an open access article distributed under the terms and conditions of the Creative Commons Attribution (CC BY) license (
Share and Cite
MDPI and ACS Style
Yeh, P.-J.; Wu, R.-C.; Chen, C.-M.; Chiu, C.-T.; Lai, M.-W.; Chen, C.-C.; Kuo, C.-J.; Hsu, J.-T.; Su, M.-Y.; Le, P.-H. Risk Factors, Clinical and Endoscopic Features, and Clinical Outcomes in Patients with Cytomegalovirus Esophagitis. J. Clin. Med. 2022, 11, 1583.
AMA Style
Yeh P-J, Wu R-C, Chen C-M, Chiu C-T, Lai M-W, Chen C-C, Kuo C-J, Hsu J-T, Su M-Y, Le P-H. Risk Factors, Clinical and Endoscopic Features, and Clinical Outcomes in Patients with Cytomegalovirus Esophagitis. Journal of Clinical Medicine. 2022; 11(6):1583.
Chicago/Turabian Style
Yeh, Pai-Jui, Ren-Chin Wu, Chien-Ming Chen, Cheng-Tang Chiu, Ming-Wei Lai, Chien-Chang Chen, Chia-Jung Kuo, Jun-Te Hsu, Ming-Yao Su, and Puo-Hsien Le. 2022. "Risk Factors, Clinical and Endoscopic Features, and Clinical Outcomes in Patients with Cytomegalovirus Esophagitis" Journal of Clinical Medicine 11, no. 6: 1583.
APA Style
Yeh, P.-J., Wu, R.-C., Chen, C.-M., Chiu, C.-T., Lai, M.-W., Chen, C.-C., Kuo, C.-J., Hsu, J.-T., Su, M.-Y., & Le, P.-H. (2022). Risk Factors, Clinical and Endoscopic Features, and Clinical Outcomes in Patients with Cytomegalovirus Esophagitis. Journal of Clinical Medicine, 11(6), 1583.
Note that from the first issue of 2016, this journal uses article numbers instead of page numbers. See further details here.
Article Metrics
No
Article Access Statistics
For more information on the journal statistics, click here.
Multiple requests from the same IP address are counted as one view.
Zoom | Orient | As Lines | As Sticks | As Cartoon | As Surface | Previous Scene | Next Scene
Export citation file: BibTeX)
MDPI and ACS Style
Yeh, P.-J.; Wu, R.-C.; Chen, C.-M.; Chiu, C.-T.; Lai, M.-W.; Chen, C.-C.; Kuo, C.-J.; Hsu, J.-T.; Su, M.-Y.; Le, P.-H. Risk Factors, Clinical and Endoscopic Features, and Clinical Outcomes in Patients with Cytomegalovirus Esophagitis. J. Clin. Med. 2022, 11, 1583.
AMA Style
Yeh P-J, Wu R-C, Chen C-M, Chiu C-T, Lai M-W, Chen C-C, Kuo C-J, Hsu J-T, Su M-Y, Le P-H. Risk Factors, Clinical and Endoscopic Features, and Clinical Outcomes in Patients with Cytomegalovirus Esophagitis. Journal of Clinical Medicine. 2022; 11(6):1583.
Chicago/Turabian Style
Yeh, Pai-Jui, Ren-Chin Wu, Chien-Ming Chen, Cheng-Tang Chiu, Ming-Wei Lai, Chien-Chang Chen, Chia-Jung Kuo, Jun-Te Hsu, Ming-Yao Su, and Puo-Hsien Le. 2022. "Risk Factors, Clinical and Endoscopic Features, and Clinical Outcomes in Patients with Cytomegalovirus Esophagitis" Journal of Clinical Medicine 11, no. 6: 1583.
APA Style
Yeh, P.-J., Wu, R.-C., Chen, C.-M., Chiu, C.-T., Lai, M.-W., Chen, C.-C., Kuo, C.-J., Hsu, J.-T., Su, M.-Y., & Le, P.-H. (2022). Risk Factors, Clinical and Endoscopic Features, and Clinical Outcomes in Patients with Cytomegalovirus Esophagitis. Journal of Clinical Medicine, 11(6), 1583.
Note that from the first issue of 2016, this journal uses article numbers instead of page numbers. See further details here.
clear
J. Clin. Med., EISSN 2077-0383, Published by MDPI
RSS Content Alert
Further Information
Article Processing Charges Pay an Invoice Open Access Policy Contact MDPI Jobs at MDPI
Guidelines
For Authors For Reviewers For Editors For Librarians For Publishers For Societies For Conference Organizers
MDPI Initiatives
Sciforum MDPI Books Preprints.org Scilit SciProfiles Encyclopedia JAMS Proceedings Series
Follow MDPI
LinkedIn Facebook X
© 1996-2025 MDPI (Basel, Switzerland) unless otherwise stated
Disclaimer
Disclaimer/Publisher’s Note: The statements, opinions and data contained in all publications are solely those of the individual author(s) and contributor(s) and not of MDPI and/or the editor(s). MDPI and/or the editor(s) disclaim responsibility for any injury to people or property resulting from any ideas, methods, instructions or products referred to in the content.
Terms and Conditions Privacy Policy
We use cookies on our website to ensure you get the best experience.Read more about our cookies here.
Share Link
clear
clear
Back to TopTop |
189075 | https://www.tuhh.de/ti3/paper/rump/Ru19.pdf | Error bounds for computer arithmetics This research was partially supported by CREST, Japan Science and Technology Agency.
Siegfried M. Rump Institute for Reliable Computing, Hamburg University of Technology, Am Schwarzenberg-Campus 3, 21071 Hamburg, Germany, and Visiting Professor at Waseda University, Faculty of Science and Engineering, 3–4–1 Okubo, Shinjuku-ku, Tokyo 169–8555, Japan.
Email: rump@tuhh.de Abstract—This note summarizes recent progress in error bounds for compound operations performed in some computer arithmetic. Given a general set of real numbers together with some operations satisfying the first standard model, we identify three types A, B, and C of weak sufficient assumptions implying new results and sharper error estimates. Those include linearized error estimates in the number of operations, faithfully rounded and reproducible results. All types of assumptions are satisfied for an IEEE-754 p-digit base-β floating-point arithmetic.
I. INTRODUCTION In this note we summarize some recent results on IEEE-754 floating-point and on a more general computer arithmetic. Sev-eral findings are presented unifying and generalizing previous ones. At few places short proofs are given rather than lengthy explanations if we think it is the easier way to understand the matter.
An introduction and basic properties of floating-point and of more general computer arithmetics can be found in Higham’s ASNA , see also , and in particular in the excellent books by Muller et al. and by Brent and Zimmermann.
We start with the most popular model of computer arith-metic, namely IEEE-754 p-digit floating-point arithmetic to some base β. For this model we show that traditional estimates using the relative rounding error unit u “ 1 2β1´p can be replaced by optimal bounds for individual opera-tions. Moreover, it was a surprise that the traditional bound γk “ ku{p1 ´ kuq for compound operations, for example for summation, dot products and also some matrix factorizations, can be linearized to ku.
Next, we consider an arbitrary set A of real numbers and a computer arithmetic on A with the sole requirement to satisfy the first standard model , i.e., the relative error of the computed result to the exact result is bounded by some constant. Then we ask what additional assumptions are necessary to prove error estimates such as the linearized bounds mentioned above. This approach is in some way opposite to fixing the set A to p-digit base-β numbers and precisely defining the operations as in IEEE-754 arithmetic.
We identify three mutually different additional assumptions.
For a computer arithmetic and a, b P A denote by a ‘ b the computed result of a b. The first Assumption A is that the absolute error of an addition or subtraction is bounded by the minimum absolute value of the operands, that is Assumption A: |a ‘ b ´ pa bq| ď minp|a|, |b|q.
That suffices to prove the linearized error bounds mentioned above. Assumption A is satisfied for any IEEE-754 p-digit base-β floating-point arithmetic with some nearest rounding.
On the one hand, Assumption A is very weak [the error of 2 ‘ 3 to 5 is bounded by 2], on the other hand Assumption A excludes directed rounding [adding an arbitrarily small number to 1 results in one of the neighbors of 1].
In IEEE-754 arithmetic the relative error of an operation aeb is not only bounded by u|a˝b|, but in fact by u¨ufppa˝bq, where ufppxq denotes the largest power of β less than or equal to |x|. That is the kernel of the alternative second Assumption B, namely that errors are bounded relative to specific numbers “near” the actual result rather than to the real result itself.
That suffices to prove the linear error estimates. For IEEE-754 p-digit base-β floating-point arithmetic such numbers are powers of β, and that second Assumption B is satisfied for any rounding, including the directed and faithful ones, the latter meaning that there is no other floating-point number between the real result and the computed result.
Another interpretation of the fact that in IEEE-754 errors are bounded by u ¨ ufppa ˝ bq is that those numbers are a constant times a power of β. This is the kernel of the third Assumption C, namely that errors grow by b¨βk for some constants b, β, k.
With Assumption C it follows that errors grow linearly with the height of a summation tree. For IEEE-754 p-digit base-β floating-point arithmetic the third Assumption C is satisfied for any rounding, including directed or faithful ones.
The conclusions by Assumption A are true without re-striction on the number of summands. The second and third Assumptions B and C are targeted to prove results for a general rounding, including directed, faithful but also nearest. Such results require a mandatory but weak restriction on the size of the problem. Therefore Assumptions B and C are more general than Assumption A, but imply such a restriction.
The first standard model together with Assumptions A and B leads to the optimal error bound for summation, namely that ku can be replaced by ku{p1kuq, see Table I. A mandatory restriction on k of size u´1 applies.
Depending on the assumption, the error bounds for sum-mation are as follows. Let a set A Ď R with a computer arithmetic according to the first standard model with relative rounding error unit u be given. For p1, . . . , pn P A denote by ˆ s the computed sum in any order. Depending on the additional assumption, the error of summation satisfies |ˆ s ´ řn i“1 pi| ď Φ řn i“1 |pi| with constants Φ according to Table I. The bounds depend on the number of summands n except the third one depending on the height h of the summation tree.
TABLE I: Error bounds for summation.
Assumption Φ rounding condition A pn ´ 1qu nearest none B pn ´ 1qu any n ď 1 β´1 2 u´1 C hu any h ď u´1{2 ´ 1 A and B pn´1qu 1pn´1qu nearest n ď 1 β´1 2 u´1 The motivation of the improved error estimates is not only a matter of beauty, but it often suffices to show similar linearized estimates for other types of compound operations such as dot products, blocked summation or sums of products.
Next we reduce the arithmetic to the first standard model, not requiring any of the Assumptions A, B or C. Based on that we introduce a simplified pair arithmetic producing a faithfully rounded result under precisely specified conditions. In the world of IEEE-754 arithmetic this widens the applicability because existing pair arithmetics rely on error-free transforma-tions. Those, however, do not exist in case of directed rounding because the error of an approximate operation needs not to be representable.
Finally we close the circle and return to IEEE-754 binary arithmetic with rounding to nearest. For that we discuss how to obtain efficiently a reproducible result for summation.
II. OPTIMAL ERROR BOUNDS FOR THE TWO STANDARD MODELS We start with the most popular model of computer arith-metic, namely IEEE-754 p-digit base-β floating-point arith-metic. For simplicity, we assume the set F of floating-point numbers to have no restriction on the exponent range: F “ t0u Y tM ¨ βe : M, e P Z, βp´1 ď |M| ă βpu.
(1) Let fl: R Ñ F denote a round-to-nearest function, that is |t ´ flptq| “ min fPF |t ´ f|, t P R.
(2) To carry out rounding error analysis of algorithms, frequently the first or second standard model of computer arithmetic is used. That means that, according to Table II, the relative error of a floating-point operation ˝ P t, ´, ˆ, {u shall be bounded with respect to the true result for the first, and with respect to the computed result for the second standard model.
The relative error of the first and second standard model for TABLE II: First and second standard model for x, y P F.
standard model property I flpx ˝ yq “ px ˝ yqp1 δq, |δ| ď u II flpx ˝ yq “ x˝y 1δ , |δ| ď u rounding t P R is E1ptq “ |t ´ flptq| |t| and E2ptq “ |t ´ flptq| |flptq| , respectively, with the convention 0{0 “ 0. It is well known that E1ptq ď v :“ u 1u and E2ptq ď u for the relative rounding error unit u :“ 1 2β1´p.
For arithmetic operations x e y :“ flpx ˝ yq, however, the upper bound for E1 or E2 is not always attained. The maximum value for E1 is achieved if and only if there exist x, y P F with x ˝ y “ 1 u, the number half-way between 1 and its successor 1 2u. The same is true for E2 if ties are rounded to even.
The worst case bounds for the individual operations were proved by Jeannerod et al. in as given in Table III. The TABLE III: Optimal relative error bounds for various inputs t and x, y P F for IEEE-754 p-digit base-β floating-point arithmetic.
t bound on E1ptq bound on E2ptq real number u 1u u x ˘ y u 1u u xy u 1u u x{y # u ´ 2u2 if β “ 2, u 1u if β ą 2 # u´2u2 1u´2u2 if β “ 2, u if β ą 2 ?x 1 ´ 1 ?12u ?1 2u ´ 1 bounds are optimal, possibly under some mild (necessary and sufficient) conditions on β and p outlined in . Specifically, for addition, subtraction, and multiplication the condition for optimality is that β is even, and in the case of multiplication in base 2 it requires that 2p1 is not a Fermat prime. In most practical situations such conditions are satisfied.
A rounding function fl: R Ñ F is defined by its “switch-ing points” (called rounding boundary in ). In IEEE-754 these are the midpoints, i.e., the arithmetic mean of adjacent floating-point numbers, thus minimizing the maximum relative error E1ptq of the first standard model.
We may ask for switching points minimizing the maximum relative error for a nearest rounding of the second, or of both standard models. These have been identified in as by Table IV. This is not only true for the grid of p-digit base-β floating-point numbers, but for general sets of real numbers.
TABLE IV: Optimal switching points within adjacent elements of F.
Minimizing E1ptq Minimizing E2ptq Minimizing maxpE1ptq, E2ptqq Arithmetic mean Harmonic mean Geometric mean III. LINEARLY BOUNDED ERROR ESTIMATES FOR COMPOUND OPERATIONS In this section we still assume an IEEE-754 p-digit base-β floating-point arithmetic with some nearest rounding. Here “some” means that ties may be rounded in any way.
The most simple and important compound operation is the sum of floating-point numbers. Let n numbers p1, . . . , pn P F be given, set ˆ s1 :“ p1 and define recursively ˆ si :“ ˆ si´1 ‘ pi for i P t2, . . . , nu.
(3) Since ˆ si´1 ‘ pi “ pˆ si´1 piqp1 δiq for some |δi| ď u, a straightforward and standard computation yields |ˆ sn ´ n ÿ i“1 pi| ď p1 uqn´1 ´ 1 ˘ n ÿ i“1 |pi|.
(4) This is the classical Wilkinson-type bound . To cover higher order terms, the unwieldy factor on the right-hand side is often replaced by γn´1 “ pn´1qu 1´pn´1qu, provided that pn ´ 1qu ă 1 (see ).
It was a surprise when it was proved in that, at least for recursive summation (3), the factor γn´1 can be replaced by pn ´ 1qu without restriction on n: | ˆ sn ´ n ÿ i“1 pi | ď pn ´ 1qu n ÿ i“1 |pi| .
(5) That theoretical estimate was supplemented in by the computable estimate | ˆ sn ´ n ÿ i“0 pi | ď pn ´ 1qu ¨ ufpp ˆ Snq, where ˆ Sn is obtained by (3) when replacing pi by |pi|. The unit in the first place ufppxq of x P R denotes the value of the leading digit in the β-adic representation of x (with ufpp0q :“ 0). It was introduced in and proved to be useful to transform complicated rounding error analyses into inequalities. At least for β “ 2 it can be computed, without branch, using three floating-point operations and one absolute value, see [34, Algorithm 3.5] and .
OPEN PROBLEM 1. Design an algorithm of similar com-plexity and without branch to compute ufppxq for β ą 2.
After that first linear error estimate, the race began. For the remainder of this section, in order to show the historical progress, we still restrict our attention to a computer arithmetic in base β ě 2 with p ě 2 mantissa digits following the IEEE-754 standard, so that u “ 1 2β1´p.
The next target was dot products. Rather than treating sums of products of floating-point numbers, the error of a sum of real numbers was estimated. To our knowledge that was the first time to take that more general perspective. More precisely, consider a real vector x1, . . . , xn P R and suppose the sum of flpxiq is evaluated in floating-point arithmetic with result ˆ r.
Then it was shown by Jeannerod in that, no matter what the order of evaluation of the floating-point sum, |ˆ r ´ n ÿ i“1 xi| ď nu n ÿ i“1 |xi|.
(6) The result is true without any restriction of n. For floating-point vectors a, b P Fn it follows as a corollary that the result ˆ r of the floating-point dot product, no matter what the order of evaluation and barring underflow, satisfies |ˆ r ´ aT b| ď nu|aT ||b|.
(7) As a consequence, the error of the floating-point product of two matrices A, B with inner dimension k is bounded by ku|A||B|. In it was also shown that the bound for summation (5) is true in IEEE-754 p-digit base-β floating-point arithmetic for any order of evaluation; in the next section we will see that (5), (6) and (7) are true for any computer arithmetic satisfying the first error model and |a‘b´pabq| ď minp|a|, |b|q. The latter is Assumption A.
Given a vector p P Fn, let ˆ r be the value of the Euclidean norm }p}2 calculated in floating-point arithmetic in any order of evaluation. The standard error estimate ˇ ˇ ˆ r ´ }p}2 ˇ ˇ ď ´ p1 uqn{21 ´ 1 ¯ }p}2 was, without restriction on n, improved in to ˇ ˇ ˆ r ´ }p}2 ˇ ˇ ď pn{2 1qu}p}2.
Next consider the product of floating-point numbers. First, Graillat et al. proved in that for a P F, β “ 2, the result ˆ r of the power ak1 computed by successive multiplications and barring over- and underflow satisfies |ˆ r ´ ak1| ď ku|ak1| if k 1 ď a 21{3 ´ 1 ¨ u´1{2. (8) More generally, in the product of real numbers was treated. For x0, x1, . . . , xk P R with ℓof them in F, denote by ˆ r the floating-point product of all the flpxiq in any order of evaluation, and set K :“ 2k 1 ´ ℓ and ω :“ # 1 if β is odd, 2 if β is even.
(9) Then, in the absence of underflow and overflow, ˇ ˇˆ r ´ k ź i“0 xi ˇ ˇ ď Ku ˇ ˇ k ź i“0 xi ˇ ˇ if K ă cω β u´1{2.
(10) Note that the index i starts from 0. In particular, if β “ 2 and all the xi are in F, then pK, ωq “ pk, 2q and (10) becomes ˇ ˇˆ r ´ k ź i“0 xi ˇ ˇ ď ku ˇ ˇ k ź i“0 xi ˇ ˇ if k ă u´1{2.
(11) For β “ 2 and p ě 4, the constraint in (11) cannot be replaced by k ă 12u´1{2.
OPEN PROBLEM 2. Assume IEEE-754 p-digit base-β arith-metic. Let T be a binary tree with k 1 leaves, where each inner node represents a division. Associate to each leaf a floating point number, denote by r the value of the root for real division {, and by ˆ r for floating-point division { . Is |ˆ r ´ r| ď ku|r| if k ă u´1{2 true? Is it also true for mixed multiplications and divisions?
If yes and assuming the first standard model, what are the necessary assumptions on the computer arithmetic?
As another consequence, the classical factor γ2n for Horner’s scheme was improved as well in .
Let x, a0, a1, . . . , an P F be given and let ˆ r be the approximation to řn i“0 aixi produced by Horner’s scheme. Then, using ω defined in (9) and in the absence of underflow and overflow, ˇ ˇ ˇ ˇ ˇˆ r ´ n ÿ i“0 aixi ˇ ˇ ˇ ˇ ˇ ď 2nu n ÿ i“0 |aixi| if n ă 1 2 ˆcω β u´1{2 ´ 1 ˙ .
(12) Finally, it was shown in that the concept of linearizing bounds by replacing γk by ku is also true for some standard numerical linear algebra algorithms. If for some A P Fmˆn with m ě n Gaussian elimination runs to completion, then the computed factors ˆ L and ˆ U satisfy (comparison and absolute value to be understood entrywise) ˆ L ˆ U “ A ∆A, |∆A| ď nu|ˆ L|| ˆ U|.
(13a) If for symmetric A P Fnˆn the Cholesky decomposition runs to completion, then the computed factor ˆ R satisfies ˆ RT ˆ R “ A ∆A, |∆A| ď pn 1qu| ˆ RT || ˆ R|.
(13b) If Tx “ b is solved by substitution for b P Fn and nonsingular triangular T P Fnˆn, then the computed solution ˆ x satisfies pT ∆Tqˆ x “ b, |∆T| ď nu|T|.
(13c) Each of these bounds improves upon the corresponding clas-sical ones, that is, γn|ˆ L|| ˆ U|, γn1| ˆ RT || ˆ R|, γn|T|.
In contrast to the classical ones, the new bounds are valid without restriction on n.
IV. GENERAL COMPUTER ARITHMETIC Up to now we actively assumed to use an IEEE-754 p-digit base-β floating-point arithmetic. Then, for example, lin-earized error estimates follow. Next, for a computer arithmetic satisfying the first standard model, we will passively identify sufficient additional assumptions to prove certain results, for example linearized error estimates. In particular we move away from a specified grid, working instead on an arbitrary set of real numbers.
Let a set A of real numbers together with operations e : A ˆ A Ñ A for ˝ P t, ´, ˆ, {u be given satisfying the first standard model @x, y P A: x e y “ px ˝ yqp1 δq, |δ| ď u (14) for a constant u. That is our minimum accuracy requirement bounding the relative error of x e y P A with respect to the real result x ˝ y P R.
Most types of computer arithmetic used for numerical computations satisfy the first standard model, chief amongst IEEE-754-type arithmetics. That includes flush-to-zero models [no gradual underflow] if the underflow range is excluded.
The first and second standard model are not satisfied for multiplication and division in fixed-point arithmetic (although addition and subtraction is always exact). They are also not satisfied for an arithmetic without guard digit [13, Section 2.4] and which has been used in the early days of computers.1 The standard model (14) leaves much freedom for the actual definition of the computer arithmetic, it neither implies xey “ x ˝ y if x ˝ y P A, nor a e b “ c e d if a ˝ b “ c ˝ d.
Moreover, there is quite a gap between the active “best ap-proximation” property (2) and the mere accuracy requirement (14) as by the following example.
Example 1: Consider a 3-digit decimal arithmetic, and xy for x “ 4.96 and y “ 5. Then x y “ 9.96 is representable in 3 decimal digits, and x e y “ x y “ 9.96 is the best approximation in the sense of (2). However, any choice of xey P t 9.92, 9.93, 9.94, 9.95, 9.96, 9.97, 9.98, 9.99, 10.0 u satisfies the standard model (14) for u “ 1 2β1´p “ 0.005.
Consider the sum of p1, . . . , pn P A. The first standard model (14) suffices to prove the standard estimate (4), but without additional assumptions, the factor p1 uqn´1 ´ 1 cannot be replaced by pn ´ 1qu.
Example 2: Consider a logarithmic number system F :“ t0u Y t˘ck : k P Zu for 1 ă c P R with rounding upwards.
Then u “ c´1 c1 and, for sufficiently small e P F, p1‘eq‘e “ c2 but c2 ´ p1 2eq ą 2 c´1 c1p1 2eq “ 2up1 2eq. The reason is that an arbitrarily small summand e causes a relative error of almost size u.
A. The standard model together with Assumption A Surprisingly, for the improved and linearized bound (5) the following Assumption A suffices: @a, b P A: | pa ‘ bq ´ pa bq | ď minp|a|, |b|q.
(15) For a computer arithmetic with some nearest rounding (2) that follows by | pa ‘ bq ´ pa bq | ď minp|a ´ pa bq|, |b ´ pa bq|q “ minp|a|, |b|q.
(16) It is not true for a directed rounding, but it is true for Dekker’s truncated rounding [5, Definition 3.5], i.e., a faithful rounding such that a nonzero error pa ‘ bq ´ pa bq and the summand of smallest absolute value have opposite signs.
To demonstrate how the first standard model and Assump-tion A, that is (14) and (15) interplay, we repeat the proof in of the linearized estimate (5). Given p P An and proceeding by induction we set sn :“ ˆ sn´1 pn, so that the induction hypothesis implies ∆:“ | ˆ sn ´ řn i“1 pi | “ |ˆ sn ´ sn ˆ sn´1 ´ řn´1 i“1 pi| ď |ˆ sn ´ sn| pn ´ 2qu řn´1 i“1 |pi| .
(17) 1Note that still today almost all cheap decimal pocket calculators without exponent have no guard digit so that completely wrong results may be produced.
We distinguish two cases. First, assume |pn| ď u řn´1 i“1 |pi|.
Then (15) implies |ˆ sn ´ sn| “ |pˆ sn´1 ‘ pnq ´ pˆ sn´1 pnq| ď |pn| ď u n´1 ÿ i“1 |pi| , (18) and inserting into (17) finishes this part of the proof. Hence-forth, assume u řn´1 i“1 |pi| ă |pn|. Then (14) gives |ˆ sn ´ sn| ď u|sn| “ u|ˆ sn´1 ´ n´1 ÿ i“1 pi n ÿ i“1 pi| , so that (17) and the induction hypothesis yield ∆ ď u “ pn ´ 2qu řn´1 i“1 |pi| řn i“1 |pi| ‰pn ´ 2qu řn´1 i“1 |pi| ă u “ pn ´ 2q|pn| |pn| řn´1 i“1 |pi| ‰ pn ´ 2qu řn´1 i“1 |pi| “ pn ´ 1qu řn i“1 |pi| .
l By Table III, u can be replaced by u 1u for an IEEE-754 p-digit base-β floating-point arithmetic.
Assumption A, that is (15) is the key to the proof. In fact, the weaker assumption (20) in the following Theorem 4.1 suffices to prove more: in it is shown that, for any order of evaluation, both the linearized error estimates (5) for the sum and (7) for the dot product remain true for an arithmetic according to the first standard model (14) with the additional assumption (20). The proof is more involved.
Theorem 4.1: Let a binary tree T with root r be given.
For a node j of T, denote the set of inner nodes of the subtree with root j by Nj, and the set of its leaves by Lj. To each leaf i P Lr associate a real number xi, and let to each inner node j P Nr a real number εj be associated. Define sj :“ # xj if j P Lr psleftpjq srightpjqqp1 εjq if j P Nr, where leftpjq and rightpjq denote the left and right child of an inner node j, respectively. Furthermore, define for all inner nodes j δj :“ sj ´ sleftpjq ´ srightpjq (19) as well as, with the convention 0 0 :“ 0, ξj :“ |δj| ř iPLj |si| ř iPNjztju |δi|.
Suppose |δj| ď min kPtleftpjq,rightpjqu ␣ |sk| ÿ iPNjzNk ξi ÿ iPLk |si| ( (20) is true for all inner nodes j. Then ∆r :“ sr´ř iPLr si satisfies |∆r| ď ÿ iPNr |δi| ď ÿ iPNr ξi ÿ iPLr |si| ď ÿ iPNr |εi| ÿ iPLr |si|.
(21) The estimate is sharp in the sense that for arbitrary εj P r0, 1s there exists a tree T such that (20) is satisfied and there are equalities in (21).
Here sj is the computed value of sleftpjq srightpjq with error δj for an inner node j. Therefore, the only assumption (20) in Theorem 4.1 is a trivial consequence of Assumption A.
For the specific case of p-digit arithmetic to base β it follows ř ξi ď ř |εi| ď pn ´ 1qv with v :“ u{p1 uq according to Table III.
The replacement of u by v was used in for a simple proof of the linearized estimate (7) for dot products.
Example 3: For real xj, denote the floating-point sum of all flpxjq by ˆ r, so that |ˆ r´řn j“1 flpxjq| ď pn´1qv řn j“1 |flpxjq|.
Rounding to nearest implies |flpxjq´xj| ď v|xj|, so that |ˆ r ´ řn j“1 xj| ď ˇ ˇˆ r ´ řn j“1 flpxjq ˇ ˇ řn j“1 ˇ ˇxj ´ flpxjq ˇ ˇ ď pv pn ´ 1qvp1 vqq řn j“1 |xj|, and v pn ´ 1qvp1 vq ď nvp1 vq ď n v 1´v “ nu proves the desired estimate (7).
We will use the same concept later by improving pn ´ 1qu into the optimal factor pn´1qu 1pn´1qu for summation to show linearized estimates for other compound operations such as blocked summation or sums of products.
B. The standard model together with Assumption B Next we are interested in linear estimates for other types of rounding, for example directed or some faithful rounding.
Following the first standard model (14) one might think just to replace u by 2u to obtain similar results. That is not true because Assumption A (and also the weaker (20)) is not satisfied: adding an arbitrarily small positive e P F to 1 in rounding upwards results in the successor of 1 with an error larger than e.
For the specific case of rounding upwards and IEEE-754 binary floating-point arithmetic, the following estimate, similar to (5), with adapted relative rounding error unit was shown for β “ 2 in . Let x P Fn be given, and denote by ˆ r the sum computed in any order of evaluation and with all additions in rounding upwards. Then in it was shown that |ˆ r ´ r| ď 2pn ´ 1qu n ÿ i“1 |xi| provided that 4nu ď 1.
(22) The restriction on n is necessary for any base β: for x1 “ 1 and x2...n arbitrarily small positive numbers the sum increases each intermediate result to the next successor in F. Up to n ď βp ´ βp´1 summands the error is 2u, but passing the intermediate result β increases the error to βu, eventually spoiling the estimate (22).
Estimate (22) holds true for rounding upwards. We improve that by showing that under Assumption B, that is (26) for the εk defined in (25), a linearized error bound is true for any directed or faithful rounding. In terms of IEEE-754 p-digit base-β floating-point arithmetic this assumption amounts to the fact that bounds on the maximum relative error in the intervals ˘rβm, βm1s are constant and with respect to βm, namely uβm. The mathematical formulation is Assumption B, that is (26) with respect to (25). It implies the mandatory restriction on n.
Theorem 4.2: Let a binary tree T with n leaves be given. To each leaf associate a real number xi and to each inner node associate a real number sk forming vectors x P Rn and s P Rn´1. Denote by σk the sum of the values associated with the children of an inner node k, and define δk :“ sk ´ σk for 1 ď k ď n ´ 1.
(23) Let nonnegative real numbers λ, µ be given such that λ ď n ÿ i“1 |xi| ă µ.
(24) Define for 1 ď k ď n ´ 1 εk :“ # |δk| λ if |σk| ă µ, |δk| µ otherwise, (25) with the convention 0 0 :“ 0. Assume n´1 ÿ i“1 εi ď µ ´ λ λ .
(26) Then |σk| ă µ2{λ for 1 ď k ď n ´ 1, and for r denoting the root of T, ˇ ˇsr ´ n ÿ i“1 xi ˇ ˇ ď n´1 ÿ i“1 |δi| ď n´1 ÿ i“1 εi n ÿ i“1 |xi|.
(27) The interpretation, in particular the role of λ and µ for Assumption B, becomes clear from the proof of the following corollary.
Corollary 4.3: Let F be a p-digit floating-point number system in base β, and denote by s the result of a floating-point summation of x1, . . . , xn P F using some faithful addition. If n ď 1 β´1 2 u´1, then ˇ ˇs ´ n ÿ j“1 xj ˇ ˇ ď 2pn ´ 1qu n ÿ j“1 |xj|.
PROOF.
Let m P Z such that λ :“ βm ď řn j“1 |xj| ă βm1 “: µ. Let σk be as in Theorem 4.2, and denote by ufppσkq the largest power of β being less than or equal to |σk|. If |σk| ă µ, then ufppσkq ď λ and (25) implies εk “ |δk| λ ď |δk| ufppσkq “ |flpσkq ´ σk| ufppσkq ď 2u, and otherwise µ ď |σk| ă µ2{λ “ βµ shows ufppσkq “ µ and εk “ |δk| µ “ |δk| ufppσkq “ |flpσkq ´ σk| ufppσkq ď 2u.
Thus, all εk are bounded by 2u. Additionally, the limit on n implies n´1 ÿ j“1 εj ď pn ´ 1q2u ď β ´ 1 “ µ ´ λ λ , (28) so that the assumption (26) in Theorem 4.2 is satisfied. Thus, ˇ ˇs ´ n ÿ j“1 xj ˇ ˇ ď n´1 ÿ j“1 |δj| ď pn ´ 1q ¨ 2u n ÿ j“1 |xj|.
l Theorem 4.2 is tailored to the fact that the maximum abso-lute error of a floating-point operation is uniformly bounded by 2u{λ in the interval rλ, µs with λ “ βm and µ “ βm1, that is Assumption B. As we have seen, some restriction on n is mandatory; here the maximal number of such errors in rλ, µs, see (28), bounds the number of summands n to 1 β´1 2 u´1.
The application of the theorem to faithful rounding is just an example, it applies to nearest rounding with replacing 2u by u and other roundings as well.
C. The standard model together with Assumption C So far we saw that given x1, . . . , xn the error of the sum in some computer arithmetic is bounded by pn ´ 1qu ř |xi| no matter what the order of evaluation, and for a nearest rounding without restriction on n. The corresponding traditional con-stant p1uqn´1´1 in the estimate is straightforward to prove, whereas for the linearized bounds some effort is necessary.
An interesting generalization concerns error bounds depend-ing on the height of a summation tree. For an addition tree of height h, the traditional Wilkinson-type constant p1 uqh ´ 1 in the estimate follows straightforwardly. That bound can be linearized to hu for a standard model together with an Assumption C. For a balanced tree in binary64 that restricts the number of summands to n ď 294,906,264.
In terms of IEEE-754 p-digit base-β floating-point arith-metic Assumption C amounts to the fact that bounds on the maximum relative error increase from the interval ˘rβm, βm1s to the interval ˘rβm1, βm2s by a constant factor β. That factor β is associated with the arithmetic model. The mathematical formulation of Assumption C are assumptions (29) and (30) in Theorem 4.4. Note that β is some real number, not necessarily related to some grid.
Theorem 4.4: Let an α-ary tree T with root r and height h be given. For an inner node j of T, denote the set of leaves of the corresponding subtree by Lj and the set of all its inner nodes including j by Nj. To each leaf i of T associate a real number xi. Moreover, let positive real numbers b, ε as well as β ě α be given, and let two numbers δj P R and bj P t0u Y tβmb | m P Zu (29) be assigned to each inner node j of T. Suppose that for each inner node j |δj| ď bj ď ε ˜ ÿ iPLj |xi| ÿ iPNjztju |δi| ¸ .
(30) If h is restricted by h ď 2 b chε´1 ´ 1 with ch :“ # β´1 ´ β´2 if α “ β 1 ´ αβ´1 otherwise, (31) then ÿ iPNr |δi| ď hε ÿ iPLr |xi|.
(32) The following corollary formulates the result for IEEE-754 p-digit base-β floating-point arithmetic. From the proof the previous interpretation of Assumption C becomes clear. It is satisfied for any nearest as well as for any directed or faithfully rounded summation.
Corollary 4.5: For an IEEE-754 p-digit base-β floating-point arithmetic, let s be the result of a floating-point summa-tion of p1, . . . , pn P F in some nearest addition in any order.
If the height h of the corresponding binary summation tree satisfies h ď # u´ 1 2 ´ 1 if β “ 2 a 4 ´ 8β´1u´ 1 2 ´ 1 otherwise, (33) then ˇ ˇ ˇ ˇ ˇs ´ n ÿ j“1 pj ˇ ˇ ˇ ˇ ˇ ď hu n ÿ j“1 |pj|.
(34) The result remains valid for any faithful addition when replac-ing the error constant u by 2u in (33) and (34).
PROOF.
Let T denote a binary summation tree, where to each inner node j of T the respective intermediate summation result sj including the perturbations δi is associated. Using the notation as in Theorem 4.4 it follows sj “ ř iPLj xi ř iPNj δi. Furthermore, let b “ ε “ η, where η “ u in case of a nearest addition, and η “ 2u in case of faithful addition. Define bj :“ η ¨ ufppsj ´ δjq for all inner nodes j. This definition of bj complies with assumption (29), i.e., bj P t0u Y tβmη | m P Zu. Moreover, |δj| ď bj ď η|sj ´ δj| ď η ˜ ÿ iPLj |xi| ÿ iPNjztju |δi| ¸ validates the assumption (30). Finally, for α “ 2, h ď $ ’ ’ & ’ ’ % η´ 1 2 ´ 1 “ 2 a pβ´1 ´ β´2qη´1 ´ 1 if β “ α a 4 ´ 8β´1 η´ 1 2 ´ 1 “ 2 a p1 ´ αβ´1qη´1 ´ 1 otherwise shows the equivalence of (31) and (33). Thus (34) follows.˝ Similar to Example 3, the result extends to dot products.
Denote by s the result of a floating-point dot product of a, b P Fn in some rounding to nearest. Let the height h of the corresponding binary evaluation tree satisfy (33). Then, barring over- and underflow, ˇ ˇs ´ aT b ˇ ˇ ď hu n ÿ i“1 |aibi|.
(35) For faithful rounding the result is true when replacing the error constant u by 2u.
Given some computer arithmetic satisfying Assumption B or C, there is any freedom for the corresponding subset of R of representable numbers. Unless that set is fairly exotic, we can expect that Assumption C implies Assumption B.
D. The standard model together with Assumptions A and B Even the linearized bounds presented so far are not optimal for an IEEE-754 p-digit base-β floating-point arithmetic. We guess everybody thinking about a worst case error for recursive summation quickly constructs x “ p1, u, . . . , uq. The result ˆ r of a nearest addition with rounding ties to even is 1, so that the error is pn ´ 1qu and satisfies |ˆ r ´ n ÿ i“1 xi| “ pn ´ 1qu 1 pn ´ 1qu n ÿ i“1 |xi|.
(36) By means of explicit examples (cf. ) it is easy to see that some restriction on n is mandatory for (36). But although it was common belief that this is the worst case, it could not be proved.
The first result in this direction can be found in , where Mascarenhas introduces a new concept of using continuous mathematics to analyze floating-point arithmetic to prove (36) for recursive summation provided that n ď 1 20u´1.
However, despite the comparatively small upper bound on n and the restriction to recursive summation, the given proof is rather complicated and longish. In [24, Theorem 5] a more general result was proved using fairly simple arguments.
The assumptions are in fact more general than Assumptions A and B together. However, the mathematical statement is technical, so we state only the following corollary (which is [24, Proposition 1]) for IEEE-754 p-digit base-β arithmetic.
Note that this arithmetic satisfies Assumptions A and B.
Theorem 4.6: Let a p-digit floating-point arithmetic to base β be given. Let ˆ r be the result of a floating-point summation of p1, . . . , pn P F in some nearest addition in arbitrary order.
Then ˇ ˇ ˇ ˇ ˇˆ r ´ n ÿ j“1 pj ˇ ˇ ˇ ˇ ˇ ď pn ´ 1qu 1 pn ´ 1qu n ÿ j“1 |pj| if n ď 1 β ´ 1 2 u´1.
(37) As has been mentioned, the result holds true for a more general computer arithmetic as well. In that case the mandatory restriction on n has to be re-computed.
The upper bound on n is almost sharp. For rounding ties away from zero it cannot be replaced by the next larger integer; for rounding ties to even, p ě 3 mantissa digits and even β the upper bound cannot be increased by 2 β 2 , see .
OPEN PROBLEM 3. Devise a sharp error estimate for dot products in the spirit of (37).
E. Some applications As exploited in , this Theorem 4.4 has a number of consequences. Denote by s “ floatpexpressionq the result of an expression with each operation replaced by the correspond-ing floating-point operation in some nearest rounding. The evaluation may be in any order but, if applicable, respecting parentheses. First, consider a sum of products s :“ n ÿ i“1 m ź j“1 pij for pij P F.
(38) Provided that pn m ´ 2qu ă 1, the standard Wilkinson-type error estimate gives ˇ ˇ ˇ ˇ ˇfloat n ÿ i“1 m ź j“1 pij ˘ ´ s ˇ ˇ ˇ ˇ ˇ ď γnm´2 n ÿ i“1 m ź j“1 |pij|.
Corollary 4.5 and barring over- and underflow implies the linearized estimate ˇ ˇ ˇ ˇ ˇfloat n ÿ i“1 m ź j“1 pij ˘ ´ s ˇ ˇ ˇ ˇ ˇ ď pn m ´ 2qu n ÿ i“1 m ź j“1 |pij| (39) provided that m ď β´ 1 2 u´ 1 2 , n ď 1 β ´ 1 2 u´1, and m ď n.
For binary floating-point numbers, assuming m ď u´ 1 2 suffices for (39) to hold true. The proof is very similar to that in Example 3 for dot products, where the improved error bound v “ u 1u instead of u was sufficient to obtain the error bound nu; now the optimal error bound (37) is used instead of pn ´ 1qu.
Another direct application is a bound on the error of a Vandermonde matrix times a vector. Let Vij “ αi j for 0 ď i, j ď n for given αj P F. Then pV xqi “ n ř j“0 αi jxj, so that for a vector x P Fn1, starting with index 0, we obtain |floatpV xq ´ V x| ď diag pnu, nu u, . . . , 2nuq |V | |x| ď 2nu |V | |x|.
Another application is an error estimate for blocked summa-tion. Let a vector p P Fmn be given and consider s :“ float ˜ n ÿ i“1 ˆ m ÿ j“1 pij ˙¸ .
(40) Then |s ´ ř ij pij| ď γnm´2 ř ij |pij|, the standard Wilkinson-type error estimate, improves to ˇ ˇ ˇ ˇs ´ ÿ ij pij ˇ ˇ ˇ ˇ ď pn m ´ 2qu ÿ ij |pij| (41) provided that maxpm, nq ď 1 β´1 2 u´1.
V. FAITHFUL RESULTS BY A SIMPLIFIED PAIR ARITHMETIC In the previous sections error bounds for single or com-pound operations were shown, either for an actively given arithmetic such as IEEE-754, or for a computer arithmetic pas-sively satisfying the first error model (14) with some additional weak assumptions. In this section, accuracy estimates related to the condition number of the problem will be investigated, in particular methods to achieve a faithfully rounded result.
Common methods to improve the accuracy are compensated algorithms. Prominent examples are Kahan’s and Shewchuk’s summation algorithms , for which a small backward error2 of size 2u follows. The doubly compensated summation 2The computed result is the true result for a small perturbation of the input data .
by Priest requires ordering of the summands and proves a forward error of size 2u.
A notable exception to the many algorithms is Neumaier’s summation which he found as a master student in 1974.
Obviously without knowing, he uses what we call today “error-free transformations”, a term I coined in . For example, consider function [x,y] = FastTwoSum(a,b) x = a + b; y = a - (x - b); For any two floating-point numbers a, b P F with |a| ě |b| it holds a b “ x y for a nearest rounding and β ď 3. Similar algorithms for addition without constraint on the ordering of the summands (TwoSum) and for products (TwoProduct) are known , , . Note that for IEEE-754 precision-p base-β arithmetic it is necessary that a b is computed in rounding to nearest, otherwise the error a ‘ b ´ pa bq need not be representable.
Algorithm TwoSum implies an error-free vector transfor-mation. Given a vector p P Fn, the call q = VecSum(p) of function p = VecSum(p) for i=2:n [p(i),p(i-1)] = TwoSum(p(i),p(i-1)) produces a vector q P Fn with ř pi “ ř qi and qn “ floatpř piq. Summing the vector q in floating-point after a single call of VecSum is Algorithm Sum2 in , which is identical to Neumaier’s fourth algorithm in . The accuracy of the result depends on the condition number k “ př |pi|q{| ř pi|. The results in imply that basically for k À 1{p2n2uq the result of Sum2 is faithfully rounded, so that there is no other floating-point number between the true and the computed result. A similar result holds true for the dot product algorithm Dot2 in .
Using similar techniques, a number of algorithms with faithfully rounded result have been developed for several standard problems in numerical analysis. For example, Graillat gave in a compensated scheme for śn i“1 xi with faithfully rounded result provided that n ă ?1 ´ u ?4 2u 2 a p1 ´ uqu u´1{2 ; Boldo and Mu˜ noz showed in a compensated polynomial evaluation to be faithful provided that k :“ řn i“0 |pi||xi| | řn i“0 pixi| ă p1 ´ uqp1 ´ 2nuq2 4n2up2 uq ; in algorithm Sum2 is shown to be faithful if pn ´ 2qpn ´ 1q p1 ´ pn ´ 2quqp1 ´ pn ´ 1quq ď 1 2ku.
In binary64 the assumptions read n ă 47, 453, 132, and, for n “ 1000, k ă 1.13 ¨ 109 and k ă 4.52 ¨ 109, respectively.
Sometimes restrictions apply, for example the latter result for summation supposes recursive summation in binary, and all results suppose that transformation algorithms such as TwoSum are indeed error-free, i.e., produce x y “ a b.
Another approach to compute a faithfully rounded result, also based on error-free transformations, is Bailey’s double-double arithmetic . Here numbers are represented as an unevaluated sum of two elements of F.
The double-double arithmetic is analyzed in . For addi-tion, for example, two algorithms are given. The first algorithm [18, Algorithm 5], called “sloppy addition”, was already given by Dekker as add2 in . However, the result may have no Function (c, g) = SloppyDWPlusDW(a, e, b, f) rc, ts “ TwoSumpa, bq s “ flpe fq g “ flpt sq rc, gs “ FastTwoSumpc, gq significance at all.
Alternatively, an accurate algorithm AccurateDWPlusDW is analyzed with relative error not larger than 3u2 1´4u. The double-double arithmetic is based on IEEE-754 binary arithmetic and error-free transformations.
The target for this section is to introduce a new and simpli-fied pair arithmetic with the goal to give conditions for which the final result is faithful. That applies to general arithmetic expressions comprising of, ´, ˆ, {, ?¨. As we will see this includes all methods mentioned at the beginning of this section. Another target is to require as weak assumptions on the arithmetic as possible, but nevertheless guaranteeing a faithfully rounded result under specified conditions.
Our pair arithmetic is more general than previous ap-proaches in several aspects. First, we require only an arithmetic following the first standard model, neither of the previous Assumptions A, B or C has to be satisfied. Hence a situation as in Example 1 may occur.
Second, to estimate the error of an individual operation only an approximation of the residual is needed, for example of a ‘ b ´ pa bq for addition. Again that approximation is only required to satisfy the first standard model. In particular, “error-free transformations” are no longer needed but replaced by “approximate transformations”.
Third, for a pair pc, gq, no relation between c and g is required. Fourth, the pair operations are simplified requiring less operations compared to double-double. Fifth, for every pair pc, gq the first part c is equal to the result when computing in the given computer arithmetic.
As a special example, all of the following results are true for an IEEE-754 p-digit base-β arithmetic and any round-ing scheme. As has been mentioned, for directed rounding error-free transformations are not possible because the error a ‘ b ´ pa bq need not be representable. Nevertheless, our arithmetical model remains applicable.
Let A be an arbitrary discrete set of real numbers. For a given positive constant v ă 1 define the working set of A by W :“ tr P R | Df P A: |f ´ r| ď v|r|u.
(42) Consider a real function g: Rn Ñ R and let x P An be such that gpxq P W. The left-hand side of c Ð gpxq for c P A is called an A-arithmetic approximation if c “ gpxqp1 εq with |ε| ď v, (43) abbreviated by c Ð gpxq. We choose the notation “Д rather than “flp¨q” to indicate that only the error estimate (43) has to be satisfied, i.e., a relation rather than a function. Our general assumption on the arithmetic in A is as follows.
Assumption 5.7: For a, b P A and ˝ P t, ´, ˆ, {u, let ˆ c :“ a ˝ b. If ˆ c P W, we assume that c Ð a ˝ b can be evaluated and satisfies |c ´ ˆ c| ď v|ˆ c| according to (43). A similar statement is true for the square root. Moreover, assume that for t Ð a ˝ b ´ c if c Ð a ˝ b for ˝ P t, ´, ˆu, t Ð a ´ bc if c Ð a{b, t Ð a ´ c2 if c Ð ?a (44) a method to evaluate t is available satisfying the estimate in (43) with appropriate interpretation.
For the special case of IEEE-754 binary64 floating-point arithmetic with rounding to nearest, Assumption 5.7 is satisfied if the real result ˆ c does not cause over- or underflow by setting v :“ u{p1 uq for u :“ 2´53, by replacing c Ð a ˝ b by c “ flpa ˝ bq and evaluating the expressions in (44) by appropriate error-free transformations3, possibly using the fused multiply-add operation FMA.
Next we define our pair arithmetic . An algorithm for subtraction follows directly from addition, and all results hold true mutatis mutandis. To ease the exposition we omit sub-traction by the technical assumption A “ ´A. The comments “//TwoSum or Add3”, “FMA or TwoProduct” etc. in the following algorithms refer to a possible implementation when using IEEE-754 arithmetic; they are not mandatory.
Function (c,g) = CPairSum((a, e),(b, f)) c Ð a b t Ð a b ´ c // TwoSum or Add3 s Ð e f g Ð t s The first part c of the result of our pair operations is always equal to the one computed in the given arithmetic.
That property is spoiled by the final normalization step in SloppyDWPlusDW, also used in . Technically, the double-double SloppyDWPlusDW and our CPairSum are identical up to the final normalization; however, the assumptions of our pair 3Error-free transformations require the absence of over- and underflow not only for the results but also for all intermediate values; see Boldo et al. .
Function (c,g) = CPairProd((a, e),(b, f)) c Ð ab t Ð ab ´ c // FMA or TwoProduct q Ð af r Ð be s Ð q r g Ð t s Function (c,g) = CPairDiv((a, e),(b, f)) c Ð a{b t Ð a ´ bc // FMA p Ð t e q Ð cf r Ð p ´ q s Ð b f g Ð r{s arithmetic are far less. The flop counts for the pair addition algorithms are as follows.
CPairSum 8 flops SloppyDWPlusDW 11 flops AccurateDWPlusDW 20 flops Compared to double-double arithmetic, our other pair opera-tions have a smaller flop count as well.
In turn, the results of the double-double arithmetic are usu-ally more accurate than those of our pair arithmetic. However, one target was to derive provable conditions for a faithful result with as weak assumptions on the arithmetic as possible.
Consider an arbitrary arithmetic expression represented by a binary tree. For given input data we henceforth assume that all intermediate operations are well defined with result in the working set W. That is in particular satisfied if, when using IEEE-754 p-digit base-β floating-point arithmetic, no intermediate over- or underflow occur.
To formulate the conditions for a faithfully rounded result, we need to define the condition number of an arithmetic expression. An essential ingredient is the notation of the No Inaccurate Cancellation (NIC) principle. Demmel et al. used that in to identify algorithms computing accurate results basically independent of the condition number.4 4A famous example is to treat Hilbert matrices as Cauchy matrices allowing to faithfully compute the inverse or smallest singular value up to about dimension 108 solely in binary64.
Function (c,g) = CPairSqrt((a, e)) c Ð ?a t Ð a ´ c2 // FMA r Ð t e s Ð c c g Ð r{s Definition 5.8: Let T be an evaluation tree with input data p (the values at the leaves), and inner nodes consisting of operations from the set t, ˆ, {, ?¨u. If no sum with at least one addend not being input data is performed on numbers with opposite signs, then pT, pq complies with the No Inaccurate Cancellation (NIC) principle.
The rationale is to avoid catastrophic cancellation. If an arithmetic expression does not satisfy the NIC principle, for example xy´x, then for large positive x and small positive y cancellation and a large relative error to the true result y occurs. That cannot happen for an arithmetic expression satis-fying the NIC principle, the relative error of every intermediate to the corresponding true result may grow, but very slowly.
Definition 5.9: Consider an evaluation tree T with input data p P An and inner nodes consisting of operations from the set t, ˆ, {, ?¨u. Let any pair of input numbers pi and pj that is added in T with negative result be replaced by p1 i :“ ´pi and p1 j :“ ´pj, respectively. Moreover, let all other input numbers pk be replaced by their absolute value p1 k :“ |pk|. The so obtained data p1 is called NIC remodeled input data to pT, pq.
The rationale behind this definition is as follows. Let a compound operation be given depending on x P Rn. Then the error of an approximation is usually estimated relative to the maximal possible value S for all possible sign combinations of the xi. Examples are S “ ř |xi| for summation (4) or S “ řn i“0 |aixi| for Horner’s scheme (12). In those examples, the ratio between S and the true value for the original xi is the condition number, and that is the result of the problem with NIC remodeled data5.
Suppose evaluating an arithmetic expression by our pair arithmetic results in pc, gq. In order to obtain a faithfully rounded result, an element of A, we need to add c and g approximately. Here assuming the first standard model is not sufficient as by Example 1; this (and only this) final addition has to be done in some nearest rounding, otherwise the result cannot be guaranteed to be faithful. To be precise, we say ˆ c P A is a nearest A-approximation to c P R if @a P A: |ˆ c ´ c| ď |a ´ c|.
(45) Finally, we need a measurement for the minimum relative distance between two adjacent numbers in A, namely η :“ inf "|s ´ t| |s t| : s, t P A, s ‰ t , (46) For a p-digit base-β floating-point arithmetic it follows η “ u β´u ą u{β.
Based on that we can state our result for general arithmetic expressions.
Theorem 5.10: Let an arithmetic expression be given by an evaluation tree T with n leaves, where to each inner node j an operation ˝j out of t, ˆ, {, ?¨u is assigned. Moreover, to 5That is not always true, for example when divisions occur; however, for our condition to prove that a result is faithful it is sufficient.
every node j, inner node or leaf, let an integer kj be assigned according to kj :“ $ ’ ’ ’ ’ ’ ’ & ’ ’ ’ ’ ’ ’ % 0 if j is leaf maxtkleftpjq, krightpjqu 1 if ˝j “ kleftpjq krightpjq 1 if ˝j “ ˆ kleftpjq krightpjq 2 if ˝j “ { P 4 5kchildpjq 5 4 T if ˝j “ ?¨, (47) where leftpjq, rightpjq, and childpjq are denoting the left, right, and only child of j, respectively. For given input data p P An, let ppi, 0q be the pairs at the leaves of T, and denote by pc, gq the result evaluated at root r using our pair arithmetic.
Furthermore, let ˆ c be the true result of the expression for input data p, and let ˆ C be the true result for the NIC remodeled input data p1.
Suppose that all denominators and all expressions below a square root comply with the NIC principle. Furthermore, sup-pose that kj is not larger than u´ 1 2 for any node j comprising of division or square root. Otherwise kj is unbounded.
Assume the pair arithmetic produces pc, gq P A2 as a final result with final k according to (47), and define k :“ ˆ C |ˆ c| with the convention 0 0 :“ 1.
(48) Let η be defined as in (46). If k is restricted via k ď c mintη, uu ku2 ´ 2, (49) then a nearest A-approximation of cg according to (45) is a faithful rounding of ˆ c. For an expression complying with the NIC principle condition (49) reduces to k ď a mintη, uu u ´ 2.
That theorem yields conditions on k and the length of input data for all examples mentioned at the beginning of this section, and more.
For simplicity, assume an IEEE-754 p-digit base-β floating-point arithmetic. Then η ą u{β reduces the condition on k to k ď b 1 βku ´ 2 for general expressions, and the same with k “ 1 for expressions complying with the NIC principle. The error estimates in Table V follow by calculating the maximum value of k for the input expression.
For all examples mentioned at the beginning of this section, a faithfully rounded result is computed with as many or fewer operations and with weaker condition on n, but for any order of evaluation, and for any base β.
In the third example “binary” refers to binary summation.
The sixth example “mixed ˆ, {” means that for n input num-bers xi a tree is evaluated with each node being multiplication or division. The final k satisfies k ď 2pn ´ 1q and the tree complies with the NIC principle, so that k “ 1 and the bound on n for a faithfully rounded result follows.
Note in particular for polynomial interpolation, the second last problem, all denominators satisfy the NIC principle so that TABLE V: Faithfully rounded results by the pair arithmetic.
Problem k bound on n s :“ řn i“0 pixi řn i“0 |pixi| |s| n ď 1 2?βku ´ 1 s :“ řn i“1 xi řn i“1 |xi| |s| n ď 1 ?βku ´ 1 s :“ řn i“1 xi binary řn i“1 |xi| |s| rlog2pnqs ď 1 ?βku ´ 2 s :“ řn i“1 xiyi řn i“1 |xiyi| |s| n ď 1 ?βku ´ 2 śn i“1 xi 1 n ď 1 ?βu ´ 1 mixed ˆ, { 1 n ď 1 2?βu s :“ řn i“0 ś j‰i x´xj ś j‰i xi´xj yi “: řn i“0 Θipxq yi řn i“0 |Θipxqyi|{|s| n ď 1 5 pβkuq´ 1 2 ´ 3 5 s :“ }x}2 1 n ď 1 ?βu ´ 4 Theorem 5.10 is applicable. With adapted constants that holds also true for the faster approach R :“ n ź j“0 px ´ xjq, ppxq “ n ÿ i“0 R px ´ xiq ś j‰ipxi ´ xjq yi.
For the last example in Table V, the Euclidean norm of a vector, double-double arithmetic adapted to nonnegative summands is used by Graillat et al. in requiring 13n 1 operations compared to 10n 1 for our pair arithmetic. They prove, however, that for recursive summation in binary arith-metic the result is faithful for considerably larger maximum vector length n ď 1 24uu2 ´ 3. That is useful for binary32, lifting the bound in Table V on n from 2, 892 to 699, 047; for binary64 the bound n ď 67, 108, 860 from Table V may be sufficient.
VI. FAITHFULLY ROUNDED AND REPRODUCIBLE RESULTS The results in the previous section are proved to be faithfully rounded for not too large condition number. However, the condition number is, in general, not known. Besides, for small condition number likely the nearest approximation is computed, but not always.
Recently, so-called “reproducible” results became popular.
In this section we restrict our attention to summation. Then the true sum is a real number, and reproducibility means to produce exactly the same floating-point approximation no matter what the order of evaluation. That implies addition to become associative, a property which is outside the scope of traditional floating-point algorithms. In addition to “always exactly the same result” we add some not explicitly specified accuracy requirement such as backward stability in a certain sense. Such a requirement is often forgotten in the literature.
For the remaining of this note we assume IEEE-754 p-digit binary arithmetic with the nearest rounding tie-to-even. It is not difficult to extend the methods to general base β.
Following we discuss two approaches producing a faithfully rounded and/or reproducible result, independent of the condi-tion number and guaranteeing associativity. First, a limited exponent range allows to compute the exact sum, for example using a long accumulator as popularized by Kulisch , or Malcolm’s adding by exponents which is based on Wolfe’s approach in 1964. In either case the exact value of the sum can be extracted and rounded faithfully and/or reproducibly.
The second approach splits the bits of a given vector into slices which can be regarded as scaled integers. For example, in binary64 corresponding to 53-bit mantissa (including the implicit 1), suppose each slice is 53 ´ M bits wide. Then at least 2M numbers [scaled integers representable in at most 53 ´ M bits] within a slice can be added without error, see Figure 1.
This idea is due to Zielke and Drygalla who developed it to improve the accuracy of summation and dot products. No analysis is given, and they use a splitting by integer scaling in a way that the exponent range of the input is severely limited.
In the shortcomings are removed, and a complete anal-ysis is given to achieve a faithfully rounded result. Recently, that method has been used for reproducible summation and popularized by Demmel et al. , .
53 M bits 53 M bits 53 M bits Fig. 1: Splitting bits of a vector by Zielke/Drygalla .
The Wolfe/Malcolm methods adjoins each input to a fixed exponent, whereas the slices by Zielke/Drygalla are created depending on the actual input data, a static versus dynamic approach. As a consequence, the first method generates an array of variables over the whole exponent range, independent of the input data, whereas the second method uses just as many slices as necessary to produce a faithfully rounded or reproducible result. In that is used to compute a correctly rounded approximation of a sum with arbitrary precision.
A key point is an efficient way to extract the input into slices . The extraction is relative to some σ so that p “ qp1 and the bits of q and p1 do not overlap as by ExtractVector.
A choice for σ is a power of 2 larger than ř |pi|.
The bit patterns of Algorithm ExtractVector is outlined in Figure 2. The main property of the extraction is the error-free transformation řn i“1 pi “ τřn i“1 p1 i, see for details.
Algorithm 1: Error-free vector transformation extract-ing high order part.
function rτ, p1s “ ExtractVectorpσ, pq τ “ 0 for i “ 1 : n qi “ flpflpσ piq ´ σq p1 i “ flppi ´ qiq τ “ flpτ qiq end for Here τ is the error-free sum of the first slice, comprised of the leading bits of the vector entries pi belonging to this slice [the bold lines in Figure 2]. Note in particular that τ is a sum of scaled integers, thus the computation is associative and error-free.
input p output p´ = 2k u = 2k-53 2k-M bold parts sum to Fig. 2: ExtractVector: error-free transformation ř pi “ τ ř p1 i.
Next, this process is applied to the vector p iteratively, resulting in Algorithm Transform. Here realmin denotes the smallest positive normalized floating-point number. In it Algorithm 2: Error-free transformation of a vector pp0q of length n.
function rτ1, τ2, ppmq, σs “ Transformppp0qq µ “ maxp|pp0q i |q if µ “ 0, τ1 “ τ2 “ ppmq “ σ “ 0, return, end if M “ P log2pn 2q T σ0 “ 2Mrlog2pµqs tp0q “ 0, m “ 0 repeat m “ m 1 rτ pmq, ppmqs “ ExtractVectorpσm´1, ppm´1qq tpmq “ flptpm´1q τ pmqq σm “ flp2Muσm´1q until |tpmq| ě flp22Muσm´1q or σm´1 ď realmin σ “ σm´1 rτ1, τ2s “ FastTwoSumptpm´1q, τ pmqq is shown that this algorithm stops, and for each intermediate m between 1 and its final value řn i“1 pp0q i “ tpm´1q τ pmq řn i“1 ppmq i and max |ppmq i | ď p2Muqmσ0 (50) is satisfied. Moreover it is shown that, denoting the final ppmq by p1, floatpτ1 pτ2 přn i“1 p1 iqqq is a faithful approximation of the exact sum.
That algorithm offers a convenient way to compute a reproducible result by choosing a fixed maximum number of extractions m and return res = tpm´1q τ pmq, i.e., ignoring the remainder terms ppmq i . By the nature of the algorithm and (50), the quantities tpm´1q and τ pmq are uniquely determined and, as sums of scaled integers, independent of the order of evaluation in ExtractVector and Transform.
The quantity σ0 is of the order ř |xi|, so that the omitted summands ppmq i are bounded by about pnuqm ř |xi|. Hence for a condition number up to about upnuq´m the result res is faithful, where for larger condition number the accuracy decreases.
For any value of m the result is reproducible. In binary64 and n “ 1000, the result is faithful up to condition number k ď 9 ¨ 109 for m “ 2, and up to k ď 8 ¨ 1022 for m “ 3.
The equality in (50) implies that for any m the exact sum is available, so a rounded to nearest or K-fold, i.e. an unevaluated sum of K numbers, result can be computed as well .
VII. SUMMARY We first assumed an IEEE-754 p-digit binary floating-point arithmetic, for which the general bound for the first and second standard model is u{p1 uq and u, respectively. Optimal bounds were given, in particular improved for division and square root.
Next we considered a general computer arithmetic satisfying the first standard model, where “rounding” mutated to an arbitrary perturbation of the true real result. With the additional Assumption A, that is | pa ‘ bq ´ pa bq | ď minp|a|, |b|q, standard error estimates for summation, dot products and others of type γk “ ku{p1 ´ kuq reduce to ku for any order of evaluation and without restriction on k. Assumption A is not satisfied for directed or faithful rounding.
For the first standard model together with Assumption B, the same holds true for general perturbations of the true result including directed or faithful rounding with adapted relative rounding error unit u. A mandatory but weak restriction of order u´1 on the number of operations applies.
For the first standard model together with Assumption C instead, the results were extended to replacing γh by hu for h denoting the height of a tree. Again, this is true for any rounding and a mandatory weak restriction on the height.
For a first standard error model together with the additional Assumptions A and B, an optimal error bound ku{p1kuq was shown for k 1 summands provided that k ď β´1 2 u´1. That implies linearized error bounds for other compound operations such as blocked summation or sums of products.
Next we assumed nothing but the first standard model, in particular none of the Assumptions A, B or C. Based on that a pair arithmetic was introduced, simpler and more generally applicable than existing ones. In particular, the often used error-free transformations were weakened by not assuming equality in the transformation. That opens this approach, for example, for IEEE-754 to any base β and any rounding scheme. Sufficient conditions were given that the computed result is a faithfully rounded exact result for arbitrary expres-sions consisting of , ´, ˆ, {, ?¨.
Finally, for IEEE-754 binary arithmetic, a summation method was introduced for computing a result that is guaran-teed to be faithfully rounded or rounded to nearest independent of the condition number of the sum. The method computes slices of the result and is an efficient way to compute repro-ducible results.
ACKNOWLEDGMENT The author is indebted to Claude-Pierre Jeannerod, Marko Lange and Paul Zimmermann for their thorough reading and many valuable comments, including advices on the structure of the paper. Moreover many thanks to Sylvie Boldo, Florian B¨ unger, Stef Graillat, Christoph Lauter, Jean-Michel Muller and Takeshi Ogita for very fruitful suggestions and remarks.
REFERENCES D.H. Bailey. A Fortran-90 based multiprecision system. ACM Trans.
Math. Software, 21(4):379–387, 1995.
S. Boldo, S. Graillat, and J.-M. Muller. On the robustness of the 2Sum and Fast2Sum algorithms. ACM Trans. Math. Softw., 44(1):1–14, 2017.
S. Boldo and C. Mu˜ noz. Provably faithful evaluation of polynomials. In Proceedings of the 2006 ACM Symposium on Applied Computing, SAC ’06, 1328–1332, 2006.
R. Brent and P. Zimmermann. Modern Computer Arithmetic. Cambridge University Press, New York, NY, USA, 2010.
T.J. Dekker.
A floating-point technique for extending the available precision. Numerische Mathematik, 18:224–242, 1971.
J. Demmel, I. Dumitriu, O. Holtz, and P. Koev. Accurate and efficient expression evaluation and linear algebra. Acta Numerica, 2008:87–145, 2008.
J. Demmel and H.D. Nguyen. Fast reproducible floating-point summa-tion. 163–172. Proc. 21st IEEE Symposium on Computer Arithmetic, Austin, Texas, 2013.
J. Demmel and H.D. Nguyen. Parallel reproducible summation. IEEE Trans. Comp., 64(7):2060–2070, 2015.
D. Goldberg. What every computer scientist should know about floating-point arithmetic. ACM Computaing Surveys, 23(1):5–47, 1991.
S. Graillat. Accurate floating-point product and exponentiation. IEEE Trans. Comp., 58(7):994–1000, 2009.
S. Graillat, C. Lauter, P. Tang, N. Yamanaka, and S. Oishi. Efficient calculations of faithfully rounded l2-norms of n-vectors. ACM TOMS, 41(4):24:1–20, 2015.
S. Graillat, V. Lef evre, and J.-M. Muller. On the maximum relative error when computing integer powers by iterated multiplications in floating-point arithmetic. Numerical Algorithms, 70(3):653–667, 2015.
N. J. Higham. Accuracy and Stability of Numerical Algorithms. SIAM Publications, Philadelphia, 2nd edition, 2002.
IEEE, New York. ANSI/IEEE 754-2008: IEEE Standard for Floating-Point Arithmetic, 2008.
C.-P. Jeannerod and S.M. Rump.
Improved error bounds for inner products in floating-point arithmetic.
SIAM J. Matrix Anal. Appl.
(SIMAX), 34(2):338–344, 2013.
C.-P. Jeannerod and S.M. Rump. On relative errors of floating-point operations: optimal bounds and applications. Math. Comp., 87:803–819, 2017.
C.-P. Jeannerod, J.-M. Muller, and P. Zimmermann. On various ways to split a floating-point number. In ARITH 2018 - 25th IEEE Symposium on Computer Arithmetic, pages 53–60. IEEE, June 2018.
M. Joldes, J.-M. Muller, and V. Popescu. Tight and rigorous error bounds for basic building blocks of double-word arithmetic. ACM Trans. Math.
Softw., 44(2):1–27, 2017.
W. M. Kahan. A survey of error analysis. In Proceedings of the IFIP Congress, Ljubljana, Information Processing 71, 1214–1239. North– Holland, Amsterdam, 1972.
D.E. Knuth.
The Art of Computer Programming: Seminumerical Algorithms, volume 2. Addison Wesley, Reading, Massachusetts, 3rd edition, 1998.
U. Kulisch and W.L. Miranker.
Computer Arithmetic in Theory and Practice. Academic Press, New York, 1981.
M. Lange and S.M. Rump. Error estimates for the summation of real numbers with application to floating-point summation. BIT, 57:927–941, 2017.
M. Lange and S.M. Rump. Faithfully rounded floating-point operations.
ACM Trans. Math. Softw., to appear, 2019.
M. Lange and S.M. Rump. Sharp estimates for perturbation errors in summations. Math.Comp., 88:349–368, 2019.
P. Langlois and N. Louvet. More instruction aevel parallelism explains the actual efficiency of compensated algorithms. Technical report, 2007.
V. Lefevre. Correctly rounded arbitrary-precision floating-point summa-tion. IEEE Transactions on Computers, 66(12):14, 2017.
M. Malcolm.
On accurate floating-point summation.
Comm. ACM, 14(11):731–736, 1971.
W.F.
Mascarenhas.
Floating-point numbers are real numbers.
ArXi:1605:09202, 2016.
J.-M. Muller, N. Brunie, F. de Dinechin, C.-P. Jeannerod, M. Joldes, V.
Lef evre, G. Melquiond, N. Revol, and S. Torres. Handbook of Floating-Point Arithmetic. Birkh¨ auser Boston, 2nd edition, 2018.
A. Neumaier. Rundungsfehleranalyse einiger Verfahren zur Summation endlicher Summen. Zeitschrift f¨ ur Angew. Math. Mech. (ZAMM), 54:39– 51, 1974.
T. Ogita, S.M. Rump, and S. Oishi. Accurate sum and dot product.
SIAM Journal on Scientific Computing (SISC), 26(6):1955–1988, 2005.
K. Ozaki, T. Ogita, F. B¨ unger, and S. Oishi. Accelerating interval matrix multiplication by mixed precision arithmetic. Nonlinear Theory and Its Applications, IEICE, 6(3):364–376, 2015.
D.M. Priest.
On properties of floating-point arithmetics: numer-ical stability and the cost of accurate computations.
PhD the-sis, Mathematics Department, University of California at Berkeley, CA, 1992.
ftp://ftp.icsi.berkeley.edu/pub/theory/ priest-thesis.ps.Z.
S.M. Rump.
Error estimation of floating-point summation and dot product. BIT, 52(1):201–220, 2012.
S.M. Rump, F. B¨ unger, and C.-P. Jeannerod. Improved error bounds for floating-point products and Horner’s scheme. BIT, 56(1):293–307, 2015.
S.M. Rump and C.-P. Jeannerod. Improved backward error bounds for LU and Cholesky factorizations. SIAM J. Matrix Anal. Appl. (SIMAX), 35(2):684–698, 2014.
S.M. Rump and M. Lange. On the definition of unit roundoff. BIT, 56(1):309–317, 2015.
S.M. Rump, T. Ogita, and S. Oishi. Accurate floating-point summation part I: faithful rounding. SIAM J. Sci. Comput. (SISC), 31(1):189–224, 2008.
S.M. Rump, T. Ogita, and S. Oishi. Accurate floating-point summation part II: sign, K-fold faithful and rounding to nearest.
Siam J. Sci.
Comput. (SISC), 31(2):1269–1302, 2008.
J.R. Shewchuk.
Adaptive precision floating-point arithmetic and fast robust geometric predicates. Discrete Comput. Geom., 18(3):305–363, 1997.
J.H. Wilkinson.
Error analysis of floating-point computation.
Nu-merische Mathematik, 2:319–340, 1960.
J.M. Wolfe. Reducing truncation errors by programming. Comm. ACM, 7(6):355–356, 1964.
G. Zielke and V. Drygalla. Genaue L¨ osung linearer Gleichungssysteme.
GAMM Mitt. Ges. Angew. Math. Mech., 26:7–108, 2003. |
189076 | https://www.mbl.edu/news/why-do-spiders-have-8-legs-live-science | Why Do Spiders Have 8 Legs? | Live Science | Marine Biological Laboratory
Skip to main content
The University of Chicago Marine Biological Laboratory
MENU
Utility
Visit The MBL
Contact
Careers
Give
About Expand Submenu
BACK About the MBL
Leadership
Advisory Bodies Advisory Bodies Expand Submenu
BACKAdvisory Bodies - Overview
MBL Society MBL Society Expand Submenu
BACKMBL Society - Overview
Anne Hart
MBL Society Members
MBL Society Membership Criteria and Procedures
MBL Society Council Charter
MBL Society Steering Committee
MBL Community MBL Community Expand Submenu
BACKMBL Community - Overview
Land Acknowledgement Statement
MBL Community Committee
University of Chicago
Financial Statements
Careers & Internships Careers & Internships Expand Submenu
BACKCareers & Internships - Overview
Our Community
Benefits Summary
Equal Employment Opportunity Equal Employment Opportunity Expand Submenu
BACKEqual Employment Opportunity - Overview
Accommodation of Individuals with Disabilities
Drug-Free Workplace
Harassment-Free Workplace
Notice of Nondiscrimination
Staff Toolbox
Electronic Posters
Contact HR
History & Archives History & Archives Expand Submenu
BACKHistory & Archives - Overview
Nobel Laureates Affiliated with the MBL
Legacy of Leadership Legacy of Leadership Expand Submenu
BACKLegacy of Leadership - Overview
Cornelia Clapp
Jewel Plummer Cobb
Jean Clark Dan and Katsuma Dan
Ernest Everett Just
Stephen W. Kuffler
Joseph L. Martinez, Jr. and James G. Townsel
Lilian Vaughan and Thomas Hunt Morgan
Seeing Life Exhibit Seeing Life Exhibit Expand Submenu
BACKSeeing Life Exhibit - Overview
Seeing Life at MBL
Bronze Memorial Tablet
Charles Otis Whitman
Edmund Beecher Wilson
Thomas Hunt Morgan
Cornelia Clapp
Ernest Everett Just
Ethel Browne Harvey
Albert Szent-Györgyi
Jean Clark Dan
Katsuma Dan
"Last One to Go"
Shinya Inoué
Polarized Light Microscopy
Seeing Spindles
Video Microscopy
Advertisements from the Collecting Net
Exhibit Videos
Alumni Alumni Expand Submenu
BACKAlumni - Overview
Alumni Spotlight
Alumni Bookshelf
Class Photos Class Photos Expand Submenu
BACKClass Photos - Overview
Biology of Aging Class Photos Biology of Aging Class Photos Expand Submenu
BACKBiology of Aging Class Photos - Overview
Biology of Aging 2020s
Gene Regulatory Networks for Development Class Photos Gene Regulatory Networks for Development Class Photos Expand Submenu
BACKGene Regulatory Networks for Development Class Photos - Overview
Gene Regulatory Networks for Development 2000s
Gene Regulatory Networks for Development 2010s
Biology of Parasitism Class Photos Biology of Parasitism Class Photos Expand Submenu
BACKBiology of Parasitism Class Photos - Overview
Biology of Parasitism 1980s
Biology of Parasitism 1990s
Biology of Parasitism 2000s
Biology of Parasitism 2010s
Biology of Parasitism 2020s
Brains, Minds and Machines Class Photos Brains, Minds and Machines Class Photos Expand Submenu
BACKBrains, Minds and Machines Class Photos - Overview
Brains, Minds and Machines 2010s
Brains, Minds and Machines 2020s
Embryology Class Photos Embryology Class Photos Expand Submenu
BACKEmbryology Class Photos - Overview
Embryology 1880s
Embryology 1940s
Embryology 1950s
Embryology 1960s
Embryology 1970s
Embryology 1980s
Embryology 1990s
Embryology 2000s
Embryology 2010s
Embryology 2020s
Endocrine-Disrupting Chemicals: Hazards and Opportunities Class Photos Endocrine-Disrupting Chemicals: Hazards and Opportunities Class Photos Expand Submenu
BACKEndocrine-Disrupting Chemicals: Hazards and Opportunities Class Photos - Overview
Endocrine-Disrupting Chemicals: Hazards and Opportunities 2020s
Frontiers in Reproduction Class Photos Frontiers in Reproduction Class Photos Expand Submenu
BACKFrontiers in Reproduction Class Photos - Overview
Frontiers in Reproduction 1990s
Frontiers in Reproduction 2000s
Frontiers in Reproduction 2010s
Frontiers in Reproduction 2020s
Frontiers in Stem Cells and Regeneration Class Photos Frontiers in Stem Cells and Regeneration Class Photos Expand Submenu
BACKFrontiers in Stem Cells and Regeneration Class Photos - Overview
Frontiers in Stem Cells and Regeneration 2010s
Frontiers in Stem Cells and Regeneration 2020s
History of Biology Class Photos History of Biology Class Photos Expand Submenu
BACKHistory of Biology Class Photos - Overview
History of Biology 1980s
History of Biology 1990s
History of Biology 2000s
History of Biology 2010s
History of Biology 2020s
Marine Ecology Class Photos Marine Ecology Class Photos Expand Submenu
BACKMarine Ecology Class Photos - Overview
Marine Ecology 1960s
Marine Ecology 1970s
Marine Ecology 1980s
Marine Ecology 1990s
Medical Informatics Class Photos Medical Informatics Class Photos Expand Submenu
BACKMedical Informatics Class Photos - Overview
Medical Informatics 1990s
Medical Informatics 2000s
Microbial Diversity Class Photos Microbial Diversity Class Photos Expand Submenu
BACKMicrobial Diversity Class Photos - Overview
Microbial Diversity 1970s
Microbial Diversity 1980s
Microbial Diversity 1990s
Microbial Diversity 2000s
Microbial Diversity 2010s
Molecular and Cell Biology of Symbiosis Class Photos Molecular and Cell Biology of Symbiosis Class Photos Expand Submenu
BACKMolecular and Cell Biology of Symbiosis Class Photos - Overview
Molecular and Cell Biology of Symbiosis 2020s
Molecular Evolution Class Photos Molecular Evolution Class Photos Expand Submenu
BACKMolecular Evolution Class Photos - Overview
Molecular Evolution 2000s
Molecular Evolution 2010s
Molecular Mycology Class Photos Molecular Mycology Class Photos Expand Submenu
BACKMolecular Mycology Class Photos - Overview
Molecular Mycology 1990s
Molecular Mycology 2000s
Molecular Mycology 2010s
Molecular Mycology 2020s
Neural Systems & Behavior Class Photos Neural Systems & Behavior Class Photos Expand Submenu
BACKNeural Systems & Behavior Class Photos - Overview
Neural Systems & Behavior 1970s
Neural Systems & Behavior 1980s
Neural Systems & Behavior 1990s
Neural Systems & Behavior 2000s
Neural Systems & Behavior 2010s
Neural Systems & Behavior 2020s
Neurobiology Class Photos Neurobiology Class Photos Expand Submenu
BACKNeurobiology Class Photos - Overview
Neurobiology 1970s
Neurobiology 1980s
Neurobiology 1990s
Neurobiology 2000s
Neurobiology 2010s
Neurobiology 2020s
Physiology Class Photos Physiology Class Photos Expand Submenu
BACKPhysiology Class Photos - Overview
Physiology 1890s
Physiology 1950s
Physiology 1960s
Physiology 1970s
Physiology 1980s
Physiology 1990s
Physiology 2000s
Physiology 2010s
Physiology 2020s
Semester in Environmental Science Class Photos Semester in Environmental Science Class Photos Expand Submenu
BACKSemester in Environmental Science Class Photos - Overview
SES 2000s
SES 2010s
SES 2020s
Summer Program in Neuroscience, Excellence and Success Class Photos Summer Program in Neuroscience, Excellence and Success Class Photos Expand Submenu
BACKSummer Program in Neuroscience, Excellence and Success Class Photos - Overview
SPINES 2020s
Zebrafish Class Photos Zebrafish Class Photos Expand Submenu
BACKZebrafish Class Photos - Overview
Zebrafish 2020s
ARTC Reunions ARTC Reunions Expand Submenu
BACKARTC Reunions - Overview
Embryology 130th Anniversary Celebration
SES 25th Anniversary Celebration and SBD Welcome Symposium
Neurobiology 50th Anniversary Symposium
Microbial Diversity 50th Anniversary Symposium
Support the Alumni Fund
Contact Alumni Staff
Campus Facilities & Resources Campus Facilities & Resources Expand Submenu
BACKCampus Facilities & Resources - Overview
Housing Housing Expand Submenu
BACKHousing - Overview
Apply for Housing
Campus Info for Guests
Reservation Process
General Information and Rates
Housing Services
Housing Policies
Housing Buildings
Local Rentals
Contact Housing Staff
Dining Dining Expand Submenu
BACKDining - Overview
Dining Venues
Dietary Accommodations
Swope Café Weekly Specials
Dining Programs
MBL Dining Hall Meals
Sodexo Sustainability @ MBL
Contact Dining Staff
Campus Maps
Parking, Transportation and Grounds Parking, Transportation and Grounds Expand Submenu
BACKParking, Transportation and Grounds - Overview
Day Visitor Parking Request
MBL Shuttle Service MBL Shuttle Service Expand Submenu
BACKMBL Shuttle Service - Overview
MBL Policy For Shuttle Van Use
Contact Transportation & Grounds Staff
Information Technology Information Technology Expand Submenu
BACKInformation Technology - Overview
Multimedia Services
Telecommunications
Shipping and Receiving
Security and Campus Emergencies
Supplier User Guide
Summer Camp & Children's Programs
Bicycle Rentals
Boating Club
Campus Accessibility
Single User Restrooms
Local Area Information Local Area Information Expand Submenu
BACKLocal Area Information - Overview
Travel Information
Local Hotels
Sculptures on the MBL Campus
Sustainability Sustainability Expand Submenu
BACKSustainability - Overview
Recycling
Composting
Electric Vehicles
Nitrogen and Carbon Footprint Tracking
Water Kiosk and Bottle Refill Stations
Friends of the MBL/Get Involved
Production Studio
Campus Health
Research Expand Submenu
BACK Groundbreaking Research at the MBL
Research Centers Research Centers Expand Submenu
BACKResearch Centers - Overview
Ecosystems Center Ecosystems Center Expand Submenu
BACKEcosystems Center - Overview
About The Ecosystems Center
Ecosystems Center 50th Anniversary Celebration Ecosystems Center 50th Anniversary Celebration Expand Submenu
BACKEcosystems Center 50th Anniversary Celebration - Overview
50th Anniversary Program
50th Anniversary Speakers
Ecosystems Education Info
Employment/Internships
Research Projects Research Projects Expand Submenu
BACKResearch Projects - Overview
Long-Term Ecological Research (LTER)
MBL Greenhouse
Oceanic Flux Program (OFP) Oceanic Flux Program (OFP) Expand Submenu
BACKOceanic Flux Program (OFP) - Overview
OFP Collaborations & Outreach
OFP Bibliography
Microscopic Photos
OFP Cruise Photos
OFP Staff
Contact Us
Terrestrial Isotope Biogeochemistry
Ecosystem Modeling Ecosystem Modeling Expand Submenu
BACKEcosystem Modeling - Overview
Frontiers Model
Hydra Model
General Ecosystem Model (GEM) General Ecosystem Model (GEM) Expand Submenu
BACKGeneral Ecosystem Model (GEM) - Overview
GEM Download Page
Multiple Element Limitation (MEL) Multiple Element Limitation (MEL) Expand Submenu
BACKMultiple Element Limitation (MEL) - Overview
MBL MEL IV: Eight resources
MBL MEL V: Annual model
MEL Download Page
MEL II: Two elemental resources
Modelshell
Non-Equilibrium, Stable-Isotope Simulator (NESIS)
Terrestrial Ecosystem Model (TEM)
Tropical Forest Regrowth (TFR)
Center Reports
Stable Isotope Laboratory Stable Isotope Laboratory Expand Submenu
BACKStable Isotope Laboratory - Overview
Our Equipment
Sample Requirements
Services & Charges
Shipping & Turnaround
Media from the Lab
Ecosystems Publications
Ecosystems Center Faculty & Staff
Eugene Bell Center Eugene Bell Center Expand Submenu
BACKEugene Bell Center - Overview
About the Bell Center
Bell Center Faculty and Staff
Featured Projects
Bell Center Publications Bell Center Publications Expand Submenu
BACKBell Center Publications - Overview
Bell Center Publications 2024
Bell Center Publications 2023
Bell Center Publications 2022
Bell Center Publications 2021
Contact Bell Center
Josephine Bay Paul Center Josephine Bay Paul Center Expand Submenu
BACKJosephine Bay Paul Center - Overview
About the JPBC
Bay Paul Publications Bay Paul Publications Expand Submenu
BACKBay Paul Publications - Overview
JBPC Publications 2024
JBPC Publications 2023
JBPC Publications 2022
JBPC Publications 2021
JBPC Publications: 2011 - 2020
JBPC Publications: 2000 - 2010
JBPC Publications: 1996 - 1999
The Keck Facility
Mobile Genetic Element Cluster
2024 CGUE Conference on Unicellular Eukaryotes 2024 CGUE Conference on Unicellular Eukaryotes Expand Submenu
BACK2024 CGUE Conference on Unicellular Eukaryotes - Overview
CGUE Registration and Travel Information
CGUE Speakers
CGUE Early Career Events and More
CGUE Program
Bay Paul Center Faculty and Staff
Whitman Center Whitman Center Expand Submenu
BACKWhitman Center - Overview
About the Whitman Center
Registration and Housing
Whitman Directory Whitman Directory Expand Submenu
BACKWhitman Directory - Overview
2025 Whitman Fellows
2025 Whitman Scientists
Pre-arrival, Arrival & Departure Pre-arrival, Arrival & Departure Expand Submenu
BACKPre-arrival, Arrival & Departure - Overview
Getting Started
Labs, Equipment, and Facilities
Responsible Conduct
Shut Down and Winter Storage
Faculty Directory
Whitman Directory
Researchers and Support Staff
Strategic Initiatives Strategic Initiatives Expand Submenu
BACKStrategic Initiatives - Overview
Imaging Innovation Initiative
New Research Organisms
Research Organisms Research Organisms Expand Submenu
BACKResearch Organisms - Overview
African Clawed Frog
Atlantic Horseshoe Crab
Axolotl
Bat Sea Star
California Two-Spot Octopus
Comb Jelly
Common Cuttlefish
Flamboyant Cuttlefish
Green Algae
Hummingbird Bobtail Squid
Lesser Pacific Striped Octopus
Little Skate
Longfin Squid
Northern Star Coral
Parhyale
Rotifer
Sea Lamprey
Sea Robin
Striped Bass
MBL/UChicago PhD Graduate Program
Scientific Publications Scientific Publications Expand Submenu
BACKScientific Publications - Overview
2024 Scientific Publications
2023 Scientific Publications
Who is at the MBL
MBLWHOI Library
Resources & Research Facilities Resources & Research Facilities Expand Submenu
BACKResources & Research Facilities - Overview
Imaging Resources Imaging Resources Expand Submenu
BACKImaging Resources - Overview
Microscopes in Development
Central Microscopy Facility Central Microscopy Facility Expand Submenu
BACKCentral Microscopy Facility - Overview
CMF Equipment
Scheduling Time and Trainings
MLSC Imaging Community
Contact CMF
Laboratory Operations Laboratory Operations Expand Submenu
BACKLaboratory Operations - Overview
Reserve Equipment
General Use Rooms
Submersible Heaters
Stockroom Information
Marine Resources Center Marine Resources Center Expand Submenu
BACKMarine Resources Center - Overview
About the MRC
MRC Services MRC Services Expand Submenu
BACKMRC Services - Overview
On-Campus Tank Space and Specimens Form
Small Boat Program
Research at the MRC
Cephalopods Cephalopods Expand Submenu
BACKCephalopods - Overview
Cephalopod Breeding Center
Gemma Tracker
Contact MRC Staff
Zebrafish Facility Zebrafish Facility Expand Submenu
BACKZebrafish Facility - Overview
Zebrafish Use
The National Xenopus Resource The National Xenopus Resource Expand Submenu
BACKThe National Xenopus Resource - Overview
About the NXR
Ordering and Pricing
Protocols
3D Printing Files
Mutant Frog Stocks
Transgenic Tropicalis
Transgenic Laevis
Wild Type Frog Stocks
Research Facility Resource
Workshops and Meetings
Contact The NXR
Diving at MBL Diving at MBL Expand Submenu
BACKDiving at MBL - Overview
Dive Manual
10X Genomics Chromium Controller
Live Organism Use Form
Research Administration Research Administration Expand Submenu
BACKResearch Administration - Overview
Contact Research Staff
Office of Sponsored Programs Office of Sponsored Programs Expand Submenu
BACKOffice of Sponsored Programs - Overview
Sponsored Programs Sponsored Programs Expand Submenu
BACKSponsored Programs - Overview
Types of Agreements & Roles
Gifts vs. Grants
Uniform Guidance Information Uniform Guidance Information Expand Submenu
BACKUniform Guidance Information - Overview
Federal Agency Guidelines and Procurement
Subawards
Costs and Fees
Admin Salaries and Conferences
PI Absence from Project
Closeout
Grant & Contract Administration Grant & Contract Administration Expand Submenu
BACKGrant & Contract Administration - Overview
Conflict of Interest and PHS (Public Health Science) Sponsored Specific Requirements
Effort Reporting: Semiannual Payroll Verification
Federal Policies & OSP Procedures
Funding & Sponsor Information
NIH and NSF Registration Requests
NSF Off-Campus/Off-Site Research Safe and Harassment-free Fieldwork (SAHF) Plan
OSP Forms
All OSP News
Contact OSP
For Foreign Nationals For Foreign Nationals Expand Submenu
BACKFor Foreign Nationals - Overview
Exchange Visitors Program Exchange Visitors Program Expand Submenu
BACKExchange Visitors Program - Overview
J-1 Visa Instructions
English Proficiency Policy for International Participants
Resources
Tax Information for Recipients of Scholarships, Fellowships, Stipends and Honoraria Tax Information for Recipients of Scholarships, Fellowships, Stipends and Honoraria Expand Submenu
BACKTax Information for Recipients of Scholarships, Fellowships, Stipends and Honoraria - Overview
Applying for SSN or Individual Taxpayer ID
Foreign National Information System (FNIS)
Visas, Employment Restrictions, and Payments
Types of Visas
Contact Foreign Nationals Staff
Compliance and Ethics Compliance and Ethics Expand Submenu
BACKCompliance and Ethics - Overview
IACUC IACUC Expand Submenu
BACKIACUC - Overview
IACUC Charter
Reporting Animal Care Concerns
Committee and Regulations
Animal Use Guidelines and Purchasing
Independent Review Board
Intellectual Property
Policies
Resources
Training
Contact HR Director
MBL Toolbox
Research Registration Forms
Funding Opportunities
Health & Safety Health & Safety Expand Submenu
BACKHealth & Safety - Overview
Biological Safety Biological Safety Expand Submenu
BACKBiological Safety - Overview
Biosafety Committee
IBC Application
IBC Meeting Minutes
IBC Charter
Chemical Safety Chemical Safety Expand Submenu
BACKChemical Safety - Overview
Acutely Hazardous Chemicals
Chemical Inventory Instructions
Chemical Storage and Segregation
Peroxide Forming Chemicals (PFC) Guidelines
Safety Data Sheet (SDS) Information
DEA Controlled Substances
Occupational Health and Safety Occupational Health and Safety Expand Submenu
BACKOccupational Health and Safety - Overview
Electrical Safety
Fall Protection Program
Fire Safety Policy
Forklift Safety
Hearing Conservation
Ladder Safety
Lockout/Tagout Program
Machine Guarding
Occupational Health Program for Animal Handlers
Occupational Health and Safety Program
Workstation Ergonomics
Radiation and Laser Safety Radiation and Laser Safety Expand Submenu
BACKRadiation and Laser Safety - Overview
Radiation Safety Committee Charter
Radiation Safety Committee
Research Safety Questionnaire Form
Waste Management
Contact EHS Staff
Education Expand Submenu
BACK Educational Opportunities
Advanced Research Training Courses Advanced Research Training Courses Expand Submenu
BACKAdvanced Research Training Courses - Overview
Course Offerings Course Offerings Expand Submenu
BACKCourse Offerings - Overview
Analytical and Quantitative Light Microscopy Analytical and Quantitative Light Microscopy Expand Submenu
BACKAnalytical and Quantitative Light Microscopy - Overview
Course Faculty and Lecturers
Financial and Course Support
2025 AQLM Schedule
Biology of Aging Biology of Aging Expand Submenu
BACKBiology of Aging - Overview
Course Faculty and Lecturers
Financial and Course Support
2023 Biology of Aging Schedule
Biology of Parasitism: Modern Approaches Biology of Parasitism: Modern Approaches Expand Submenu
BACKBiology of Parasitism: Modern Approaches - Overview
Course Faculty and Lecturers
Financial and Course Support
2025 BoP Schedule
Student Testimonials
Blout Lectureship
Englund Lectureship
Biology of the Inner Ear: Experimental and Analytical Approaches Biology of the Inner Ear: Experimental and Analytical Approaches Expand Submenu
BACKBiology of the Inner Ear: Experimental and Analytical Approaches - Overview
Course Faculty and Lecturers
Financial and Course Support
BIE Schedule
BIE Protocols
Brains, Minds and Machines Brains, Minds and Machines Expand Submenu
BACKBrains, Minds and Machines - Overview
Course Faculty and Lecturers
Financial and Course Support
2024 BMM Schedule
AI@MBL: Machine Learning for Microscopy Image Analysis AI@MBL: Machine Learning for Microscopy Image Analysis Expand Submenu
BACKAI@MBL: Machine Learning for Microscopy Image Analysis - Overview
Course Faculty and Lecturers
Financial and Course Support
2025 AI@MBL Schedule
Embryology: Concepts and Techniques in Modern Developmental Biology Embryology: Concepts and Techniques in Modern Developmental Biology Expand Submenu
BACKEmbryology: Concepts and Techniques in Modern Developmental Biology - Overview
Course Faculty and Lecturers
Financial and Course Support
2025 Embryology Schedule
Course Structure and Modules
Course Objectives
Course History
Named Lectureships
Student Testimonials
Embryology Publications
Endocrine-Disrupting Chemicals: Hazards and Opportunities Endocrine-Disrupting Chemicals: Hazards and Opportunities Expand Submenu
BACKEndocrine-Disrupting Chemicals: Hazards and Opportunities - Overview
Course Faculty and Lecturers
Financial and Course Support
2025 ECHO Schedule
Frontiers in Reproduction Frontiers in Reproduction Expand Submenu
BACKFrontiers in Reproduction - Overview
Course Faculty and Lecturers
Financial and Course Support
2025 FIR Schedule
Annual FIR Symposium
Course History
Student Testimonials
Awards
Frontiers in Stem Cells & Regeneration Frontiers in Stem Cells & Regeneration Expand Submenu
BACKFrontiers in Stem Cells & Regeneration - Overview
Course Faculty and Lecturers
2024 SCARE Schedule
Financial and Course Support
Gene Regulatory Networks for Development Gene Regulatory Networks for Development Expand Submenu
BACKGene Regulatory Networks for Development - Overview
Course Faculty and Lecturers
Financial and Course Support
2025 GERN Schedule
Methods in Computational Neuroscience Methods in Computational Neuroscience Expand Submenu
BACKMethods in Computational Neuroscience - Overview
Course Faculty and Lecturers
Financial and Course Support
2025 MCN Schedule
Course Materials
Lecture Videos
Microbial Diversity Microbial Diversity Expand Submenu
BACKMicrobial Diversity - Overview
Course Faculty and Lecturers
Financial and Course Support
2025 Microbial Diversity Schedule
Student Testimonials
Publications Publications Expand Submenu
BACKPublications - Overview
Microbial Diversity Publications 1996-2019
Molecular Mycology: Current Approaches to Fungal Pathogenesis Molecular Mycology: Current Approaches to Fungal Pathogenesis Expand Submenu
BACKMolecular Mycology: Current Approaches to Fungal Pathogenesis - Overview
Course Faculty and Lecturers
Financial and Course Support
2024 MOMY Schedule
The Deborah Hogan and Andrew Alspaugh Endowed Lecture
The Jack Edwards and Aaron Mitchell Endowed Lecture
Molecular and Cell Biology of Symbiosis Molecular and Cell Biology of Symbiosis Expand Submenu
BACKMolecular and Cell Biology of Symbiosis - Overview
Financial and Course Support
2025 Molecular and Cell Biology of Symbiosis Schedule
Neural Systems & Behavior Neural Systems & Behavior Expand Submenu
BACKNeural Systems & Behavior - Overview
Course Faculty and Lecturers
Financial and Course Support
Course Structure and Modules
2025 NS&B Calendar
2025 NS&B Students
NS&B Endowed Lectures
Publications
Video Lectures
Video Tutorials
Course History
Student Testimonials
Neurobiology: Mechanisms & Advanced Approaches Neurobiology: Mechanisms & Advanced Approaches Expand Submenu
BACKNeurobiology: Mechanisms & Advanced Approaches - Overview
Course Faculty and Lecturers
Financial and Course Support
2025 Neurobiology Schedule
Publications
Student Testimonials
Optical Microscopy & Imaging in the Biomedical Sciences Optical Microscopy & Imaging in the Biomedical Sciences Expand Submenu
BACKOptical Microscopy & Imaging in the Biomedical Sciences - Overview
Course Faculty and Lecturers
Financial and Course Support
2025 OMIBS Schedule
Physiology: Modern Cell Biology Using Microscopic, Biochemical and Computational Approaches Physiology: Modern Cell Biology Using Microscopic, Biochemical and Computational Approaches Expand Submenu
BACKPhysiology: Modern Cell Biology Using Microscopic, Biochemical and Computational Approaches - Overview
Course Faculty and Lecturers
Financial and Course Support
About Physiology
2025 Physiology Schedule
Student Testimonials
Strategies and Techniques for Analyzing Microbial Population Structures Strategies and Techniques for Analyzing Microbial Population Structures Expand Submenu
BACKStrategies and Techniques for Analyzing Microbial Population Structures - Overview
Course Faculty and Lecturers
Financial and Course Support
2025 STAMPS Schedule
Summer Program in Neuroscience, Excellence and Success Summer Program in Neuroscience, Excellence and Success Expand Submenu
BACKSummer Program in Neuroscience, Excellence and Success - Overview
Course Faculty and Lecturers
Financial and Course Support
2025 SPINES Schedule
Visual Neuroscience Visual Neuroscience Expand Submenu
BACKVisual Neuroscience - Overview
Course Faculty and Lecturers
Financial and Course Support
2025 VN Schedule
VN Modules
2025 VN Students
Workshop on Molecular Evolution Workshop on Molecular Evolution Expand Submenu
BACKWorkshop on Molecular Evolution - Overview
Course Faculty and Lecturers
Financial and Course Support
2025 Molecular Evolution Schedule
Zebrafish Development and Genetics Zebrafish Development and Genetics Expand Submenu
BACKZebrafish Development and Genetics - Overview
Course Faculty and Lecturers
Financial and Course Support
2025 Zebrafish Schedule
Student Awards Presentation
Faculty Opportunities in MBL Courses
Before You Apply
Apply
Courses Calendar
Lecture Schedules
Info for Students Info for Students Expand Submenu
BACKInfo for Students - Overview
Immigration and Tax Guidance
Building Innovation: A faculty enhancement program
MBL Alumni ROCS MBL Alumni ROCS Expand Submenu
BACKMBL Alumni ROCS - Overview
2023-24 Alumni ROCS Awarded Projects
2022-23 Alumni ROCS Awarded Projects
2021-2022 Alumni ROCS Awarded Projects
Info for Course Faculty and Staff Info for Course Faculty and Staff Expand Submenu
BACKInfo for Course Faculty and Staff - Overview
Info for Foreign National Course Faculty
Education Travel Information Education Travel Information Expand Submenu
BACKEducation Travel Information - Overview
Travel Policies
MBL Light Microscopy Bootcamp
MBL/UChicago Graduate Research Fellowship Program
Undergraduate Programs Undergraduate Programs Expand Submenu
BACKUndergraduate Programs - Overview
Semester in Environmental Science Semester in Environmental Science Expand Submenu
BACKSemester in Environmental Science - Overview
SES Courses SES Courses Expand Submenu
BACKSES Courses - Overview
SES Faculty
2025 SES Calendar
2025 SES Syllabus
Distinguished Scientist Seminar Series
SES Research Symposium
SES Application Instructions
SES Tuition and Aid
SES FAQ
SES Program Credits
Participating Colleges
Student Life
SES Alumni SES Alumni Expand Submenu
BACKSES Alumni - Overview
SES Students & Projects
SES Alumni List by College
Participation in Conferences
SES 25th Anniversary Celebration and SBD Welcome Symposium SES 25th Anniversary Celebration and SBD Welcome Symposium Expand Submenu
BACKSES 25th Anniversary Celebration and SBD Welcome Symposium - Overview
SES/SBD Event Gallery
SES Multimedia
Support SES
SES Contacts
Semester in Biological Discovery (SBD) Semester in Biological Discovery (SBD) Expand Submenu
BACKSemester in Biological Discovery (SBD) - Overview
2026 SBD Course Offerings 2026 SBD Course Offerings Expand Submenu
BACK2026 SBD Course Offerings - Overview
Biodiversity: Exploring the Marine Diversity of Woods Hole Using Molecular Tools Biodiversity: Exploring the Marine Diversity of Woods Hole Using Molecular Tools Expand Submenu
BACKBiodiversity: Exploring the Marine Diversity of Woods Hole Using Molecular Tools - Overview
Biodiversity: Exploring the Marine Diversity of Woods Hole Using Molecular Tools Schedule
Meet the Faculty: Andrew Gillis
Dynamic Camouflage: Behavior, Visual Perception and Neural Skin Patterning in Cephalopods Dynamic Camouflage: Behavior, Visual Perception and Neural Skin Patterning in Cephalopods Expand Submenu
BACKDynamic Camouflage: Behavior, Visual Perception and Neural Skin Patterning in Cephalopods - Overview
Dynamic Camouflage: Behavior, Visual Perception and Neural Skin Patterning in Cephalopods Schedule
Meet the Faculty: Roger Hanlon
Embryology Embryology Expand Submenu
BACKEmbryology - Overview
Embryology Schedule
Fundamentals of Synapses Fundamentals of Synapses Expand Submenu
BACKFundamentals of Synapses - Overview
Fundamentals of Synapses Schedule
Imaging for Biological Research
Introduction to Parasitology Introduction to Parasitology Expand Submenu
BACKIntroduction to Parasitology - Overview
Introduction to Parasitology Schedule
Microbes Across Environments
Science Writing
Stem Cells and Regeneration: From Aquatic Research Organisms to Mammals Stem Cells and Regeneration: From Aquatic Research Organisms to Mammals Expand Submenu
BACKStem Cells and Regeneration: From Aquatic Research Organisms to Mammals - Overview
Stem Cells and Regeneration Schedule
Meet the Faculty: Karen Echeverri
SBD Application Instructions
SBD Tuition and Aid
SBD FAQ
SBD Program Credits
Student Life
SBD Contacts
UChicago Undergraduate Programs
High School Science Discovery Program High School Science Discovery Program Expand Submenu
BACKHigh School Science Discovery Program - Overview
Frequently Asked Questions
Participating Schools
Sample Course Schedule
Student Experiences
Internships Internships Expand Submenu
BACKInternships - Overview
NSF REU Biological Discovery in Woods Hole NSF REU Biological Discovery in Woods Hole Expand Submenu
BACKNSF REU Biological Discovery in Woods Hole - Overview
NSF REU Program Details
NSF REU Program FAQ
NSF REU Abstracts and Videos
NSF REU – Beaufort Lagoon Ecosystems LTER
REU – Oceanic Flux Program
NSF REU – Plum Island LTER
Jeff Metcalf Summer for Undergraduate Research Fellows (SURF) Jeff Metcalf Summer for Undergraduate Research Fellows (SURF) Expand Submenu
BACKJeff Metcalf Summer for Undergraduate Research Fellows (SURF) - Overview
Metcalf Mentors
2019 Metcalf Abstracts
2018 Metcalf Abstracts
2017 Metcalf Abstracts
2016 Metcalf Abstracts
Dartmouth College E.E. Just Internship Dartmouth College E.E. Just Internship Expand Submenu
BACKDartmouth College E.E. Just Internship - Overview
Dartmouth College E.E. Just Internship Mentors
The Blue Economy Internship Program (BEIP)
Marine Aquarist and Research Internship
Ecosystems Center Paid Internship
Logan Science Journalism Program Logan Science Journalism Program Expand Submenu
BACKLogan Science Journalism Program - Overview
2025 SJP Fellows
Apply for SJP
SJP Eligibility
Biomedical Course Biomedical Course Expand Submenu
BACKBiomedical Course - Overview
2025 Schedule
Environmental Course Environmental Course Expand Submenu
BACKEnvironmental Course - Overview
2025 Schedule
Alumni and Testimonials Alumni and Testimonials Expand Submenu
BACKAlumni and Testimonials - Overview
2024 SJP Fellows
2023 SJP Fellows
2022 SJP Fellows
2021 SJP Fellows
Faculty and Staff Faculty and Staff Expand Submenu
BACKFaculty and Staff - Overview
Emeritus Directors
Funding and Support
Contact SJP Staff
Comparative Developmental Biology Comparative Developmental Biology Expand Submenu
BACKComparative Developmental Biology - Overview
CDB Schedule
Student Accessibility Services Student Accessibility Services Expand Submenu
BACKStudent Accessibility Services - Overview
Accommodation Request Form
Scientific Vendor Partnerships
MBL Toolbox
Education Registration Forms
Contact Education Staff
Events Expand Submenu
BACK What's Happening
Falmouth Forum Falmouth Forum Expand Submenu
BACKFalmouth Forum - Overview
2024-25 Falmouth Forum
2023-24 Falmouth Forum
2022-23 Falmouth Forum
2021-22 Falmouth Forum
2020-21 Falmouth Forum
2019-20 Falmouth Forum
2018-19 Falmouth Forum
2017-18 Falmouth Forum
2016-17 Falmouth Forum
2015-16 Falmouth Forum
2014-15 Falmouth Forum
Falmouth Forum Archives
Friday Evening Lectures Friday Evening Lectures Expand Submenu
BACKFriday Evening Lectures - Overview
2024 Friday Evening Lectures
2023 Friday Evening Lectures
2022 Friday Evening Lectures
2021 Friday Evening Lectures
2020 Friday Evening Lectures
2019 Friday Evening Lectures
2018 Friday Evening Lectures
2017 Friday Evening Lectures
2016 Friday Evening Lectures
2015 Friday Evening Lectures
Nikon Small World Exhibit
Exhibit: Plankton in Our Midst
MBL Video Center
Public Events & Lectures
For Scientists & Students For Scientists & Students Expand Submenu
BACKFor Scientists & Students - Overview
Endowed Lectures
Flash Talk Presentations
Monday Night Neuroscience Seminars
Tuesday Evening Seminar Series on Cell Division and the Cytoskeleton
Wednesday Evening Neural Systems & Behavior Seminars
Thursday Evening NeuroImaging Seminars
Conferences Conferences Expand Submenu
BACKConferences - Overview
General Information & FAQs
Conference Policies
For Conference Organizers For Conference Organizers Expand Submenu
BACKFor Conference Organizers - Overview
Meeting Spaces and Other Facilities Meeting Spaces and Other Facilities Expand Submenu
BACKMeeting Spaces and Other Facilities - Overview
Auditoriums
Classrooms and Conference Space
Multi-Function and Event Space
Setup Options
Conference Rates
Housing and Meals
Conference Add-ons
Upcoming Conferences at the MBL
Contact Conference Staff
MBL Virtual Backgrounds
News Expand Submenu
BACK All MBL News
Newsletter
Recent Grants Recent Grants Expand Submenu
BACKRecent Grants - Overview
Grant Awards 2022-23
MBL Accolades
MBL Publications
Media Information
Science Communication Resources for Scientists
Obituaries
Keyword
Type
Back to NewsNewsOverview
Newsletter
Recent Grants
Overview
Grant Awards 2022-23
MBL Accolades
MBL Publications
Media Information
Science Communication Resources for Scientists
Obituaries
Home
All MBL News
Why Do Spiders Have 8 Legs? | Live Science
Bold jumper spider (Phidippus audax). Credit: Phil Myers, cc-by-nc-sa-3.0
By Cameron Duke | Live Science July 24, 2023
Share Articles
Share Marine Biological Laboratory | Why Do Spiders Have 8 Legs? | Live Science on Facebook
Share Marine Biological Laboratory | Why Do Spiders Have 8 Legs? | Live Science on Twitter
Share Marine Biological Laboratory | Why Do Spiders Have 8 Legs? | Live Science on LinkedIn
Related Scientists
Full Name Nipam Patel
Categories
Media Mentions
MBL Director Nipam Patel and Research Associate Heather Bruce are quoted in this story.
There seems to be no ideal number of legs. Humans have two, dogs have four, insects have six and millipedes can have over 1,000. So what made spiders settle for eight legs?
"I think the best answer and the simplest answer is that spiders have eight legs because their parents did," Thomas Hegna, an assistant professor of invertebrate paleontology at the State University of New York at Fredonia, told Live Science. "But then that gets into sort of a regress, and somewhere this all had to start."
If we follow the succession of eight-legged spider parents back to about 500 million years ago, during the middle Cambrian Period, we arrive at the root of the chelicerate lineage, the group of arthropods that contains spiders. If we go even further back, to 541 million years ago, we find the ocean-dwelling lobopods, the ancestors of all arthropods. Read rest of the story here.
Source: Why Do Spiders Have 8 Legs? | Live Science
Related Tags
Crustaceans
Evo-Devo
Evolution
Share Articles
Share Marine Biological Laboratory | Why Do Spiders Have 8 Legs? | Live Science on Facebook
Share Marine Biological Laboratory | Why Do Spiders Have 8 Legs? | Live Science on Twitter
Share Marine Biological Laboratory | Why Do Spiders Have 8 Legs? | Live Science on LinkedIn
The University of Chicago Marine Biological Laboratory
MBL In Your Inbox
Sign up to receive “The Collecting Net,” our bi-weekly newsletter.
Sign up for the Collecting Net
MBL Toolbox
Quickly access resources and logistical information for students, instructors and researchers.
Prepare Your Lab
Housing Information
Imaging Resources
Order an Organism
What's Happening at the MBL?
Who Is At The MBL?
Access the MBL Toolbox
Connect
Facebook
Instagram
Vimeo
Youtube
LinkedIn
Twitter
Main navigation
About About the MBL
Leadership
Advisory Bodies
MBL Community
University of Chicago
Financial Statements
Careers & Internships
History & Archives
Alumni
Campus Facilities & Resources
Local Area Information
Sustainability
Friends of the MBL/Get Involved
Production Studio
Campus Health
Research Groundbreaking Research at the MBL
Research Centers
Faculty Directory
Whitman Directory
Researchers and Support Staff
Strategic Initiatives
Research Organisms
MBL/UChicago PhD Graduate Program
Scientific Publications
Who is at the MBL
MBLWHOI Library
Resources & Research Facilities
Research Administration
MBL Toolbox
Research Registration Forms
Funding Opportunities
Health & Safety
Education Educational Opportunities
Advanced Research Training Courses
MBL Light Microscopy Bootcamp
MBL/UChicago Graduate Research Fellowship Program
Undergraduate Programs
High School Science Discovery Program
Internships
Logan Science Journalism Program
Comparative Developmental Biology
Student Accessibility Services
Scientific Vendor Partnerships
MBL Toolbox
Education Registration Forms
Contact Education Staff
Events What's Happening
Falmouth Forum
Friday Evening Lectures
Nikon Small World Exhibit
Exhibit: Plankton in Our Midst
MBL Video Center
Public Events & Lectures
For Scientists & Students
Conferences
MBL Virtual Backgrounds
News All MBL News
Newsletter
Recent Grants
MBL Accolades
MBL Publications
Media Information
Science Communication Resources for Scientists
Obituaries
Footer Menu 2
MBL Gift Shop
MBL Intranet
Stay For The Season
Visit The MBL
© 1996-2025, The Marine Biological Laboratory, MARINE BIOLOGICAL LABORATORY, MBL, and the 1888 logo are registered trademarks and service marks of The Marine Biological Laboratory.
Subfooter
Privacy Policy
Accessibility |
189077 | https://www.cliffsnotes.com/study-guides/calculus/calculus/applications-of-the-derivative/increasing-decreasing-functions | Do Not Sell or Share My Personal Information
When you visit our website, we store cookies on your browser to collect information. The information collected might relate to you, your preferences or your device, and is mostly used to make the site work as you expect it to and to provide a more personalized web experience. However, you can choose not to allow certain types of cookies, which may impact your experience of the site and the services we are able to offer. Click on the different category headings to find out more and change our default settings according to your preference. You cannot opt-out of our First Party Strictly Necessary Cookies as they are deployed in order to ensure the proper functioning of our website (such as prompting the cookie banner and remembering your settings, to log into your account, to redirect you when you log out, etc.). For more information about the First and Third Party Cookies used please follow this link.
More information
Manage Consent Preferences
Essential Cookies
Always Active
Essential Cookies are required for providing you with features or services that you have requested. For example, certain Cookies enable you to log into secure areas of our Services.
Share or Sale of Personal Data
Always Active
Under the California Consumer Privacy Act, you have the right to opt-out of the sale of your personal information to third parties. These cookies collect information for analytics and to personalize your experience with targeted ads. You may exercise your right to opt out of the sale of personal information by using this toggle switch. If you opt out we will not be able to offer you personalised ads and will not hand over your personal information to any third parties. Additionally, you may contact our legal department for further clarification about your rights as a California consumer by using this Exercise My Rights link.
If you have enabled privacy controls on your browser (such as a plugin), we have to take that as a valid request to opt-out. Therefore we would not be able to track your activity through the web. This may affect our ability to personalize ads according to your preferences.
Advertising Cookies
Always Active
Advertising Cookies collect data about your online activity and identify your interests so that we can provide advertising that we believe is relevant to you. Advertising Cookies may include Retargeting Cookies.
Analytics Cookies
Always Active
Analytics Cookies allow us to understand how visitors use our Services. They do this by collecting information about the number of visitors to the Services, what pages visitors view on our Services and how long visitors are viewing pages on the Services.
Analytics Cookies also help us measure the performance of our advertising campaigns in order to help us improve our campaigns and the Services’ content for those who engage with our advertising.
Cookie List
Consent Leg.Interest
label
label
label |
189078 | https://sscchemistry.weebly.com/monoatomic-and-polyatomic-ions.html | Monoatomic and Polyatomic Ions - SSC Chemistry
SSC Chemistryanya.covarrubias@ssccardinals.org anna.ahrens@ssccardinals.org
Home
Lab Introduction
Quantitative Basics
The Atom and Periodic Table
Electrons and Light
1st Term Review
Ionic Bonding
Covalent Bonding
Chemical Reactions
The Mole and Stiocheometry
2nd Term Review
Monoatomic Ions and Oxidation Numbers
LEARNING TARGET: STUDENT CAN IDENTIFY MONATOMIC AND POLYATOMIC IONS.
For Monoatomic Ions, Go to pages 208-209
For Polytomic Ions, Go to Pages 184,190,218,275Monoatomic ions are the elements that have either lost or gained electrons by bonding with another element. The Periodic Table helps us with all the monoatmoic ions in the "A groups" (the s and p-block), but it doesn't give us any information about the "B group" (the d-block). To be able to identify the charge of the B group we have to use the oxidation number. An oxidation number IS ALWAYS GIVEN IN THE NAME. We will discuss this more when we start naming ionic compounds. Below is a chart with the most common monoatomic ions, including some from the A group that we can find on the Periodic Table and others from the B group that we cannot. Notice that oxidation numbers are a Roman numeral. The Roman numeral tells the CHARGE of the monoatomic ion. So copper (I) has a positive one charge. ALL NUMBERS THAT ARE OXIDATION NUMBERS ARE POSITIVE. We will only see oxidation numbers in ionic bonding.
Polyatomic Ions
Polyatomic ions are a group of elements that are COVALENTLY BONDED, but are still not happy in terms of their electrons. Because of this, the molecule tries to give and take electrons just like an unhappy element. In the illustration below Nitrate is one nitrogen and three oxygen. However, the electrons have multiple places they can be located and therefore move giving the opportunity for other ions to bond with the molecule. An example of how nitrate would bond with sodium would be NaNO3. You must memorize the polyatomic ions! It becomes VERY important to know these when we start naming ionic and covalent bonds. Even though polyatomic ions are bonded covalently, we will only see polyatomic ions in ionic bonds.
Ion and Formula WS
Memorizing the Polyatomic Ions (Quiz Wednesday)
Quiz tomorrow over the following
PP for Polyatomic Ions to Study
electron dot structure
charges of monoatomic ions in A group
formula writing for the elements
polyatomic ion (name, charges, and formulas) use the chart below that was on the video
Acids and Bases
Titration Lab >
Lab Book Requirements
Lab Analysis
Sci Not and Sig Figs
Dimensional Analysis
Density and Volume
% Yeild & % Error
Test Review
Atomic Review >
Isotopes AND % Abundance
Periodic Table Review >
Periodic Trends
History of the Atom
Test Review
Atomic Orbitals and Aufbau
Properties of a Wave and Photons
Electromagnetic Spectrum and Light
Test Review
Monoatomic and Polyatomic Ions
Ionic Naming and Formula Practice
Ionic Properties and Acids and Bases
Acid Naming
Test Review
VSPER Theory and Calculations
Covalent Naming >
Artificial Flavor Lab
Polarity/Dipole and Hybridization
Resonance and Exceptions
Metallic Bonding
Test Review
Parts/Types of Chemical RXN
Identifying Types and Practice
Net Ionic Equations and Solubility >
Circle of Copper Lab
Test Review
Mole Conversions and Calculation
Limiting Reactant
% Composition and Empirical/Molecular Formulas
Lab Final-Making Salt
Powered by Create your own unique website with customizable templates.Get Started |
189079 | https://www.bill.com/learning/investment-management | Home
/
Learning Center
/
Investment management: What it is and why it matters
Investment management: What it is and why it matters
Brendan Tuytel
Contributor
Table of contents
What is investment management?Types of investment managementThe importance of investment managementAdvantages and disadvantages of investment managementInvestment management processAbout investment management firmsThe Future of Investment ManagementHow investment management firms bill in 2025
Get more from BILL
Subscribe to finance insights and thought leadership content delivered straight to your inbox.
Thank you! Your submission has been received!
Oops! Something went wrong while submitting the form.
By continuing, you agree to BILL's Terms of Service and Privacy Notice.
The term “investment management” creates images of people in suits on the stock exchange frantically finalizing trades to try and maximize their returns. But the truth is far from that.
After a drop off in 2008, investment activity has returned to its usual amount and in 2024, 62% of adults in the United States were invested in stocks. This means that 62% of adults are participating in investment management.
For the new adopters, practiced vets, and trained professionals, investment management is at the center of generating money from savings. And here’s what you need to know about the practice and industry that surrounds it.
Key takeaways
Investment management involves handling a portfolio to meet investment goals, including strategy, buying assets, and managing taxes.
There are different types of investment management, like active vs. passive and discretionary vs. non-discretionary.
Technology, automation, and sustainable investing are shaping the future of investment management, making it more accessible and efficient.
What is investment management?
Investment management is the process of handling an investment portfolio or set of assets to successfully achieve an investment goal.
What falls under investment management generally includes creating an investment strategy, planning out a portfolio breakdown, buying and selling assets, and planning for the eventual taxation of any capital gains.
Almost any type of investment falls under investment management. Whether it’s choosing a single mutual fund or picking and choosing individual stocks and bonds, you may be doing investment management without even knowing it.
What are investment management services?
The term investment management services is most often used when discussing asset management firms or wealth management services. These are businesses whose sole responsibility is managing the money of their clients to maximize the return on their assets.
Instead of an individual managing their investments, a third-party does it all for them. They’re responsible for staying on top of shifts in the markets and making decisions on the fly to achieve the goals of their clients.
Investment management vs. financial planning
While both investment management and financial planning involve overseeing money usage, they serve different purposes.
Financial planning is a larger umbrella which includes budgeting and day-to-day money management. For example, a financial planner could help their client figure out their typical monthly earnings and bills, outline an amount that could be put in investments, and help plan for year-end taxes given all of the client’s financial activity.
Investment management hones in on solely the investment activity. They’re not concerned with what a client is doing day-to-day with their money so long as the investments are performing well.
Types of investment management
Investment management is as involved or hands-free as a process as you want it to be. Here’s how investment management services are categorized in the industry.
Active vs. passive investment management
Investors have the option of making individual investments (e.g. stocks and bonds) or indexed investments (e.g. mutual funds).
The act of buying individual investments is called active investment. This process is more time-intensive and requires an eye for detail as the portfolio is likely made up of a greater volume of investments, each with a higher volatility.
Passive investing favors mutual funds or ETFs over individual investments. This classification of investments includes pre-bundled individual investments, which helps curb the volatility. As a result, it’s less time-consuming and more hands-off, hence the name “passive” investing.
Discretionary vs. non-discretionary management
Depending on the client, they may want to approve every single purchase and sale of an investment, or they may trust their investment manager completely. This difference refers to discretionary and non-discretionary investment management.
In discretionary management, the investment manager has the authority to buy and sell without getting the approval of the client. However, this doesn’t give them free reign to do whatever they’d like as they have to abide by the investment strategy and uphold their fiduciary duty.
In non-discretionary management, the investment manager cannot make any changes without the explicit approval of the client. They may advise the client and suggest moves, but they cannot execute on the plan without the client’s approval.
Alternative investment strategies
While investment most often refers to the management of an investment portfolio composed of stocks and bonds, there are alternatives that can similarly be managed by individuals or professionals.
Some examples of alternative investment strategies include:
Real estate
Private debt
Foreign currency
Shorting
Private markets
Hedge funds
Venture capital
With each of these options, it’s best to work with a specialist who’s familiar with the unique ins-and-outs of the specific investment strategy to get the best return.
The importance of investment management
Generally speaking, the purpose of investment management is to maximize the money earned on a portfolio of assets. But this means different things to different investors.
The goals of an investor will vary based on their risk tolerance, timeline, and willingness to put in the time and effort to manage a portfolio.
What this means is that “maximizing the return on an investment” depends heavily on the individual. And their investment management approach or investment management services should reflect that.
Think of investment management like planning out a route for a road trip. It starts with questions like where do you want to go, how long do you have to get there, and are you willing to risk something off the beaten path on the journey, then creates a plan built around these factors.
In the case of investing, similar questions are asked that impact the distribution of funds. The distribution will have a unique diversification, level of risk, and asset allocation that reflects the person at the heart of it all.
Streamline your clients' wealth management process and grow your firm.
Get started
Advantages and disadvantages of investment management
For some, investment management is a no-brainer decision, while others may still be on the fence about trusting their money. These are the advantages and disadvantages considered when making the decision.
Advantages
A knowledgeable resource: Investment managers do more than manage money. They help their clients understand the dynamics of investing so they know exactly what’s going on with their money and for what purpose.
Understanding market trends: Someone with a history of playing the markets recognizes when something is trending in a good or bad direction. With their know-how, investment managers can potentially outperform the market.
Preparing for taxation: How money gets invested affects the taxes on capital gains. Savvy investment minimizes the tax bill while maximizing the return.
Disadvantages
Paying management fees: Management fees are typically based on a percentage of the managed assets. These fees add up on large portfolios and must be paid regardless of whether the investments were successful or not.
Performance still hinges on the market: Investment firms can only take advantage of the opportunities in front of them. If the market as a whole is in a downturn, their options are going to be limited.
Establishing trust: Giving up savings to a stranger is daunting and requires trust. That trust is required to weather any low periods and achieve the client’s long-term goals.
Investment management process
An investment management engagement can be broken down into 7 simple steps.
1. Collecting information
At the start of an engagement is a conversation where the investment manager gets familiar with their client’s goals. Some factors they’ll dig into are:
What is the purpose of the investment?
What is the level of expected return?
When are they expecting to see that return?
How much risk are they willing to take on to achieve the result?
Are there any legal restrictions on what they can invest in?
Are there any industries or business types they would or wouldn’t want to invest in?
Once the investment manager has what they need, they can get to work.
2. Creating an investment policy statement (IPS)
How the investment manager will service the client is formally documented in an investment policy statement. These are the guidelines on what they can invest in, how the money will be distributed, and how decisions will be made.
The IPS plays an essential role in establishing their fiduciary duty: if any decisions are not in line with this document, they are not in the client’s best interest.
3. Making an investment allocation plan
Once the IPS is finalized, the investment manager has what they need to start planning how to distribute the money. Using models and metrics, they’ll experiment with different potential allocations before determining the best approach.
4. Buying the investments
Investments are purchased only once a plan has been finalized. If the investment management is discretionary, purchases will only be made if the client approves of them. In a non-discretionary investment agreement, the client remains hands-off.
5. Monitoring and revision
The portfolio is monitored to ensure all purchases went through smoothly and that all cash is allocated. There may be revisions made at this point but unless the investment manager is taking an active, non-discretionary role, it’s most likely that the portfolio will be left to accrue data before changes are made.
6. Performance reporting
Traditionally, performance reports and statements are provided to the client monthly. This includes an overview of how the entire portfolio is performing and a breakdown of the individual investments.
7. Continued communication and adjustment
As performance is monitored, the investment manager will communicate the results and make suggestions regarding new opportunities or adjustments that could bolster returns. Managers are available to help their clients understand the whats, whys, and hows of their portfolio so they can feel confident about their money.
Automate your financial operations—demo BILL today.
Demo
About investment management firms
An investment management firm is composed of certified professional managers who handle managing assets as well as report on their performance to their clients through monthly reports and meetings.
Each professional manager would oversee the portfolios of a selection of clients. They’re responsible for asking the right questions, then creating and executing an investment strategy based on the answers.
The management firm makes their money off of management fees, a percentage of the value invested by their clients ranging from 0.5% to 4% of the portfolio. This payment structure incentivizes firms to make money for their clients—the more money they generate, the larger the assets, the greater the fee.
To prevent any malpractice, registered investment management firms face penalties if they fail to act as a fiduciary of their clients. They must always act in the best interests of their clients or face a penalty, including potential jail time.
The Future of Investment Management
Investment management has been around for hundreds of years, and will continue to be around for the foreseeable future. But how it looks is having drastic changes that reflect the changing landscape and desires of clients.
Technology and automation
The introduction of AI and automated portfolio management is putting investment power back in the hands of its users through “robo advisors.” These innovations are more accessible and cost-efficient then using traditional investment management.
Looking forward, investment management firms need to hone in on how they service their clients. They need to think about the value they provide their clients beyond the results.
At the same time, adopting some of these tools could help investment managers with forecasting, modeling, and improving the composition of a portfolio. Adopting this technology would free up time that could be used on providing a personalized service.
The rise of sustainable investing
Environmental, social, and governance (ESG) investing has grown with an increased demand for investments that further social causes. This is sometimes referred to as “ethical investing.”
Part of ESG investing involves incorporating social impact into the financial management and portfolio allocation. In short, it factors in social benefit into its decision making, which would be formally outlined in the investment policy statement (ISP).
With this in mind, it’s prudent to get familiar with investments in renewable energy, climate projects, ecotech, and green bonds.
Changing regulations and security
In the current technological landscape, transparency and security are prioritized. Governments and financial regulators aim to protect the clients with their money on the line by making firms disclose privacy policies, security policies, and their usage of AI technology in their decision-making.
In the United States, this is done by the Division of Investment Management, a section of the Securities and Exchange Commission (SEC).
But regulations can only move so fast. There’s still work to be done with regards to cryptocurrency and decentralized finance to guarantee money is only being used with the best interests of the client in mind.
How investment management firms bill in 2025
Investment management firms care about the money their clients trust them with, but they need to also care about how they’re getting paid. To get paid promptly and securely, they turn to BILL.
“It comes down to three key benefits: enhanced service, security, and time savings.” - Laura Blaire, COO, CCO, and Managing Partner of JFS Wealth Advisors.
BILL leverages automation to save time invoicing clients while maintaining the high level of security needed, keeping the information of both the firm and the client safe. And with overnight payments, payment times are cut down so money is in the bank sooner.
Reach out for a demo to see how BILL could save you both time and money on billing clients.
Start using BILL today.
Sign up
Author
Brendan Tuytel
Contributor
Brendan Tuytel is a freelance writer, who writes content for BILL. He draws from his studies of economics and multiple years of bookkeeping experience where he helped businesses understand and measure their financial health.
Author
Brendan Tuytel
Contributor
Brendan Tuytel is a freelance writer, who writes content for BILL. He draws from his studies of economics and multiple years of bookkeeping experience where he helped businesses understand and measure their financial health.
Get more from BILL
Subscribe to finance insights and thought leadership content delivered straight to your inbox.
Thank you! Your submission has been received!
Oops! Something went wrong while submitting the form.
By continuing, you agree to BILL's Terms of Service and Privacy Notice.
Accounting
Continue learning with BILL
What is accrued interest? Definition and example
What is lean accounting? Definition & how it works
What is grant accounting? A comprehensive guide for nonprofits
Cost-to-cost method: Definition, formulas & examples
What is SaaS accounting? Different types and metrics to track
What is robotic accounting? Benefits & how it works
What is WealthTech? Technology for firms & advisors
BILL and its affiliates do not provide tax, legal or accounting advice. This material has been prepared for informational purposes only, and is not intended to provide, and should not be relied on, for tax, legal or accounting advice. You should consult your own tax, legal and accounting advisors before engaging in any transaction. BILL assumes no responsibility for any inaccuracies or inconsistencies in the content. While we have made every attempt to ensure that the information contained in this site has been obtained from reliable sources, BILL is not responsible for any errors or omissions, or for the results obtained from the use of this information. All information in this site is provided “as is”, with no guarantee of completeness, accuracy, timeliness or of the results obtained from the use of this information, and without warranty of any kind, express or implied. In no event shall BILL, its affiliates or parent company, or the directors, officers, agents or employees thereof, be liable to you or anyone else for any decision made or action taken in reliance on the information in this site or for any consequential, special or similar damages, even if advised of the possibility of such damages. Certain links in this site connect to other websites maintained by third parties over whom BILL has no control. BILL makes no representations as to the accuracy or any other aspect of information contained in other websites. |
189080 | https://en.wikipedia.org/wiki/Messenger_RNA | Jump to content
Search
Contents
1 Synthesis
1.1 Transcription
1.2 Uracil substitution for thymine
1.3 Eukaryotic pre-mRNA processing
1.3.1 Splicing
1.3.2 5' cap addition
1.3.3 Editing
1.3.4 Polyadenylation
1.4 Transport
1.5 Translation
2 Structure
2.1 Coding regions
2.2 Untranslated regions
2.3 Poly(A) tail
2.4 Monocistronic versus polycistronic mRNA
2.5 mRNA circularization
3 Degradation
3.1 Prokaryotic mRNA degradation
3.2 Eukaryotic mRNA turnover
3.3 AU-rich element decay
3.4 Nonsense-mediated decay
3.5 Small interfering RNA (siRNA)
3.6 MicroRNA (miRNA)
3.7 Other decay mechanisms
4 Applications
5 History
6 See also
7 References
7.1 Further reading
8 External links
Messenger RNA
Afrikaans
العربية
Azərbaycanca
বাংলা
閩南語 / Bn-lm-gí
Български
Bosanski
Català
Čeština
Dansk
Deutsch
Eesti
Ελληνικά
Español
Euskara
فارسی
Français
Galego
한국어
Հայերեն
हिन्दी
Hrvatski
Bahasa Indonesia
Italiano
עברית
ქართული
Қазақша
Kurdî
Latina
Latviešu
Lietuvių
Magyar
Македонски
മലയാളം
მარგალური
مصرى
Bahasa Melayu
Nederlands
日本語
Norsk bokmål
Norsk nynorsk
Occitan
Polski
Português
Romnă
Русский
Scots
Sicilianu
Slovenčina
Slovenščina
کوردی
Српски / srpski
Srpskohrvatski / српскохрватски
Suomi
Svenska
தமிழ்
ไทย
Türkçe
Українська
اردو
Tiếng Việt
吴语
中文
Edit links
Article
Talk
Read
Edit
View history
Tools
Actions
Read
Edit
View history
General
What links here
Related changes
Upload file
Permanent link
Page information
Cite this page
Get shortened URL
Download QR code
Print/export
Download as PDF
Printable version
In other projects
Wikimedia Commons
Wikidata item
Appearance
From Wikipedia, the free encyclopedia
| |
| This article needs additional citations for verification. Please help improve this article by adding citations to reliable sources. Unsourced material may be challenged and removed.Find sources: "Messenger RNA" – news · newspapers · books · scholar · JSTOR (May 2025) (Learn how and when to remove this message) |
RNA that is read by the ribosome to produce a protein
"MRNA" redirects here. For other uses, see MRNA (disambiguation).
Not to be confused with modRNA.
In molecular biology, messenger ribonucleic acid (mRNA) is a single-stranded molecule of RNA that corresponds to the genetic sequence of a gene, and is read by a ribosome in the process of synthesizing a protein.
mRNA is created during the process of transcription, where an enzyme (RNA polymerase) converts the gene into primary transcript mRNA (also known as pre-mRNA). This pre-mRNA usually still contains introns, regions that will not go on to code for the final amino acid sequence. These are removed in the process of RNA splicing, leaving only exons, regions that will encode the protein. This exon sequence constitutes mature mRNA. Mature mRNA is then read by the ribosome, and the ribosome creates the protein utilizing amino acids carried by transfer RNA (tRNA). This process is known as translation. All of these processes form part of the central dogma of molecular biology, which describes the flow of genetic information in a biological system.
As in DNA, genetic information in mRNA is contained in the sequence of nucleotides, which are arranged into codons consisting of three ribonucleotides each. Each codon codes for a specific amino acid, except the stop codons, which terminate protein synthesis. The translation of codons into amino acids requires two other types of RNA: transfer RNA, which recognizes the codon and provides the corresponding amino acid, and ribosomal RNA (rRNA), the central component of the ribosome's protein-manufacturing machinery.
The concept of mRNA was developed by Sydney Brenner and Francis Crick in 1960 during a conversation with François Jacob. In 1961, mRNA was identified and described independently by one team consisting of Brenner, Jacob, and Matthew Meselson, and another team led by James Watson. While analyzing the data in preparation for publication, Jacob and Jacques Monod coined the name "messenger RNA".
Synthesis
[edit]
The brief existence of an mRNA molecule begins with transcription, and ultimately ends in degradation. During its life, an mRNA molecule may also be processed, edited, and transported prior to translation. Eukaryotic mRNA molecules often require extensive processing and transport, while prokaryotic mRNA molecules do not. A molecule of eukaryotic mRNA and the proteins surrounding it are together called a messenger RNP.[citation needed]
Transcription
[edit]
Main article: Transcription (genetics)
Transcription is when RNA is copied from DNA in the nucleus. During transcription, RNA polymerase makes a copy of a gene from the DNA to mRNA as needed. This process differs slightly in eukaryotes and prokaryotes. One notable difference is that prokaryotic RNA polymerase associates with DNA-processing enzymes during transcription so that processing can proceed during transcription. Therefore, this causes the new mRNA strand to become double stranded by producing a complementary strand known as the tRNA strand, which when combined are unable to form structures from base-pairing. Moreover, the template for mRNA is the complementary strand of tRNA, which is identical in sequence to the anticodon sequence that the DNA binds to. The short-lived, unprocessed or partially processed product is termed precursor mRNA, or pre-mRNA; once completely processed, it is termed mature mRNA.[citation needed]
Uracil substitution for thymine
[edit]
mRNA uses uracil (U) instead of thymine (T) in DNA. uracil (U) is the complementary base to adenine (A) during transcription instead of thymine (T). Thus, when using a template strand of DNA to build RNA, thymine is replaced with uracil. This substitution allows the mRNA to carry the appropriate genetic information from DNA to the ribosome for translation. Regarding the natural history, uracil came first then thymine; evidence suggests that RNA came before DNA in evolution. The RNA World hypothesis proposes that life began with RNA molecules, before the emergence of DNA genomes and coded proteins. In DNA, the evolutionary substitution of thymine for uracil may have increased DNA stability and improved the efficiency of DNA replication.
Eukaryotic pre-mRNA processing
[edit]
Main article: Post-transcriptional modification
Processing of mRNA differs greatly among eukaryotes, bacteria, and archaea. Non-eukaryotic mRNA is, in essence, mature upon transcription and requires no processing, except in rare cases. Eukaryotic pre-mRNA, however, requires several processing steps before its transport to the cytoplasm and its translation by the ribosome.
Splicing
[edit]
Main article: RNA splicing
The extensive processing of eukaryotic pre-mRNA that leads to the mature mRNA is the RNA splicing, a mechanism by which introns or outrons (non-coding regions) are removed and exons (coding regions) are joined.
5' cap addition
[edit]
Main article: 5' cap
A 5' cap (also termed an RNA cap, an RNA 7-methylguanosine cap, or an RNA m7G cap) is a modified guanine nucleotide that has been added to the "front" or 5' end of a eukaryotic messenger RNA shortly after the start of transcription. The 5' cap consists of a terminal 7-methylguanosine residue that is linked through a 5'-5'-triphosphate bond to the first transcribed nucleotide. Its presence is critical for recognition by the ribosome and protection from RNases.[citation needed]
Cap addition is coupled to transcription, and occurs co-transcriptionally, such that each influences the other. Shortly after the start of transcription, the 5' end of the mRNA being synthesized is bound by a cap-synthesizing complex associated with RNA polymerase. This enzymatic complex catalyzes the chemical reactions that are required for mRNA capping. Synthesis proceeds as a multi-step biochemical reaction.[citation needed]
Editing
[edit]
In some instances, an mRNA will be edited, changing the nucleotide composition of that mRNA. An example in humans is the apolipoprotein B mRNA, which is edited in some tissues, but not others. The editing creates an early stop codon, which, upon translation, produces a shorter protein. Another well-defined example is A-to-I (adenosine to inosine) editing, which is carried out by double-strand specific adenosine-to inosine editing (ADAR) enzymes. This can occur in both the open reading frame and untranslated regions, altering the structural properties of the mRNA. Although essential for development, the exact role of this editing is not fully understood
Polyadenylation
[edit]
Main article: Polyadenylation
Polyadenylation is the covalent linkage of a polyadenylyl moiety to a messenger RNA molecule. In eukaryotic organisms most messenger RNA (mRNA) molecules are polyadenylated at the 3' end, but recent studies have shown that short stretches of uridine (oligouridylation) are also common. The poly(A) tail and the protein bound to it aid in protecting mRNA from degradation by exonucleases. Polyadenylation is also important for transcription termination, export of the mRNA from the nucleus, and translation. mRNA can also be polyadenylated in prokaryotic organisms, where poly(A) tails act to facilitate, rather than impede, exonucleolytic degradation. Polyadenylation occurs during and/or immediately after transcription of DNA into RNA. After transcription has been terminated, the mRNA chain is cleaved through the action of an endonuclease complex associated with RNA polymerase. After the mRNA has been cleaved, around 250 adenosine residues are added to the free 3' end at the cleavage site. This reaction is catalyzed by polyadenylate polymerase. Just as in alternative splicing, there can be more than one polyadenylation variant of an mRNA.
Polyadenylation site mutations also occur. The primary RNA transcript of a gene is cleaved at the poly-A addition site, and 100–200 A's are added to the 3' end of the RNA. If this site is altered, an abnormally long and unstable mRNA construct will be formed.
Transport
[edit]
Another difference between eukaryotes and prokaryotes is mRNA transport. Because eukaryotic transcription and translation is compartmentally separated, eukaryotic mRNAs must be exported from the nucleus to the cytoplasm—a process that may be regulated by different signaling pathways. Mature mRNAs are recognized by their processed modifications and then exported through the nuclear pore by binding to the cap-binding proteins CBP20 and CBP80, as well as the transcription/export complex (TREX). Multiple mRNA export pathways have been identified in eukaryotes.
In spatially complex cells, some mRNAs are transported to particular subcellular destinations. In mature neurons, certain mRNA are transported from the soma to dendrites. One site of mRNA translation is at polyribosomes selectively localized beneath synapses. The mRNA for Arc/Arg3.1 is induced by synaptic activity and localizes selectively near active synapses based on signals generated by NMDA receptors. Other mRNAs also move into dendrites in response to external stimuli, such as β-actin mRNA. For export from the nucleus, actin mRNA associates with ZBP1 and later with 40S subunit. The complex is bound by a motor protein and is transported to the target location (neurite extension) along the cytoskeleton. Eventually ZBP1 is phosphorylated by Src in order for translation to be initiated. In developing neurons, mRNAs are also transported into growing axons and especially growth cones. Many mRNAs are marked with so-called "zip codes", which target their transport to a specific location. mRNAs can also transfer between mammalian cells through structures called tunneling nanotubes.
Translation
[edit]
Main article: Translation (biology)
Because prokaryotic mRNA does not need to be processed or transported, translation by the ribosome can begin immediately after the end of transcription. Therefore, it can be said that prokaryotic translation is coupled to transcription and occurs co-transcriptionally.
Eukaryotic mRNA that has been processed and transported to the cytoplasm (i.e., mature mRNA) can then be translated by the ribosome. Translation may occur at ribosomes free-floating in the cytoplasm, or directed to the endoplasmic reticulum by the signal recognition particle. Therefore, unlike in prokaryotes, eukaryotic translation is not directly coupled to transcription. It is even possible in some contexts that reduced mRNA levels are accompanied by increased protein levels, as has been observed for mRNA/protein levels of EEF1A1 in breast cancer.[non-primary source needed]
Structure
[edit]
Coding regions
[edit]
Main article: Coding region
Coding regions are composed of codons, which are decoded and translated into proteins by the ribosome; in eukaryotes usually into one and in prokaryotes usually into several. Coding regions begin with the start codon and end with a stop codon. In general, the start codon is an AUG triplet and the stop codon is UAG ("amber"), UAA ("ochre"), or UGA ("opal"). The coding regions tend to be stabilised by internal base pairs; this impedes degradation. In addition to being protein-coding, portions of coding regions may serve as regulatory sequences in the pre-mRNA as exonic splicing enhancers or exonic splicing silencers.
Untranslated regions
[edit]
Main articles: 5' UTR and 3' UTR
Untranslated regions (UTRs) are sections of the mRNA before the start codon and after the stop codon that are not translated, termed the five prime untranslated region (5' UTR) and three prime untranslated region (3' UTR), respectively. These regions are transcribed with the coding region and thus are exonic as they are present in the mature mRNA. Several roles in gene expression have been attributed to the untranslated regions, including mRNA stability, mRNA localization, and translational efficiency. The ability of a UTR to perform these functions depends on the sequence of the UTR and can differ between mRNAs. Genetic variants in 3' UTR have also been implicated in disease susceptibility because of the change in RNA structure and protein translation.
The stability of mRNAs may be controlled by the 5' UTR and/or 3' UTR due to varying affinity for RNA degrading enzymes called ribonucleases and for ancillary proteins that can promote or inhibit RNA degradation. (See also, C-rich stability element.)
Translational efficiency, including sometimes the complete inhibition of translation, can be controlled by UTRs. Proteins that bind to either the 3' or 5' UTR may affect translation by influencing the ribosome's ability to bind to the mRNA. MicroRNAs bound to the 3' UTR also may affect translational efficiency or mRNA stability.
Cytoplasmic localization of mRNA is thought to be a function of the 3' UTR. Proteins that are needed in a particular region of the cell can also be translated there; in such a case, the 3' UTR may contain sequences that allow the transcript to be localized to this region for translation.
Some of the elements contained in untranslated regions form a characteristic secondary structure when transcribed into RNA. These structural mRNA elements are involved in regulating the mRNA. Some, such as the SECIS element, are targets for proteins to bind. One class of mRNA element, the riboswitches, directly bind small molecules, changing their fold to modify levels of transcription or translation. In these cases, the mRNA regulates itself.
Poly(A) tail
[edit]
Main article: Polyadenylation
The 3' poly(A) tail is a long sequence of adenine nucleotides (often several hundred) added to the 3' end of the pre-mRNA. This tail promotes export from the nucleus and translation, and protects the mRNA from degradation.
Monocistronic versus polycistronic mRNA
[edit]
See also: Cistron
An mRNA molecule is said to be monocistronic when it contains the genetic information to translate only a single protein chain (polypeptide). This is the case for most of the eukaryotic mRNAs. On the other hand, polycistronic mRNA carries several open reading frames (ORFs), each of which is translated into a polypeptide. These polypeptides usually have a related function (they often are the subunits composing a final complex protein) and their coding sequence is grouped and regulated together in a regulatory region, containing a promoter and an operator. Most of the mRNA found in bacteria and archaea is polycistronic, as is the human mitochondrial genome. Dicistronic or bicistronic mRNA encodes only two proteins.
mRNA circularization
[edit]
In eukaryotes mRNA molecules form circular structures due to an interaction between the eIF4E and poly(A)-binding protein, which both bind to eIF4G, forming an mRNA-protein-mRNA bridge. Circularization is thought to promote cycling of ribosomes on the mRNA leading to time-efficient translation, and may also function to ensure only intact mRNA are translated (partially degraded mRNA characteristically have no m7G cap, or no poly-A tail).
Other mechanisms for circularization exist, particularly in virus mRNA. Poliovirus mRNA uses a cloverleaf section towards its 5' end to bind PCBP2, which binds poly(A)-binding protein, forming the familiar mRNA-protein-mRNA circle. Barley yellow dwarf virus has binding between mRNA segments on its 5' end and 3' end (called kissing stem loops), circularizing the mRNA without any proteins involved.
RNA virus genomes (the + strands of which are translated as mRNA) are also commonly circularized. During genome replication the circularization acts to enhance genome replication speeds, cycling viral RNA-dependent RNA polymerase much the same as the ribosome is hypothesized to cycle.
Degradation
[edit]
Different mRNAs within the same cell have distinct lifetimes (stabilities). In bacterial cells, individual mRNAs can survive from seconds to more than an hour. However, the lifetime averages between 1 and 3 minutes, making bacterial mRNA much less stable than eukaryotic mRNA. In mammalian cells, mRNA lifetimes range from several minutes to days. The greater the stability of an mRNA the more protein may be produced from that mRNA. The limited lifetime of mRNA enables a cell to alter protein synthesis rapidly in response to its changing needs. There are many mechanisms that lead to the destruction of an mRNA, some of which are described below.
Prokaryotic mRNA degradation
[edit]
In general, in prokaryotes the lifetime of mRNA is much shorter than in eukaryotes. Prokaryotes degrade messages by using a combination of ribonucleases, including endonucleases, 3' exonucleases, and 5' exonucleases. In some instances, small RNA molecules (sRNA) tens to hundreds of nucleotides long can stimulate the degradation of specific mRNAs by base-pairing with complementary sequences and facilitating ribonuclease cleavage by RNase III. It was recently shown that bacteria also have a sort of 5' cap consisting of a triphosphate on the 5' end. Removal of two of the phosphates leaves a 5' monophosphate, causing the message to be destroyed by the exonuclease RNase J, which degrades 5' to 3'.
Eukaryotic mRNA turnover
[edit]
Inside eukaryotic cells, there is a balance between the processes of translation and mRNA decay. Messages that are being actively translated are bound by ribosomes, the eukaryotic initiation factors eIF-4E and eIF-4G, and poly(A)-binding protein. eIF-4E and eIF-4G block the decapping enzyme (DCP2), and poly(A)-binding protein blocks the exosome complex, protecting the ends of the message. The balance between translation and decay is reflected in the size and abundance of cytoplasmic structures known as P-bodies. The poly(A) tail of the mRNA is shortened by specialized exonucleases that are targeted to specific messenger RNAs by a combination of cis-regulatory sequences on the RNA and trans-acting RNA-binding proteins. Poly(A) tail removal is thought to disrupt the circular structure of the message and destabilize the cap binding complex. The message is then subject to degradation by either the exosome complex or the decapping complex. In this way, translationally inactive messages can be destroyed quickly, while active messages remain intact. The mechanism by which translation stops and the message is handed-off to decay complexes is not understood in detail. The majority of mRNA decay was believed to be cytoplasmic; however, recently, a novel mRNA decay pathway was described, which starts in the nucleus.
AU-rich element decay
[edit]
The presence of AU-rich elements in some mammalian mRNAs tends to destabilize those transcripts through the action of cellular proteins that bind these sequences and stimulate poly(A) tail removal. Loss of the poly(A) tail is thought to promote mRNA degradation by facilitating attack by both the exosome complex and the decapping complex. Rapid mRNA degradation via AU-rich elements is a critical mechanism for preventing the overproduction of potent cytokines such as tumor necrosis factor (TNF) and granulocyte-macrophage colony stimulating factor (GM-CSF). AU-rich elements also regulate the biosynthesis of proto-oncogenic transcription factors like c-Jun and c-Fos.
Nonsense-mediated decay
[edit]
Main article: Nonsense-mediated decay
Eukaryotic messages are subject to surveillance by nonsense-mediated decay (NMD), which checks for the presence of premature stop codons (nonsense codons) in the message. These can arise via incomplete splicing, V(D)J recombination in the adaptive immune system, mutations in DNA, transcription errors, leaky scanning by the ribosome causing a frame shift, and other causes. Detection of a premature stop codon triggers mRNA degradation by 5' decapping, 3' poly(A) tail removal, or endonucleolytic cleavage.
Small interfering RNA (siRNA)
[edit]
Main article: siRNA
In metazoans, small interfering RNAs (siRNAs) processed by Dicer are incorporated into a complex known as the RNA-induced silencing complex or RISC. This complex contains an endonuclease that cleaves perfectly complementary messages to which the siRNA binds. The resulting mRNA fragments are then destroyed by exonucleases. siRNA is commonly used in laboratories to block the function of genes in cell culture. It is thought to be part of the innate immune system as a defense against double-stranded RNA viruses.
MicroRNA (miRNA)
[edit]
Main article: microRNA
MicroRNAs (miRNAs) are small RNAs that typically are partially complementary to sequences in metazoan messenger RNAs. Binding of a miRNA to a message can repress translation of that message and accelerate poly(A) tail removal, thereby hastening mRNA degradation. The mechanism of action of miRNAs is the subject of active research.
Other decay mechanisms
[edit]
There are other ways by which messages can be degraded, including non-stop decay and silencing by Piwi-interacting RNA (piRNA), among others.
Applications
[edit]
See also: mRNA vaccine and RNA therapeutics
The administration of a nucleoside-modified messenger RNA sequence can cause a cell to make a protein, which in turn could directly treat a disease or could function as a vaccine; more indirectly the protein could drive an endogenous stem cell to differentiate in a desired way.
The primary challenges of RNA therapy center on delivering the RNA to the appropriate cells. Challenges include the fact that naked RNA sequences naturally degrade after preparation; they may trigger the body's immune system to attack them as an invader; and they are impermeable to the cell membrane. Once within the cell, they must then leave the cell's transport mechanism to take action within the cytoplasm, which houses the necessary ribosomes.
Overcoming these challenges, mRNA as a therapeutic was first put forward in 1989 "after the development of a broadly applicable in vitro transfection technique." In the 1990s, mRNA vaccines for personalized cancer have been developed, relying on non-nucleoside modified mRNA. mRNA based therapies continue to be investigated as a method of treatment or therapy for both cancer as well as auto-immune, metabolic, and respiratory inflammatory diseases. Gene editing therapies such as CRISPR may also benefit from using mRNA to induce cells to make the desired Cas protein.
Since the 2010s, RNA vaccines and other RNA therapeutics have been considered to be "a new class of drugs". The first mRNA-based vaccines received restricted authorization and were rolled out across the world during the COVID-19 pandemic by Pfizer–BioNTech COVID-19 vaccine and Moderna, for example. The 2023 Nobel Prize in Physiology or Medicine was awarded to Katalin Karikó and Drew Weissman for the development of effective mRNA vaccines against COVID-19. New approaches to modulate RNA levels as a therapeutics include the use of antisense oligonucleotides, including for neurodevelopment diseases associated with high mortality.
History
[edit]
Several molecular biology studies during the 1950s indicated that RNA played some kind of role in protein synthesis, but that role was not clearly understood. For instance, in one of the earliest reports, Jacques Monod and his team showed that RNA synthesis was necessary for protein synthesis, specifically during the production of the enzyme β-galactosidase in the bacterium E. coli. Arthur Pardee also found similar RNA accumulation in 1954. In 1953, Alfred Hershey, June Dixon, and Martha Chase described a certain cytosine-containing DNA (indicating it was RNA) that disappeared quickly after its synthesis in E. coli. In hindsight, this may have been one of the first observations of the existence of mRNA but it was not recognized at the time as such.
The idea of mRNA was first conceived by Sydney Brenner and Francis Crick on 15 April 1960 at King's College, Cambridge, while François Jacob was telling them about a recent experiment conducted by Arthur Pardee, himself, and Monod (the so-called PaJaMo experiment, which did not prove mRNA existed but suggested the possibility of its existence). With Crick's encouragement, Brenner and Jacob immediately set out to test this new hypothesis, and they contacted Matthew Meselson at the California Institute of Technology for assistance. During the summer of 1960, Brenner, Jacob, and Meselson conducted an experiment in Meselson's laboratory at Caltech which was the first to prove the existence of mRNA. That fall, Jacob and Monod coined the name "messenger RNA" and developed the first theoretical framework to explain its function.
In February 1961, James Watson revealed that his Harvard-based research group had been right behind them with a series of experiments whose results pointed in roughly the same direction. Brenner and the others agreed to Watson's request to delay publication of their research findings. As a result, the Brenner and Watson articles were published simultaneously in the same issue of Nature in May 1961, while that same month, Jacob and Monod published their theoretical framework for mRNA in the Journal of Molecular Biology.
See also
[edit]
Extension Poly(A) Test
GeneCalling, an mRNA profiling technology
Missense mRNA
mRNA display
mRNA surveillance
Prokaryotic mRNA degradation
Transcriptome, the sum of all RNA in a cell
modRNA Nucleoside-modified messenger RNA
References
[edit]
^ "The Information in DNA Is Decoded by Transcription | Learn Science at Scitable". www.nature.com. Retrieved 2024-05-03.
^ "RNA world (article) | Natural selection". Khan Academy. Retrieved 2024-05-03.
^ "The presence of thymine the place of uracil also confers additional stability to DNA. How?". Toppr Ask. Retrieved 2024-05-04.
^ Watson JD (February 22, 2013). Molecular Biology of the Gene, 7th edition. Pearson Higher Ed USA. ISBN 9780321851499.
^ Marasco LE, Kornblihtt AR (2023). "The physiology of alternative splicing". Nature Reviews Molecular Cell Biology. 24 (4): 242–254. doi:10.1038/s41580-022-00545-z. PMID 36229538.
^ Rogalska ME, Vivori C, Valcárcel J (2023). "Regulation of pre-mRNA splicing: Roles in physiology and disease, and therapeutic prospects". Nature Reviews Genetics. 24 (4): 251–269. doi:10.1038/s41576-022-00556-8. PMID 36526860.
^ Slotkin W, Nishikura K (2013). "Adenosine-to-inosine RNA editing and human disease". Genome Medicine. 5 (11): 105. doi:10.1186/gm508. PMC 3979043. PMID 24289319.
^ Choi YS, Patena W, Leavitt AD, McManus MT (March 2012). "Widespread RNA 3'-end oligouridylation in mammals". RNA. 18 (3): 394–401. doi:10.1261/rna.029306.111. PMC 3285928. PMID 22291204.
^ Hajnsdorf E, Kaberdin VR (2018). "RNA polyadenylation and its consequences in prokaryotes". Philosophical Transactions of the Royal Society B: Biological Sciences. 373 (1762). doi:10.1098/rstb.2018.0166. PMC 6232592. PMID 30397102.
^ Quaresma AJ, Sievert R, Nickerson JA (April 2013). "Regulation of mRNA export by the PI3 kinase/AKT signal transduction pathway". Molecular Biology of the Cell. 24 (8): 1208–1221. doi:10.1091/mbc.E12-06-0450. PMC 3623641. PMID 23427269.
^ Kierzkowski D, Kmieciak M, Piontek P, Wojtaszek P, Szweykowska-Kulinska Z, Jarmolowski A (September 2009). "The Arabidopsis CBP20 targets the cap-binding complex to the nucleus, and is stabilized by CBP80". The Plant Journal. 59 (5): 814–825. doi:10.1111/j.1365-313X.2009.03915.x. PMID 19453442.
^ Strässer K, Masuda S, Mason P, Pfannstiel J, Oppizzi M, Rodriguez-Navarro S, Rondón AG, Aguilera A, Struhl K, Reed R, Hurt E (May 2002). "TREX is a conserved complex coupling transcription with messenger RNA export". Nature. 417 (6886): 304–308. Bibcode:2002Natur.417..304S. doi:10.1038/nature746. PMID 11979277. S2CID 1112194.
^ Katahira J, Yoneda Y (27 October 2014). "Roles of the TREX complex in nuclear export of mRNA". RNA Biology. 6 (2): 149–152. doi:10.4161/rna.6.2.8046. PMID 19229134.
^ Cenik C, Chua HN, Zhang H, Tarnawsky SP, Akef A, Derti A, Tasan M, Moore MJ, Palazzo AF, Roth FP (April 2011). "Genome analysis reveals interplay between 5'UTR introns and nuclear mRNA export for secretory and mitochondrial genes". PLOS Genetics. 7 (4) e1001366. doi:10.1371/journal.pgen.1001366. PMC 3077370. PMID 21533221.
^ Steward O, Levy WB (March 1982). "Preferential localization of polyribosomes under the base of dendritic spines in granule cells of the dentate gyrus". The Journal of Neuroscience. 2 (3): 284–291. doi:10.1523/JNEUROSCI.02-03-00284.1982. PMC 6564334. PMID 7062109.
^ Steward O, Worley PF (April 2001). "Selective targeting of newly synthesized Arc mRNA to active synapses requires NMDA receptor activation". Neuron. 30 (1): 227–240. doi:10.1016/s0896-6273(01)00275-6. PMID 11343657. S2CID 13395819.
^ Job C, Eberwine J (December 2001). "Localization and translation of mRNA in dendrites and axons". Nature Reviews. Neuroscience. 2 (12): 889–898. doi:10.1038/35104069. PMID 11733796. S2CID 5275219.
^ Oleynikov Y, Singer RH (February 2003). "Real-time visualization of ZBP1 association with beta-actin mRNA during transcription and localization". Current Biology. 13 (3): 199–207. Bibcode:2003CBio...13..199O. doi:10.1016/s0960-9822(03)00044-7. PMC 4765734. PMID 12573215.
^ Hüttelmaier S, Zenklusen D, Lederer M, Dictenberg J, Lorenz M, Meng X, et al. (November 2005). "Spatial regulation of beta-actin translation by Src-dependent phosphorylation of ZBP1". Nature. 438 (7067): 512–515. Bibcode:2005Natur.438..512H. doi:10.1038/nature04115. PMID 16306994. S2CID 2453397.
^ Oleynikov Y, Singer RH (October 1998). "RNA localization: different zipcodes, same postman?". Trends in Cell Biology. 8 (10): 381–383. doi:10.1016/s0962-8924(98)01348-8. PMC 2136761. PMID 9789325.
^ Ainger K, Avossa D, Diana AS, Barry C, Barbarese E, Carson JH (September 1997). "Transport and localization elements in myelin basic protein mRNA". The Journal of Cell Biology. 138 (5): 1077–1087. doi:10.1083/jcb.138.5.1077. PMC 2136761. PMID 9281585.
^ Haimovich G, Ecker CM, Dunagin MC, Eggan E, Raj A, Gerst JE, Singer RH (November 2017). "Intercellular mRNA trafficking via membrane nanotube-like extensions in mammalian cells". Proceedings of the National Academy of Sciences of the United States of America. 114 (46): E9873 – E9882. Bibcode:2017PNAS..114E9873H. doi:10.1073/pnas.1706365114. PMC 5699038. PMID 29078295.
^ Haimovich G, Dasgupta S, Gerst JE (February 2021). "RNA transfer through tunneling nanotubes". Biochemical Society Transactions. 49 (1): 145–160. doi:10.1042/BST20200113. PMID 33367488. S2CID 229689880.
^ Irastortza-Olaziregi M, Amster-Choder O (2021). "Coupled Transcription-Translation in Prokaryotes: An Old Couple with New Surprises". Frontiers in Microbiology. 11 624830. doi:10.3389/fmicb.2020.624830. PMC 7858274. PMID 33552035.
^ Lin CY, Beattie A, Baradaran B, Dray E, Duijf PH (September 2018). "Contradictory mRNA and protein misexpression of EEF1A1 in ductal breast carcinoma due to cell cycle regulation and cellular stress". Scientific Reports. 8 (1) 13904. Bibcode:2018NatSR...813904L. doi:10.1038/s41598-018-32272-x. PMC 6141510. PMID 30224719.
^ Shabalina SA, Ogurtsov AY, Spiridonov NA (2006). "A periodic pattern of mRNA secondary structure created by the genetic code". Nucleic Acids Research. 34 (8): 2428–2437. doi:10.1093/nar/gkl287. PMC 1458515. PMID 16682450.
^ Katz L, Burge CB (September 2003). "Widespread selection for local RNA secondary structure in coding regions of bacterial genes". Genome Research. 13 (9): 2042–2051. doi:10.1101/gr.1257503. PMC 403678. PMID 12952875.
^ Lu YF, Mauger DM, Goldstein DB, Urban TJ, Weeks KM, Bradrick SS (November 2015). "IFNL3 mRNA structure is remodeled by a functional non-coding polymorphism associated with hepatitis C virus clearance". Scientific Reports. 5 16037. Bibcode:2015NatSR...516037L. doi:10.1038/srep16037. PMC 4631997. PMID 26531896.
^ a b Kozak M (March 1983). "Comparison of initiation of protein synthesis in procaryotes, eucaryotes, and organelles". Microbiological Reviews. 47 (1): 1–45. doi:10.1128/MMBR.47.1.1-45.1983. PMC 281560. PMID 6343825.
^ Niehrs C, Pollet N (December 1999). "Synexpression groups in eukaryotes". Nature. 402 (6761): 483–487. Bibcode:1999Natur.402..483N. doi:10.1038/990025. PMID 10591207. S2CID 4349134.
^ Mercer TR, Neph S, Dinger ME, Crawford J, Smith MA, Shearwood AM, Haugen E, Bracken CP, Rackham O, Stamatoyannopoulos JA, Filipovska A, Mattick JS (August 2011). "The human mitochondrial transcriptome". Cell. 146 (4): 645–658. doi:10.1016/j.cell.2011.06.051. PMC 3160626. PMID 21854988.
^ Wells SE, Hillner PE, Vale RD, Sachs AB (July 1998). "Circularization of mRNA by eukaryotic translation initiation factors". Molecular Cell. 2 (1): 135–140. CiteSeerX 10.1.1.320.5704. doi:10.1016/S1097-2765(00)80122-7. PMID 9702200.
^ López-Lastra M, Rivas A, Barría MI (2005). "Protein synthesis in eukaryotes: the growing biological relevance of cap-independent translation initiation". Biological Research. 38 (2–3): 121–146. doi:10.4067/S0716-97602005000200003. hdl:10533/176032. PMID 16238092.
^ Zhang X, Liang Z, Wang C, Shen Z, Sun S, Gong C, Hu X (2022). "Viral Circular RNAs and Their Possible Roles in Virus-Host Interaction". Frontiers in Immunology. 13 939768. doi:10.3389/fimmu.2022.939768. PMC 9247149. PMID 35784275.
^ Lewin B, Krebs JE, Kilpatrick ST, Goldstein ES, eds. (2011). Lewin's genes X (10th ed.). Sudbury, Mass.: Jones and Bartlett. ISBN 9780763766320. OCLC 456641931.
^ Yu J, Russell JE (September 2001). "Structural and functional analysis of an mRNP complex that mediates the high stability of human beta-globin mRNA". Molecular and Cellular Biology. 21 (17): 5879–5888. doi:10.1128/mcb.21.17.5879-5888.2001. PMC 87307. PMID 11486027.
^ Deana A, Celesnik H, Belasco JG (January 2008). "The bacterial enzyme RppH triggers messenger RNA degradation by 5' pyrophosphate removal". Nature. 451 (7176): 355–358. Bibcode:2008Natur.451..355D. doi:10.1038/nature06475. PMID 18202662. S2CID 4321451.
^ Parker R, Sheth U (March 2007). "P bodies and the control of mRNA translation and degradation". Molecular Cell. 25 (5): 635–646. doi:10.1016/j.molcel.2007.02.011. PMID 17349952.
^ Chattopadhyay S, Garcia Martinez J, Haimovich G, Fischer J, Khwaja A, Barkai O, Chuartzman SG, Schuldiner M, Elran R, Rosenberg M, Urim S, Deshmukh S, Bohnsack K, Bohnsack M, Perez Ortin J, Choder M (November 2022). "RNA-controlled nucleocytoplasmic shuttling of mRNA decay factors regulates mRNA synthesis and a novel mRNA decay pathway". Nature Communications. 13(1): 7184 (1) 7184. Bibcode:2022NatCo..13.7184C. doi:10.1038/s41467-022-34417-z. PMC 9684461. PMID 36418294.
^ Chen CY, Gherzi R, Ong SE, Chan EL, Raijmakers R, Pruijn GJ, Stoecklin G, Moroni C, Mann M, Karin M (November 2001). "AU binding proteins recruit the exosome to degrade ARE-containing mRNAs". Cell. 107 (4): 451–464. doi:10.1016/S0092-8674(01)00578-5. PMID 11719186. S2CID 14817671.
^ Fenger-Grøn M, Fillman C, Norrild B, Lykke-Andersen J (December 2005). "Multiple processing body factors and the ARE binding protein TTP activate mRNA decapping". Molecular Cell. 20 (6): 905–915. doi:10.1016/j.molcel.2005.10.031. PMID 16364915.
^ Shaw G, Kamen R (August 1986). "A conserved AU sequence from the 3' untranslated region of GM-CSF mRNA mediates selective mRNA degradation". Cell. 46 (5): 659–667. doi:10.1016/0092-8674(86)90341-7. PMID 3488815. S2CID 40332253.
^ Chen CY, Shyu AB (November 1995). "AU-rich elements: characterization and importance in mRNA degradation". Trends in Biochemical Sciences. 20 (11): 465–470. doi:10.1016/S0968-0004(00)89102-1. PMID 8578590.
^ Isken O, Maquat LE (August 2007). "Quality control of eukaryotic mRNA: safeguarding cells from abnormal mRNA function". Genes & Development. 21 (15): 1833–1856. doi:10.1101/gad.1566807. PMID 17671086.
^ Obbard DJ, Gordon KH, Buck AH, Jiggins FM (January 2009). "The evolution of RNAi as a defence against viruses and transposable elements". Philosophical Transactions of the Royal Society of London. Series B, Biological Sciences. 364 (1513): 99–115. doi:10.1098/rstb.2008.0168. PMC 2592633. PMID 18926973.
^ Robert E. Farrell, Jr. RNA Methodologies, 5th Edition. Academic Press, 2017
^ Brennecke J, Stark A, Russell RB, Cohen SM (March 2005). "Principles of microRNA-target recognition". PLOS Biology. 3 (3): e85. doi:10.1371/journal.pbio.0030085. PMC 1043860. PMID 15723116.
^ Tasuku Honjo, Michael Reth, Andreas Radbruch, Frederick Alt. Molecular Biology of B Cells, 2nd Edition. Academic Press, 2014 (including "updated research on microRNAs")
^ Eulalio A, Huntzinger E, Nishihara T, Rehwinkel J, Fauser M, Izaurralde E (January 2009). "Deadenylation is a widespread effect of miRNA regulation". RNA. 15 (1): 21–32. doi:10.1261/rna.1399509. PMC 2612776. PMID 19029310.
^ a b Hajj KA, Whitehead KA (12 September 2017). "Tools for translation: non-viral materials for therapeutic mRNA delivery". Nature Reviews Materials. 2 (10) 17056. Bibcode:2017NatRM...217056H. doi:10.1038/natrevmats.2017.56.
^ a b Gousseinov E, Kozlov M, Scanlan C (September 15, 2015). "RNA-Based Therapeutics and Vaccines". Genetic Engineering News.
^ Kaczmarek JC, Kowalski PS, Anderson DG (June 2017). "Advances in the delivery of RNA therapeutics: from concept to clinical reality". Genome Medicine. 9 (1) 60. doi:10.1186/s13073-017-0450-0. PMC 5485616. PMID 28655327.
^ Schlake T, Thess A, Fotin-Mleczek M, Kallen KJ (November 2012). "Developing mRNA-vaccine technologies". RNA Biology. 9 (11): 1319–30. doi:10.4161/rna.22269. PMC 3597572. PMID 23064118.
^ Haridi R (2021-04-23). "The mRNA revolution: How COVID-19 hit fast-forward on an experimental technology". New Atlas. Retrieved 2021-04-26.
^ Kowalska J, Wypijewska del Nogal A, Darzynkiewicz ZM, Buck J, Nicola C, Kuhn AN, Lukaszewicz M, Zuberek J, Strenkowska M, Ziemniak M, Maciejczyk M, Bojarska E, Rhoads RE, Darzynkiewicz E, Sahin U, Jemielity J (2014), "mRNA-based therapeutics–developing a new class of drugs.", Nature Reviews Drug Discovery, vol. 13, no. 10, pp. 759–780, doi:10.1093/nar/gku757, PMC 4176373, PMID 25150148
^ Barbier AJ, Jiang AY, Zhang P, Wooster R, Anderson DG (June 2022). "The clinical progress of mRNA vaccines and immunotherapies". Nature Biotechnology. 40 (6): 840–854. doi:10.1038/s41587-022-01294-2. PMID 35534554. S2CID 248667843.
^ "The Nobel Prize in Physiology or Medicine 2023". NobelPrize.org. Retrieved 2023-10-03.
^ "Hungarian and US scientists win Nobel for COVID-19 vaccine discoveries". Reuters. 2023-10-02. Retrieved 2023-10-03.
^ "The Nobel Prize in Physiology or Medicine 2023". NobelPrize.org. Retrieved 2023-10-03.
^ Golinski SR, Soriano K, Briegel AC, Burke MC, Yu TW, Nakayama T, Hu R, Smith RS (24 October 2024). "Gene therapy for targeting a prenatally enriched potassium channel associated with severe childhood epilepsy and premature death". bioRxiv 10.1101/2024.10.24.620125.
^ Monod J, Pappenheimer AM, Cohen-Bazire G (1952). "La cinétique de la biosynthèse de la β-galactosidase chez E. coli considérée comme fonction de la croissance". Biochimica et Biophysica Acta (in French). 9 (6): 648–660. doi:10.1016/0006-3002(52)90227-8. PMID 13032175.
^ Pardee AB (May 1954). "Nucleic Acid Precursors and Protein Synthesis". Proceedings of the National Academy of Sciences of the United States of America. 40 (5): 263–270. Bibcode:1954PNAS...40..263P. doi:10.1073/pnas.40.5.263. PMC 534118. PMID 16589470.
^ Hershey AD, Dixon J, Chase M (July 1953). "Nucleic acid economy in bacteria infected with bacteriophage T2. I. Purine and pyrimidine composition". The Journal of General Physiology. 36 (6): 777–789. doi:10.1085/jgp.36.6.777. PMC 2147416. PMID 13069681.
^ a b c Cobb M (29 June 2015). "Who discovered messenger RNA?". Current Biology. 25 (13): R526 – R532. Bibcode:2015CBio...25.R526C. doi:10.1016/j.cub.2015.05.032. PMID 26126273.
Further reading
[edit]
Alsaweed M, Lai CT, Hartmann PE, Geddes DT, Kakulas F (February 2016). "Human milk miRNAs primarily originate from the mammary gland resulting in unique miRNA profiles of fractionated milk". Scientific Reports. 6 (1) 20680. Bibcode:2016NatSR...620680A. doi:10.1038/srep20680. PMC 4745068. PMID 26854194.
Lillycrop KA, Burdge GC (October 2012). "Epigenetic mechanisms linking early nutrition to long term health". Best Practice & Research. Clinical Endocrinology & Metabolism. 26 (5): 667–676. doi:10.1016/j.beem.2012.03.009. PMID 22980048.
Melnik BC, Kakulas F, Geddes DT, Hartmann PE, John SM, Carrera-Bastos P, Cordain L, Schmitz G (21 June 2016). "Milk miRNAs: simple nutrients or systemic functional regulators?". Nutrition & Metabolism. 13 (1) 42. doi:10.1186/s12986-016-0101-2. PMC 4915038. PMID 27330539.
Vickers MH (June 2014). "Early life nutrition, epigenetics and programming of later life disease". Nutrients. 6 (6): 2165–2178. doi:10.3390/nu6062165. PMC 4073141. PMID 24892374.
Zhou Q, Li M, Wang X, Li Q, Wang T, Zhu Q, Zhou X, Wang X, Gao X, Li X (2012). "Immune-related microRNAs are abundant in breast milk exosomes". International Journal of Biological Sciences. 8 (1): 118–123. doi:10.7150/ijbs.8.118. PMC 3248653. PMID 22211110.
Krause W (2023). "mRNA — From COVID-19 Treatment to Cancer Immunotherapy". Biomedicines. 11 (2): 308. doi:10.3390/biomedicines11020308. PMC 9953480. PMID 36830845.
External links
[edit]
Scholia has a profile for messenger RNA (Q188928).
Wikimedia Commons has media related to mRNA.
RNAi Atlas: a database of RNAi libraries and their target analysis results
miRSearch Archived 2012-12-04 at the Wayback Machine: Tool for finding microRNAs that target mRNA
How mRNA is coded?: YouTube video
What is mRNA?: theconversation.com
| v t e Types of nucleic acids |
| Constituents | Nucleobases Nucleosides Nucleotides Deoxynucleotides |
| Ribonucleic acids (coding, non-coding) | | | | --- | | Translational | Messenger + precursor, heterogenous nuclear modified Messenger Transfer Ribosomal Transfer-messenger | | Regulatory | Interferential + Micro + Small interfering + Piwi-interacting Antisense Processual + Small nuclear + Small nucleolar + Small Cajal Body RNAs + Y RNA Enhancer RNAs | | Others | Guide Ribozyme Small hairpin Small temporal Trans-acting small interfering Subgenomic messenger | |
| Deoxyribonucleic acids | Organellar + Chloroplast + Mitochondrial Complementary Deoxyribozyme Genomic Hachimoji Multicopy single-stranded |
| Analogues | Xeno + Glycol + Threose + Hexose + Locked + Peptide + Morpholino Phosphorothioate |
| Cloning vectors | Phagemid Plasmid Lambda phage Cosmid Fosmid Artificial chromosomes + P1-derived + Bacterial + Yeast + Human |
| |
| v t e Gene expression |
| Introductionto genetics | Genetic code Central dogma + DNA → RNA → Protein Special transfers + RNA→RNA + RNA→DNA + Protein→Protein |
| Transcription | | | | --- | | Types | Bacterial Archaeal Eukaryotic | | Key elements | Transcription factor RNA polymerase Promoter | | Post-transcription | Precursor mRNA (pre-mRNA / hnRNA) 5' capping Splicing Polyadenylation Histone acetylation and deacetylation | |
| Translation | | | | --- | | Types | Bacterial Archaeal Eukaryotic | | Key elements | Ribosome Transfer RNA (tRNA) Ribosome-nascent chain complex (RNC) Post-translational modification | |
| Regulation | Epigenetic + imprinting Transcriptional + Gene regulatory network + cis-regulatory element lac operon Post-transcriptional + sequestration (P-bodies) + alternative splicing + microRNA Translational Post-translational + reversible + irreversible |
| Influential people | François Jacob Jacques Monod |
Retrieved from "
Categories:
RNA
Gene expression
Protein biosynthesis
Molecular genetics
Spliceosome
RNA splicing
Life sciences industry
Hidden categories:
CS1: long volume value
CS1 French-language sources (fr)
Articles needing additional references from May 2025
All articles needing additional references
Articles with short description
Short description is different from Wikidata
All articles with unsourced statements
Articles with unsourced statements from February 2024
All pages needing factual verification
Wikipedia articles needing factual verification from June 2021
Commons category link is on Wikidata
Webarchive template wayback links
Messenger RNA
Add topic |
189081 | https://www.mit.edu/~hsmui/files/handouts/mathcircle/diophantine.pdf | Diophantine Equations Holden Mui Name: Date: A Diophantine equation is an algebraic equation in which the solutions of interest are those for which all variables are integers. There are two main techniques for solving Diophantine equations.
• Factoring a Diophantine equation can reduce it to a finite case check.
• Bounding a Diophantine equation restricts the size of a variable, thus reducing the equation to several small cases.
Example 1. There is a prime number p such that 16p + 1 is the cube of a positive integer. Find p.
Diophantine Equations Page 2 Example 2. Find all positive integers x such that there exists a positive integer y satisfying 1 x + 1 y = 1 7.
Example 3. Find the sum of all positive integers n for which n2 −19n + 99 is a perfect square.
Example 4. Find all triples of positive integers (x, y, z) for which x3 + y3 + z3 −3xyz = 11.
Diophantine Equations Page 3 Problems Problem 1. Find all pairs of positive integers (x, y) such that x2 −y2 = 23.
Problem 2. Let A, M, and C be digits with (100A + 10M + C)(A + M + C) = 2005 What is A?
Problem 3. Find the prime number p such that 71p + 1 is a perfect square.
Diophantine Equations Page 4 Problem 4. There exist unique positive integers x and y that satisfy the equation x2 + 84x + 2008 = y2. Find x + y.
Problem 5. Find all triples (a, b, c) of positive integers which satisfy the simultaneous equations ab + bc = 44 ac + bc = 23 Problem 6. Find all pairs of integers (x, y) for which 4y −615 = x2.
Diophantine Equations Page 5 Problem 7. Find all pairs of positive integers (a, b) for which a2 + b exceeds a + b2 by 36.
Problem 8. Find n such that n −76 and n + 76 are both cubes of positive integers.
Problem 9. Find all pairs of positive integers (m, n) for which n4 −7n2 + 1 = m2.
Diophantine Equations Page 6 Problem 10. Find 3x2y2 if x and y are integers such that y2 + 3x2y2 = 30x2 + 517.
Problem 11. If p, q, and r are primes with pqr = 7(p + q + r), find p + q + r.
Problem 12. Find all right triangles with integer side lengths whose perimeter and area are equal.
Diophantine Equations Page 7 Problem 13. Find all triples of positive integers (p, q, r) for which a regular p-gon, a regular q-gon, and a regular r-gon can meet at a corner such that they don’t overlap and there are no gaps.
Problem 14. Find all pairs of integers (x, y) that satisfy the equation 2(x2 + y2) + x + y = 5xy.
Diophantine Equations Page 8 Challenge Problems Challenge 1. Show that there is no triple of positive integers (a, b, c) for which a2 + b + c, b2 + c + a, and c2 + a + b are all square.
Challenge 2. Solve in integers the equation x2 + xy + y2 = x + y 3 + 1 3 .
Diophantine Equations Page 9 Challenge 3. Find all sets {a1, a2, a3, a4} of four distinct positive integers such that there are four pairs (i, j) with 1 ≤i < j ≤4 for which ai + aj divides a1 + a2 + a3 + a4.
Challenge 4. Determine all pairs (x, y) of positive integers such that 3 p 7x2 −13xy + 7y2 = |x −y| + 1.
Diophantine Equations Page 10 Challenge 5. Find all triples (a, b, c) of positive integers such that a3 + b3 + c3 = (abc)2. |
189082 | https://www.mimuw.edu.pl/~ados/publications/papers/possquare.pdf | SHARP INEQUALITIES FOR THE SQUARE FUNCTION OF A NONNEGATIVE MARTINGALE ADAM OSE ¸KOWSKI Abstract. We determine the optimal constants Cp and C∗ p such the following holds: if f is a nonnegative martingale and S(f), f∗denote its square and maximal function, then ||S(f)||p ≤Cp||f||p, p < 1, and ||S(f)||p ≤C∗ p||f∗||p, p ≤1.
1. Introduction Square-function inequalities play an important role in harmonic analysis, classi-cal and noncommutative probability theory and other areas of mathematics. The reader is referred to, for example, the works of Stein , , Delacherie and Meyer , Pisier and Xu and Randrianantoanina . The purpose of this paper is to provide some new sharp bounds for the moments of a square function under the assumption that the martingale is nonnegative.
Let us start with some definitions. Throughout the paper, (Ω, F, P) will be a nonatomic probability space, filtered by a nondecreasing family (Fn)∞ n=0 of sub-σ-fields of F. Let f = (fn) be a real-valued martingale adapted to (Fn) and let d f = (d fn) stand for its difference sequence: d f0 = f0, d fn = fn −fn−1, n = 1, 2, . . . .
A martingale f is called simple, if for any n = 0, 1, 2, . . . the random variable fn takes only a finite number of values and there exists an integer m such that fn = fm almost surely for n > m.
For any nonnegative integer n, let Sn(f) and f ∗ n be given by Sn(f) = n X k=0 |d fk|2 !1/2 and f ∗ n = max 0≤k≤n |fk|.
Then one defines the square function S(f) and the maximal function f ∗by S(f) = lim n→∞Sn(f) and f ∗= lim n→∞f ∗ n.
In the paper we are interested in the inequalities between the moments of S(f), f and f ∗. For p ∈R, let ||f||p = sup n ||fn||p = (E|fn|p)1/p , if p ̸= 0, 2000 Mathematics Subject Classification. Primary: 60G42. Secondary: 60G44.
Key words and phrases. Martingale, square function, maximal function, moment inequality.
Partially supported by MEiN Grant 1 PO3A 012 29 and Foundation for Polish Science.
1 2 ADAM OSE ¸KOWSKI and ||f||0 = sup n ||fn||0 = sup n exp E log |fn| , with the convention that if p ≤0 and P(|X| = 0) > 0, then ||X||p = 0.
Let us mention here some related results from the literature. An excellent source of information is the survey by Burkholder (see also the references therein). The inequality (1.1) cp||f||p ≤||S(f)||p ≤Cp||f||p, if 1 < p < ∞, valid for all martingales, was proved by Burkholder in . Later, Burkholder refined his proof and shown that (cf. ) the inequality holds with c−1 p = Cp = p∗−1, where p∗= max{p, p/(p −1)}. Furthermore, the constant cp is optimal for p ≥2, Cp is the best for 1 < p ≤2 and the proof carries over to the case of martingales taking values in a separable Hilbert space. The right inequality (1.1) does not hold for general martingales if p ≤1 and nor does the left one if p < 1. It was shown by the author in that c1 = 1/2 is the best. In the remaining cases the optimal constants cp and Cp are not known.
Let us now turn to a related maximal inequality. If p > 1, then the estimate (1.1) and Doob’s maximal inequality imply the existence of some finite c∗ p, C∗ p such that for any martingale f, (1.2) c∗ p||f ∗||p ≤||S(f)||p ≤C∗ p||f ∗||p.
On the other hand, neither of the inequalities holds for p < 1 without additional assumptions on f. The limit case p = 1 was studied by Davis , who proved the validity of the estimate using a clever decomposition of the martingale f. Then Burkholder proved in that the optimal choice for the constant C∗ 1 is √ 3. In the other cases (except for p = 2, when c∗ 2 = 1/2 and C∗ 2 = 1) the optimal values of c∗ p and C∗ p are not known.
In the paper we study the square function inequalities for the case p < 1 under an additional assumption that the martingale f is nonnegative. The main results of the paper are summarized in the theorem below. For p < 1, let Cp = Z ∞ 1 (1 + t2)p/2 dt t2 1/p , if p ̸= 0, C0 = lim p→0 Cp = exp Z ∞ 1 1 2 log(1 + t2)dt t2 .
Theorem 1.1. Assume f is a nonnegative martingale.
(i) We have (1.3) ||f||p ≤||S(f)||p ≤Cp||f||p, if p < 1, and the inequality is sharp.
(ii) We have (1.4) ||S(f)||p ≤ √ 2||f ∗||p, if p ≤1, and the constant √ 2 is the best possible.
The paper is organized as follows. In the next section we describe the technique invented by Burkholder to study the inequalities involving a martingale, its square and maximal function and present its extension, which is needed to establish (1.4).
SHARP SQUARE FUNCTION INEQUALITIES 3 Section 3 is devoted to the proofs of the inequalities (1.3) and (1.4), while in Sec-tion 4 it is shown that these estimates are sharp. Finally, in the last section we present a different proof of the inequality (1.4) in the case p = 1.
2. On Burkholder’s method The inequalities (1.3) and (1.4) will be established using Burkholder’s technique, which reduces the problem of proving of a given martingale inequality to a finding certain special function. Here is the version of Theorem 2.1 in , modified in such a way that it works for the estimates involving a positive martingale and its square function. The analogous proof is omitted; however, see the proof of Theorem 2.2 below.
Theorem 2.1. Suppose that U and V are functions from (0, ∞)2 into R satisfying (2.1) V (x, y) ≤U(x, y) and the further condition that if d is a simple F-measurable function with Ed = 0 and P(x + d > 0) = 1, then (2.2) EU(x + d, p y2 + d2) ≤U(x, y).
Under these two conditions, we have (2.3) EV (fn, Sn(f)) ≤EU(f0, f0), for all nonnegative integers n and simple positive martingales f.
The condition (2.2) follows immediately from the following inequality, which is a bit easier to check: for any positive x and any number d > −x, (2.2′) U(x + d, p y2 + d2) ≤U(x, y) + Ux(x, y)d.
The inequality (1.4) may be proved using a special function involving three variables. However, this function seems to be difficult to construct and we have managed to find it only in the case p = 1 (see Section 5 below). To overcome this problem, we need an extension of Burkholder’s method allowing to work with other operators: we will establish a stronger result (2.4) ||T(f)||p ≤ √ 2||f ∗||p, if p ≤1.
Here, given a martingale f, we define a sequence (Tn(f)) by T0(f) = |f0|, Tn+1(f) = (T 2 n(f) + d f 2 n+1)1/2 ∨f ∗ n+1, n = 0, 1, 2, . . . , and T(f) = limn→∞Tn(f). Observe that Tn(f) ≥Sn(f) for all n, which can be easily proved by induction. Thus (2.4) implies (1.4).
Theorem 2.2. Suppose that U and V are functions from {(x, y, z) ∈(0, ∞)3 : y ≥ x ∨z} into R satisfying (2.5) V (x, y, z) ≤U(x, y, z), (2.6) U(x, y, z) = U(x, y, x ∨z) and the further condition that if 0 < x ≤z ≤y and d is a simple F-measurable function with Ed = 0 and P(x + d > 0) = 1, then (2.7) EU(x + d, p y2 + d2 ∨(x + d), z) ≤U(x, y, z).
4 ADAM OSE ¸KOWSKI Under these three conditions, we have (2.8) EV (fn, Tn(f), f ∗ n) ≤EU(f0, f0, f0), for all nonnegative integers n and simple positive martingales f.
Proof. By (2.5), it suffices to show that EU(fn, Tn(f), f ∗ n) ≤EU(f0, f0, f0), for all nonnegative integers n and simple positive martingales f.
To this end, we will prove that the process (Xn)∞ n=1, given by Xn = U(fn, Tn(f), f ∗ n), is a supermartingale. Observe that Tn+1(f) = (T 2 n(f) + d f 2 n+1)1/2 ∨fn+1 for any n = 0, 1, 2, . . .. Hence we have, by (2.6), E U(fn+1, Tn+1(f), f ∗ n+1)|Fn = E U(fn + d fn+1, (T 2 n(f) + d f 2 n+1)1/2 ∨(fn + d fn+1), f ∗ n)|Fn .
Using the condition (2.7) conditionally on Fn, this can be bounded from above by U(fn, Tn(f), f ∗ n).
□ Again we replace the property (2.7), this time with the following stronger con-dition: for any 0 < x ≤z ≤y and any d > −x, (2.7′) U(x + d, p y2 + d2 ∨(x + d), z) ≤U(x, y, z) + Ad, where A = A(x, y, z) = ( Ux(x, y, z), if x < z, limt↑z Ux(t, y, z) if x = z.
3. The proofs of the inequalities Let us start with some reductions. By standard approximation, it is enough to establish the inequalities (1.3) and (1.4) for simple and positive martingales only. The next observation is that, by Jensen’s inequality, we have ||f||p = ||f0||p.
Therefore, all we need is to show the following ,,local” versions: for n = 0, 1, 2, . . ., (3.1) ||f0||p ≤||Sn(f)||p ≤Cp||f0||p, if p < 1, and (3.2) ||Tn(f)||p ≤ √ 2||f ∗ n||p, if p ≤1.
Finally, we will be done if we establish the inequalities (3.1) and (3.2) for p ̸= 0; the case p = 0 follows then by passing to the limit. Hence, till the end of this section, we assume p ̸= 0.
3.1. The proof of (3.1). First note that the left inequality is obvious, since ||f0||p = ||S0(f)||p ≤||Sn(f)||p. Furthermore, clearly, it is sharp; hence we may restrict ourselves to the right inequality in (3.1). It is equivalent to (3.3) pESp n(f) ≤pCp pEf p 0 .
Let us introduce the functions Vp, Up : (0, ∞)2 →R by Vp(x, y) = pyp and Up(x, y) = px Z ∞ x (y2 + t2)p/2 dt t2 .
SHARP SQUARE FUNCTION INEQUALITIES 5 Now (3.3) can be stated as EVp(fn, Sn(f)) ≤EUp(f0, f0), that is, the inequality (2.3).
Therefore, by Theorem 2.1, we need to check the conditions (2.1) and (2.2’).
The inequality (2.1) follows from the identity Up(x, y) −Vp(x, y) = px Z ∞ x h (y2 + t2)p/2 −ypi dt t2 .
To check (2.2’), note that the integration by parts yields (3.4) Up(x, y) = p(y2 + x2)p/2 + p2x Z ∞ x (y2 + t2)p/2−1dt and Upx(x, y) = p Z ∞ x (y2 + t2)p/2 dt t −p(y2 + x2)p/2 x = p2 Z ∞ x (y2 + t2)p/2−1dt.
Hence we must prove that p(y2 + d2 + (x + d)2)p/2 + p2(x + d) Z ∞ x+d (y2 + d2 + t2)p/2−1dt −p(y2 + x2)p/2 −p2x Z ∞ x (y2 + t2)p/2−1dt −p2d Z ∞ x (y2 + t2)p/2−1 ≤0, or, equivalently, F(x) := p(y2 + d2 + (x + d)2)p/2 −(y2 + x2)p/2 x + d −p2 Z ∞ x (y2 + t2)p/2−1dt − Z ∞ x+d (y2 + d2 + t2)p/2−1dt ≤0.
We have F ′(x)(x + d)2 =p2(y2 + x2)p/2−1(x + d)d −p h (y2 + d2 + (x + d)2)p/2 −(y2 + x2)p/2i , (3.5) which is nonnegative due to the mean value property of the function t 7→tp/2.
Hence F(x) ≤lim s→∞F(s) = 0 and the proof is complete.
3.2. The proof of the inequality (3.2). We start with an auxiliary technical result.
Lemma 3.1. (i) If z ≥d > 0 and y > 0, then (3.6) p h (y2 + d2 + z2)p/2 −(y2 + (z −d)2)p/2i −p2z Z z z−d (y2 + t2)p/2−1dt ≤0.
(ii) If −z < d ≤0 and Y > 0, then (3.7) p(Y + (z + d)2)p/2 −(Y 2 −d2 + z2)p/2 z + d + p2 Z z z+d (Y + t2)p/2−1dt ≤0.
6 ADAM OSE ¸KOWSKI (iii) If y ≥z ≥x > 0, then (3.8) p h (y2 + x2)p/2 −2p/2zpi + p2 x2 + y2 2x Z z x (y2 + t2)p/2−1dt ≥0.
(iv) If D ≥z ≥x > 0, y ≥z, then p h (y2 + (D −x)2 + D2)p/2 −(y2 + x2)p/2 + 2p/2(zp −Dp) i −p2D Z z x (y2 + t2)p/2−1dt ≤0.
(3.9) Proof. Denote the left hand sides of (3.6) – (3.9) by F1(d), F2(d), F3(x) and F4(x), respectively. The inequalities will follow by simple analysis of the derivatives.
(i) We have F ′ 1(d) = p2d[(y2 + d2 + z2)p/2−1 −(y2 + (z −d)2)p/2−1] ≤0, as (z −d)2 ≤d2 + z2. Hence F1(d) ≤F1(0+) = 0.
(ii) The expression F ′ 2(d)(z + d)2 equals p h (Y −d2 + z2)p/2 −(Y + (z + d)2)p/2 + p 2(Y −d2 + z2)p/2−1 · 2d(z + d) i ≥0, due to the mean value property. This yields F2(d) ≤F2(0) = 0.
(iii) We have F ′ 3(x) = p2 2 1 −y2 x2 (y2 + x2)p/2−1x + Z z x (y2 + t2)p/2−1dt ≤0 and F3(x) ≥F3(z) = p[(y2 + z2)p/2 −2p/2zp] ≥0.
(iv) Finally, F ′ 4(x) = p2(D −x) h −(y2 + (D −x)2 + D2)p/2−1 + (y2 + x2)p/2−1i ≥0 and hence F4(x) ≤F4(z) = p h (y2 + (D −z)2 + D2)p/2 −(y2 + z2)p/2i −p2p/2(Dp −zp).
The right hand side decreases as y increases. Therefore F4(z) ≤p h (z2 + (D −z)2 + D2)p/2 −2p/2Dpi ≤0, as z2 + (D −z)2 + D2 ≤2D2.
□ Now we reduce the inequality (3.2) to (2.8). Let Vp(x, y, z) = p yp −2p/2(x ∨z)p and Up(x, y, z) = p2x Z x∨z x (y2 + t2)p/2−1dt + p(y2 + x2)p/2 −p2p/2(x ∨z)p.
(3.10) Now we see that (3.2) is equivalent to EVp(fn, Tn(f), f ∗ n) ≤EUp(f0, f0, f0), which is (2.8). Hence we need to check (2.5), (2.6) and (2.7’).
The property (2.5) is a consequence of the identity Up(x, y, z) −Vp(x, y, z) = p[(y2 + x2)p/2 −yp] + p2x Z x∨z x (y2 + t2)p/2−1dt.
SHARP SQUARE FUNCTION INEQUALITIES 7 The equation (2.6) follows directly from the definition of Up. All that is left is to prove the last condition. We consider two cases.
1◦The case x + d ≤z. Then (2.7’) reads p(y2 + d2 + (x + d)2)p/2 + p2(x + d) Z z x+d (y2 + d2 + t2)p/2−1dt ≤p(y2 + x2)p/2 + p2(x + d) Z z x (y2 + t2)p/2−1dt, or, in equivalent form, p(y2 + d2 + (x + d)2)p/2 −(y2 + x2)p/2 x + d −p2 Z z x (y2 + t2)p/2−1dt − Z z x+d (y2 + d2 + t2)p/2−1dt ≤0.
Denote the left hand side by F(x) and observe that (3.5) is valid; this implies F(x) ≤F((z −d) ∧z). If z −d < z, then F(z −d) ≤0, which follows from (3.6).
If conversely, z ≤z −d, then F(z) ≤0, which is a consequence of (3.7) (with Y = y2 + d2).
2◦The case x + d > z. If x + d ≥ p y2 + d2, then (2.7’) takes form p[(y2 + x2)p/2 −2p/2zp] + p2(x + d) Z z x (y2 + t2)p/2−1dt ≥0.
The left hand side is an increasing function of d, hence, if we fix all the other parameters, it suffices to show the inequality for the least d, which is determined by the condition x + d = p y2 + d2, that is, d = (y2 −x2)/(2x); however, then the estimate is exactly (3.8). Finally, assume x + d < p y2 + d2. Then (2.7’) becomes p(y2 + d2 + (x + d)2)p/2 −p2p/2(x + d)p ≤ p(y2 + x2)p/2 + p2(x + d) Z z x (y2 + t2)p/2−1dt −p2p/2zp, which is (3.9) with D = x + d.
4. Sharpness Now we will prove that the constants Cp and √ 2 in (1.3) and (1.4) can not be replaced by smaller ones. We will construct the appropriate examples on the probability space ([0, 1], B([0, 1]), |·|), a unit interval equipped with its Borel subsets and the Lebesgue measure. We will identify a set A ∈B([0, 1]) with its indicator function.
4.1. Sharpness of (1.3). Fix ε > 0 and define f by fn = (1 + nε) (0, (1 + nε)−1], n = 0, 1, 2, . . ..
Then it is easy to check that f is a nonnegative martingale, d f0 = (0, 1], d fn = ε (0, (1 + nε)−1] −(1 + (n −1)ε) ((1 + nε)−1, (1 + (n −1)ε)−1], for n = 1, 2, . . ., and S(f) = ∞ X n=0 (1 + nε2 + (1 + nε)2)1/2 ((1 + (n + 1)ε)−1, (1 + nε)−1].
8 ADAM OSE ¸KOWSKI Furthermore, for p < 1 we have ||f||p = 1 and, if p ̸= 0, ||S(f)||p p = ε ∞ X n=0 (1 + nε2 + (1 + nε)2)p/2 (1 + (n + 1)ε)(1 + nε) , which is a Riemann sum for Cp p. Finally, the case p = 0 is dealt with by passing to the limit; this is straightforward, as the martingale f does not depend on p.
4.2. Sharpness of (1.4). Fix M > 1, an integer N ≥1 and let f = f (N,M) be given by fn = M n (0, M −n], n = 0, 1, 2, . . . , N, and fN = fN+1 = fN+2 = . . . .
Then f is a nonnegative martingale, f ∗= M N(0, M −N] + N X n=1 M n−1(M −n, N −n+1], d f0 = (0, 1], d fn = (M n −M n−1) (0, M −n] −M n−1 (M n, M −n+1], for n = 1, 2, . . . , N, and d fn = 0 for n > N. Hence the square function is equal to 1 + N X k=1 (M k −M k−1)2 !1/2 = 1 + M −1 M + 1(M 2N −1) 1/2 on the interval (0, M −N], and is given by 1 + n−1 X k=1 (M k −M k−1)2 + M 2n−2 !1/2 = 1 + M −1 M + 1(M 2n−2 −1) + M 2n−2 1/2 on the set (M −n, M −n+1], for n = 1, 2, . . . , N.
Now, if M →∞, then ||S(f)||1 →1 + √ 2N and ||f||1 →1 + N, therefore, for M and N sufficiently large, the ratio ||S(f)||1/||f||1 can be made arbitrarily close to √ 2. Similarly, for p < 1, ||S(f)||p/||f||p → √ 2 as M →∞(here we may keep N fixed). Thus the constant √ 2 is the best possible.
5. On an alternative proof of (1.4) Let us present here (the sketch of) the direct proof of the inequality (1.4) in the case p = 1, without using the operators (Tn(f)). As previously, it is based on a construction of the special function; here is a modification of Theorem 2.1 from for the case of positive martingales.
Theorem 5.1. Suppose that U and V are functions from (0, ∞)3 into R satisfying (5.1) V (x, y, z) ≤U(x, y, z), (5.2) U(x, y, z) = U(x, y, x ∨z) and the further condition that if 0 < x ≤z and d is a simple F-measurable function with Ed = 0 and P(x + d > 0) = 1, then (5.3) EU(x + d, p y2 + d2, z) ≤U(x, y, z).
Under these three conditions, we have (5.4) EV (fn, Sn(f), fn) ≤EU(f0, f0, f0), for all nonnegative integers n and simple positive martingales f.
SHARP SQUARE FUNCTION INEQUALITIES 9 To show (1.4), take V (x, y, z) = y − √ 2(x ∨z) and introduce the function U(x, y, z) = 1 2 √ 2 y2 −x2 −(x ∨z)2 x ∨z .
These functions satisfy (5.1), (5.2), (5.3): the first inequality is equivalent to (y − √ 2(x ∨z))2 2 √ 2(x ∨z) ≥0, the second equation follows immediately from the definition of U. The third con-dition is a consequence of the stronger estimate U(x + d, p y2 + d2, z) ≤U(x, y, z) + Ux(x, y, z)d, valid for x, y, z > 0 and d > −x. The final observation is that U(x, x, x) ≤0 for all positive x. By the theorem above and the approximation argument (leading from simple to general martingales), (1.4) follows. The proof is complete.
References D. L. Burkholder, Martingale transforms, Ann. Math. Statist. 37 (1966), pp. 1494–1504.
, Explorations in martingale theory and its applications, Ecole d’Ete de Probabilits de Saint-Flour XIX—1989, pp. 1–66, Lecture Notes in Math., 1464, Springer, Berlin, 1991.
, The best constant in the Davis inequality for the expectation of the martingale square function, Trans. Amer. Math. Soc. 354 (2002), pp. 91–105.
B. Davis, On the integrability of the martingale square function, Israel J. Math. 8 (1970), pp. 187–190.
C. Dellacherie and P.-A. Meyer, Probabilities and potential B: Theory of martingales, North Holland, Amsterdam, 1982.
A. Os¸ ekowski, Two inequalities for the first moment of a martingale, its square and maximal function, Bull. Polish Acad. Sci. Math. 53 (2005), pp. 441-449.
G. Pisier and Q. Xu, Non-commutative martingale inequalities, Commun. Math. Phys. 189 (1997), 667–698.
N. Randrianantoanina, Square function inequalities for non-commutative martingales, Israel J. Math. 140 (2004), 333-365.
E. M. Stein, The development of square functions in the work of A. Zygmund, Bull. Amer.
Math. Soc. 7 (1982), 359–376.
, Harmonic Analysis: Real-Variable Methods, Orthogonality and Oscillatory Integrals, Princeton University Press, Princeton, NJ, 1993.
Department of Mathematics, Informatics and Mechanics, Warsaw University, Ba-nacha 2, 02-097 Warsaw, Poland E-mail address: ados@mimuw.edu.pl |
189083 | http://matchstickpuzzles.blogspot.com/ | Matchstick Puzzles
Matchstick Puzzles
A Collection
23 September 2022
662. Mathematical: 5 minus 3 is not 4
Easy Medium Hard Extreme
ADD 4 matchsticks to make the equation true.
The following rules apply:
No number is greater than 9
Negative numbers are not allowed
You may not change the equal (=) sign
Show Answer
Hide Answer
2 comments:
Email ThisBlogThis!Share to XShare to FacebookShare to Pinterest
10 October 2021
661. Reduce from 4 to 3 square matchsticks
Easy Medium Hard Extreme
The size of the layout shown above is exactly 4 square matchsticks. Make sure to understand that A + B is exactly 1 square matchstick.
MOVE 4 matchsticks to reduce the size of the shape to exactly 3 square matchsticks.
Show Answer
C + D is equal to 1 square matchstick.
Hide Answer
1 comment:
Email ThisBlogThis!Share to XShare to FacebookShare to Pinterest
25 September 2021
660. Six to five equal triangles
Easy Medium Hard Extreme
There are 6 equal triangles in the layout above. MOVE 2 matchsticks to form 5 equal triangles.
Show Answer
Hide Answer
No comments:
Email ThisBlogThis!Share to XShare to FacebookShare to Pinterest
07 September 2021
659. Triangle 3X3: Remove 6 leaving 0 triangles
Easy Medium Hard Extreme
TAKE AWAY 6 matchsticks so that there are no triangles left.
Show Answer
Hide Answer
No comments:
Email ThisBlogThis!Share to XShare to FacebookShare to Pinterest
28 August 2021
658. Mathematical: 0 minus 3 is not 7
Easy Medium Hard Extreme
ADD 3 matchsticks to make the equation true.
The following rules apply:
No number is greater than 9
Negative numbers are not allowed
You may not change the equal (=) sign
Show Answer
Hide Answer
2 comments:
Email ThisBlogThis!Share to XShare to FacebookShare to Pinterest
03 August 2021
657. Change the 2:3:4 ratio to 2:2:5
Easy Medium Hard Extreme
The above square is divided into 3 parts in the ratio 2:3:4. Change this ratio to 2:2:5 by moving only 1 matchstick.
Show Answer
Hide Answer
No comments:
Email ThisBlogThis!Share to XShare to FacebookShare to Pinterest
29 July 2021
656. Three right angles
Easy Medium Hard Extreme
Three matchsticks form 2 right angles as shown in Figure A above. By moving only 1 matchstick you can create 3 right angles as shown in Figure B. However, this is only one solution and there are 7 more where 3 right angles can be created. Can you find them all?
The rules are:
Start each time with the layout as in Figure A.
Move only 1 matchstick.
You may not move the horizontal matchstick in Figure A.
Show Answer
Hide Answer
1 comment:
Email ThisBlogThis!Share to XShare to FacebookShare to Pinterest
27 July 2021
655. Four to six squares
Easy Medium Hard Extreme
Move only 2 matchsticks to create 6 squares. The squares are allowed to differ in size.
However, squares must be linked and each matchstick must form the side of at least one square.
Show Answer
Don't miss the 2X2 square.
Hide Answer
No comments:
Email ThisBlogThis!Share to XShare to FacebookShare to Pinterest
10 July 2021
654. Increase the inner triangles
Easy Medium Hard Extreme
This puzzle is similar to but different from numbers 517 and 592.
In the figure above 9 matchsticks were placed on a triangular grid to form 7 inner triangles.
What is the maximum number of inner triangles you can create by rearranging the 9 matchsticks on the grid?
Show Answer
Hide Answer
No comments:
Email ThisBlogThis!Share to XShare to FacebookShare to Pinterest
06 July 2021
653. The farmer and his 4 sons
Easy Medium Hard Extreme
A farmer lives on the area marked "A" on his farm. He decides to divide the rest of the farm between his 4 sons with the SIZE and SHAPE of each of the 4 farms exactly the same.
Add as many matchsticks as you want to the layout to show how he has done this.
Show Answer
Hide Answer
No comments:
Email ThisBlogThis!Share to XShare to FacebookShare to Pinterest
04 July 2021
652. Three square units extreme
Easy Medium Hard Extreme
Credit: The Moscow puzzles (1972) Boris A. Kordemsky
The length of a matchstick is 1 unit, so in the layout above 12 matchsticks were used to create an area "W" of 3 square units. (see puzzle 121 and also how area W was calculated below)
There is another way to create 3 square units, but you may NOT MOVE the matchsticks which from line AC. Can you find it?
Take note of the calculations below as you might need some of the information in your solution.
The area W was calculated as:
= Total area ABC - area X - area Y - area Z
= 1/2(3 X 4) - 1 - 1 - 1
= 3 square units
Angle 𝛳 was calculated as:
= arcsine(AB/AC)
= arcsine(3/5)
= 36.869898 degrees
Show Answer
We must now proof that the area of the parallelogram is 3 square units.
To calculate the height:
sin𝛳 = height/1
height = sin(36.869898)
height = 0.6 unit
The area of the parallelogram is:
= base X height
= 5 X 0.6
= 3 square units
Hide Answer
No comments:
Email ThisBlogThis!Share to XShare to FacebookShare to Pinterest
03 July 2021
651. Three square units
Easy Medium Hard Extreme
The length of a matchstick is 1 unit. Use 12 matchsticks to create an area of 3 square units. To make your life easier, 8 of the 12 matches have already been placed in the correct positions.
Show Answer
Area A is exactly 3 square units. (4 - 1)
Hide Answer
No comments:
Email ThisBlogThis!Share to XShare to FacebookShare to Pinterest
19 June 2021
650. Three squares different in size
Easy Medium Hard Extreme
There are 5 squares in the layout above, 4 small and 1 big.
Move 2 matchsticks to create 3 squares of 3 different sizes.
Show Answer
Hide Answer
No comments:
Email ThisBlogThis!Share to XShare to FacebookShare to Pinterest
16 June 2021
649. Three-way bridge
Easy Medium Hard Extreme
There are three islands in the ocean and your challenge is to construct a bridge to connect them. Unfortunately, as can be seen with matchsticks A, B and C, the length of a matchstick is just too short to connect the islands directly.
How would you construct an interlocking bridge, that can actually carry some weight in real life, by only using matchsticks A, B and C? Also, the three matchsticks are not allowed to touch the ocean.
Show Answer
Hide Answer
No comments:
Email ThisBlogThis!Share to XShare to FacebookShare to Pinterest
08 June 2021
648. Divide into 2 parts
Easy Medium Hard Extreme
MOVE 2 matchsticks in the layout above to divide it into 2 parts equal in shape and size.
Show Answer
Hide Answer
2 comments:
Email ThisBlogThis!Share to XShare to FacebookShare to Pinterest
04 June 2021
647. Rugby
Easy Medium Hard Extreme
A rugby player missed a kick to goal by kicking the ball to the right of the posts and also lower than the crossbar.
MOVE only 2 matchsticks to make the ball go through the posts and over the crossbar. You may not move the ball.
Show Answer
Tilt your head to the right to see the solution.
Hide Answer
No comments:
Email ThisBlogThis!Share to XShare to FacebookShare to Pinterest
03 June 2021
646. Cross to squares
Easy Medium Hard Extreme
There are no squares in the layout above. Move 4 matchsticks to change the cross into 4 EQUAL squares.
Show Answer
Hide Answer
No comments:
Email ThisBlogThis!Share to XShare to FacebookShare to Pinterest
02 June 2021
645. Fix the equation
Easy Medium Hard Extreme
You would need math skills to solve this one.
The equation 7/9 = 3 is not true. Move only 2 matchsticks to make it true.
Show Answer
Hide Answer
No comments:
Email ThisBlogThis!Share to XShare to FacebookShare to Pinterest
01 June 2021
644. Polygon with 7 sides
Easy Medium Hard Extreme
Shown above is an eight sided convex polygon built from 2 squares and 6 triangles. 17 matchsticks were used to achieve this. The outside angles are all greater than 180 degrees as illustrated with angle "a" above.
Create your own convex polygon. The following rules apply:
Polygon must have 7 sides
12 matchsticks to be used
All outside angles to be greater than 180 degrees
There must be 3 equal sized triangles
There must be 2 equal sized squares
All matchsticks must be flat on the surface.
Show Answer
Hide Answer
No comments:
Email ThisBlogThis!Share to XShare to FacebookShare to Pinterest
30 May 2021
643. Cup and saucer
Easy Medium Hard Extreme
Move 4 matchsticks to convert the cup and saucer into 3 DIFFERENT in size triangles.
Show Answer
Hide Answer
No comments:
Email ThisBlogThis!Share to XShare to FacebookShare to Pinterest
29 May 2021
642. Stacked triangles
Easy Medium Hard Extreme
In the figure above are 4 equal triangles. Move 4 matchsticks to create 2 equal triangles.
Show Answer
Hide Answer
No comments:
Email ThisBlogThis!Share to XShare to FacebookShare to Pinterest
28 May 2021
641. Three different squares
Easy Medium Hard Extreme
MOVE 3 matchsticks to create only 3 squares. The 3 squares must have 3 DIFFERENT sizes.
Show Answer
Note that all matchsticks form part of a square i.e. there are no redundant matchsticks. There are also other solutions.
Hide Answer
No comments:
Email ThisBlogThis!Share to XShare to FacebookShare to Pinterest
24 May 2021
640. Big and small squares
Easy Medium Hard Extreme
There are 4 squares in the layout above. A broken matchstick is exactly half the length of a full length matchstick. Also, matchstick A is placed exactly in the middle of matchstick B.
Move 3 matchsticks to create 8 squares. All matchsticks don't have to be flat on the surface and square sizes may differ.
Show Answer
There are 5 small and 3 bigger squares.
Hide Answer
No comments:
Email ThisBlogThis!Share to XShare to FacebookShare to Pinterest
22 May 2021
639. From three to two squares
Easy Medium Hard Extreme
Reduce the number of squares from 3 to 2 by moving 3 matchsticks. Squares may differ in size.
Show Answer
Hide Answer
No comments:
Email ThisBlogThis!Share to XShare to FacebookShare to Pinterest
21 May 2021
638. Create 5 triangles
Easy Medium Hard Extreme
Move 4 matchsticks to create 5 triangles. The triangles may differ in size.
Show Answer
There are 4 small triangles and one big triangle.
Hide Answer
No comments:
Email ThisBlogThis!Share to XShare to FacebookShare to Pinterest
Older PostsHome
Subscribe to: Posts (Atom)
Translate
Select Language▼
Welcome
I've always been intrigued by these little sticks with their rounded heads. There are just so many fun things you can do with them...
I started collecting matchstick puzzles a long time ago and have decided to display them all on this blog.
~Cape Town, SA~
Search this Blog
The Puzzles
0001 Fly in Wine Glass [Hard]
0002 Square in Square [Hard]
0003 Roman and Maths [Hard]
0004 Big Square Small Square [Medium]
0005 Farmer and 4 Sons [Hard]
0006 Eight Triangles and Hexagon [Hard]
0007 Egypt [Hard]
0008 Jumping Frogs [Hard]
0009 The Pig and the Tail [Hard]
0010 Easterly Fish Southerly Fish [Medium]
0011. Square 3X3: 14 squares reduced to 5 [Easy]
0012 Crossing Sticks [Hard]
0013 Five Triangles and the Pentagram [Medium]
0014 Tower upside down moving 4 sticks [Medium]
0015 Short and Long [Hard]
0016 Add 5 to form 8 Triangles [Hard]
0017 Square 2X2: Move 4 to form 10 Squares [Hard]
0018 Square 1X1: Add 4 to create 4 Triangles and 2 Squares [Extreme]
0019 Triangle 3X3: Remove 3 leaving 7 Triangles [Medium]
0020. 5 Equal Squares: Remove 3 leaving 3 Squares [Easy]
0021 The King and his Guards [Medium]
0022 The ONE Puzzle [Easy]
0023 6 Sticks Touching [Hard]
0024 14 Squares reduced to 3 [Medium]
0025 Square 2X2 : Remove 2 leaving 2 Squares [Easy]
0026 The farmer and his 5 sons [Hard]
0027 Squares 2X2: Move 4 creating 3 equal sized Squares [Medium]
0028 Move 3 creating 4 equal sized Squares [Hard]
0029 Square 2X2: Move 2 to create 7 Squares [Medium]
0030 The farmer and his 6 sons [Hard]
0031 Square 2X2: Remove 3 and Move 2 to form 3 Squares [Hard]
0032 Square 2X2: Move 3 to create 3 squares [Easy]
0033 Square 2X2: Remove 1 and move 4 to create 11 Squares [Medium]
0034. The farmer and his 8 sons [Hard]
0035 Create 3 Groups of 8 Sticks in each Group [Hard]
0036 Star of David : Move 2 to form 6 Triangles [Medium]
0037. Square 1X1 : Add 5 creating 6 Squares [Medium]
0038. 7 Triangles reduced to 3 [Easy]
0039. 3 Triangles changed to 5 by moving 3 [Medium]
0040. Digital Mirror Image [Hard]
0041. Move 3 creating 9 Squares [Medium]
0042. 6 Squares reduced to 5 [Hard]
0043. Maze 5X5 : Move 4 creating 3 Squares [Medium]
0044. Remove 3 leaving no Triangles [Medium]
0045. Turn matchsticks in the circle [Hard]
0046. Reduce 6 Triangles to 3 [Medium]
0047. 6 Farms to 6 Farms [Extreme]
0048. 13 Squares Reduced to 8 [Medium]
0049. EAT and ATE and TEA [Medium]
0050. Vertical Horizontal and Diagonal Rows [Hard]
0051. Land Water and Island [Hard]
0052. Dividing the Sphinx [Hard]
0053. Oil and No Oil [Medium]
0054. Adding Sticks without forming a Square [Easy]
0055. Square 4X4 : How many Squares [Medium]
0056. 25 Broken matchsticks adding up to 15 [Medium]
0057. Dividing Figure and put together to form a Rectangle [Hard]
0058. Reduce 5 Squares to 4 [Extreme]
0059. Reduce 6 Squares formed by Crosses to 3 Squares [Easy]
0060. Black and White Knights [Hard]
0061. Roman numerals: 12 changed to 7 [Easy]
0062. Farmer and Neighboring Sons [Hard]
0063. House fronting West [Easy]
0064. Roman numeral: Tongue in the cheek [Easy]
0065. 3 Squares using 8 Sticks [Easy]
0066. Move 3 sticks creating 5 squares [Easy]
0067. Divide Trapezoid into 7 equal sized Areas [Hard]
0068. Crosses in Circle [Medium]
0069. Squares forming L [Medium]
0070. Triangle 3X3 : Remove 6 Matchsticks leaving no Triangles [Easy]
0071. Add 5 Matchsticks to make 9 [Easy]
0072. Triangle 2X2 : Move 2 creating 3 Triangles [Easy]
0073. Roman Numerals: 1 minus 3 is not 2 [Medium]
0074. The Letter A [Hard]
0075. 3 Rectangles converted into 6 Squares [Easy]
0076. Add 2 and Divide into 2 Equal Sized Figures [Medium]
0077. Form 8 Triangles [Hard]
0078. Change the Cross [Hard]
0079. Forming 3 Squares [Easy]
0080. Square 3X3 : Remove 11 leaving 3 Squares [Easy]
0081. Roman 13 Changed to 8 [Medium]
0082. Divide Figure into 4 Parts [Medium]
0083. Triangle 3X3 : Remove 3 Leaving no Triangles [Easy]
0084. Reduce 6 triangles to 4 triangles [Medium]
0085. Chair Fell Over [Hard]
0086. Farms and Windmills and Oil Drills and Boreholes [Hard]
0087. Peace Love in the World [Easy]
0088. Divide into 2 Equal Shapes using 4 Matchsticks [Hard]
0089. Cross on the Church [Medium]
0090. Remove 4 leaving 4 Triangles [Easy]
0091. Another fly in the wine glass [Hard]
0092. Create 3 squares with 8 sticks [Hard]
[0093. Convert 4 Triangles to 3 Squares {Medium]](
0094. Roman numeral 11 changed to 6 [Easy]
0095. Matchstick Sudoku [Medium]
0096. Subtracting Triangles [Extreme]
0097. Triangle reduced to 3 square units [Hard]
0098. Create 2 Squares Moving 2 Matchsticks [Medium]
0099. Roman numerals: 7 plus 1 is not 1 [Easy]
0100. Remove 4 leaving 4 Triangles [Easy]
0101. 4X4 Square : Remove 8 sticks leaving 6 squares [Medium]
0102. Roman numerals: 7 minus 2 = 2: Move 2 [Medium]
0103. Snake to Squares [Hard]
0104. The 8 As [Extreme]
0105. Fish swimming in opposite Direction [Hard]
0106. Change to One [Hard]
0107. Remove 6 Leaving 2 Squares [Medium]
0108. The Numbers 1 to 8 in Squares [Hard]
0109. 3X3 Squares: Remove 8 leaving 2 Squares [Easy]
0110. 1X1 Square : Add 16 creating 5 equal Squares [Hard]
0111. Ten and Eleven [Medium]
0112. Remove 8 leaving 6 squares [Medium]
0113. The Farmer Four Sons Diamonds and Gold [Hard]
0114. Forming 5 groups with 3 matchsticks in each group [Hard]
0115. 3X3 Triangle. Remove 5 sticks leaving 4 Triangles [Medium]
0116 Add 17 sticks to form 7 Squares [Extreme]
0117. The King and his Guards: Remove 2 Guards [Hard]
0118. Roman numerals : 4-2 is not 5: Move 1 [Easy]
0119. Treasure Rooms and Treasures [Hard]
0120. Numeric Puzzle : 3X3=5 : Move 1 [Medium]
0121. Create 3 square units with 12 sticks [Extreme]
0122. Broken sticks add up to 25 [Hard]
0123. The Farmer Four Sons and Oil Drills [Hard]
0124. Move 1 stick creating 6 Squares [Medium]
0125. Turn around sticks moving 2 at a time [Hard]
0126. Change cross into 4 squares [Medium]
0127 Square 3X3: Move 8 sticks to form 3 squares of different sizes [Hard]
0128 Turn around matchsticks in a circle [Hard]
0129. Rectangle 3X2 : Remove 3 leaving 4 equal squares [Easy]
0130. The King and his Guards: Remove 4 Guards
0131. Triangle 3X3 : Remove 7 leaving 5 triangles [Medium]
0132. Sticks in a row : Form 8 Squares [Hard]
0133. Triangular Prism [Extreme]
0134. Triangle 2X2 : Remove 2 leaving 2 triangles [Easy]
0135. Chair facing to the right [Hard]
0136. Roman numerals: 10 minus 1 = 1: Move 1 [Hard]
0137. Square 3X3 : Remove 11 leaving 4 equal squares [Medium]
0138. Sticks in a row : Turn 3 at a time [Hard]
0139. Another Egypt [Hard]
0140. Triangle 3X3 : Remove 3 leaving 7 Triangles [Easy]
0141. Number Puzzle : 3 plus 3=5 Move 1 [Easy]
0142. Create 5 Diamonds [Hard]
0143. Proof that 4 plus 2 is NIL [Easy]
0144. Square 4X4 : Remove 5 leaving 6 equal squares [Medium]
0145. Roman numerals: 1 and 2 and 3 = 4 : Move 1 [Hard]
0146. 5 Squares : Move 4 creating 3 squares [Hard]
0147. Three Diamonds to Four Triangles [Easy]
0148. Rectangle 4X2 : Remove 6 leaving 6 equal squares [Medium]
0149. Square 1X1 : Add 4 creating 4 triangles and 1 square [Hard]
0150. Triangle 3X3 : Move 5 creating 5 triangles [Medium]
0151. Change 5 squares to 4 by moving 2 sticks [Medium]
0152. Roman numerals: 2 plus 2 is not 1 [Easy]
0153. Triangle 3X3 : Move 1 leaving 9 triangles [Easy]
0154. Duplicate the Arrow [Easy]
0155. Remove squares to form equal areas [Hard]
0156. Roman numerals: 2 plus 1 is not 1 [Easy]
0157. Chair lying on its back [Hard]
0158. Square 3X3 : Remove 6 leaving 3 unequal squares [Medium]
0159. Small diamonds to big diamond [Easy]
0160. Triangle 2X2 : Move 2 forming 2 triangles [Easy]
0161. Move 3 to create 3 triangles [Medium]
0162. Roman numerals: 3 minus 1 is not 3 [Easy]
0163. Change 3 triangles to 4 triangles using 7 sticks [Hard]
0164. Rectangle 3X2 : Remove 4 sticks leaving 3 squares [Easy]
0165 From Wine Glasses to a House [Medium]
0166 Triangle 3X3 : Remove 1 leaving 10 triangles [Easy]
0167. The farmer and his paddocks [Medium]
0168. Squares : Move 11 sticks to change 7 squares to 10 [Medium]
0169. The farmer and his paddocks : move only 3 sticks [Medium]
0170. Change Roman Numeral 7 to 1 [Hard]
0171. Squares: Remove 10 leaving 4 equal squares [Medium]
0172. Rhombus Square and Parallelogram [Medium]
0173. All the right angles [Medium]
0174. Rhomboid and Rhombus [Medium]
0175 Four sticks in each row and column [Medium]
0176. Rectangle 3X2: Remove 5 leaving 3 equal squares [Easy]
0177. Number Puzzle: 9-8=15: Move 1 [Medium]
0178. Double-crossed matchsticks [Hard]
0179. Move 4 to form 3 rhombuses [Medium]
0180. Rectangle 4X2: Remove 7 leaving 6 squares [Easy]
0181. Forming 8 nodes [Extreme]
0182. 7 Equal squares reduced to 5 equal squares [Hard]
0183. Roman numerals: 3 minus 2 is not 5 [Easy]
0184. Key to squares [Medium]
0185. Rectangle 3X2: Remove 4 leaving 4 equal squares [Easy]
0186. Divide triangle into 3 equal parts [Hard]
0187. Changing 3 to 9 [Easy]
0188. Create 4 triangles using 6 matchsticks [Hard]
0189. Create 9 rhombuses adding 12 sticks [Hard]
0190. Count the squares [Easy]
0191. Squares : Increase 7 squares to 9 [Medium]
0192. Mathematical : 99 minus 59 is not 10 [Easy]
0193. Squares 4X4: Remove 16 leaving 2 squares of equal size [Hard]
0194. Forming 7 squares with 24 sticks [Medium]
0195. Create a 3 dimensional cube moving 3 sticks [Medium]
0196. Rectangle 4X2 : Remove 10 leaving 3 squares [Easy]
0197. Two plus one is three [Easy]
0198. Move 1 stick creating 8 squares [Medium]
0199. Create 3 parallelograms and 2 triangles [Hard]
0200. Big squares and small squares [Medium]
0201. A rhombus and a triangle [Easy]
0202. No squares to 2 squares [Medium]
0203. Rectangle 4X2 : Remove 10 leaving 5 squares [Easy]
0204. Mathematical : 1 plus 2 minus 3 is not 199 [Extreme]
0205. Move 2 creating 4 rectangles and 2 squares [Medium]
0206. Square 4X4: Remove sticks leaving no squares [Medium]
0207. Star to 6 rhombuses [Easy]
0208. Move 3 creating 4 equal squares [Hard]
0209. Tree to fish [Easy]
0210. Square 3X3: Move 2 creating 4 squares [Easy]
0211. Turn 7 sticks pointing downwards [Hard]
0212. Divide the Factory [Extreme]
0213. Change the L into 4 squares [Medium]
0214. Triangle 3X3: Remove 4 leaving 7 triangles [Medium]
0215. Squares in squares [Medium]
0216. Create 6 rhombuses and 6 triangles [Hard]
0217. Rearrange 24 matchsticks [Medium]
0218. Square 4X4: Remove 9 leaving 7 rectangles [Hard]
0219. Move 2 to create 6 squares [Easy]
0220. Mathematical : 9 plus 2 is not equal to 5 plus 1 [Medium]
0221. Move 2 and create 2 squares [Easy]
0222. Move 3 creating 3 squares [Medium]
0223. Triangle 3X3: Remove 5 leaving 5 equal triangles [Easy]
0224. Form a 3X5 rectangle [Hard]
0225. Move 1 to create 5 rectangles [Medium]
0226. Move 4 sticks to create 4 equal squares [Hard]
0227. Square 3X3: Move 3 creating 7 equal squares [Medium]
0228. Remove 7 sticks leaving 4 equal squares [Easy]
0229. The 23 matchsticks [Hard]
0230. Move 4 to form 5 squares [Easy]
0231. Move 4 to create 7 squares [Medium]
0232. Remove 6 leaving 3 equal squares [Easy]
0233. Create 6 rhombuses [Hard]
0234. One equals Seven [Hard]
0235. Move 2 creating 6 squares [Easy]
0236. Africa [Hard]
0237. Move 4 to create 7 squares [Medium]
0238. Square 3X3: Move 2 creating 7 squares [Medium]
0239. Move 6 and create 6 squares [Hard]
0240. Create 3 rhombuses from 3 triangles [Easy]
0241. Value of ? [Hard]
0242. Move 4 sticks to create 9 squares [Easy]
0243. Route map: The hexagon [Medium]
0244. Square 5X5: count the squares [Medium]
0245. Move 4 creating 3 squares [Medium]
0246. Roman numerals: 10 plus 10 is not equal to 10 [Easy]
0247. Move 2 sticks to create 6 unequal triangles [Medium]
0248. Turn the snake into squares [Easy]
0249. Move 3 to create 8 squares [Easy]
0250. Small and big squares [Medium]
0251. Convert the key into 5 squares [Hard]
0252. Add 5 sticks to create 8 squares [Easy]
0253. Matchstick Nodes: Three sticks [Hard]
0254. Form the correct equation [Hard]
0255. Remove 1 to leave 3 squares [Easy]
0256. Letters between the lines [Hard]
0257. Move 4 to create 4 squares [Medium]
0258. Square 2X2: Move 4 creating 2 squares [Easy]
0259. Northerly Fish Southerly Fish [Hard]
0260. Remove 3 sticks leaving 5 squares [Easy]
0261. Remove 22 sticks leaving 4 equal squares [Medium]
0262. Triangle 3X3 : Remove 6 leaving 3 triangles [Easy]
0263. Move 1 stick to create 8 squares [Medium]
0264. Square 4X4: Remove 4 Leaving 14 squares [Easy]
0265. Move 4 to create 5 triangles [Easy]
0266. Create 6 rhombuses and 1 triangle [Hard]
0267. Move 5 to create 7 squares [Medium]
0268. Roman Numerals: 5 minus 5 is not 5 [Easy]
0269. Move 6 to create 18 parallelograms [Medium]
0270. Double the size [Easy]
0271. Mathematical: 1 plus 1 is 2 [Hard]
0272. Move 2 to create 4 squares [Easy]
0273. Triangle 3X3: Remove 7 leaving 3 triangles [Easy]
0274. Divide into 2 equal halves [Hard]
0275. Move 6 to create 4 equal squares [Medium]
0276. Move 2 to create 2 squares [Medium]
0277. Move 8 to create 18 squares [Easy]
0278. Create 2 squares and 5 rectangles [Hard]
0279. Remove 16 leaving 13 squares [Medium]
0280. Moving adjacent matchsticks [Hard]
0281. Remove 4 leaving 7 squares [Easy]
0282. Mathematical: 8 plus 3 is not 7 [Medium]
0283. Move 4 to create 17 squares [Easy]
0284. Move 3 to create 3 triangles [Medium]
0285. Thirty minutes past four [Hard]
0286. Triangle 3X3: Remove 8 leaving 2 triangles [Easy]
0287. Move 2 creating 4 squares [Easy]
0288. Create TWO by removing 5 matchsticks [Easy]
0289. Move 4 to create 9 squares [Medium]
0290. Move 1 to create 5 squares [Easy]
0291. Mathematical: 3 plus 8 is not zero [Medium]
0292. Move 6 to create 14 squares [Medium]
0293. Move 4 to create 4 equal squares [Medium]
0294. Mathematical: Move 1 and remove 1 [Easy]
0295. Move 4 to create 6 triangles [Easy]
0296. The funny structure [Extreme]
0297. House fronting East [Easy]
0298. Move 2 leaving no squares [Medium]
0299. Mathematical: 6 minus 6 is not equal to 8 [Hard]
0300. The Donkey [Hard]
0301. Move 2 to create 8 squares [Easy]
0302. The swimming pool and the trees [Hard]
0303. Six to nil [Easy]
0304. School of Fish [Hard]
0305. Divide the square [Hard]
0306. Reduce the squares [Easy]
0307. Move 4 to create 4 triangles [Easy]
0308. Reduce the cubes [Hard]
0309. All the parallelograms [Medium]
0310. Move 4 to create 6 squares [Easy]
0311. Roman numerals: 10 plus 10 is not equal to 5 [Easy]
0312. Move 8 to form 10 squares [Hard]
0313. Rectangle 4X2: Move 3 to create 6 squares [Medium]
0314. Roman numerals: 5 plus 5 is not equal to 9 [Medium]
0315. Letters to structures [Hard]
0316. Square 3X3: Remove 6 leaving 3 squares [Easy]
0317. Move 2 to create 6 triangles [Easy]
0318. Move 3 to create 3 triangles [Easy]
0319. Move 6 to create 17 squares [Extreme]
0320. Mathematical: Move 2 and statement is still true [Medium]
0321. Divide the triangle [Medium]
0322. Move 4 to create 12 squares [Easy]
0323. Move 1 to create 2 squares [Easy]
0324. Triangle 3X3: Remove 6 leaving 5 triangles [Easy]
0325. Rectangle 5X2: Remove 3 leaving 7 squares [Medium]
0326. Mathematical: 14 plus 7 minus 4 is not 11 [Hard]
0327. Triangles to no triangles [Hard]
0328. Move 1 to create 4 [Easy]
0329. Move 2 to create 10 squares [Easy]
0330. Move 2 to create 4 triangles [Medium]
0331. Move 3 to fix the equation [Hard]
0332. Move 2 to create 11 squares [Easy]
0333. Mathematical: 9 times 2 is not 7 plus 8 [Medium]
0334. Move 4 to create 9 squares [Easy]
0335. The coffee table and the chair [Medium]
0336. Parallelograms in abundance [Medium]
0337. Remove 2 leaving 6 squares [Easy]
0338. Route map: 5 Triangles [Medium]
0339. Reduce the squares from 4 to 2 [Hard]
0340. Move 4 and create 10 triangles [Hard]
0341.The 5 trapezoids [Medium]
0342. Move 7 to create 7 rhombuses [Hard]
0343. Get 4 triangles [Hard]
0344. Cube to triangles [Easy]
0345. Divide the land into 4 parts [Hard]
0346. Unbalanced Structure [Hard]
0347. Move 6 to create 10 squares [Medium]
0348. Mathematical: Take away 4 to make equation true [Medium]
0349. Ice cream cone [Hard]
0350. Odd one out [Medium]
0351. Move 2 to create 7 squares [Easy]
0352. Move 5 to create 3 rhombuses [Medium]
0353. Roman numerals: 6 plus 2 is not 5 [Easy]
0354. Divide the figure into 2 equal areas [Medium]
0355. Move 4 to create 5 rhombuses [Hard]
0356. Move 1 to make equation true [Easy]
0357. Staggered triangles [Hard]
0358. Move 1 to create 3 squares [Easy]
0359. Change the arrows [Medium]
0360. Creating a circle [Hard]
0361. Noughts and crosses board [Easy]
0362. Move 3 to create 4 rhombuses [Hard]
0363. Moving geometric shapes [Hard]
0364. Reduce the area to 5 square units [Easy]
0365. The farmer and his 4 sons and windmills [Medium]
0366. Digital Clock: 6H27 am [Medium]
[0367. Mathematical: 3 minus 2 is not 9 {Medium]](
0368. Move 4 to create 6 squares [Medium]
0369. Reduce area to 5/9 of original [Hard]
0370. Remove 4 to create 5 squares [Medium]
0371. Mathematical: 7 plus 1 is not 0 [Hard]
0372. Create 4 triangles [Hard]
0373. Move 4 to create 6 squares [Medium]
0374. Pentomino shapes in a row [Easy]
0375. Cocktail to house [Medium]
0376. Missing triangles in a row [Extreme]
0377. Triangle 3X3: Remove 6 leaving 3 rhombuses [Easy]
0378. Garden spade and fork [Medium]
0379. Rhombus and square [Medium]
0380. Big and small square [Easy]
0381. Move 6 to drive a Mitshubishi [Easy]
0382. Divide into 6 shapes [Hard]
0383. Roman Numerals: 6 minus 4 is not 9 [Easy]
0384. Move 2 to create 6 triangles [Medium]
0385. Matchstick somersault [Medium]
0386. Reduce the area to 6 square units [Easy]
0387. Series of matchsticks [Extreme]
0388. Snake to 2 squares [Medium]
0389. The 5 smaller farms [Medium]
0390 Transparent 98 [Hard]
0391. Square 3X3: Move 5 to create 4 squares [Hard]
0392. Vertical Symmetry [Hard]
0393. Rectangle 4X2: Remove 4 to create 5 squares [Easy]
0394. Mathematical: 5 minus 2 is not 8 [Medium]
0395. Triangle 4X4: Move 4 to create 2 triangles [Easy]
0396. The cross [Medium]
0397. Route map: 4 squares [Easy]
0398. Two glasses [Medium]
0399. Remove 3 to create 7 equal squares [Medium]
0400. Add 4 to create 18 squares [Hard]
0401. Create 2 squares [Medium]
0402. Reduce from 5 to 4 square matchsticks [Hard]
0403. House from a different angle [Easy]
0404. Divide into 2 equal pieces [Extreme]
0405. Move 3 to create 5 squares [Medium]
0406. The hidden square [Medium]
0407. Rectangle 4X2: Remove 5 to create 4 squares [Easy]
0408. Move 3 to create 2 squares [Medium]
0409. Mathematical: 7 minus 6 is not 4 [Hard]
0410. Square 3X3: Remove 8 to form 3 squares [Medium]
0411. Triangle 3X3: Reduce size to one third [Medium]
0412. Reduce from 4 to 3 square matchsticks [Extreme]
0413. Break the ladder [Easy]
0414. Add 4 to create 10 squares [Medium]
0415. Triangle 3X3: Remove 7 leaving 4 triangles [Easy]
0416. The 100 metres [Hard]
0417. Double the area [Easy]
0418. Two small and one big triangle [Medium]
0419. Move 2 to create 3 triangles [Medium]
0420. Mathematical: 7 minus 9 is not 3 [Easy]
0421. Move 3 to create 6 squares [Medium]
0422. Square 3X3: Remove 4 leaving 6 squares [Easy]
0423. Mathematical: 6 minus 9 is not 6 [Medium]
0424. Reduce from 4 to 2 square matchsticks [Extreme]
0425. Convert the snake into 3 squares [Hard]
0426. Convert the snake into 2 squares [Hard]
0427. Form 8 rows of 5 matchsticks [Hard]
0428. Mathematical: 1 minus 1 is not 3 [Hard]
0429. Rectangle 4X2: Take away 5 to create 4 squares [Easy]
0430. Move 3 to create 5 squares [Easy]
0431. Dominoes in a rectangle [Medium]
0432. Triangle 3X3: Take away 2 leaving 8 triangles [Easy]
0433. Mathematical: 8 plus 9 is not 4 [Medium]
0434. Matchstick heads only [Easy]
0435. Move 4 to create 4 unequal squares [Medium]
0436. Bridge over river [Hard]
0437. Triangle 3X3: Remove 9 leaving 2 unequal triangles [Medium]
0438. Move 2 to create 6 squares [Easy]
0439. Move 4 to create 2 equal shapes [Medium]
0440. Square 3X3: Take away 8 to leave 4 squares [Easy]
0441. Bridge over water [Hard]
0442. Empty portrait frames [Medium]
0443. Find the 3 whole numbers [Medium]
0444. Mathematical: 3 plus 3 is not 8 [Easy]
0445. House to squares [Hard]
0446. Next square in the row [Hard]
0447. Add 2 to create 5 triangles [Easy]
0448. Mathematical: 3 plus 7 is not 8 [Medium]
0449. Move 4 and create 11 triangles [Medium]
0450. Remove 6 to leave 6 squares [Easy]
0451. Create 3 equal squares[Medium]
0452. Divide into 2 equal shapes [Hard]
0453. All the arrows [Easy]
0454. Square 3X3: Take away 4 leaving 9 squares [Medium]
0455. Skiing out of control [Easy]
0456. Mathematical: 4 minus 4 is not 1 [Hard]
0457. Triangle 3X3: Remove 4 leaving 4 triangles [Easy]
0458. Move 2 to create 6 squares [Easy]
0459. The Horse [Easy]
0460. Spokes to triangles [Easy]
0461. Create 4 equal shapes [Medium]
0462. Move 6 to create 4 squares [Medium]
0463. Change to equal ratio [Medium]
0464. Mathematical: 4 plus 3 is not 6 [Easy]
0465. Remove 12 to form 3 squares [Easy]
0466. From 0 to 7 triangles [Hard]
0467. Building with blocks [Medium]
0468. Move 4 to create 2 squares [Medium]
0469. Broken matchstick migration [Medium]
0470. Roman numerals: 5 minus 4 is not 4 [Hard]
0471. Mathematical: 6 minus 4 is not 7 [Easy]
0472. Rectangle 3X2: Move 3 to create 5 squares [Medium]
0473. The matching pair [Easy]
0474. Polygon with 7 sides [Hard]
0475. Rectangle 4X2: Take away 7 [Easy]
0476. Increase the crossings [Medium]
0477. Ten Triangles [Hard]
0478. Divide into 3 shapes [Medium]
0479. Reduce the area to 8 [Medium]
0480. Triangle 3X3: Remove 3 to leave 9 [Easy]
0481. Move 4 to create 5 triangles [Medium]
0482. Mathematical: 7 minus 0 is not 6 [Hard]
0483. Odd one out [Extreme]
0484. The 12 matchstick challenge [Hard]
0485. Rectangle 4X2: Take away 9 to leave 4 squares [Easy]
0486. Divide into 2 equal parts [Medium]
0487. The next number [Extreme]
0488. Four equal squares [Easy]
0489. Mathematical: 1 plus 3 is not 2 [Medium]
0490. Move 3 to form 9 squares [Medium]
0491. Triangle 3X3: Remove 6 leaving 3 triangles [Medium]
0492. Pathway with 7 matchsticks [Medium]
0493. Divide the grid into the ratio 1:4:4 [Easy]
0494. Move 3 to create 14 squares [Medium]
0495. Rectangle 4X2: Remove 8 leaving 4 squares [Easy]
0496. Triangle 4X4: Reduce to one quarter [Hard]
0497. Mathematical: 1 plus 4 is not 6 [Medium]
0498. Link between rows [Extreme]
0499. Rectangle 3X2: Move 3 to create 5 squares [Medium]
0500. Matchstick Grid: Count the heads [Medium]
0501. Remove 1 leaving 9 squares [Easy]
0502. Pathway with 19 matchsticks [Hard]
0503. Relationship of 4 figures [Hard]
0504. Move 2 to create 9 squares [Easy]
0505. Mathematical: 8 minus 5 is not 7 [Easy]
0506. Triangle 3X3: Take away 6 [Easy]
0507. Divide square into 4 equal parts [Hard]
0508. Move 2 to create 6 squares and 4 rectangles [Hard]
0509. Mathematical: 3 minus 0 is not 5 [Easy]
0510. Take away 4 to leave 5 squares [Easy]
0511. Only 1 head in each node [Medium]
0512. The 7 small triangles [Hard]
0513. Rectangle 4X3: Remove 4 to create 8 squares [Medium]
0514. Divide square into 4 equal parts [Hard]
0515. Mathematical: 4 minus 7 is not 6 [Medium]
0516. Triangle 3X3: Take away 3 to leave 6 triangles [Easy]
0517. Increase the inner triangles [Medium]
0518. Rectangle 4X3: Remove 3 to leave 9 squares [Medium]
0519. Move 1 to create 2 squares [Easy]
0520. Equal to two [Easy]
0521. Three numerical rows [Hard]
0522. Mathematical: 3 minus 6 is not 1 [Medium]
0523. Balance the scale: Add 1 [Medium]
0524. Invalid Roman numeral [Medium]
0525. Move 3 to create 9 squares [Medium]
0526. Move to create 2 squares [Easy]
0527. Windmill to 8 triangles [Easy]
0528. Mathematical: 8 minus 2 is not 4 [Hard]
0529. Take away 4 to leave 7 squares [Easy]
0530. Square 3X3: Remove 6 leaving 5 squares [Medium]
0531. Move 4 to create 6 squares [Medium]
0532. The 3 square matchsticks [Hard]
0533. Reversed Tic-Tac-Toe [Easy]
0534. Count the triangles [Easy]
0535. Smallest number [Medium]
0536. Seven sections [Medium]
0537. Reduce the area to 8 [Medium]
0538. Mathematical: 5 plus 1 is not 2 [Hard]
0539. Rectangle 4X3: Remove 4 to leave 10 squares [Easy]
0540. Two triangles [Easy]
0541. The TOWN named OE [Medium]
0542. One third [Hard]
0543. Four to two triangles [Hard]
0544. Move 3 to form 2 squares [Easy]
0545. Divide into two identical pieces [Medium]
0546. Change to two [Easy]
0547. Pathway with 14 matchsticks [Hard]
0548. Mathematical: 9 minus 0 is not 1 [Easy]
0549. Divide the square into 2 parts [Extreme]
0550. Move 3 to form 4 squares [Hard]
0551. Route map: 1- and 2-way roads [Medium]
0552. Change the ratios [Easy]
0553. Mathematical: 5 minus 9 is not 7 [Medium]
0554. Count the 4-sided figures [Medium]
0555. Change the 3:4:5 ratio to 1:3:8 [Medium]
0556. Letters and Numbers [Hard]
0557. Turn the direction [Medium]
0558. Move 1 and still have 5 squares [Easy]
0559. Divide into 4 areas [Medium]
0560. Find the next number [Hard]
0561. Pathway with 7 matchsticks revisited [Medium]
0562. Mathematical: 5 minus 1 is not 1 [Easy]
0563. Square 3X3: Take away 8 [Easy]
0564. Triangle 3X3: Remove 6 [Easy]
0565. Move 2 to form 2 triangles [Easy]
0566. Matchstick Grid: Diagonal heads [Medium]
0567. The wood log problem [Extreme]
0568. Change to numerical equation [Hard]
0569. Divide into 4 areas [Hard]
0570. The biggest number [Easy]
0571. Change the 2:3:4 ratio to 1:1:1 [Easy]
0572. Mathematical: 4 minus 7 is not 8 [Medium]
0573. The third grader puzzle [Easy]
0574. Triangle 3X3: Remove 4 leaving 6 triangles [Easy]
0575. Three matchstick node notations [Hard]
0576. Divide into 4 areas [Hard]
0577. One square [Easy]
0578. Remove 3 leaving 3 triangles [Easy]
0579. Mathematical: 6 plus 0 is not 7 [Medium]
0580. Divide the rectangle in ratio 3:4:5 [Medium]
0581. Remove 6 leaving 3 squares [Easy]
0582. Mathematical: 4 plus 5 is not 5 [Hard]
0583. Divide the square [Hard]
0584. The four rectangles [Hard]
0585. Four-sided figures [Medium]
0586. Little square and rectangle [Extreme]
0587. Mathematical: 0 minus 5 is not 1 [Easy]
0588. Route Map: Change one direction [Medium]
0589. Missing number [Extreme]
0590. Balance the scale: Move 1 [Hard]
0591. Mathematical: 8 plus 7 is not 5 [Medium]
0592. Increase the inner triangles [Medium]
0593. Divide into 3 shapes [Extreme]
0594. Mathematical: 8 minus 3 is not 3 [Easy]
0595. Polygon with 8 sides [Hard]
0596. M for Matchstick [Medium]
0597. Not Roman numerals [Easy]
0598. Three squares and four rectangles [Medium]
[0599. Mathematical: 3 minus 9 is not 9 Hard}
0600. The clock and the number [Hard]
0601. Rectangle 3X2: Remove 6 [Easy]
0602. Divide into 3 equal parts [Medium]
0603. Move 3 to create 2 rhombuses [Medium]
0604. Mathematical: 5 minus 4 is not 3 [Easy]
0605. Five matchsticks in a row [Medium]
0606. Nodes: 4 matchsticks [Medium]
0607. Mathematical: 4 - 0 is not 7 [Medium]
0608. Spot the twins [Easy]
0609. Five to two squares [Medium]
0610. Mathematical: 9 minus 6 is not 4 [Hard]
0611. Heads and squares [Hard]
0612. Ship to Triangles [Medium]
0613. Five to six squares [Medium]
0614. Divide the area [Easy]
0615. Mathematical: 2 minus 2 is not 5 [Medium]
0616. Five to three squares [Medium]
0617. Change the matchstick heads [Medium]
0618. Mathematical: 4 minus 0 is not 5 [Medium]
0619. Mathematical: 7 minus 2 is not 8 [Easy]
0620. Divide into two parts [Medium]
0621. Divide into two parts [Easy]
0622. The uncompleted square [Medium]
0623. Four to three equal triangles [Easy]
0624. One minus one is not one [Medium]
0625. Take away two [Easy]
0626. Create six equal squares [Hard]
0627. The upside down CAT [Medium]
0628. The kneeling man [Easy]
0629. Remove four to leave no triangles [Medium]
0630. Triangle reduced to 5 square units [Easy]
0631. Mathematical: 4 plus 0 is not 0 [Hard]
0632. Move two to create two squares [Easy]
0633. Move three to create three [Hard]
0634. Triangle reduced to 4 square units [Hard]
0635. Two is not equal to six [Medium]
0636. Move two to create two squares [Easy]
0637. The sand timer [Hard]
0638. Create 5 triangles [Medium]
0639. From three to two squares [Easy]
0640. Big and small squares [Medium]
0641. Three different squares [Hard]
[0642. Stacked triangles Medium}
0643. Cup and saucer [Medium]
0644. Polygon with 7 sides [Hard]
0645. Fix the equation [Extreme]
0646. Cross to squares [Hard]
0647. Rugby [Medium]
0648. Divide into 2 parts [Easy]
0649. Three-way bridge [Hard]
0650. Three squares different in size [Medium]
0651. Three square units [Medium]
0652. Three square units extreme [Extreme]
0653. The farmer and his 4 sons [Medium]
0654. Increase the inner triangles [Hard]
0655. Four to six squares [Easy]
0656. Three right angles [Medium]
0657. Change the 2:3:4 ratio to 2:2:5 [Easy]
0658. Mathematical: 0 minus 3 is not 7 [Medium]
0659. Triangle 3X3: Remove 6 leaving 0 triangles [Easy]
0660. Six to five equal triangles [Medium]
0661. Reduce from 4 to 3 square matchsticks [Hard]
0662. Mathematical: 5 minus 3 is not 4 [Medium]
Awesome Inc. theme. Powered by Blogger.
Original text
Rate this translation
Your feedback will be used to help improve Google Translate |
189084 | https://crypto.stackexchange.com/questions/113250/finding-lagrange-interpolation-polynomial-modulo-p | secret sharing - Finding (lagrange) interpolation polynomial modulo p - Cryptography Stack Exchange
Join Cryptography
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…
1.
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
2.
3.
current community
Cryptography helpchat
Cryptography Meta
your communities
Sign up or log in to customize your list.
more stack exchange communities
company blog
5. Log in
6. Sign up
1. 1. Home
2. Questions
3. Unanswered
4. AI Assist Labs
5. Tags
7. Chat
8. Users
10. Companies
2. 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
Finding (lagrange) interpolation polynomial modulo p
Ask Question
Asked 10 months ago
Modified10 months ago
Viewed 113 times
This question shows research effort; it is useful and clear
-1
Save this question.
Show activity on this post.
I'm trying to find an interpolation polynomial degree t t less than the number of n n points for modulo prime p p. Additionally, the results of this polynomial should be in small intervals (i.e. in [0,3][0,3]). I will give an example from Shamir Secret Sharing. Let secret is 23 23, I have 6 6 points (p(1)=2,p(2)=0,p(3)=1,p(4)=3,p(5)=0,p(6)=1 p(1)=2,p(2)=0,p(3)=1,p(4)=3,p(5)=0,p(6)=1). How can I find a a and b b from polynomial p(x)=a x 2+b x+23 p(x)=a x 2+b x+23?
secret-sharing
polynomial
Share
Share a link to this question
Copy linkCC BY-SA 4.0
Improve this question
Follow
Follow this question to receive notifications
edited Oct 18, 2024 at 12:19
Mahesh S R
1,796 1 1 gold badge 5 5 silver badges 22 22 bronze badges
asked Oct 16, 2024 at 21:09
SemiramisSemiramis
31 4 4 bronze badges
2
You are overloading the meaning of p and p p. p is a prime number while p p denotes a polynomial. Please edit your question to clarify what you mean. –Dilip Sarwate Commented Oct 17, 2024 at 13:40
Your data are inconsistent. A quadratic polynomial with zeroes at 2 2 and 5 5 _must_ satisfy symmetry requirements p(3)=p(4)p(3)=p(4) and p(1)=p(6)p(1)=p(6) which what you have written down fails to do. –Dilip Sarwate Commented Oct 17, 2024 at 16:25
Add a comment|
2 Answers 2
Sorted by: Reset to default
This answer is useful
1
Save this answer.
Show activity on this post.
p(2)=p(5)=0 p(2)=p(5)=0 implies that the OP's p(x)p(x) is of the form
c(x)(x−2)(x−5)=c(x)(x 2−7 x+10).c(x)(x−2)(x−5)=c(x)(x 2−7 x+10).
Since the OP insists that p(x)p(x) is a quadratic polynomial, it must be that c(x)c(x) is a constant, say α α. So,
a x 2+b(x)+23=α(x 2−7 x+10)a x 2+b(x)+23=α(x 2−7 x+10)
showing 10 α=23 10 α=23, i.e.,
α=((10−1 mod q)×23)mod q α=((10−1 mod q)×23)mod q
where q q is the size of the prime field that is being used. Note that Shamir secret-sharing requires that q>23 q>23. Once the value of α α has been determined, it can be used to find a a and b b.
Note: While the p(x)p(x) thus found satisfies p(2)=p(5)=0 p(2)=p(5)=0, it does not satisfy the other constraints (p(1)=2 p(1)=2, etc) which appear to be written down at random. A quadratic polynomial with zeroes at 2 2 and 5 5 _must_ satisfy symmetry requirements p(3)=p(4)p(3)=p(4) and p(1)=p(6)p(1)=p(6), and the data that the OP has chosen to provide does not meet these requirements.
Share
Share a link to this answer
Copy linkCC BY-SA 4.0
Improve this answer
Follow
Follow this answer to receive notifications
edited Oct 17, 2024 at 16:20
answered Oct 17, 2024 at 14:10
Dilip SarwateDilip Sarwate
2,801 18 18 silver badges 25 25 bronze badges
Add a comment|
This answer is useful
1
Save this answer.
Show activity on this post.
In general you cannot. A polynomial of degree d d requires d+1 d+1 equations to uniquely determine it. On the flip side, unless there are dependencies in the k k values that are given (i.e., they lie in a subspace) the polynomial interpolating at those points will usually be degree k−1.k−1.
An example of this would be if your values were
p(i)=i 2+i+2(mod 29),i=1,…,6 p(i)=i 2+i+2(mod 29),i=1,…,6
for example, i.e., they happened to be the values of a degree 2 polynomial, then Lagrange interpolation would recover this polynomial.
_Note:_ For crypto applications points and values are randomly chosen for security and in general result in a full degree interpolating polynomial.
Share
Share a link to this answer
Copy linkCC BY-SA 4.0
Improve this answer
Follow
Follow this answer to receive notifications
edited Oct 17, 2024 at 21:31
answered Oct 16, 2024 at 21:46
kodlukodlu
25.3k 2 2 gold badges 30 30 silver badges 64 64 bronze badges
1
1 Since 23 23 is the secret, it should be smaller that the field size, no? –Dilip Sarwate Commented Oct 17, 2024 at 13:43
Add a comment|
Your Answer
Thanks for contributing an answer to Cryptography Stack Exchange!
Please be sure to _answer the question_. Provide details and share your research!
But _avoid_ …
Asking for help, clarification, or responding to other answers.
Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Draft saved
Draft discarded
Sign up or log in
Sign up using Google
Sign up using Email and Password
Submit
Post as a guest
Name
Email
Required, but never shown
Post Your Answer Discard
By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.
Start asking to get answers
Find the answer to your question by asking.
Ask question
Explore related questions
secret-sharing
polynomial
See similar questions with these tags.
The Overflow Blog
Robots in the skies (and they use Transformer models)
Research roadmap update, August 2025
Report this ad
Related
16Necessity for finite field arithmetic and the prime number p in Shamir's Secret Sharing Scheme
1Lagrange Interpolation for finite field GF(2^8), for Secret Reconstruction
8Does Runge phenomenon affect Shamir's secret sharing scheme?
4Shamir Interpolation Problem (Recovering secret with fewer parts than the threshold)
6Simplified Lagrange interpolation formula inside Shamir's secret sharing scheme
4What is the reason for Shamir scheme to use modulo prime?
1Gap in Shamir's Secret Sharing Scheme security proof
1Is there supposed to be a unique polynomial when given t shares for (t,n) Shamir Secret Sharing scheme?
Hot Network Questions
How can I get my security door to stay shut in place as to not obstruct the deadbolt?
Why do these two lines have the same probability of intersecting the circle?
Confusion about infinity in gravitational potential energy (GPE)
Excessive flame on one burner
How are the crosswords made for mass-produced puzzle books?
Do strong winds also cause a hovering creature to fall?
Why didn't Alexander sail his Army across Gulf instead of Gedrodsian desert?
how should one interpret a cert who's Issuer is different from the DirName of the X509v3 Authority Key Identifier extension?
Splitting a long introduction into collapsible elements easier to read and translate
What is the principle of non-liability in the European Union wrt the US Section 230 and social media?
How to balance research and teaching responsibilities?
Simple arithmetic goes wrong?
"can" or "could"... which one should I use in this phrase?
Should 1 Peter 1:5 read ‘a’ or ‘the’ before ‘salvation’, or should it just read ‘salvation’?
Why wrap fruitcake in grease-proof paper?
Why does Wittgenstein use long, step-by-step chains of reasoning in his works?
Need help understanding the value of this infinite continued fraction intuitively
Which passport to use at immigration counter
Fewest cages for unique Killer Kropki Sudoku
Using adverbs of frequency in the Future Simple and Present Continuous tenses
Why does bible claim Asa was perfect all his days when he didn't rely on God in the latter years
Will NASA send Juno to investigate 3i/Atlas?
How to get code of a program that decrypts itself?
Are trills only for high-pitched notes?
Question feed
Subscribe to RSS
Question feed
To subscribe to this RSS feed, copy and paste this URL into your RSS reader.
Why are you flagging this comment?
It contains harassment, bigotry or abuse.
This comment attacks a person or group. Learn more in our Code of Conduct.
It's unfriendly or unkind.
This comment is rude or condescending. Learn more in our Code of Conduct.
Not needed.
This comment is not relevant to the post.
Enter at least 6 characters
Something else.
A problem not listed above. Try to be as specific as possible.
Enter at least 6 characters
Flag comment Cancel
You have 0 flags left today
Cryptography
Tour
Help
Chat
Contact
Feedback
Company
Stack Overflow
Teams
Advertising
Talent
About
Press
Legal
Privacy Policy
Terms of Service
Your Privacy Choices
Cookie Policy
Stack Exchange Network
Technology
Culture & recreation
Life & arts
Science
Professional
Business
API
Data
Blog
Facebook
Twitter
LinkedIn
Instagram
Site design / logo © 2025 Stack Exchange Inc; user contributions licensed under CC BY-SA. rev 2025.8.21.33163
By clicking “Accept all cookies”, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy.
Accept all cookies Necessary cookies only
Customize settings
Cookie Consent Preference Center
When you visit any of our websites, 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 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 manage your preferences. Please note, blocking some types of cookies may impact your experience of the site and the services we are able to offer.
Cookie Policy
Accept all cookies
Manage Consent Preferences
Strictly Necessary Cookies
Always Active
These cookies are necessary for the website to function and cannot be switched off in our systems. They are usually only set in response to actions made by you which amount to a request for services, such as setting your privacy preferences, logging in or filling in forms. You can set your browser to block or alert you about these cookies, but some parts of the site will not then work. These cookies do not store any personally identifiable information.
Cookies Details
Performance Cookies
[x] Performance Cookies
These cookies allow us to count visits and traffic sources so we can measure and improve the performance of our site. They help us to know which pages are the most and least popular and see how visitors move around the site. All information these cookies collect is aggregated and therefore anonymous. If you do not allow these cookies we will not know when you have visited our site, and will not be able to monitor its performance.
Cookies Details
Functional Cookies
[x] Functional Cookies
These cookies enable the website to provide enhanced functionality and personalisation. They may be set by us or by third party providers whose services we have added to our pages. If you do not allow these cookies then some or all of these services may not function properly.
Cookies Details
Targeting Cookies
[x] Targeting Cookies
These cookies are used to make advertising messages more relevant to you and may be set through our site by us or by our advertising partners. They may be used to build a profile of your interests and show you relevant advertising on our site or on other sites. They do not store directly personal information, but are based on uniquely identifying your browser and internet device.
Cookies Details
Cookie List
Clear
- [x] checkbox label label
Apply Cancel
Consent Leg.Interest
- [x] checkbox label label
- [x] checkbox label label
- [x] checkbox label label
Necessary cookies only Confirm my choices |
189085 | https://guidesly.com/fishing/fish-species/oyster-toadfish | Latest
Blog
Fish Species
Places
Guide Articles
Search
Oyster Toadfish
Species Details
Opsanus Tau
Batrachoididae
Batrachoidiformes
Wrecks, reefs
3 - 5 lbs.
12" - 16"
Oyster Toadfish (Opsanus tau) Fish Description
The Oyster Toadfish are also commonly known as the ugly toad, oyster cracker, oyster catcher, and bar dog. They appear principally yellowish with an olive-brown back along with a pattern of brown blotches or sloping lines, and a pale belly.
They are scaleless and have a compressed, flattened body that can grow 12 inches lengthwise. Oyster Toadfish have “whiskers” or paunchy flaps on their cheeks and jaws. On top of their large, flat head are bulging large eyes accompanied by a broad mouth filled with strong, rounded teeth. They very much resemble a toad, as they are named.
But careful, Oyster Toadfish has a venomous spine on their first dorsal fin. Venom from the Oyster Toadfish has been compared to the pain levels of a bee or wasp sting.
Oyster Toadfish Diet and Size
Oyster Toadfish are omnivores. Their common prey includes the following: crustaceans, mollusks, amphipods, squid, and other smaller fish.
The length of the Oyster toadfish is about 17 inches and rarely do they grow longer than 15 inches.
Interesting Facts About The Oyster Toadfish
The name of the Oyster Toadfish is from their likeness to toads, with their scaleless brown skin covered with thick mucus and sometimes warts.
In comparison to any Chesapeake Bay fish, the Oyster Toadfish lays the largest eggs in the area.
Oyster toadfish can live in rather poor conditions, being able to tolerate litter and polluted water, and can survive out of water for a lengthy period.
Oyster toadfish are edible! But they are rarely eaten because of their off putting appearance.
NASA, back in 1998, has sent the oyster toadfish to space in the investigation of the effects of microgravity on the development of otolithic organs.
Oyster Toadfish Fishing Techniques
When fishing for Oyster Toadfish, they can and will take on an angler's bait rather quickly. Baits like pieces of squid, clams, bloodworms, and peeler crabs would be good to lure them in. Though most anglers despised the toadfish, as they are often caught accidentally, you mustn’t throw caution out in the wind when out fishing for this one, because the Oyster Toadfish has powerful, snapping jaws and sharp, venomous spines on its dorsal fin. Not quite the gamefish you’d expect, are they?
Oyster Toadfish Habitat and Distribution
Bottom-dwelling fish, similar species such as the Oyster Toadfish are found around oyster reefs, but also inhabit areas among wreck, debris, rocks, and other dark secluded places in the shallow waters during the warmer months, but deeper in the channels in the winter.
These types of fish are in abundance throughout the year by the Chesapeake Bay; They are frequently found in the middle and lower Bay areas.
Oyster Toadfish can be also be located near the shore from Maine to Florida.
Also Known As:
Ugly toad, oyster cracker, oyster catcher, bar dog
Loading...
Guides
Guidesly Pro AppGuidesly PlusBecome a Guide
Access Your Guide Admin
FAQs
Anglers
Book a GuideGuidesly AppWhat is GuideslyAll GuidesAll Destinations
Sign Up
Articles
LatestBlogPlacesGuide ArticlesNymphingFish SpeciesUS FishingTechniques
Corporate |
189086 | https://www.youtube.com/watch?v=FgvuD4yhK_M | Understanding Function Behavior with Derivatives: Increasing, Decreasing, or Neither
Concepts and Solutions by Sabih Siddiqui
443 subscribers
9 likes
Description
107 views
Posted: 15 Apr 2024
In this detailed tutorial, we'll unravel the concept of determining whether a function is increasing, decreasing, or neither using its derivative. 📈📉 Many students find this topic challenging, but fear not! We'll break it down step-by-step to make it crystal clear. Learn how to analyze the behavior of a function by examining the sign of its derivative. Discover the critical points and intervals of increase or decrease. We'll work through examples and provide intuitive explanations to help you grasp this fundamental concept in calculus.
Whether you're preparing for exams or seeking a deeper understanding of calculus, this video will equip you with the skills to confidently analyze function behavior. Subscribe now and conquer the mysteries of function derivatives!
FunctionBehavior #Derivatives #IncreasingFunction #DecreasingFunction #CalculusTutorial #MathTutorials #StudentSuccess #MathematicsEducation #functions #CAIE #ASLevel #ALevel #Pure #Mathematics #exam
Transcript:
so the concept that we'll be looking at for functions is how to determine that whether a function is an increasing function or a decreasing function or okay what is the function function we have an input and see similarly we have an output right function for example let's take the common ones that are in your the function for example yals to very basic quadratic function x² say now if I have a look at this function for for these values what do you see as I'm going down not forget about going down because when we are studying function we talk about the X values being our input right so I am going write what is happening to the values of X they are what is happening to the values of x x is the first one - 2.09 - 2.06 - 2.0 the values of X I'm going from left to right what happens to the values of X they are increasing but as the values of X are increasing the value of y is decreasing so when I'm increasing the input the output is decreasing for any of these portions up to this point if I find the gradient at any of these points will the gradient be positive or negative negative gradient so the function is decreasing for this domain when X is less than zero what about when X is greater than Z now if go from left to right what happens to the values of X the values of X increase I'm increasing my input my output is what is happening to the output the output is also increasing input output so the function is increasing for this domain for the domain X less than zero the function is decreasing for the domain X is greater than zero the function is decreasing but if you generally ask me function increasing decre say neither because function single is this clear or not we have to decide and we do this through the gradient how do we find the gradient of a [Music] function how do we find the gradient of a function the derivative of x the gradient right so if that is positive I'm sure that the derivative is positive then I call the function increasing and what about decreasing call the gradient is negative f- this now let's have a look at a few of the P questions now what is the question asking the function f is defined by this page number one in anybody Book C FX is 1 over 3x + 2 + x² for X is less than -1 determine whether f is increasing function and decreasing function or neither so what will we have to do first find the derivative or derivative f of x is1 DX + 22 f of x is - e upon x + 2 that right now what can I say about this derivative what is the domain the question that provid us with the domain it's telling us that X is less than netive one okay X input or X is being input over here and over here is that right if I input X over here no matter what values of X I input over here once this squares it will become positive so can you say for all values of X this denominator will be positive and what about the entire fraction negative is that right minus 3 if you divide a positive value the entire fraction will be negative and what about this part if I inut the values of X less than1 then what happens to plus 2X be negative so part be negative this is negative for all values of X this + 2x is negative for the given domain is that right so I know part be negative be negative the two negatives will add up but the final answer will still be the two negatives will add up but the final answer will still be negative - 3 -2 example to Minus 5 is that right so can I say that for the given domain F der of X is less than zero so the function is what can I say about the function decreasing function is this clear let's have a look at another question this concept is the function f is defined by FX is 1 / 3 2x - 1^ 3 / 2 - 2x4 X is between half and a it is given that f is a f is a decreasing function find the maximum possible value of the constant a so the if the question is telling me that it is the decreasing function that what can then what can I say for the function just derivative is negative derivative is always negative right so first I need to find the derivative how do we find the derivative for this one 3 mtip 1 3 2x - 1 power 2 - 2 mtip then f of x is 3 three cancel and I'm left 2x - 1 to the^ halfus is that right right now what do I know that F derivative of x is less than 0 so I can say 2x - 1^ half - 2 is less than zero negative is that right because the question is telling us that it is a decreasing function half is less than two can I Square both sides now so 2x - 1 is less than 4 2x is less than 5 x is less than 5 so X has to be less than 5/ the question already told us that it is greater than half why the question told us that it is greater than under Ro under Ro this value has to be positive this value cannot be negative if I want to take the under root under root positive has to be greater than one half sorry X has to be greater than half and we found out that X has to be less than 5/2 so the maximum value of a is this multiply what can I say the maximum value is where makes sense let's have a look at another sech question booklet page number 209 209 this one function f is defined by this determine whether f is an increasing function a decreasing function or nice and what is the domain that the question is provided X is an element of real number X is any real number that means from negative Infinity to positive Infinity can take any particular value so first we find the Dera what will be the derivative to the power then plus now part one may let me just clear this part one as well because it is relatable to now okay let's have a look at part A this was part here was a hint as to what we need to do Express this and this completed Square form how do we do this five common here Y 2 - 6 y + 50 y - + 3² + 50 y - [Music] 3² 5 braet y - 3² - 45 is that right now for this one we found the derivative now how will we figure out that this derivative positive negative how do we do this now your this function and this function what is the difference Y what you what do you have in place of Y x square is this right or wrong now what can you tell me about the derivative or X is an element of real what about once it comes over here Square okay all values of X will become positive right and what is the domain x infinity to positive Infinity minus 3 the number can either be positive or negative but once I Square it positive right so you Square value will become positive is that right value positive or positive here finally the derivative is always positive for the given domain or for all values the domain is all values of X so I can say that x² Square x² - 3 sare greater than equal to Z is that right f of x greater than or equal to 5 derivative is always five or greater than five that means the gradient is five or greater than five that means the gradient is always positive so the function is a is inreasing there make sure you write it down L it let's do just one more part this is page number 303 from the new book the function this is defined by X is 2 - 3 / 4x - 3 and the domain is X is greater than P / 4 find F derivative of x and hence determine whether f is an increasing function a decreasing function or neither how do we find the derivative of f first let's just reite 4x - P ^1 then Dera is 0 - 3 -1 it becomes + 3 4 4 x - 2 yes now squares are never negative so the denominator will always be positive the denominator is positive and the numerator is 12 the final answer is bound to be positive so the function is can we surely say that no matter what domain we have right so what can I say that greater than Z let's see for the given domain yeah all Val so function is is this clear cannot claim the function to be increasing or decreasing I should be extremely sure about the function being increasing or decreasing |
189087 | https://www.sciencedirect.com/science/article/abs/pii/S0096300312000057 | Existence and uniqueness of solutions of functional equations arising in dynamic programming - ScienceDirect
Typesetting math: 100%
Skip to main contentSkip to article
Journals & Books
Access throughyour organization
Purchase PDF
Search ScienceDirect
Article preview
Abstract
Introduction
Section snippets
References (9)
Cited by (6)
Applied Mathematics and Computation
Volume 218, Issue 13, 1 March 2012, Pages 7221-7230
Existence and uniqueness of solutions of functional equations arising in dynamic programming
Author links open overlay panel H.K.Pathak, Deepmala
Show more
Add to Mendeley
Share
Cite
rights and content
Abstract
In this paper, we study solvability of two functional equations arising in dynamic programming of multistage decision processes. By using Boyd and Wong fixed point theorem, some existence and uniqueness theorems of solutions and iterative approximation for solving these class of functional equations are established. The results presented here extend, improve and unify the corresponding results due to Bellman, Bhakta and Mitra, Bhakta and Choudhary, Liu and Kang, Liu et al., Jiang et al. and others. We also discuss some illustrative examples to highlight the realized improvements.
Introduction
It is well known that the existence problems of solutions of various functional equations arising in dynamic programming are both theoretical and practical interest. Bellman initiated and studied the existence of solutions for some class of functional equations arising in dynamic programming. The basic form of functional equation of dynamic programming given by Bellman and Lee is f(x)=max y∈D H(x,y,f(T(x,y))),x∈S.This paper deals with solvability of functional equations and system of functional equations arising in dynamic programming of multistage decision processes as follows:f(x)=opt y∈D opt{u(x,y),p i(x,y)+A i(x,y,f(a i(x,y))):i=1,2},∀x∈S,and f(x)=opt y∈D opt{u(x,y),p i(x,y)+q i(x,y)f(a i(x,y)):i=1,2,3},∀x∈S,where “opt” denotes the “sup” or “inf”, x and y stands for the state and decision vectors respectively, a i represents the transformation of the processes, f(x) denotes the optimal return function with initial state x.
It is clear that (1), (2) include many functional equations as special cases, respectively. For instance, the following functional equations f(x)=inf y∈D max{r(x,y),s(x,y),f(b(x,y))},∀x∈S,f(x)=inf y∈D max{r(x,y),f(b(x,y))},∀x∈S,f(x)=sup y∈D{A(x,y,f(a(x,y)))},∀x∈S,f(x)=sup y∈D{p(x,y)+A(x,y,f(a(x,y)))},∀x∈S,f(x)=sup y∈D{p(x,y)+f(a(x,y))},∀x∈S,f(x)=inf y∈D A(x,y,f),∀x∈S,f(x)=opt y∈D{p(x,y)+A(x,y,f(a(x,y)))},x∈S,f(x)=inf y∈D max{p(x,y),f(a(x,y)),q(x,y)+f(b(x,y))},∀x∈S,f(x)=opt y∈D opt{p(x,y),q(x,y)f(a(x,y)),r(x,y)f(b(x,y)),s(x,y)f(c(x,y))},∀x∈S,studied by Bellman , Bellman and Lee , Bhakta and Mitra , Bhakta and Choudhary , Liu et al. , Liu and Kang , Jiang et al. , and others , respectively.
In Section 2, we recall some basic concepts, notations and lemmas. In Section 3, we utilize the fixed point theorem due to Boyd and Wong to establish the existence, uniqueness and iterative approximation of solution for the functional equation (1) in Banach spaces BC(S) and B(S), respectively. In Section 4, we obtain the existence, uniqueness and iterative approximations of solutions for the functional equation (2) in the complete metric space BB(S). Several examples which dwell upon the importance of our results are also included. The results presented here generalize, improve and extend the results of Bellman , Bhakta and Mitra , Bhakta and Choudhury , Liu et al. , Liu and Kang , Jiang et al. and Liu .
Access through your organization
Check access to the full text by signing in through your organization.
Access through your organization
Section snippets
Preliminaries
Throughout this paper, we assume that R=(-∞,+∞),R+=[0,+∞) and R-=(-∞,0]. For any t∈R,[t] denotes the largest integer not exceeding t and (X,‖·‖) and (Y,‖·‖′) be real Banach spaces. S⊆X be the state space and D⊆Y be decision space. Define Φ 1={φ:φ:R+→R+is right continuous at t=0},Φ 2={φ:φ:R+→R+is non decreasing},Φ 3={φ:φ∈Φ 1 and φ(0)=0},Φ 4={φ:φ∈Φ 1∩Φ 2 and φ(t)0},Φ 5={φ:φ∈Φ 2 for φ(t)0},Φ 6={(φ,ψ):φ,ψ∈Φ 2,ψ(t)>0 and∑n=0∞ψ(φ n(t))<∞for t>0},Φ 7={(φ,ψ):φ,ψ∈Φ 2,ψ(t)>0 and lim n→∞ψ(φ n(t))=0 for t>0},B(S)={f:f:S→R
Properties of solutions for functional equation in BC(S) and B(S)
Now we prove the existence, uniqueness and iterative approximations of solutions for the functional equation (1) in BC(S) and B(S), respectively, by using Boyd and Wong fixed point theorem.
Theorem 3.1
Let S is compact, u,p i:S×D→R,a i:S×D→S and A i:S×D×R→R be mappings for i = 1, 2. Let φ∈Φ 3 and ψ∈Φ 4 such that
(C 1)u,p i and A i are bounded for i = 1, 2.
(C 2)for each x 0∈S,u(x,y)→u(x 0,y),p i(x,y)→p i(x 0,y)and a i(x,y)→a i(x 0,y)as x→x 0 uniformly for y∈D and i = 1, 2.
(C 3)max{|A i(x,y,z)-A i(x 0,y,z)|:i=1,2}⩽φ(‖x-x 0‖)for x,x 0∈S,y∈D,z∈R .
(C 4)max{|A i(
Properties of solutions for functional equation in BB(S)
Theorem 4.1
Let u,p i,q i:S×D→R and a i:S×D→S for i=1,2,3 , and satisfy the following conditions
(B 1)u and p i are bounded on B¯(0,k)×D for k⩾1 and i=1,2,3 .
(B 2)max{‖a i(x,y)‖:i=1,2,3}⩽‖x‖, for(x,y)∈S×D .
(B 3)there exists a constant α such that sup y∈D max{|q i(x,y)|:i=1,2,3}⩽α<1 ,
then the functional equation (2) possesses a unique solution w∈BB(S)and{H n h}n⩾1 converges to w for each h∈BB(S), where H is defined by Hh(x)=opt y∈D opt{u(x,y),p i(x,y)+q i(x,y)h(a i(x,y)):i=1,2,3},∀x∈S.
Proof
It follows from (B 1) and (B 2) that for each k⩾1 and h∈BB
Acknowledgments
The research of the second author is supported by the Department of Science and Technology, Government of India under INSPIRE Fellowship Program No. DST/INSPIRE Fellowship/2010/.
Recommended articles
References (9)
P.C. Bhakta et al.
Some existence theorems for functional equations arising in dynamic programming
J. Math. Anal. Appl. (1984)
P.C. Bhakta et al.
Some existence theorems for functional equations arising in dynamic programming II
J. Math. Anal. Appl. (1988)
Z. Liu et al.
Solutions to two functional equations arising in dynamic programming
J. Comput. Appl. Math. (2006)
Z. Liu
Existence theorems of solutions for certain classes of functional equations arising in dynamic programming
J. Math. Anal. Appl. (2001)
There are more references available in the full text version of this article.
Cited by (6)
Common fixed point theorem of six self-mappings in Menger spaces using (CLRST) property
2018, Open Mathematics
### Common fixed point theorems with an application in dynamic programming
2018, Miskolc Mathematical Notes
### Ultimate numerical bound estimation of chaotic dynamical finance model
2016, Springer Proceedings in Mathematics and Statistics
### Existence and uniqueness of solutions for certain functional equations and system of functional equations arising in dynamic programming
2016, Analele Stiintifice Ale Universitatii Ovidius Constanta Seria Matematica
### On solvability for certain functional equations arising in dynamic programming
2015, Springer Proceedings in Mathematics and Statistics
### Existence theorems for solvability of a functional equation arising in dynamic programming
2014, International Journal of Mathematics and Mathematical Sciences
View full text
Copyright © 2012 Elsevier Inc. All rights reserved.
Recommended articles
Sufficient conditions for the solvability of a Sylvester-like absolute value matrix equation
Applied Mathematics Letters, Volume 112, 2021, Article 106818 Behnam Hashemi
### An efficient Newton-type matrix splitting algorithm for solving generalized absolute value equations with application to ridge regression problems
Journal of Computational and Applied Mathematics, Volume 457, 2025, Article 116329 Xuehua Li, Cairong Chen
### Two efficient iteration methods for solving the absolute value equations
Applied Numerical Mathematics, Volume 208, Part B, 2025, pp. 148-159 Xiaohui Yu, Qingbiao 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. We may also use additional cookies to analyze, improve, and personalize our content and your digital experience. You can manage your cookie preferences using the “Cookie Settings” link. For more information, see ourCookie Policy
Cookie Settings Accept all cookies
Cookie Preference Center
We use cookies which are necessary to make our site work. We may also use additional cookies to analyse, improve and personalise our content and your digital experience. For more information, see our Cookie Policy and the list of Google Ad-Tech Vendors.
You may choose not to allow some types of cookies. However, blocking some types may impact your experience of our site and the services we are able to offer. See the different category headings below to find out more or change your settings.
You may also be able to exercise your privacy choices as described in our Privacy Policy
Allow all
Manage Consent Preferences
Strictly Necessary Cookies
Always active
These cookies are necessary for the website to function and cannot be switched off in our systems. They are usually only set in response to actions made by you which amount to a request for services, such as setting your privacy preferences, logging in or filling in forms. You can set your browser to block or alert you about these cookies, but some parts of the site will not then work.
Cookie Details List
Performance Cookies
[x] Performance Cookies
These cookies allow us to count visits and traffic sources so we can measure and improve the performance of our site. They help us to know which pages are the most and least popular and see how visitors move around the site.
Cookie Details List
Contextual Advertising Cookies
[x] Contextual Advertising Cookies
These cookies are used for properly showing banner advertisements on our site and associated functions such as limiting the number of times ads are shown to each user.
Cookie Details List
Cookie List
Clear
[x] checkbox label label
Apply Cancel
Consent Leg.Interest
[x] checkbox label label
[x] checkbox label label
[x] checkbox label label
Confirm my choices
×
Read strategically, not sequentially
ScienceDirect AI extracts key findings from full-text articles, helping you quickly assess the article's relevance to your research.
Unlock your access |
189088 | https://phys.org/news/2025-09-camouflage-caution-anti-predator-strategies.html | Camouflage or caution? How anti-predator strategies have evolved
Topics
Week's top
Latest news
Unread news
Subscribe
Science X Account
Sign In
Sign in with
Forget Password?
Not a member? Sign up
Learn more
Nanotechnology
Physics
Earth
Astronomy & Space
Chemistry
Biology
Other Sciences
Medicine
Technology
Discover more
Science
Sciences
science
scientific
Science studies courses
Health report subscription
Technology policy books
astronomy telescopes online purchase
Science magazine subscription
purchase ecological survey services
share this!
39
Tweet
Share
Email
Home
Biology
Evolution
Home
Biology
Ecology
September 25, 2025
The GIST
Camouflage or caution? How anti-predator strategies have evolved
by University of Melbourne
edited by Robert Egan
Robert Egan
associate editor
Meet our editorial team
Behind our editorial process
Editors' notes
This article has been reviewed according to Science X's editorial process and policies. Editors have highlighted the following attributes while ensuring the content's credibility:
fact-checked
peer-reviewed publication
trusted source
proofread
Credit: Pixabay/CC0 Public Domain
Predators and the environment determine why some animals use camouflage to avoid being eaten, while others use bright colors to warn them off, new research reveals. Published today in the journal Science, the findings help explain the evolution and global distribution of the most common color strategies used by insects to avoid predators.
The global study took place across six continents and involved over 50 scientific collaborators.
Using the same experiment, researchers deployed more than 15,000 artificial prey with three different colors to investigate which strategy works best to deter predators: a classic warning pattern of orange and black, a dull brown that blends in, and an unusual bright blue and black.
The study's lead author, University of Melbourne's Dr Iliana Medina Guzman, said the answer to why some animals use camouflage over warning colors to deter predators turned out to be more complex than expected.
"Our findings showed there is no single best color strategy to deter predators, but that context is critical," Dr Medina Guzman, from the School of BioSciences, said.
"The different characteristics of the predator and prey communities, as well as habitat in that part of the globe, heavily decide which strategy performs better in each place.
"This makes sense when we see animals employing so many varying camouflage and warning color strategies as defense systems all over the world."
Predators had the biggest influence on which color strategy was most successful for prey, the study revealed.
"In environments where predators are competing intensely for food, they are more likely to risk attacking prey that might be dangerous or distasteful. Hence, we saw that camouflage worked best in areas with lots of predation," Dr Medina Guzman said.
Discover more
Sciences
science
scientific
Science
buy biotechnology products
Robotics components
purchase genome sequencing services
purchase quantum physics books
Robotics kits
physics books for sale
"Whereas, in places where cryptic prey (insects who use camouflage) are abundant, hiding becomes less effective, as predators are better at looking for those types of animals."
The findings help scientists understand why some species, such as the cryptic bogong moth or the brightly colored harlequin bug, have evolved their strategies against predators.
Dr William Allen, an evolutionary ecologist at Swansea University, in the UK, was the senior author on the research.
"For a long time, scientists have wondered why some animals use one defense strategy over the other and our study sheds important information on how animal communities and the environment influence this," Dr Allen said.
"We hope our findings can help build a better understanding of the evolution and global distribution of the most common antipredator color strategies in animals."
More information: Iliana Medina et al, Global selection on insect antipredator coloration, Science (2025). DOI: 10.1126/science.adr7368. www.science.org/doi/10.1126/science.adr7368
Journal information:Science
Provided by University of Melbourne
Citation: Camouflage or caution? How anti-predator strategies have evolved (2025, September 25) retrieved 28 September 2025 from
This document is subject to copyright. Apart from any fair dealing for the purpose of private study or research, no part may be reproduced without the written permission. The content is provided for information purposes only.
Explore further
Bugs resort to several colours to protect themselves from predators
39 shares
Facebook
Twitter
Email
Feedback to editors
Featured
Last Comments
Popular
Scientists achieve electrically driven perovskite laser using dual-cavity design
11 hours ago
0
3,300 year-old Egyptian bone whistle discovered at 18th Dynasty city of Akhenaten
12 hours ago
0
Saturday Citations: Epiphanies and brain states; a baffling skull find; achieving well-being in old age
Sep 27, 2025
0
Termite observations reveal their sophisticated technique to prevent contamination in fungal crop
Sep 26, 2025
0
Intense groundwater flow destabilizes ice in North America's Great Lakes, simulations show
Sep 26, 2025
0
##### Mushrooms may have been part of early human diets: Primate study explores who eats what, and when 6 hours ago ##### Scientists achieve electrically driven perovskite laser using dual-cavity design 11 hours ago ##### Study of extreme Indian rainfall upends conventional wisdom 12 hours ago ##### 3,300 year-old Egyptian bone whistle discovered at 18th Dynasty city of Akhenaten 12 hours ago ##### Physicists demonstrate 3,000 quantum-bit system capable of continuous operation 12 hours ago ##### Saturday Citations: Epiphanies and brain states; a baffling skull find; achieving well-being in old age Sep 27, 2025 ##### Icy planetesimal with high nitrogen and water content discovered in white dwarf's atmosphere Sep 27, 2025 ##### New adaptive optics system promises sharper gravitational-wave observations Sep 27, 2025 ##### AggreBots: Tiny living robots made from lung cells could one day deliver medicine inside the body Sep 27, 2025 ##### A 3000-year-old copper smelting site could be key to understanding the origins of iron Sep 26, 2025
Relevant PhysicsForums posts
Body dysmorphic disorder (BDD)
Sep 20, 2025
Warning of new opiod drug (nitazene) on the streets: Question re strength...
Sep 19, 2025
Chagas disease now established in US
Sep 17, 2025
Flesh-eating New World Screwworm re-occurrence in US
Sep 17, 2025
Here is a weird moth larva
Sep 17, 2025
Mold in Plastic Water Bottles? What does it eat?
Sep 16, 2025
More from Biology and Medical
1
2
Related Stories ##### Bugs resort to several colours to protect themselves from predators Jun 25, 2020 ##### Aposematism: Standing out from the crowd can be a survival mechanism Jul 2, 2025 ##### Frogs use brains or camouflage to evade predators Aug 17, 2022 ##### Cuttlefish 'mesmerize' their prey with a moving skin pattern, study finds Mar 26, 2025 ##### Comparing effectiveness of camouflage in different animals Sep 14, 2022 ##### Young frogs may camouflage selves as animal poo: study Oct 12, 2023
Recommended for you ##### Mushrooms may have been part of early human diets: Primate study explores who eats what, and when 6 hours ago ##### Termite observations reveal their sophisticated technique to prevent contamination in fungal crop Sep 26, 2025 ##### Million-year-old skull could change human evolution timeline Sep 26, 2025 ##### What noise does a fish make? New underwater tool lets ecologists ID fish from their sounds Sep 26, 2025 ##### Global project unearths new species of wildflowers Sep 26, 2025 ##### Baltic diatoms remained genetically stable for millennia—then humans came into play Sep 26, 2025
Load comments (0)
Get Instant Summarized Text (Gist)
The effectiveness of camouflage versus warning coloration in prey depends on local predator behavior, prey community traits, and habitat. Camouflage is more successful in areas with intense predation, while warning colors are favored where cryptic prey are common. No single color strategy is universally superior; environmental context shapes the evolution and distribution of these anti-predator defenses.
This summary was automatically generated using LLM. Full disclaimer
Let us know if there is a problem with our content
Use this form if you have come across a typo, inaccuracy or would like to send an edit request for the content on this page. For general inquiries, please use our contact form. For general feedback, use the public comments section below (please adhere to guidelines).
Please select the most appropriate category to facilitate processing of your request
Your message to the editors
Your email (optional, only if you'd like a response)
Send Feedback
Thank you for taking time to provide your feedback to the editors.
Your feedback is important to us. However, we do not guarantee individual replies due to the high volume of messages.
E-mail the story
Camouflage or caution? How anti-predator strategies have evolved
Your friend's email
Your email
[x] I would like to subscribe to Science X Newsletter. Learn more
Your name
Note
Your email address is used only to let the recipient know who sent the email. Neither your address nor the recipient's address will be used for any other purpose. The information you enter will appear in your e-mail message and is not retained by Phys.org in any form.
Your message
Send
Newsletter sign up
Get weekly and/or daily updates delivered to your inbox. You can unsubscribe at any time and we'll never share your details to third parties.
Subscribe
More informationPrivacy policy
Donate and enjoy an ad-free experience
We keep our content available to everyone. Consider supporting Science X's mission by getting a premium account.
Remove ads
Maybe later
### Medical Xpress Medical research advances and health news
### Tech Xplore The latest engineering, electronics and technology advances
### Science X The most comprehensive sci-tech news coverage on the web
Newsletters
Subscribe
Science X Daily and the Weekly Email Newsletter are free features that allow you to receive your favorite sci-tech news updates in your email inbox
Follow us
Top
Home
Search
Mobile version
Help
FAQ
About
Contact
Support us
Science X Account
Archive
News wire
Android app
iOS app
RSS feeds
Push notification
© Phys.org 2003 - 2025 powered by Science X Network
Privacy policyTerms of use
E-mail newsletter
Subscribe
Follow us
It appears that you are currently using Ad Blocking software. What are the consequences?× |
189089 | https://www.senate.mn/committees/2025-2026/3124_Committee_on_Higher_Education/UMN-Enrollments-2015-2024.pdf | University of Minnesota 1/13/2025 Official Enrollment Statistics Fall 2015 - Fall 2024 Undergraduate Fall 2015 Fall 2016 Fall 2017 Fall 2018 Fall 2019 Fall 2020 Fall 2021 Fall 2022 Fall 2023 Fall 2024 % Change Fall 2015 - Fall 2024 Crookston 1874 1821 1797 1834 1839 1754 1574 1489 1650 1729 -7.7% Duluth 8929 9051 9199 9109 8841 8351 8084 7754 7475 7336 -17.8% Morris 1741 1680 1554 1488 1400 1243 1189 1024 980 936 -46.2% Rochester 402 424 459 519 554 614 631 617 568 568 41.3% Twin Cities 30511 30975 31535 31455 31367 30907 30734 30560 30469 31855 4.4% TOTAL 43457 43951 44544 44405 44001 42869 42212 41444 41142 42424 -2.4% Rochester hosts an additional 118 undergraduates as part their UMN System Partner programs, that are officially counted in Twin Cities colleges.
Graduate and Professional Fall 2015 Fall 2016 Fall 2017 Fall 2018 Fall 2019 Fall 2020 Fall 2021 Fall 2022 Fall 2023 Fall 2024 % Change Fall 2015 - Fall 2024 Crookston 0 0 0 0 0 0 0 0 0 0 Duluth 1007 1021 1021 1033 982 945 833 839 801 809 -19.7% Morris 0 0 0 0 0 0 0 0 0 0 Rochester 0 0 0 0 0 0 0 0 0 0 Twin Cities 16294 16389 16122 16038 15958 15768 15947 15488 15137 15141 -7.1% TOTAL 17301 17410 17143 17071 16940 16713 16780 16327 15938 15950 -7.8% Rochester hosts 242 graduate and professional students as part their UMN System Partner programs, that are officially counted in Twin Cities colleges.
Total Degree Seeking Fall 2015 Fall 2016 Fall 2017 Fall 2018 Fall 2019 Fall 2020 Fall 2021 Fall 2022 Fall 2023 Fall 2024 % Change Fall 2015 - Fall 2024 Crookston 1874 1821 1797 1834 1839 1754 1574 1489 1650 1729 -7.7% Duluth 9936 10072 10220 10142 9823 9296 8917 8593 8276 8145 -18.0% Morris 1741 1680 1554 1488 1400 1243 1189 1024 980 936 -46.2% Rochester 402 424 459 519 554 614 631 617 568 568 41.3% Twin Cities 46805 47364 47657 47493 47325 46675 46681 46048 45606 46996 0.4% TOTAL 60758 61361 61687 61476 60941 59582 58992 57771 57080 58374 -3.9% Rochester hosts a total of 360 students as part their UMN System Partner programs, that are officially counted in Twin Cities colleges.
Source: University of Minnesota Institutional Analysis. |
189090 | https://brilliant.org/wiki/graphs/ | Graphs
Thaddeus Abiy,
Beakal Tiliksew,
Christopher Williams,
and
Karleigh Moore
Arron Kau
Josh Silverman
Pi Han Goh
Josh Silverman
Sam Solomon
Jimin Khim
Blake Farrow
Suyeon Khim
contributed
There are many systems in the natural world and in society that are amenable to mathematical and computational modeling. However, not everything is easily codified as a system of particles with coordinates and momenta. Some systems and problems such as social networks, ecologies, and genetic regulatory schemes are intrinsically divorced from spacetime descriptions, and instead are more naturally expressed as graphs that reflect their topological properties. At their simplest, graphs are simply collections of nodes – representing some class of objects like people, corporate boards, proteins, or destinations on the globe – and edges, which serve to represent connections like friendships, bridges, or molecular binding interactions.
Contents
What is a graph?
Representation of Graphs
Breadth-first Search
Depth-first Search
Contrasting Traversals
Additional Problems
What is a graph?
Consider the highway system of the eastern coast of the United States. A road inspector is given the task of writing reports about the current condition of each highway. What would be the most economical way for him to traverse all the cities? The problem can be modeled as a graph.
In fact, since graphs are dots and lines , they look like road maps. The dots are called vertices or nodes and the lines are called edges. They may have a value assigned to them (weighted) or they may just be a mere indicator of the existence of a path (unweighted). More formally, a graph can be defined as follows:
A graph G consists of a set of V of vertices (or nodes) and a set E of edges (or arcs) such that each edge e∈E is associated with a pair of vertices ∈V. A graph G with vertices V and edges E is written as G=(V,E).
Because graphs are so pervasive, it is useful to define different types of graphs. The following are the most common types of graphs:
Undirected graph: An undirected graph is one in which its edges have no orientations, i.e. no direction is associated with any edge. The edges (x,y) and (y,x) are equivalent.
Directed graph: A directed graph or digraph G consists of a set V of vertices (or nodes) and a set of edges (or arcs) such that each edge e∈E is associated with an ordered pair of vertices. If there is an edge (x,y), it is completely distinct from the edge (y,x).
Undirected graphs are typically represented by a line with no arrows, which implies a bidirectional relationship between node A and node B. Directed graphs use an arrow to show the relationship from A to B.
Directed acyclic graph: A directed acyclic graph (commonly abbreviated as DAG) is a directed graph with no directed cycles. A cycle is any path {A1,…,An} such that the edges A1→A2, A2→A3, …, and An→A1 all exist, thus forming a loop. A DAG is a graph without a single cycle.
List all the edges and vertices of the undirected graph G in the figure above.
The graph G consists of the set of vertices V = {Massachusetts, Maine, Connecticut, New York, Maryland, New Jersey}.
Its edges are E= {(Maine,Massachusetts) , (Massachusetts, Connecticut) ,
(Connecticut,New York), (New York,Maine), (New York,Massachusetts), (New Jersey, Maine),(Maryland, New York), (Maine, Maryland)}.
Note that since the graph is undirected, the order of the tuples in denoting the edges is unimportant. □
A
D
B
C
Government surveillance agencies have a tendency to accumulate strange new powers during times of panic. The US National Security Agency (NSA) now has the ability to monitor the communications of suspected individuals as well as the communications of people within some number of hops of any suspect. In the communication network above, which person is connected to the greatest number of people through 1 hop or less?
Details and Assumptions:
Each dot represents a person.
Each line represents communication between the people on either end.
If X communicates with Y, and Y communicates with Z, we say that X and Z have a 1-hop connection, and that X has a 0-hop connection with Y.
The correct answer is: C
Representation of Graphs
Above we represented a graph by drawing it. To represent it in a computer, however, we need more formal ways of representing graphs. Here we discuss the two most common ways of representing a graph: the adjacency matrix and the adjacency list.
The adjacency matrix
Represent the graph above using an adjacency matrix.
To obtain the adjacency matrix of the graph, we first label the rows and columns with the corresponding ordered vertices. If there exists an edge between two vertices i and j, then their corresponding cell in the matrix will be assigned 1. If there does not exist an edge, then the cell will be assigned the value 0. The adjacency matrix for the graph above is thus
abcdea01101b10000c10011d00100e10100. □
Adjacency list
An adjacency list representation of a graph is a way of associating each vertex (or node) in the graph with its respective list of neighboring vertices. A common way to do this is to create a Hash table. This table will contain each vertice as a key and the list of adjacent vertices of that vertices as a value.
For our example above, the adjacency list representation will look as follows:
We can see that the adjacency list is much less expensive on memory as the adjacency matrix is very sparse.
Most graph algorithms involve visiting each vertex in V, starting from a root node v0. There are several ways of achieving this. The two most common traversal algorithms are breadth-first search and depth-first search.
Breadth-first Search
In a breadth-first search, we start with the start node, followed by its adjacent nodes, then all nodes that can be reached by a path from the start node containing two edges, three edges, and so on. Formally the BFS algorithm visits all vertices in a graph G, that are k edges away from the source vertex s before visiting any vertex k+1 edges away. This is done until no more vertices are reachable from s. The image below demonstrates exactly how this traversal proceeds:
For a graph G=(V,E) and a source vertex v, breadth-first search traverses the edges of G to find all reachable vertices from v. It also computes the shortest distance to any reachable vertex. Any path between two points in a breadth-first search tree corresponds to the shortest path from the root v to any other node s.
| | |
--- |
| 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 | BFS(v){ {add v to queue and mark it} Add(Q, v) Mark v as visited while (not IsEmpty(Q)) do begin w = QueueFront(Q) Remove(Q) {loop invariant : there is a path from vertex w to every vertex in the queue Q} for each unvisited vertex u adjacent to w do begin Mark u as visited Add(Q , u) end { for } end{ while } |
We may think of three types of vertices in BFS as tree verties, those that have been taken of the data structure. fringe vertices, those adjacent to tree vertices but not yet visited, and undiscovered vertices, those that we have not encountered yet. If each visited vertex is connected to the edge that caused it to be added to the data structure, then these edges form a tree.
To search a connected component of a graph systematically, we begin with one vertex on the fringe, all others unseen, and perform the following step until all vertices have been visited: "move one vertex x from the fringe to the tree, and put any unseen vertices adjacent to x on the fringe." Graph traversal methods differ in how it is decided which vertex should be moved from the fringe to the tree. For breadth-first search we want to choose the vertex from the fringe that was least recently encountered; this corresponds to using a queue to hold vertices on the fringe.
What is the state of the queue at each iteration of BFS, if it is called from node 'a'?
PP
The table below shows the contents of the queue as the procedure. BFS visits vertices in the graph above. BFS will visit the same vertices as DFS. In this example all of them.
Node Visiteda b f i ce g d h Queuea(empty)bb fb f if if i cf i c e i c e i c e g c e g e g e g d g d d (empty) h (empty)
Depth-first Search
Depth-first search explores edges out of the most recently discovered vertex s that still has unexplored edges leaving it. Once all of ’s edges have been explored, the search “backtracks” to explore edges leaving the vertex from which was discovered. This process continues until we have discovered all the vertices that are reachable from the original source vertex. If any undiscovered vertices remain, then depth-first search selects one of them as a new source, and it repeats the search from that source. The algorithm repeats this entire process until it has discovered every vertex:
Visit a vertex s.
Mark s as visited.
Recursively visit each unvisited vertex attached to s.
A recursive implementation of DFS:
| | |
--- |
| 1 2 3 4 5 | procedure DFS(G,v): label v as discovered for all edges from v to w in G.adjacentEdges(v) do if vertex w is not labeled as discovered then recursively call DFS(G,w) |
A non-recursive implementation of DFS, it delays whether a vertex has been discovered until the vertex has been popped from the stack.
| | |
--- |
| 1 2 3 4 5 6 7 8 9 | procedure DFS-iterative(G,v): let S be a stack S.push(v) while S is not empty v = S.pop() if v is not labeled as discovered: label v as discovered for all edges from v to w in G.adjacentEdges(v) do S.push(w) |
Contrasting Traversals
Similar to tree traversal, the code for breadth-first search is slightly different from depth-first search. The most commonly mentioned difference is that BFS uses a queue to store alternative choices instead of a stack. This small change however leads to a classical graph traversal algorithm. Depth-first search goes down a single path until the path leads to the goal or we reach a limit. When a path is completely explored we back track. BFS however explores all paths from the starting location at the same time.
As we increase the size of our graph, the contrast between depth-first and breadth-first search is quite evident. Depth-first search explores the graph by looking for new vertices far away from the start point, taking closer vertices only when dead ends are encountered; breadth-first search completely covers the area close to the starting point, moving farther away only when everything close has been looked at. Again, the order in which the nodes are visited depends largely upon the effects of this ordering on the order in which vertices appear on the adjacency lists.
Additional Problems
John lives in the Trees of Ten Houses, and it is a most ideal and idyllic place for him and the other dwellers up in the canopy. They have invested a tremendous amount of time in engineering these houses, and to ensure no house felt isolated from the others, they built a fresh, finely crafted bridge between each and every house!
Unfortunately, the Trees of Ten Houses were not immune to thunderstorms, nor were the bridges well engineered. The night was treacherous, howling with wind and freezing with rain, so the odds for the bridges were not good--each bridge seemed just as likely to survive as to be shattered!
Fortunately, as there were so very many bridges in the Trees of Ten Houses, when John did wake the following morning, he found he was able to make his way to each and every house using only the existing bridges, though round-about routes may have been necessary. As they began rebuilding, John became curious... what were the chances that they'd all be so lucky?
More formally, if P is the probability that, after the storm, John is able to traverse to each and every house, what is ⌊1010P⌋?
Details and Assumptions:
The Trees of Ten Houses do, in fact, contain precisely 10 houses.
Before the storm, there exists a single bridge between each and every unique pair of houses.
The storm destroys each bridge with independent probability 21.
John is allowed to traverse through others' houses to try to reach all of them, but he must only use the surviving bridges to get there. No vine swinging allowed.
Tagged under #ComputerScience as this problem is quite tedious to do without it, though not impossible.
Image Credit:
Cite as:
Graphs.
Brilliant.org.
Retrieved
03:13, September 28, 2025,
from |
189091 | https://www.savemyexams.com/a-level/physics/aqa/17/revision-notes/11-engineering-physics/11-1-rotational-dynamics/11-1-8-rotational-work-and-power/ | Rotational Work & Power | AQA A Level Physics Revision Notes 2015
Home
Start studying
Study tools
Search
No subjects found
Show all subjects
Log in
Join now for free
GCSE
Biology
AQA
Biology
Exam Questions
Revision Notes
Flashcards
Target Test New
Mock Exams
Past Papers
Edexcel
Biology
Exam Questions
Revision Notes
Flashcards
Target Test New
Past Papers
OCR
Biology A (Gateway)
Exam Questions
Revision Notes
Target Test New
Past Papers
WJEC
Biology
Revision Notes
Past Papers
Chemistry
AQA
Chemistry
Exam Questions
Revision Notes
Flashcards
Target Test New
Past Papers
Edexcel
Chemistry
Exam Questions
Revision Notes
Flashcards
Target Test New
Past Papers
OCR
Chemistry A (Gateway)
Exam Questions
Revision Notes
Target Test New
Past Papers
WJEC
Chemistry
Exam Questions
Revision Notes
Target Test New
Past Papers
Physics
AQA
Physics
Exam Questions
Revision Notes
Target Test New
Past Papers
Edexcel
Physics
Exam Questions
Revision Notes
Target Test New
Past Papers
OCR
Physics A (Gateway)
Exam Questions
Revision Notes
Target Test New
Past Papers
WJEC
Physics
Revision Notes
Past Papers
Maths
AQA
Maths Foundation
Exam Questions
Revision Notes
Target Test New
Mock Exams
Past Papers
Maths Higher
Exam Questions
Revision Notes
Flashcards
Target Test New
Mock Exams
Past Papers
Further Maths
Exam Questions
Revision Notes
Target Test New
Past Papers
Edexcel
Maths Foundation
Exam Questions
Revision Notes
Target Test New
Mock Exams
Past Papers
Maths Higher
Exam Questions
Revision Notes
Flashcards
Target Test New
Mock Exams
Past Papers
OCR
Maths Foundation
Exam Questions
Revision Notes
Target Test New
Past Papers
Maths Higher
Exam Questions
Revision Notes
Target Test New
Past Papers
WJEC
Maths
Past Papers
Maths Numeracy
Past Papers
WJEC Eduqas
Maths Foundation
Revision Notes
Past Papers
Maths Higher
Revision Notes
Past Papers
English Literature
AQA
English Literature
Exam Questions
Revision Notes
Flashcards
Target Test New
Mock Exams
Past Papers
Edexcel
English Literature
Revision Notes
Past Papers
OCR
English Literature
Revision Notes
Past Papers
WJEC
English Literature
Past Papers
WJEC Eduqas
English Literature
Revision Notes
Past Papers
Business
AQA
Business
Exam Questions
Revision Notes
Target Test New
Past Papers
Edexcel
Business
Exam Questions
Revision Notes
Target Test New
Past Papers
OCR
Business
Revision Notes
Past Papers
WJEC Eduqas
Business
Past Papers
Science
AQA
Biology Combined Science: Trilogy
Exam Questions
Revision Notes
Target Test New
Past Papers
Chemistry Combined Science: Trilogy
Exam Questions
Revision Notes
Target Test New
Past Papers
Physics Combined Science: Trilogy
Exam Questions
Revision Notes
Target Test New
Past Papers
Edexcel
Biology Combined Science
Exam Questions
Revision Notes
Target Test New
Past Papers
Chemistry Combined Science
Exam Questions
Revision Notes
Target Test New
Past Papers
Physics Combined Science
Exam Questions
Revision Notes
Target Test New
Past Papers
OCR
Biology Combined Science A (Gateway)
Exam Questions
Revision Notes
Target Test New
Past Papers
Chemistry Combined Science A (Gateway)
Exam Questions
Revision Notes
Target Test New
Past Papers
Physics Combined Science A (Gateway)
Exam Questions
Revision Notes
Target Test New
Past Papers
WJEC
Biology Science (Double Award)
Revision Notes
Past Papers
Chemistry Science (Double Award)
Revision Notes
Past Papers
Physics Science (Double Award)
Revision Notes
Past Papers
Computer Science
AQA
Computer Science
Exam Questions
Revision Notes
Target Test New
Past Papers
Edexcel
Computer Science
Revision Notes
Past Papers
OCR
Computer Science
Exam Questions
Revision Notes
Flashcards
Target Test New
Past Papers
Economics
AQA
Economics
Past Papers
OCR
Economics
Past Papers
English Language
AQA
English Language
Exam Questions
Revision Notes
Target Test New
Mock Exams
Practice Papers
Past Papers
Edexcel
English Language
Revision Notes
Past Papers
English Language 2.0
Past Papers
OCR
English Language
Past Papers
WJEC
English Language
Past Papers
WJEC Eduqas
English Language Modular
Paper 1
Paper 2
Past Papers
Food & Nutrition
AQA
Food Preparation & Nutrition
Revision Notes
Past Papers
French
AQA
French
Past Papers
Edexcel
French
Past Papers
Geography
AQA
Geography
Exam Questions
Revision Notes
Flashcards
Target Test New
Past Papers
Edexcel
Geography A
Revision Notes
Flashcards
Past Papers
Geography B
Exam Questions
Revision Notes
Target Test New
Past Papers
OCR
Geography A
Past Papers
Geography B
Exam Questions
Revision Notes
Target Test New
Past Papers
WJEC
Geography
Past Papers
WJEC Eduqas
Geography A
Past Papers
Geography B
Past Papers
History
AQA
History
The Period Study
Wider World Depth Study
Thematic Study
British Depth Study
Past Papers
Edexcel
History
The Thematic & Historic Environment
Period Study
British Depth Study
Modern Depth Study
Past Papers
OCR
History A
Past Papers
History B (Schools History Project)
Past Papers
WJEC
History
Past Papers
Physical Education
AQA
Physical Education
Revision Notes
Past Papers
Edexcel
Physical Education
Past Papers
OCR
Physical Education
Past Papers
WJEC
Physical Education
Past Papers
Psychology
AQA
Psychology
Revision Notes
Past Papers
Edexcel
Psychology
Past Papers
OCR
Psychology
Past Papers
Religious Studies
AQA
Religious Studies A
Exam Questions
Revision Notes
Target Test New
Past Papers
Religious Studies B
Past Papers
Religious Studies Short Course
Revision Notes
Past Papers
Edexcel
Religious Studies A
Past Papers
Religious Studies B
Exam Questions
Revision Notes
Target Test New
Past Papers
OCR
Religious Studies Short Course
Past Papers
WJEC
Religious Studies
Past Papers
WJEC Eduqas
Religious Studies
Past Papers
Sociology
AQA
Sociology
Revision Notes
Past Papers
Spanish
AQA
Spanish
Past Papers
Edexcel
Spanish
Past Papers
Statistics
AQA
Statistics
Past Papers
Edexcel
Statistics Foundation
Revision Notes
Past Papers
Statistics Higher
Exam Questions
Revision Notes
Target Test New
Past Papers
Other Subjects
Arabic
Edexcel
Arabic
Past Papers
Art & Design
AQA
Art, Craft & Design
Past Papers
Fine Art
Past Papers
Graphic Communication
Past Papers
Photography
Past Papers
Textile Design
Past Papers
3D Design
Past Papers
Edexcel
Art, Craft & Design
Past Papers
OCR
Fine Art
Past Papers
Citizenship Studies
AQA
Citizenship Studies
Past Papers
Edexcel
Citizenship Studies
Past Papers
Dance
AQA
Dance
Past Papers
Design & Technology
AQA
Design & Technology
Past Papers
Edexcel
Design & Technology
Past Papers
OCR
Design & Technology
Past Papers
Drama
AQA
Drama
Past Papers
Edexcel
Drama
Past Papers
OCR
Drama
Past Papers
German
AQA
German
Past Papers
Edexcel
German
Past Papers
ICT
WJEC
Digital Technology
Past Papers
Latin
OCR
Latin
Past Papers
Media Studies
AQA
Media Studies
Past Papers
Music
AQA
Music
Past Papers
Edexcel
Music
Past Papers
OCR
Music
Past Papers
Polish
AQA
Polish
Past Papers
Welsh Language
WJEC
Welsh Language
Past Papers
Welsh Second Language
Past Papers
IGCSE
Biology
Cambridge (CIE)
Biology
Exam Questions
Revision Notes
Flashcards
Target Test New
Mock Exams
Past Papers
Edexcel
Biology
Exam Questions
Revision Notes
Flashcards
Target Test New
Mock Exams
Past Papers
Biology Modular
Unit 1
Unit 2
Past Papers
Oxford AQA
Biology
Past Papers
Chemistry
Cambridge (CIE)
Chemistry
Exam Questions
Revision Notes
Flashcards
Target Test New
Mock Exams
Past Papers
Edexcel
Chemistry
Exam Questions
Revision Notes
Flashcards
Target Test New
Past Papers
Chemistry Modular
Unit 1
Unit 2
Past Papers
Oxford AQA
Chemistry
Revision Notes
Past Papers
Physics
Cambridge (CIE)
Physics
Exam Questions
Revision Notes
Flashcards
Target Test New
Mock Exams
Past Papers
Edexcel
Physics
Exam Questions
Revision Notes
Flashcards
Target Test New
Past Papers
Physics Modular
Unit 1
Unit 2
Past Papers
Oxford AQA
Physics
Revision Notes
Past Papers
Maths
Cambridge (CIE)
Maths (0580)Core
Exam Questions
Revision Notes
Target Test New
Mock Exams
Past Papers
Maths (0580)Extended
Exam Questions
Revision Notes
Flashcards
Target Test New
Mock Exams
Past Papers
International Maths (0607)Core
Exam Questions
Revision Notes
Target Test New
Past Papers
International Maths (0607)Extended
Exam Questions
Revision Notes
Target Test New
Past Papers
Additional Maths
Exam Questions
Revision Notes
Target Test New
Past Papers
Edexcel
Maths A Foundation
Exam Questions
Revision Notes
Target Test New
Past Papers
Maths A Higher
Exam Questions
Revision Notes
Flashcards
Target Test New
Past Papers
Maths A Modular
Higher Unit 1
Higher Unit 2
Past Papers
Maths B
Past Papers
Further Maths
Revision Notes
Past Papers
Oxford AQA
Maths
Past Papers
Computer Science
Cambridge (CIE)
Computer Science
Exam Questions
Revision Notes
Flashcards
Target Test New
Mock Exams
Past Papers
Edexcel
Computer Science
Past Papers
Accounting
Cambridge (CIE)
Accounting
Exam Questions
Revision Notes
Target Test New
Mock Exams
Past Papers
Edexcel
Accounting
Introduction to Bookkeeping & Accounting
Financial Statements
Past Papers
Business
Cambridge (CIE)
Business Last exams 2026
Exam Questions
Revision Notes
Target Test New
Past Papers
Business First exams 2027
Exam Questions
Revision Notes
Flashcards
Target Test New
Mock Exams
Past Papers
Edexcel
Business
Revision Notes
Past Papers
Oxford AQA
Business
Past Papers
Science
Cambridge (CIE)
Biology Co-ordinated Sciences (Double Award)
Exam Questions
Revision Notes
Target Test New
Past Papers
Chemistry Co-ordinated Sciences (Double Award)
Exam Questions
Revision Notes
Target Test New
Past Papers
Physics Co-ordinated Sciences (Double Award)
Exam Questions
Revision Notes
Target Test New
Past Papers
Edexcel
Biology Science (Double Award)
Exam Questions
Revision Notes
Target Test New
Past Papers
Chemistry Science (Double Award)
Exam Questions
Revision Notes
Target Test New
Past Papers
Physics Science (Double Award)
Exam Questions
Revision Notes
Target Test New
Past Papers
Oxford AQA
Biology Combined Science (Double Award)
Past Papers
Chemistry Combined Science (Double Award)
Revision Notes
Past Papers
Physics Combined Science (Double Award)
Revision Notes
Past Papers
Economics
Cambridge (CIE)
Economics Last exams 2025
Exam Questions
Revision Notes
Flashcards
Target Test New
Past Papers
Economics First exams 2027
Exam Questions
Revision Notes
Flashcards
Target Test New
Mock Exams
Past Papers
Edexcel
Economics
Revision Notes
Past Papers
English Language
Cambridge (CIE)
English Language
Revision Notes
Practice Papers
Past Papers
First Language English (US)
Past Papers
English as a Second Language (Speaking Endorsement)
Past Papers
Edexcel
English Language A
Paper 1 (Non-fiction Texts and Transactional Writing)
Paper 2 (Poetry and Prose Texts and Imaginative Writing)
Paper 3 (Coursework)
Past Papers
English Language B
Past Papers
English as a Second Language
Past Papers
Oxford AQA
English Language
Past Papers
English as a Second Language
Past Papers
English Literature
Cambridge (CIE)
Literature in English
Revision Notes
Past Papers
Literature (US)
Past Papers
Edexcel
English Literature
Exam Questions
Revision Notes
Target Test New
Past Papers
Oxford AQA
English Literature
Past Papers
French
Cambridge (CIE)
French
Past Papers
Edexcel
French
Past Papers
Geography
Cambridge (CIE)
Geography Last exams 2026
Exam Questions
Revision Notes
Flashcards
Target Test New
Past Papers
Geography First exams 2027
Exam Questions
Revision Notes
Target Test New
Past Papers
Edexcel
Geography
Exam Questions
Revision Notes
Target Test New
Past Papers
Oxford AQA
Geography
Past Papers
German
Cambridge (CIE)
German
Past Papers
Edexcel
German
Past Papers
History
Cambridge (CIE)
History
Revision Notes
Past Papers
Edexcel
History
Past Papers
ICT
Cambridge (CIE)
ICT
Exam Questions
Revision Notes
Target Test New
Past Papers
Edexcel
ICT
Revision Notes
Past Papers
Religious Studies
Cambridge (CIE)
Religious Studies
Revision Notes
Past Papers
Edexcel
Religious Studies
Past Papers
Spanish
Cambridge (CIE)
Spanish
Past Papers
Edexcel
Spanish
Past Papers
Other Subjects
Chinese
Edexcel
Chinese
Past Papers
Environmental Management
Cambridge (CIE)
Environmental Management
Past Papers
Marine Science
Cambridge (CIE)
Marine Science
Past Papers
Sociology
Cambridge (CIE)
Sociology
Past Papers
AS
Biology
AQA
Biology
Exam Questions
Revision Notes
Target Test New
Past Papers
Edexcel
Biology A (SNAB)
Exam Questions
Revision Notes
Target Test New
Past Papers
OCR
Biology
Exam Questions
Revision Notes
Target Test New
Past Papers
Cambridge (CIE)
Biology First exams 2025
Exam Questions
Revision Notes
Target Test New
Past Papers
Edexcel International
Biology
Exam Questions
Revision Notes
Target Test New
Past Papers
Oxford AQA
Biology
Past Papers
Chemistry
AQA
Chemistry
Exam Questions
Revision Notes
Target Test New
Past Papers
Edexcel
Chemistry
Exam Questions
Revision Notes
Target Test New
Past Papers
OCR
Chemistry
Exam Questions
Revision Notes
Target Test New
Past Papers
Cambridge (CIE)
Chemistry First exams 2025
Exam Questions
Revision Notes
Target Test New
Past Papers
Edexcel International
Chemistry
Exam Questions
Revision Notes
Target Test New
Past Papers
Physics
AQA
Physics
Exam Questions
Revision Notes
Target Test New
Past Papers
Edexcel
Physics
Exam Questions
Revision Notes
Target Test New
Past Papers
OCR
Physics
Exam Questions
Revision Notes
Target Test New
Past Papers
Cambridge (CIE)
Physics First exams 2025
Exam Questions
Revision Notes
Target Test New
Past Papers
Edexcel International
Physics
Exam Questions
Revision Notes
Target Test New
Past Papers
Oxford AQA
Physics
Past Papers
Maths
AQA
Maths
Pure
Mechanics
Statistics
Past Papers
Edexcel
Maths
Pure
Mechanics
Statistics
Past Papers
OCR
Maths
Pure
Mechanics
Statistics
Past Papers
Cambridge (CIE)
Maths
Pure 1
Pure 2
Mechanics
Probability & Statistics 1
Mock Exams
Past Papers & Practice Papers
Edexcel International
Maths
Pure 1
Pure 2
Mechanics 1
Statistics 1
Decision 1
Mock Exams
Past Papers & Practice Papers
Oxford AQA
Maths
Past Papers
Psychology
AQA
Psychology Last exams 2025
Revision Notes
Past Papers
Psychology First exams 2026
Revision Notes
Flashcards
Past Papers
Edexcel
Psychology
Past Papers
OCR
Psychology
Past Papers
Oxford AQA
Psychology
Past Papers
Business
AQA
Business
Revision Notes
Past Papers
Cambridge (CIE)
Business
Revision Notes
Past Papers
Edexcel
Business
Past Papers
OCR
Business
Past Papers
Oxford AQA
Business
Past Papers
Computer Science
AQA
Computer Science
Past Papers
CIE
Computer Science
Revision Notes
Past Papers
OCR
Computer Science
Revision Notes
Past Papers
Economics
AQA
Economics
Past Papers
Edexcel
Economics
Past Papers
OCR
Economics
Past Papers
Oxford AQA
Economics
Past Papers
English
Cambridge (CIE)
English General Paper
Past Papers
English Language
AQA
English Language
Past Papers
Edexcel
English Language
Past Papers
OCR
English Language
Past Papers
Oxford AQA
English Language
Past Papers
English Literature
AQA
English Literature
Past Papers
Edexcel
English Literature
Past Papers
OCR
English Literature
Past Papers
Oxford AQA
English Literature
Past Papers
Environmental Management
Cambridge (CIE)
Environmental Management
Revision Notes
Past Papers
French
AQA
French
Past Papers
Further Maths
AQA
Further Maths
Past Papers
Edexcel
Further Maths
Past Papers
OCR
Further Maths
Past Papers
Cambridge (CIE)
Further Maths
Past Papers
Edexcel International
Further Maths
Further Pure 1
Past Papers
Oxford AQA
Further Maths
Past Papers
Geography
AQA
Geography
Revision Notes
Past Papers
Edexcel
Geography
Revision Notes
Past Papers
OCR
Geography
Past Papers
Oxford AQA
Geography
Past Papers
History
AQA
History
Past Papers
Edexcel
History
Past Papers
OCR
History
Past Papers
Religious Studies
Edexcel
Religious Studies
Past Papers
OCR
Religious Studies
Past Papers
Other Subjects
German
AQA
German
Past Papers
Sociology
AQA
Sociology
Revision Notes
Past Papers
OCR
Sociology
Past Papers
A Level
Biology
AQA
Biology
Exam Questions
Revision Notes
Target Test New
Mock Exams
Past Papers
Edexcel
Biology A (SNAB)
Exam Questions
Revision Notes
Target Test New
Past Papers
OCR
Biology
Exam Questions
Revision Notes
Target Test New
Mock Exams
Past Papers
Cambridge (CIE)
Biology First exams 2025
Exam Questions
Revision Notes
Target Test New
Past Papers
Edexcel International
Biology
Exam Questions
Revision Notes
Target Test New
Past Papers
Oxford AQA
Biology
Past Papers
Chemistry
AQA
Chemistry
Exam Questions
Revision Notes
Target Test New
Mock Exams
Past Papers
Edexcel
Chemistry
Exam Questions
Revision Notes
Target Test New
Past Papers
OCR
Chemistry
Exam Questions
Revision Notes
Target Test New
Past Papers
Cambridge (CIE)
Chemistry First exams 2025
Exam Questions
Revision Notes
Target Test New
Past Papers
Edexcel International
Chemistry
Exam Questions
Revision Notes
Target Test New
Past Papers
Oxford AQA
Chemistry
Revision Notes
Past Papers
Physics
AQA
Physics
Exam Questions
Revision Notes
Target Test New
Mock Exams
Past Papers
Edexcel
Physics
Exam Questions
Revision Notes
Target Test New
Past Papers
OCR
Physics
Exam Questions
Revision Notes
Target Test New
Past Papers
Cambridge (CIE)
Physics First exams 2025
Exam Questions
Revision Notes
Target Test New
Past Papers
Edexcel International
Physics
Exam Questions
Revision Notes
Target Test New
Past Papers
Oxford AQA
Physics
Past Papers
Maths
AQA
Maths
Pure
Mechanics
Statistics
Mock Exams
Past Papers & Practice Papers
Edexcel
Maths
Pure
Mechanics
Statistics
Mock Exams
Past Papers & Practice Papers
Solution Banks
OCR
Maths A
Pure
Mechanics
Statistics
Mock Exams
Past Papers & Practice Papers
Maths B (MEI)
Past Papers
Cambridge (CIE)
Maths
Pure 1
Pure 3
Mechanics
Probability & Statistics 1
Probability & Statistics 2
Mock Exams
Past Papers & Practice Papers
Edexcel International
Maths
Pure 1
Pure 2
Pure 3
Pure 4
Mechanics 1
Mechanics 2
Statistics 1
Statistics 2
Decision 1
Mock Exams
Past Papers & Practice Papers
Oxford AQA
Maths
Past Papers
Psychology
AQA
Psychology Last exams 2026
Exam Questions
Revision Notes
Flashcards
Target Test New
Mock Exams
Past Papers
Psychology First exams 2027
Exam Questions
Revision Notes
Flashcards
Target Test New
Past Papers
Edexcel
Psychology
Past Papers
OCR
Psychology
Past Papers
Cambridge (CIE)
Psychology
Past Papers
Edexcel International
Psychology
Past Papers
Oxford AQA
Psychology
Past Papers
WJEC
Psychology
Past Papers
WJEC Eduqas
Psychology
Past Papers
Business
AQA
Business
Revision Notes
Past Papers
Edexcel
Business
Exam Questions
Revision Notes
Flashcards
Target Test New
Mock Exams
Past Papers
OCR
Business
Past Papers
Cambridge (CIE)
Business
Revision Notes
Past Papers
Oxford AQA
Business
Past Papers
Computer Science
AQA
Computer Science
Past Papers
Cambridge (CIE)
Computer Science
Revision Notes
Past Papers
OCR
Computer Science
Exam Questions
Revision Notes
Target Test New
Mock Exams
Past Papers
Economics
AQA
Economics
Exam Questions
Revision Notes
Target Test New
Past Papers
Edexcel
Economics A
Exam Questions
Revision Notes
Flashcards
Target Test New
Past Papers
OCR
Economics
Past Papers
Cambridge (CIE)
Economics
Past Papers
Oxford AQA
Economics
Past Papers
English Language
AQA
English Language
Past Papers
Edexcel
English Language
Past Papers
OCR
English Language
Past Papers
Cambridge (CIE)
English Language
Past Papers
Edexcel International
English Language
Past Papers
Oxford AQA
English Language
Past Papers
English Literature
AQA
English Literature A
Past Papers
English Literature B
Past Papers
Edexcel
English Literature
Past Papers
OCR
English Literature
Revision Notes
Past Papers
Cambridge (CIE)
English Literature
Past Papers
Edexcel International
English Literature
Past Papers
Oxford AQA
English Literature
Past Papers
English Language & Literature
AQA
English Language & Literature
Past Papers
French
AQA
French
Past Papers
Edexcel
French
Past Papers
Cambridge (CIE)
French Language & Literature
Past Papers
Further Maths
AQA
Further Maths
Past Papers
Edexcel
Further Maths
Core Pure
Further Statistics 1
Further Mechanics 1
Decision 1
Solution Banks
OCR
Further Maths
Past Papers
Cambridge (CIE)
Further Maths
Past Papers
Edexcel International
Further Maths
Further Pure 1
Past Papers
Oxford AQA
Further Maths
Past Papers
Geography
AQA
Geography
Exam Questions
Revision Notes
Flashcards
Target Test New
Past Papers
Edexcel
Geography
Exam Questions
Revision Notes
Target Test New
Past Papers
OCR
Geography
Past Papers
Cambridge (CIE)
Geography
Past Papers
Edexcel International
Geography
Past Papers
Oxford AQA
Geography
Past Papers
German
AQA
German
Past Papers
Edexcel
German
Past Papers
Cambridge (CIE)
German Language & Literature
Past Papers
History
AQA
History
Past Papers
Edexcel
History
Past Papers
OCR
History
Past Papers
Cambridge (CIE)
History
Past Papers
Religious Studies
AQA
Religious Studies
Past Papers
Edexcel
Religious Studies
Past Papers
OCR
Religious Studies
Past Papers
Sociology
AQA
Sociology
Revision Notes
Past Papers
OCR
Sociology
Past Papers
Cambridge (CIE)
Sociology
Past Papers
Spanish
AQA
Spanish
Past Papers
Edexcel
Spanish
Past Papers
Cambridge (CIE)
Spanish Language & Literature
Past Papers
Other Subjects
Accounting
AQA
Accounting
Past Papers
Cambridge (CIE)
Accounting
Past Papers
Art & Design
AQA
Fine Art
Past Papers
Photography
Past Papers
Product Design
Past Papers
ICT
Cambridge (CIE)
Information Technology
Past Papers
Law
AQA
Law
Past Papers
OCR
Law
Past Papers
Cambridge (CIE)
Law
Past Papers
Marine Science
Cambridge (CIE)
Marine Science
Past Papers
Physical Education
AQA
Physical Education
Past Papers
OCR
Physical Education
Past Papers
Politics
AQA
Politics
Past Papers
Edexcel
Politics
Past Papers
IB
Biology
Biology SL
Exam Questions
Revision Notes
Flashcards
Target Test New
Mock Exams
Practice Papers
Biology HL
Exam Questions
Revision Notes
Flashcards
Target Test New
Mock Exams
Practice Papers
Chemistry
Chemistry SL
Exam Questions
Revision Notes
Flashcards
Target Test New
Mock Exams
Practice Papers
Chemistry HL
Exam Questions
Revision Notes
Flashcards
Target Test New
Mock Exams
Practice Papers
Physics
Physics SL
Exam Questions
Revision Notes
Flashcards
Target Test New
Mock Exams
Practice Papers
Physics HL
Exam Questions
Revision Notes
Flashcards
Target Test New
Mock Exams
Practice Papers
Maths
Analysis & Approaches (AA)SL
Exam Questions
Revision Notes
Flashcards
Target Test New
Mock Exams
Practice Papers
Analysis & Approaches (AA)HL
Exam Questions
Revision Notes
Flashcards
Target Test New
Mock Exams
Practice Papers
Applications & Interpretation (AI)SL
Exam Questions
Revision Notes
Flashcards
Target Test New
Mock Exams
Practice Papers
Applications & Interpretation (AI)HL
Exam Questions
Revision Notes
Flashcards
Target Test New
Mock Exams
Practice Papers
Business Management
Business Management SL
Exam Questions
Revision Notes
Flashcards
Target Test New
Practice Papers
Business Management HL
Exam Questions
Revision Notes
Flashcards
Target Test New
Practice Papers
Economics
Economics SL
Revision Notes
Practice Papers
Economics HL
Exam Questions
Revision Notes
Target Test New
Practice Papers
ESS
Environmental Systems & Societies (ESS)Last exams 2025 SL
Exam Questions
Revision Notes
Target Test New
Environmental Systems & Societies (ESS)First exams 2026 SL
Exam Questions
Revision Notes
Target Test New
Environmental Systems & Societies (ESS)First exams 2026 HL
Exam Questions
Revision Notes
Target Test New
Geography
Geography SL
Revision Notes
Geography HL
Revision Notes
Psychology
Psychology SL
Revision Notes
Flashcards
Psychology HL
Revision Notes
Flashcards
O Level
Maths
Cambridge
Maths
Exam Questions
Revision Notes
Target Test New
Past Papers
Additional Maths
Exam Questions
Revision Notes
Target Test New
Past Papers
Statistics
Past Papers
Biology
Cambridge
Biology
Exam Questions
Revision Notes
Target Test New
Past Papers
Chemistry
Cambridge
Chemistry
Exam Questions
Revision Notes
Target Test New
Past Papers
Physics
Cambridge
Physics
Exam Questions
Revision Notes
Target Test New
Past Papers
Accounting
Cambridge
Accounting
Exam Questions
Revision Notes
Target Test New
Past Papers
Business Studies
Cambridge
Business Studies
Revision Notes
Past Papers
Computer Science
Cambridge
Computer Science
Exam Questions
Revision Notes
Flashcards
Target Test New
Past Papers
Economics
Cambridge
Economics
Revision Notes
Past Papers
Geography
Cambridge
Geography
Revision Notes
Past Papers
Other Subjects
Bangladesh Studies
Cambridge
Bangladesh Studies
Past Papers
English Language
Cambridge
English Language
Past Papers
English Literature
Cambridge
Literature in English
Past Papers
Environmental Management
Cambridge
Environmental Management
Past Papers
History
Cambridge
History
Revision Notes
Past Papers
Islamiyat
Cambridge
Islamiyat
Past Papers
Pakistan Studies
Cambridge
Pakistan Studies
Past Papers
Religious Studies
Cambridge
Bible Knowledge
Past Papers
AP
Biology
AP Biology
Exam Questions
Study Guides
Target Test New
Practice Tests
Past Exams
Chemistry
AP Chemistry
Exam Questions
Study Guides
Target Test New
Practice Tests
Past Exams
Physics
AP Physics 1: Algebra-Based
Exam Questions
Study Guides
Target Test New
Practice Tests
Past Exams
AP Physics 2: Algebra-Based
Past Exams
AP Physics C: Electricity & Magnetism
Past Exams
AP Physics C: Mechanics
Past Exams
Calculus
AP Calculus AB
Exam Questions
Study Guides
Target Test New
Past Exams
AP Calculus BC
Exam Questions
Study Guides
Target Test New
Past Exams
AP Precalculus
Past Exams
Economics
AP Macroeconomics
Past Exams
AP Microeconomics
Past Exams
Chinese Language
AP Chinese Language & Culture
Past Exams
Computer Science
AP Computer Science A
Past Exams
AP Computer Science Principles
Past Exams
English
AP English Language & Composition
Past Exams
AP English Literature & Composition
Past Exams
Environmental Science
AP Environmental Science
Exam Questions
Study Guides
Target Test New
Past Exams
French Language
AP French Language & Culture
Past Exams
Human Geography
AP Human Geography
Exam Questions
Study Guides
Target Test New
Past Exams
German Language
AP German Language & Culture
Past Exams
History
AP US History
Exam Questions
Study Guides
Target Test New
Past Exams
AP World History
Past Exams
AP European History
Past Exams
Latin
AP Latin
Past Exams
Research
AP Research
Past Exams
Spanish Language
AP Spanish Language & Culture
Past Exams
Statistics
AP Statistics
Exam Questions
Study Guides
Target Test New
Past Exams
Other
Level 2
AQA
Further Maths
Exam Questions
Revision Notes
Target Test New
Past Papers
WJEC
Additional Maths
Past Papers
Level 3
AQA
Mathematical Studies (Core Maths)
Paper 1
Paper 2A: Statistical Techniques
Paper 2B: Critical Path & Risk Analysis
Paper 2C: Graphical Techniques
Past Papers
Past Papers
All Past Papers
Past Papers
GCSE Past Papers
IGCSE Past Papers
AS Past Papers
A Level Past Papers
O Level Past Papers
AP Past Papers
Join now for free
Log in
Study tools
Revision Notes
Exam Questions
Smart Mark
Flashcards
Mock Exams
Target Test New
Past Papers
Test Builder New
Learning hub
Glossaries
Ambassadors
About us
Help and support
Revision NotesExam QuestionsTarget TestPast PapersMock Exams
Revision Notes
Measurements & Their Errors
2 Topics · 6 Revision Notes
1.
Use of SI Units & Their Prefixes
1. ##### SI Units
2. ##### Powers of Ten
3. ##### Estimating Physical Quantities
2.
Limitation of Physical Measurements
1. ##### Sources of Uncertainty
2. ##### Calculating Uncertainties
3. ##### Determining Uncertainties from Graphs
Particles & Radiation
5 Topics · 26 Revision Notes
1.
Atomic Structure & Decay Equations
1. ##### Atomic Structure
2. ##### Nucleon & Proton Number
3. ##### Strong Nuclear Force
4. ##### Alpha & Beta Decay
5. ##### Particles, Antiparticles & Photons
2.
Classification of Particles
1. ##### Hadrons
2. ##### Baryons
3. ##### Mesons
4. ##### Leptons
5. ##### Quarks & Antiquarks
6. ##### Strange Quarks
7. ##### Collaborative Efforts in Particle Physics
3.
Conservation Laws & Particle Interactions
1. ##### The Four Fundamental Interactions
2. ##### The Electromagnetic & Strong Force
3. ##### The Weak Interaction
4. ##### Feynman Diagrams
5. ##### Application of Conservation Laws
4.
The Photoelectric Effect
1. ##### The Electronvolt
2. ##### Threshold Frequency & Work Function
3. ##### The Photoelectric Equation
5.
Energy Levels & Photon Emission
1. ##### Collisions of Electrons with Atoms
2. ##### Energy Levels & Photon Emission
3. ##### Wave-Particle Duality
4. ##### The de Broglie Wavelength
5. ##### Diffraction Effects of Momentum
6. ##### Analysis of the Nature of Matter
Waves
5 Topics · 17 Revision Notes
1.
Longitudinal & Transverse Waves
1. ##### Progressive Waves
2. ##### Longitudinal & Transverse Waves
3. ##### Polarisation
2.
Stationary Waves
1. ##### Stationary Waves
2. ##### Formation of Stationary Waves
3. ##### Harmonics
4. ##### Required Practical: Investigating Stationary Waves
3.
Interference
1. ##### Path Difference & Coherence
2. ##### Demonstrating Interference
3. ##### Young's Double-Slit Experiment
4. ##### Developing Theories of EM Radiation
5. ##### Required Practical: Young's Slit Experiment & Diffraction Gratings
4.
Diffraction
1. ##### Single Slit Diffraction
2. ##### The Diffraction Grating
5.
Refraction
1. ##### Refraction at a Plane Surface
2. ##### Snell's Law
3. ##### Fibre Optics
Mechanics & Materials
8 Topics · 33 Revision Notes
1.
Scalars & Vectors
1. ##### Scalars & Vectors
2. ##### Resolving Vectors
2.
Moments
1. ##### Moments
2. ##### Couples
3. ##### Centre of Mass
3.
Equations of Motion
1. ##### Motion Along a Straight Line
2. ##### Motion Graphs
3. ##### SUVAT Equations
4. ##### Projectile Motion
5. ##### Drag Forces
6. ##### Terminal Velocity
7. ##### Required Practical: Determination of g
4.
Newton’s Laws of Motion
1. ##### Newton’s First Law
2. ##### Newton's Second Law
3. ##### Newton's Third Law
5.
Linear Momentum & Conservation
1. ##### Linear Momentum
2. ##### Conservation of Momentum
3. ##### Impulse
4. ##### Impulse on a Force-Time Graph
5. ##### Collisions
6.
Work, Energy & Power
1. ##### Work & Power
2. ##### Area Under a Force-Displacement Graph
3. ##### Efficiency
4. ##### Conservation of Energy
5. ##### Kinetic & Gravitational Potential Energy
7.
Bulk Properties of Solids
1. ##### Density
2. ##### Hooke's Law
3. ##### Stress & Strain
4. ##### Elastic Strain Energy
5. ##### Elastic & Plastic Behaviour
6. ##### Energy Conservation
8.
The Young Modulus
1. ##### The Young Modulus
2. ##### Required Practical: The Young Modulus
Electricity
4 Topics · 12 Revision Notes
1.
Current–Voltage Characteristics
1. ##### Basics of Electricity
2. ##### Current–Voltage Characteristics
2.
Resistance & Resistivity
1. ##### Resistivity
2. ##### Resistance in a Thermistor
3. ##### Superconductivity
4. ##### Required Practical: Investigating Resistivity
3.
Circuits & The Potential Divider
1. ##### Resistors in Series & Parallel
2. ##### Series & Parallel Circuits
3. ##### Electrical Energy & Power
4. ##### Potential Divider Circuits
4.
Electromotive Force & Internal Resistance
1. ##### Electromotive Force & Internal Resistance
2. ##### Required Practical: Investigating EMF & Internal Resistance
Further Mechanics & Thermal Physics
6 Topics · 31 Revision Notes
1.
Circular Motion
1. ##### Circular Motion
2. ##### Radians
3. ##### Angular Speed
4. ##### Centripetal Acceleration
5. ##### Centripetal Force
2.
Simple Harmonic Motion
1. ##### Conditions for Simple Harmonic Motion
2. ##### SHM Graphs
3. ##### Calculating Maximum Speed & Acceleration
4. ##### Period of Mass-Spring System
5. ##### Period of Simple Pendulum
6. ##### Harmonic Oscillators in Context
7. ##### Energy in SHM
8. ##### Required Practical: Investigating SHM
3.
Forced Vibrations & Resonance
1. ##### Damping
2. ##### Free & Forced Oscillations
3. ##### Resonance
4.
Thermal Energy Transfer
1. ##### Internal Energy
2. ##### The First Law of Thermodynamics
3. ##### Specific Heat Capacity
4. ##### Latent Heat Capacity
5.
Ideal Gases
1. ##### The Kelvin Scale & Absolute Zero
2. ##### Gas Laws
3. ##### Ideal Gas Equation
4. ##### Work Done by a Gas
5. ##### Avogadro, Molar Gas & Boltzmann Constant
6. ##### Required Practical: Investigating Gas Laws
6.
Molecular Kinetic Theory Model
1. ##### Gas Laws v Kinetic Theory
2. ##### Kinetic Theory of Gases Equation
3. ##### Average Molecular Kinetic Energy
4. ##### Brownian Motion
5. ##### Evolving Models of Gas Behaviour
Fields & Their Consequences
10 Topics · 47 Revision Notes
1.
Gravitational Fields
1. ##### Force Fields
2. ##### Gravitational Field Strength
3. ##### Representing Gravitational Fields
4. ##### Newton's Law of Gravitation
5. ##### Gravitational Field Strength in a Radial Field
2.
Gravitational Potential
1. ##### Gravitational Potential
2. ##### Calculating Gravitational Potential
3. ##### Graphical Representation of Gravitational Potential
4. ##### Work Done on a Mass
3.
Orbits of Planets & Satellites
1. ##### Circular Orbits in Gravitational Fields
2. ##### Energy of an Orbiting Satellite
3. ##### Escape Velocity
4. ##### Geostationary Orbits
4.
Electric Fields
1. ##### Coulomb's Law
2. ##### Electric Field Lines
3. ##### Electric Field Strength
4. ##### Uniform Electric Field
5. ##### Motion of Charged Particles
6. ##### Radial Electric Field
7. ##### Comparing Gravitational & Electrostatic Forces
5.
Electric Potential
1. ##### Electric Potential
2. ##### Graphical Representation of Electric Potential
3. ##### Work Done on a Charge
6.
Capacitance
1. ##### Capacitance
2. ##### Parallel Plate Capacitor
3. ##### Energy Stored by a Capacitor
7.
Capacitor Charge & Discharge
1. ##### Charge & Discharge Graphs
2. ##### The Time Constant
3. ##### Charge & Discharge Equations
4. ##### Required Practical: Charging & Discharging Capacitors
8.
Magnetic Fields
1. ##### Magnetic Flux Density
2. ##### Magnetic Force on a Current-Carrying Conductor
3. ##### Fleming's Left Hand Rule
4. ##### Force on a Moving Charge
5. ##### Circular Path of Particles
6. ##### Required Practical: Investigating Magnetic Fields in Wires
9.
Electromagnetic Induction
1. ##### Magnetic Flux
2. ##### Magnetic Flux Linkage
3. ##### Principles of Electromagnetic Induction
4. ##### Faraday's & Lenz's Laws
5. ##### Applications of EM Induction
6. ##### Required Practical: Investigating Flux Linkage on a Search Coil
10.
Alternating Currents & Transformers
1. ##### Alternating Current & Voltage
2. ##### The Operation of an Oscilloscope
3. ##### The Operation of a Transformer
4. ##### Transformer Efficiency
5. ##### Transformer Inefficiencies
Nuclear Physics
4 Topics · 25 Revision Notes
1.
Alpha, Beta & Gamma Radiation
1. ##### Rutherford Scattering
2. ##### Changing Models of the Nucleus
3. ##### Alpha, Beta & Gamma Radiation
4. ##### Inverse-Square Law of Gamma Radiation
5. ##### Background Radiation
6. ##### Radiation Safety
7. ##### Required Practical: Inverse Square-Law for Gamma Radiation
2.
Radioactive Decay
1. ##### Radioactive Decay
2. ##### Exponential Decay
3. ##### Half-Life
4. ##### Applications of Radioactivity
3.
Nuclear Instability & Radius
1. ##### Nuclear Instability
2. ##### Decay Equations
3. ##### Nuclear Excited States
4. ##### Nuclear Radius
5. ##### Closest Approach Estimate
6. ##### Nuclear Radius Equation
7. ##### Nuclear Density
4.
Nuclear Fusion & Fission
1. ##### Energy & Mass Equivalence
2. ##### Mass Difference & Binding Energy
3. ##### Nuclear Fusion & Fission
4. ##### Binding Energy
5. ##### Induced Fission
6. ##### Operation of a Nuclear Reactor
7. ##### Safety Aspects of Nuclear Reactors
Astrophysics
3 Topics · 31 Revision Notes
1.
Telescopes
1. ##### Lenses & Ray Diagrams for Telescopes
2. ##### Refracting Telescopes
3. ##### Reflecting Telescopes
4. ##### Reflecting vs Refracting Telescopes
5. ##### Resolving Power of Telescopes
6. ##### Collecting Power of Telescopes
7. ##### Radio, IR, UV & X-Ray Telescopes
8. ##### Charge-Coupled Devices (CCDs) in Astronomy
2.
Classification of Stars
1. ##### Brightness & Apparent Magnitude
2. ##### Inverse Square Law of Radiation
3. ##### Astronomical Distances
4. ##### Absolute Magnitude
5. ##### Wien’s Displacement Law
6. ##### Stefan's Law
7. ##### Emission & Absorption Spectra in Stars
8. ##### Stellar Spectral Classes
9. ##### Star Formation
10. ##### Life Cycle of a Low Mass Star
11. ##### Life Cycle of a High Mass Star
12. ##### Supernovae & Gamma Ray Bursts (GRBs)
13. ##### Supernovae as Standard Candles
14. ##### Neutron Stars & Black Holes
15. ##### The Hertzsprung-Russell Diagram
3.
Cosmology
1. ##### The Doppler Effect of Light
2. ##### Binary Star Systems
3. ##### Galactic Redshift
4. ##### Hubble's Law
5. ##### Evidence for the Big Bang
6. ##### Dark Energy
7. ##### Quasars
8. ##### Exoplanets
Medical Physics
6 Topics · 30 Revision Notes
1.
Physics of the Eye
1. ##### Converging & Diverging Lenses
2. ##### Lens Calculations
3. ##### Structure of the Eye
4. ##### Sensitivity of the Eye
5. ##### Spatial Resolution of the Eye
6. ##### Defects of Vision
2.
Physics of the Ear
1. ##### Structure of the Ear
2. ##### The Decibel Scale
3. ##### Equal Loudness Curves
4. ##### Defects of Hearing
3.
Biological Measurement
1. ##### ECG Machines
4.
Non-Ionising Imaging
1. ##### The Piezoelectric Transducer
2. ##### Generation & Detection of Ultrasound
3. ##### Ultrasound Imaging
4. ##### Applications of Ultrasound
5. ##### Fibre Optics & Endoscopy
6. ##### Magnetic Resonance Imaging
5.
X-ray Imaging
1. ##### Production of X-rays
2. ##### Rotating-Anode X-ray Tube
3. ##### Using X-rays in Medical Imaging
4. ##### X-ray Detection
5. ##### Attenuation of X-rays
6. ##### The CT Scanner
6.
Radionuclide Imaging & Therapy
1. ##### Radioactive Tracers
2. ##### The Molybdenum-Technetium Generator
3. ##### The PET Scanner
4. ##### Physical, Biological & Effective Half Life
5. ##### Gamma Camera
6. ##### Radiotherapy
7. ##### Comparing Imaging Techniques
Engineering Physics
2 Topics · 23 Revision Notes
1.
Rotational Dynamics
1. ##### Rotational Motion
2. ##### Angular Acceleration Equations
3. ##### Torque
4. ##### Moment of Inertia
5. ##### Newton’s Second Law for Rotation
6. ##### Angular Momentum
7. ##### Angular Impulse
8. ##### Rotational Work & Power
9. ##### Rotational Kinetic Energy
10. ##### Flywheels in Machines
2.
Thermodynamics & Engines
1. ##### The First Law of Thermodynamics
2. ##### p–V Diagrams
3. ##### Thermodynamic Processes
4. ##### Petrol Engine Cycle
5. ##### Diesel Engine Cycle
6. ##### Comparing Petrol & Diesel Engines
7. ##### Power Output of an Engine
8. ##### Engine Efficiency
9. ##### The Second Law of Thermodynamics
10. ##### Heat Engines
11. ##### Limitations of Real Heat Engines
12. ##### Reversed Heat Engines
13. ##### Coefficients of Performance
Turning Points in Physics
3 Topics · 27 Revision Notes
1.
The Discovery of the Electron
1. ##### Cathode Rays
2. ##### Thermionic Emission
3. ##### Specific Charge Experiments
4. ##### Significance of Thomson's Experiment
5. ##### Millikan’s Oil Drop Experiment
6. ##### Stokes' Law
2.
Wave-Particle Duality
1. ##### Theories of Light
2. ##### Young's Double Slit Interference
3. ##### Maxwell's Wave Equation
4. ##### Hertz's Discovery of Radio Waves
5. ##### Fizeau's Speed of Light Experiment
6. ##### UV Catastrophe & Black-Body Radiation
7. ##### The Discovery of Photoelectricity
8. ##### De Broglie's Hypothesis
9. ##### Electron Diffraction
10. ##### Transmission Electron Microscope
11. ##### Scanning Tunnelling Microscope
3.
Special Relativity
1. ##### The Michelson-Morley Interferometer
2. ##### The Invariance of the Speed of Light
3. ##### Inertial Frames of Reference
4. ##### Einstein's Postulates
5. ##### Time Dilation
6. ##### Muon Lifetime Experiment
7. ##### Length Contraction
8. ##### Relativistic Mass
9. ##### Relativistic Energy
10. ##### Bertozzi's Experiment
A LevelPhysicsAQARevision NotesEngineering Physics Rotational Dynamics Rotational Work & Power
Rotational Work & Power(AQA A Level Physics):Revision Note
Exam code:7408
Download PDF
Author
Ashika
Last updated
8 April 2025
Work Done & Torque
Work Done by a Rotating Object
Work has to be done on a rigid body when a torque turns in through an angle about an axis
For example, rotating cranes and fairground rides
In systems with linear acceleration, work W is the product of the force and the distance moved
Therefore, the work done for a rotating object is defined by the equation
Where:
W= work done (J)
= torque (N m)
θ= angular displacement (the angle turned through by the rotating object) (rads)
Work can also be calculated by finding thearea under a torque-angular displacement graph
Torque-angular displacement graph
The work done is the area under the torque-angular displacement graph
This is analogous to the work done being the area under a force-displacement graph
Power Output of a Rotating Object
Power is therateof doing work, and is defined by
Where:
P = power (W)
ω= angular velocity (rad s–1)
This equation is the angular version of the linear equation P =Fv
Examiner Tips and Tricks
Don't forget that θ is always in radianswhen you're doing conversions from revs s–1 or rev min–1.
Frictional Torque
In rotational mechanics, frictional forces produce a specific torque called frictional torque
This is the torque caused by the frictional force when two objects in contact move past each other
Frictional torque can be defined as:
The difference between the applied torque and the resulting net, or observed, torque
This means that thenet torque is thedifference betweenthe appliedandfrictional torque
Net torque = applied torque – frictional torque
In rotatingmachinery, power has to be expended to overcome frictional torque
This is due to resistive forces within the machinery
In most cases, frictional torque is minimised to reduce the kinetic energy losses transferred to heat and sound
The frictional force must always be subtractedfrom the torque resulting from an applied force to get thetotalornet torque in the system
Frictional torque is calculated using the same equations as torque
The only difference is F is the frictional force instead of an externally applied force
Worked Example
The figure below shows a type of circular saw. The blade is driven by an electric motor and rotates at 3100 rev min–1 when cutting a piece of wood.
A constant frictional torque of 2.7 N m acts at the bearings of the motor and axle.
A horizontal force of 45 N is needed to push a piece of wood into the saw. The force acts on the blade at an effective radius of 22 cm.
Calculate the output power of the motor when the saw is cutting the wood.
Answer:
Step 1: Calculate the torque on the saw blade
When the wood is being cut, the torque from the 45 N force is equal to the net torque of the saw blade
Step 2: Calculate the applied torque on the saw from the motor
net torque = applied torque on the saw blade – frictional torque
Step 3: Calculate the angular velocity
1 revolution = 2π radians
3100 rev min–1 = 3100 × 2π rad min–1
min–1 → sec–1 = ÷ 60
Step 4: Calculate the output power
Unlock more, it's free!
Join the 100,000+ Students that ❤️ Save My Exams
the (exam) results speak for themselves:
I would just like to say a massive thank you for putting together such a brilliant, easy to use website.I really think using this site helped me secure my top grades in science and maths. You really did save my exams! Thank you.
Beth
IGCSE Student
> This website is soooo useful and I can’t ever thank you enough for organising questions by topic like this. Furthermore, the name of the website could not have been more appropriate as it literally did SAVE MY EXAMS!
Fathima
A Level Student
> Incredible! SO worth my money, the revision notes have everything I need to know and are so easy to understand. I actually enjoy revising! It makes me feel a lot more confident for my GCSEs in a few months.
Kate
GCSE Student
> Absolutely brilliant, both my girls used it for A levels and GCSE. It's saves on paper copies, also beneficial exam questions ranked from easy to hard. It's removed a lot of stress from the exams.
Sameera
Parent
> Just to say that your resources are the best I have seen and I have been teaching chemistry at different levels for about 40 years
Mark
Chemistry Teacher
Excellent
Read more reviews
Join now for free
Test yourself
Did this page help you?
Yes No
Previous:Angular ImpulseNext:Rotational Kinetic Energy
More Exam Questions you might like
Rotational Dynamics
Thermodynamics & Engines
The Discovery of the Electron
Wave-Particle Duality
Special Relativity
Synoptic Exam Questions (Paper 1 & 2)
Practical Skills & Data Analysis (Paper 3A)
Use of SI Units & Their Prefixes
Limitation of Physical Measurements
Atomic Structure & Decay Equations
Author:Ashika
Expertise:Physics Content Creator
Ashika graduated with a first-class Physics degree from Manchester University and, having worked as a software engineer, focused on Physics education, creating engaging content to help students across all levels. Now an experienced GCSE and A Level Physics and Maths tutor, Ashika helps to grow and improve our Physics resources.
Download notes on Rotational Work & Power
Home
Resources
Home
Learning Hub
Teachers
Ambassadors
Scholarship
Join
Past Papers
Solution Banks
Sitemap
Members
Log in
Company
About us
Exam Specificity
Content Quality
Promotions
Jobs
Terms
Privacy
Cookie Policy
Help and Support
Subjects
Biology
Chemistry
Physics
Maths
Geography
English Literature
Psychology
All Subjects
TikTokInstagramFacebookTwitter
© Copyright 2015-2025 Save My Exams Ltd. All Rights Reserved.
IBO was not involved in the production of, and does not endorse, the resources created by Save My Exams. |
189092 | https://stackoverflow.com/questions/63878250/two-way-anova-using-python | pandas - Two Way Anova using Python - Stack Overflow
Join Stack Overflow
By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy.
Sign up with Google
Sign up with GitHub
OR
Email
Password
Sign up
Already have an account? Log in
Skip to main content
Stack Overflow
1. About
2. Products
3. For Teams
Stack Overflow for Teams Where developers & technologists share private knowledge with coworkers
Advertising Reach devs & technologists worldwide about your product, service or employer brand
Knowledge Solutions Data licensing offering for businesses to build and improve AI tools and models
Labs The future of collective knowledge sharing
About the companyVisit the blog
Loading…
current community
Stack Overflow helpchat
Meta Stack Overflow
your communities
Sign up or log in to customize your list.
more stack exchange communities
company blog
Log in
Sign up
Let's set up your homepage Select a few topics you're interested in:
python javascript c#reactjs java android html flutter c++node.js typescript css r php angular next.js spring-boot machine-learning sql excel ios azure docker
Or search from our full list:
javascript
python
java
c#
php
android
html
jquery
c++
css
ios
sql
mysql
r
reactjs
node.js
arrays
c
asp.net
json
python-3.x
.net
ruby-on-rails
sql-server
swift
django
angular
objective-c
excel
pandas
angularjs
regex
typescript
ruby
linux
ajax
iphone
vba
xml
laravel
spring
asp.net-mvc
database
wordpress
string
flutter
postgresql
mongodb
wpf
windows
xcode
amazon-web-services
bash
git
oracle-database
spring-boot
dataframe
azure
firebase
list
multithreading
docker
vb.net
react-native
eclipse
algorithm
powershell
macos
visual-studio
numpy
image
forms
scala
function
vue.js
performance
twitter-bootstrap
selenium
winforms
kotlin
loops
express
dart
hibernate
sqlite
matlab
python-2.7
shell
rest
apache
entity-framework
android-studio
csv
maven
linq
qt
dictionary
unit-testing
asp.net-core
facebook
apache-spark
tensorflow
file
swing
class
unity-game-engine
sorting
date
authentication
go
symfony
t-sql
opencv
matplotlib
.htaccess
google-chrome
for-loop
datetime
codeigniter
perl
http
validation
sockets
google-maps
object
uitableview
xaml
oop
visual-studio-code
if-statement
cordova
ubuntu
web-services
email
android-layout
github
spring-mvc
elasticsearch
kubernetes
selenium-webdriver
ms-access
ggplot2
user-interface
parsing
pointers
c++11
google-sheets
security
machine-learning
google-apps-script
ruby-on-rails-3
templates
flask
nginx
variables
exception
sql-server-2008
gradle
debugging
tkinter
delphi
listview
jpa
asynchronous
web-scraping
haskell
pdf
jsp
ssl
amazon-s3
google-cloud-platform
jenkins
testing
xamarin
wcf
batch-file
generics
npm
ionic-framework
network-programming
unix
recursion
google-app-engine
mongoose
visual-studio-2010
.net-core
android-fragments
assembly
animation
math
svg
session
intellij-idea
hadoop
rust
next.js
curl
join
winapi
django-models
laravel-5
url
heroku
http-redirect
tomcat
google-cloud-firestore
inheritance
webpack
image-processing
gcc
keras
swiftui
asp.net-mvc-4
logging
dom
matrix
pyspark
actionscript-3
button
post
optimization
firebase-realtime-database
web
jquery-ui
cocoa
xpath
iis
d3.js
javafx
firefox
xslt
internet-explorer
caching
select
asp.net-mvc-3
opengl
events
asp.net-web-api
plot
dplyr
encryption
magento
stored-procedures
search
amazon-ec2
ruby-on-rails-4
memory
canvas
audio
multidimensional-array
random
jsf
vector
redux
cookies
input
facebook-graph-api
flash
indexing
xamarin.forms
arraylist
ipad
cocoa-touch
data-structures
video
azure-devops
model-view-controller
apache-kafka
serialization
jdbc
woocommerce
razor
routes
awk
servlets
mod-rewrite
excel-formula
beautifulsoup
filter
docker-compose
iframe
aws-lambda
design-patterns
text
visual-c++
django-rest-framework
cakephp
mobile
android-intent
struct
react-hooks
methods
groovy
mvvm
ssh
lambda
checkbox
time
ecmascript-6
grails
google-chrome-extension
installation
cmake
sharepoint
shiny
spring-security
jakarta-ee
plsql
android-recyclerview
core-data
types
sed
meteor
android-activity
activerecord
bootstrap-4
websocket
graph
replace
scikit-learn
group-by
vim
file-upload
junit
boost
memory-management
sass
import
async-await
deep-learning
error-handling
eloquent
dynamic
soap
dependency-injection
silverlight
layout
apache-spark-sql
charts
deployment
browser
gridview
svn
while-loop
google-bigquery
vuejs2
dll
highcharts
ffmpeg
view
foreach
makefile
plugins
redis
c#-4.0
reporting-services
jupyter-notebook
unicode
merge
reflection
https
server
google-maps-api-3
twitter
oauth-2.0
extjs
terminal
axios
pip
split
cmd
pytorch
encoding
django-views
collections
database-design
hash
netbeans
automation
data-binding
ember.js
build
tcp
pdo
sqlalchemy
apache-flex
mysqli
entity-framework-core
concurrency
command-line
spring-data-jpa
printing
react-redux
java-8
lua
html-table
ansible
jestjs
neo4j
service
parameters
enums
material-ui
flexbox
module
promise
visual-studio-2012
outlook
firebase-authentication
web-applications
webview
uwp
jquery-mobile
utf-8
datatable
python-requests
parallel-processing
colors
drop-down-menu
scipy
scroll
tfs
hive
count
syntax
ms-word
twitter-bootstrap-3
ssis
fonts
rxjs
constructor
google-analytics
file-io
three.js
paypal
powerbi
graphql
cassandra
discord
graphics
compiler-errors
gwt
socket.io
react-router
solr
backbone.js
memory-leaks
url-rewriting
datatables
nlp
oauth
terraform
datagridview
drupal
oracle11g
zend-framework
knockout.js
triggers
neural-network
interface
django-forms
angular-material
casting
jmeter
google-api
linked-list
path
timer
django-templates
arduino
proxy
orm
directory
windows-phone-7
parse-platform
visual-studio-2015
cron
conditional-statements
push-notification
functional-programming
primefaces
pagination
model
jar
xamarin.android
hyperlink
uiview
visual-studio-2013
vbscript
google-cloud-functions
gitlab
azure-active-directory
jwt
download
swift3
sql-server-2005
configuration
process
rspec
pygame
properties
combobox
callback
windows-phone-8
linux-kernel
safari
scrapy
permissions
emacs
scripting
raspberry-pi
clojure
x86
scope
io
expo
azure-functions
compilation
responsive-design
mongodb-query
nhibernate
angularjs-directive
request
bluetooth
reference
binding
dns
architecture
3d
playframework
pyqt
version-control
discord.js
doctrine-orm
package
f#
rubygems
get
sql-server-2012
autocomplete
tree
openssl
datepicker
kendo-ui
jackson
yii
controller
grep
nested
xamarin.ios
static
null
statistics
transactions
active-directory
datagrid
dockerfile
uiviewcontroller
webforms
discord.py
phpmyadmin
sas
computer-vision
notifications
duplicates
mocking
youtube
pycharm
nullpointerexception
yaml
menu
blazor
sum
plotly
bitmap
asp.net-mvc-5
visual-studio-2008
yii2
electron
floating-point
css-selectors
stl
jsf-2
android-listview
time-series
cryptography
ant
hashmap
character-encoding
stream
msbuild
asp.net-core-mvc
sdk
google-drive-api
jboss
selenium-chromedriver
joomla
devise
cors
navigation
anaconda
cuda
background
frontend
binary
multiprocessing
pyqt5
camera
iterator
linq-to-sql
mariadb
onclick
android-jetpack-compose
ios7
microsoft-graph-api
rabbitmq
android-asynctask
tabs
laravel-4
environment-variables
amazon-dynamodb
insert
uicollectionview
linker
xsd
coldfusion
console
continuous-integration
upload
textview
ftp
opengl-es
macros
operating-system
mockito
localization
formatting
xml-parsing
vuejs3
json.net
type-conversion
data.table
kivy
timestamp
integer
calendar
segmentation-fault
android-ndk
prolog
drag-and-drop
char
crash
jasmine
dependencies
automated-tests
geometry
azure-pipelines
android-gradle-plugin
itext
fortran
sprite-kit
header
mfc
firebase-cloud-messaging
attributes
nosql
format
nuxt.js
odoo
db2
jquery-plugins
event-handling
jenkins-pipeline
nestjs
leaflet
julia
annotations
flutter-layout
keyboard
postman
textbox
arm
visual-studio-2017
gulp
stripe-payments
libgdx
synchronization
timezone
uikit
azure-web-app-service
dom-events
xampp
wso2
crystal-reports
namespaces
swagger
android-emulator
aggregation-framework
uiscrollview
jvm
google-sheets-formula
sequelize.js
com
chart.js
snowflake-cloud-data-platform
subprocess
geolocation
webdriver
html5-canvas
centos
garbage-collection
dialog
sql-update
widget
numbers
concatenation
qml
tuples
set
java-stream
smtp
mapreduce
ionic2
windows-10
rotation
android-edittext
modal-dialog
spring-data
nuget
doctrine
radio-button
http-headers
grid
sonarqube
lucene
xmlhttprequest
listbox
switch-statement
initialization
internationalization
components
apache-camel
boolean
google-play
serial-port
gdb
ios5
ldap
youtube-api
return
eclipse-plugin
pivot
latex
frameworks
tags
containers
github-actions
c++17
subquery
dataset
asp-classic
foreign-keys
label
embedded
uinavigationcontroller
copy
delegates
struts2
google-cloud-storage
migration
protractor
base64
queue
find
uibutton
sql-server-2008-r2
arguments
composer-php
append
jaxb
zip
stack
tailwind-css
cucumber
autolayout
ide
entity-framework-6
iteration
popup
r-markdown
windows-7
airflow
vb6
g++
ssl-certificate
hover
clang
jqgrid
range
gmail
Next You’ll be prompted to create an account to view your personalized homepage.
Home
Questions
AI Assist Labs
Tags
Challenges
Chat
Articles
Users
Jobs
Companies
Collectives
Communities for your favorite technologies. Explore all Collectives
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
Collectives™ on Stack Overflow
Find centralized, trusted content and collaborate around the technologies you use most.
Learn more about Collectives
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
Two Way Anova using Python
Ask Question
Asked 5 years ago
Modified3 years, 9 months ago
Viewed 6k times
This question shows research effort; it is useful and clear
4
Save this question.
Show activity on this post.
I am trying to do a two-way ANOVA, where I am trying to find the importance of two variables (B and M) on the classification of samples (given by the parameter C).
I am trying to reshape the data frame to make it suitable for statsmodels package. However, I have only been able to include one variable at a time (either B or M) using pd.melt.
Any suggestion on how can I use the values of both variables to perform the two-way ANOVA (in a way like the last two lines of the code given below) would be a great help.
The values of B, M and C:
```python
B : [10.,4.,4.,6.,5.]
M : [9.,6.,8.,4.,6.]
C : [1.,2.,2.,3.,1.]
import numpy as np
import pandas as pd
import statsmodels.api as sm
from statsmodels.formula.api import ols
d = pd.read_csv("/Users/Hrihaan/Desktop/Data.txt", sep="\s+")
d_melt = pd.melt(d, id_vars=['C'], value_vars=['B'])
model = ols('C ~ C(B) + C(M) + C(B):C(M)', data=d_melt).fit()
anova_table = sm.stats.anova_lm(model, typ=2)
```
python
pandas
statsmodels
anova
Share
Share a link to this question
Copy linkCC BY-SA 4.0
Improve this question
Follow
Follow this question to receive notifications
edited Sep 19, 2020 at 21:38
StupidWolf
47.1k 17 17 gold badges 49 49 silver badges 81 81 bronze badges
asked Sep 14, 2020 at 4:50
HrihaanHrihaan
285 8 8 silver badges 21 21 bronze badges
1
why would you convert B and M to categorical? And what exactly is C?StupidWolf –StupidWolf 2020-09-19 21:43:55 +00:00 Commented Sep 19, 2020 at 21:43
Add a comment|
1 Answer 1
Sorted by: Reset to default
This answer is useful
1
Save this answer.
Show activity on this post.
You were close to the answer:
```python
B = [10.,4.,4.,6.,5.]
M = [9.,6.,8.,4.,6.]
C = [1.,2.,2.,3.,1.]
import numpy as np
import pandas as pd
import statsmodels.api as sm
from statsmodels.formula.api import ols
d = pd.DataFrame()
d["B"]=B
d["M"]=M
d["C"]=C
model = ols("C ~ B + M + B:M",data = d).fit()
anova_table = sm.stats.anova_lm(model, typ=2)
```
You create a dataframe, you set your model, you perform the Anova
Share
Share a link to this answer
Copy linkCC BY-SA 4.0
Improve this answer
Follow
Follow this answer to receive notifications
answered Dec 14, 2021 at 15:15
user4624500user4624500
316 2 2 silver badges 11 11 bronze badges
1 Comment
Add a comment
Josef
JosefOver a year ago
this treats B and M as numeric and not as categorical
2021-12-14T18:14:35.807Z+00:00
0
Reply
Copy link
Your Answer
Thanks for contributing an answer to Stack Overflow!
Please be sure to answer the question. Provide details and share your research!
But avoid …
Asking for help, clarification, or responding to other answers.
Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Draft saved
Draft discarded
Sign up or log in
Sign up using Google
Sign up using Email and Password
Submit
Post as a guest
Name
Email
Required, but never shown
Post Your Answer Discard
By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.
Start asking to get answers
Find the answer to your question by asking.
Ask question
Explore related questions
python
pandas
statsmodels
anova
See similar questions with these tags.
The Overflow Blog
The history and future of software development (part 1)
Getting Backstage in front of a shifting dev experience
Featured on Meta
Spevacus has joined us as a Community Manager
Introducing a new proactive anti-spam measure
New and improved coding challenges
New comment UI experiment graduation
Policy: Generative AI (e.g., ChatGPT) is banned
Report this ad
Report this ad
15 people chatting
Python
1 hour ago - roganjosh
Related
7How can I perform a 2-way repeated measures anova using statsmodels?
2ANOVA in python
38ANOVA in python using pandas dataframe with statsmodels or scipy?
2ANOVA and HSD tests from Python dataframe
7Anova test for GLM in python
7How can I perform an n way ANOVA in python?
2stats.f_oneway Scipy Anova returns 2 arrays with 4 values
3One-way Anova loop through pandas dataframe - results in a single table
2How to obtain ANOVA table with statsmodels?
1One way Anova using STATSMODELS
Hot Network Questions
Is existence always locational?
What happens if you miss cruise ship deadline at private island?
On being a Maître de conférence (France): Importance of Postdoc
How to locate a leak in an irrigation system?
Is it ok to place components "inside" the PCB
Making sense of perturbation theory in many-body physics
Where is the first repetition in the cumulative hierarchy up to elementary equivalence?
Overfilled my oil
Riffle a list of binary functions into list of arguments to produce a result
Interpret G-code
Does the curvature engine's wake really last forever?
"Unexpected"-type comic story. Aboard a space ark/colony ship. Everyone's a vampire/werewolf
Does the mind blank spell prevent someone from creating a simulacrum of a creature using wish?
What is a "non-reversible filter"?
Why do universities push for high impact journal publications?
In the U.S., can patients receive treatment at a hospital without being logged?
Is encrypting the login keyring necessary if you have full disk encryption?
Can a cleric gain the intended benefit from the Extra Spell feat?
Proof of every Highly Abundant Number greater than 3 is Even
How do you create a no-attack area?
Why are LDS temple garments secret?
Why include unadjusted estimates in a study when reporting adjusted estimates?
What were "milk bars" in 1920s Japan?
Do we need the author's permission for reference
Question feed
Subscribe to RSS
Question feed
To subscribe to this RSS feed, copy and paste this URL into your RSS reader.
lang-py
Why are you flagging this comment?
Probable spam.
This comment promotes a product, service or website while failing to disclose the author's affiliation.
Unfriendly or contains harassment/bigotry/abuse.
This comment is unkind, insulting or attacks another person or group. Learn more in our Code of Conduct.
Not needed.
This comment is not relevant to the post.
Enter at least 6 characters
Something else.
A problem not listed above. Try to be as specific as possible.
Enter at least 6 characters
Flag comment Cancel
You have 0 flags left today
Stack Overflow
Questions
Help
Chat
Products
Teams
Advertising
Talent
Company
About
Press
Work Here
Legal
Privacy Policy
Terms of Service
Contact Us
Your Privacy Choices
Cookie Policy
Stack Exchange Network
Technology
Culture & recreation
Life & arts
Science
Professional
Business
API
Data
Blog
Facebook
Twitter
LinkedIn
Instagram
Site design / logo © 2025 Stack Exchange Inc; user contributions licensed under CC BY-SA. rev 2025.9.29.34589
By clicking “Accept all cookies”, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy.
Accept all cookies Necessary cookies only
Customize settings |
189093 | https://pressbooks.online.ucf.edu/phy2053bc/chapter/photon-energies-and-the-electromagnetic-spectrum/ | Skip to content
Chapter 29 Introduction to Quantum Physics
29.3 Photon Energies and the Electromagnetic Spectrum
Summary
Explain the relationship between the energy of a photon in joules or electron volts and its wavelength or frequency.
Calculate the number of photons per second emitted by a monochromatic source of specific wavelength and power.
Ionizing Radiation
A photon is a quantum of EM radiation. Its energy is given by and is related to the frequency and wavelength of the radiation by
where is the energy of a single photon and is the speed of light. When working with small systems, energy in eV is often useful. Note that Planck’s constant in these units is
Since many wavelengths are stated in nanometers (nm), it is also useful to know that
These will make many calculations a little easier.
All EM radiation is composed of photons. Figure 1 shows various divisions of the EM spectrum plotted against wavelength, frequency, and photon energy. Previously in this book, photon characteristics were alluded to in the discussion of some of the characteristics of UV, x rays, and rays, the first of which start with frequencies just above violet in the visible spectrum. It was noted that these types of EM radiation have characteristics much different than visible light. We can now see that such properties arise because photon energy is larger at high frequencies.
| | |
--- |
| Rotational energies of molecules | eV |
| Vibrational energies of molecules | 0.1 eV |
| Energy between outer electron shells in atoms | 1 eV |
| Binding energy of a weakly bound molecule | 1 eV |
| Energy of red light | 2 eV |
| Binding energy of a tightly bound molecule | 10 eV |
| Energy to ionize atom or molecule | 10 to 1000 eV |
| Table 1: Representative Energies for Submicroscopic Effects (Order of Magnitude Only) | |
Photons act as individual quanta and interact with individual electrons, atoms, molecules, and so on. The energy a photon carries is, thus, crucial to the effects it has. Table 1 lists representative submicroscopic energies in eV. When we compare photon energies from the EM spectrum in Figure 1 with energies in the table, we can see how effects vary with the type of EM radiation.
Gamma rays, a form of nuclear and cosmic EM radiation, can have the highest frequencies and, hence, the highest photon energies in the EM spectrum. For example, a -ray photon with has an energy . This is sufficient energy to ionize thousands of atoms and molecules, since only 10 to 1000 eV are needed per ionization. In fact, rays are one type of ionizing radiation, as are x rays and UV, because they produce ionization in materials that absorb them. Because so much ionization can be produced, a single -ray photon can cause significant damage to biological tissue, killing cells or damaging their ability to properly reproduce. When cell reproduction is disrupted, the result can be cancer, one of the known effects of exposure to ionizing radiation. Since cancer cells are rapidly reproducing, they are exceptionally sensitive to the disruption produced by ionizing radiation. This means that ionizing radiation has positive uses in cancer treatment as well as risks in producing cancer.
Figure 2. One of the first x-ray images, taken by Röentgen himself. The hand belongs to Bertha Röentgen, his wife. (credit: Wilhelm Conrad Röntgen, via Wikimedia Commons)
High photon energy also enables rays to penetrate materials, since a collision with a single atom or molecule is unlikely to absorb all the ray’s energy. This can make rays useful as a probe, and they are sometimes used in medical imaging. x rays, as you can see in Figure 1, overlap with the low-frequency end of the ray range. Since x rays have energies of keV and up, individual x-ray photons also can produce large amounts of ionization. At lower photon energies, x rays are not as penetrating as rays and are slightly less hazardous. X rays are ideal for medical imaging, their most common use, and a fact that was recognized immediately upon their discovery in 1895 by the German physicist W. C. Roentgen (1845–1923). (See Figure 2.) Within one year of their discovery, x rays (for a time called Roentgen rays) were used for medical diagnostics. Roentgen received the 1901 Nobel Prize for the discovery of x rays.
Connections: Conservation of Energy
Once again, we find that conservation of energy allows us to consider the initial and final forms that energy takes, without having to make detailed calculations of the intermediate steps. Example 1 is solved by considering only the initial and final forms of energy.
Figure 3. X rays are produced when energetic electrons strike the copper anode of this cathode ray tube (CRT). Electrons (shown here as separate particles) interact individually with the material they strike, sometimes producing photons of EM radiation.
While rays originate in nuclear decay, x rays are produced by the process shown in Figure 3. Electrons ejected by thermal agitation from a hot filament in a vacuum tube are accelerated through a high voltage, gaining kinetic energy from the electrical potential energy. When they strike the anode, the electrons convert their kinetic energy to a variety of forms, including thermal energy. But since an accelerated charge radiates EM waves, and since the electrons act individually, photons are also produced. Some of these x-ray photons obtain the kinetic energy of the electron. The accelerated electrons originate at the cathode, so such a tube is called a cathode ray tube (CRT), and various versions of them are found in older TV and computer screens as well as in x-ray machines.
Example 1: X-ray Photon Energy and X-ray Tube Voltage
Find the maximum energy in eV of an x-ray photon produced by electrons accelerated through a potential difference of 50.0 kV in a CRT like the one in Figure 3.
Strategy
Electrons can give all of their kinetic energy to a single photon when they strike the anode of a CRT. (This is something like the photoelectric effect in reverse.) The kinetic energy of the electron comes from electrical potential energy. Thus we can simply equate the maximum photon energy to the electrical potential energy—that is, . (We do not have to calculate each step from beginning to end if we know that all of the starting energy is converted to the final form .)
Solution
The maximum photon energy is , where is the charge of the electron and is the accelerating voltage. Thus,
From the definition of the electron volt, we know , where . Gathering factors and converting energy to eV yields
Discussion
This example produces a result that can be applied to many similar situations. If you accelerate a single elementary charge, like that of an electron, through a potential given in volts, then its energy in eV has the same numerical value. Thus a 50.0-kV potential generates 50.0 keV electrons, which in turn can produce photons with a maximum energy of 50 keV. Similarly, a 100-kV potential in an x-ray tube can generate up to 100-keV x-ray photons. Many x-ray tubes have adjustable voltages so that various energy x rays with differing energies, and therefore differing abilities to penetrate, can be generated.
Figure 4. X-ray spectrum obtained when energetic electrons strike a material. The smooth part of the spectrum is bremsstrahlung, while the peaks are characteristic of the anode material. Both are atomic processes that produce energetic photons known as x-ray photons.
Figure 4 shows the spectrum of x rays obtained from an x-ray tube. There are two distinct features to the spectrum. First, the smooth distribution results from electrons being decelerated in the anode material. A curve like this is obtained by detecting many photons, and it is apparent that the maximum energy is unlikely. This decelerating process produces radiation that is called bremsstrahlung (German for braking radiation). The second feature is the existence of sharp peaks in the spectrum; these are called characteristic x rays, since they are characteristic of the anode material. Characteristic x rays come from atomic excitations unique to a given type of anode material. They are akin to lines in atomic spectra, implying the energy levels of atoms are quantized. Phenomena such as discrete atomic spectra and characteristic x rays are explored further in Chapter 30 Atomic Physics.
Ultraviolet radiation (approximately 4 eV to 300 eV) overlaps with the low end of the energy range of x rays, but UV is typically lower in energy. UV comes from the de-excitation of atoms that may be part of a hot solid or gas. These atoms can be given energy that they later release as UV by numerous processes, including electric discharge, nuclear explosion, thermal agitation, and exposure to x rays. A UV photon has sufficient energy to ionize atoms and molecules, which makes its effects different from those of visible light. UV thus has some of the same biological effects as rays and x rays. For example, it can cause skin cancer and is used as a sterilizer. The major difference is that several UV photons are required to disrupt cell reproduction or kill a bacterium, whereas single -ray and X-ray photons can do the same damage. But since UV does have the energy to alter molecules, it can do what visible light cannot. One of the beneficial aspects of UV is that it triggers the production of vitamin D in the skin, whereas visible light has insufficient energy per photon to alter the molecules that trigger this production. Infantile jaundice is treated by exposing the baby to UV (with eye protection), called phototherapy, the beneficial effects of which are thought to be related to its ability to help prevent the buildup of potentially toxic bilirubin in the blood.
Example 2: Photon Energy and Effects for UV
Short-wavelength UV is sometimes called vacuum UV, because it is strongly absorbed by air and must be studied in a vacuum. Calculate the photon energy in eV for 100-nm vacuum UV, and estimate the number of molecules it could ionize or break apart.
Strategy
Using the equation and appropriate constants, we can find the photon energy and compare it with energy information in Table 1.
Solution
The energy of a photon is given by
.
Using , we find that
.
Discussion
According to Table 1, this photon energy might be able to ionize an atom or molecule, and it is about what is needed to break up a tightly bound molecule, since they are bound by approximately 10 eV. This photon energy could destroy about a dozen weakly bound molecules. Because of its high photon energy, UV disrupts atoms and molecules it interacts with. One good consequence is that all but the longest-wavelength UV is strongly absorbed and is easily blocked by sunglasses. In fact, most of the Sun’s UV is absorbed by a thin layer of ozone in the upper atmosphere, protecting sensitive organisms on Earth. Damage to our ozone layer by the addition of such chemicals as CFC’s has reduced this protection for us.
Visible Light
The range of photon energies for visible light from red to violet is 1.63 to 3.26 eV, respectively (left for this chapter’s Problems and Exercises to verify). These energies are on the order of those between outer electron shells in atoms and molecules. This means that these photons can be absorbed by atoms and molecules. A single photon can actually stimulate the retina, for example, by altering a receptor molecule that then triggers a nerve impulse. Photons can be absorbed or emitted only by atoms and molecules that have precisely the correct quantized energy step to do so. For example, if a red photon of frequency encounters a molecule that has an energy step, , equal to , then the photon can be absorbed. Violet flowers absorb red and reflect violet; this implies there is no energy step between levels in the receptor molecule equal to the violet photon’s energy, but there is an energy step for the red.
There are some noticeable differences in the characteristics of light between the two ends of the visible spectrum that are due to photon energies. Red light has insufficient photon energy to expose most black-and-white film, and it is thus used to illuminate darkrooms where such film is developed. Since violet light has a higher photon energy, dyes that absorb violet tend to fade more quickly than those that do not. (See Figure 5.) Take a look at some faded color posters in a storefront some time, and you will notice that the blues and violets are the last to fade. This is because other dyes, such as red and green dyes, absorb blue and violet photons, the higher energies of which break up their weakly bound molecules. (Complex molecules such as those in dyes and DNA tend to be weakly bound.) Blue and violet dyes reflect those colors and, therefore, do not absorb these more energetic photons, thus suffering less molecular damage.
Figure 5. Why do the reds, yellows, and greens fade before the blues and violets when exposed to the Sun, as with this poster? The answer is related to photon energy. (credit: Deb Collins, Flickr)
Transparent materials, such as some glasses, do not absorb any visible light, because there is no energy step in the atoms or molecules that could absorb the light. Since individual photons interact with individual atoms, it is nearly impossible to have two photons absorbed simultaneously to reach a large energy step. Because of its lower photon energy, visible light can sometimes pass through many kilometers of a substance, while higher frequencies like UV, x ray, and rays are absorbed, because they have sufficient photon energy to ionize the material.
Example 3: How Many Photons per Second Does a Typical Light Bulb Produce?
Assuming that 10.0% of a 100-W light bulb’s energy output is in the visible range (typical for incandescent bulbs) with an average wavelength of 580 nm, calculate the number of visible photons emitted per second.
Strategy
Power is energy per unit time, and so if we can find the energy per photon, we can determine the number of photons per second. This will best be done in joules, since power is given in watts, which are joules per second.
Solution
The power in visible light production is 10.0% of 100 W, or 10.0 J/s. The energy of the average visible photon is found by substituting the given average wavelength into the formula
.
This produces
.
The number of visible photons per second is thus
.
Discussion
This incredible number of photons per second is verification that individual photons are insignificant in ordinary human experience. It is also a verification of the correspondence principle—on the macroscopic scale, quantization becomes essentially continuous or classical. Finally, there are so many photons emitted by a 100-W lightbulb that it can be seen by the unaided eye many kilometers away.
Lower-Energy Photons
Infrared radiation (IR) has even lower photon energies than visible light and cannot significantly alter atoms and molecules. IR can be absorbed and emitted by atoms and molecules, particularly between closely spaced states. IR is extremely strongly absorbed by water, for example, because water molecules have many states separated by energies on the order of to , well within the IR and microwave energy ranges. This is why in the IR range, skin is almost jet black, with an emissivity near 1—there are many states in water molecules in the skin that can absorb a large range of IR photon energies. Not all molecules have this property. Air, for example, is nearly transparent to many IR frequencies.
Microwaves are the highest frequencies that can be produced by electronic circuits, although they are also produced naturally. Thus microwaves are similar to IR but do not extend to as high frequencies. There are states in water and other molecules that have the same frequency and energy as microwaves, typically about . This is one reason why food absorbs microwaves more strongly than many other materials, making microwave ovens an efficient way of putting energy directly into food.
Photon energies for both IR and microwaves are so low that huge numbers of photons are involved in any significant energy transfer by IR or microwaves (such as warming yourself with a heat lamp or cooking pizza in the microwave). Visible light, IR, microwaves, and all lower frequencies cannot produce ionization with single photons and do not ordinarily have the hazards of higher frequencies. When visible, IR, or microwave radiation is hazardous, such as the inducement of cataracts by microwaves, the hazard is due to huge numbers of photons acting together (not to an accumulation of photons, such as sterilization by weak UV). The negative effects of visible, IR, or microwave radiation can be thermal effects, which could be produced by any heat source. But one difference is that at very high intensity, strong electric and magnetic fields can be produced by photons acting together. Such electromagnetic fields (EMF) can actually ionize materials.
Misconception Alert: High-Voltage Power Lines
Although some people think that living near high-voltage power lines is hazardous to one’s health, ongoing studies of the transient field effects produced by these lines show their strengths to be insufficient to cause damage. Demographic studies also fail to show significant correlation of ill effects with high-voltage power lines. The American Physical Society issued a report over 10 years ago on power-line fields, which concluded that the scientific literature and reviews of panels show no consistent, significant link between cancer and power-line fields. They also felt that the “diversion of resources to eliminate a threat which has no persuasive scientific basis is disturbing.”
It is virtually impossible to detect individual photons having frequencies below microwave frequencies, because of their low photon energy. But the photons are there. A continuous EM wave can be modeled as photons. At low frequencies, EM waves are generally treated as time- and position-varying electric and magnetic fields with no discernible quantization. This is another example of the correspondence principle in situations involving huge numbers of photons.
PhET Explorations: Color Vision
Make a whole rainbow by mixing red, green, and blue light. Change the wavelength of a monochromatic beam or filter white light. View the light as a solid beam, or see the individual photons.
Figure 6. Color Vision
Section Summary
Photon energy is responsible for many characteristics of EM radiation, being particularly noticeable at high frequencies.
Photons have both wave and particle characteristics.
Conceptual Questions
1: Why are UV, x rays, and rays called ionizing radiation?
2: How can treating food with ionizing radiation help keep it from spoiling? UV is not very penetrating. What else could be used?
3: Some television tubes are CRTs. They use an approximately 30-kV accelerating potential to send electrons to the screen, where the electrons stimulate phosphors to emit the light that forms the pictures we watch. Would you expect x rays also to be created?
4: Tanning salons use “safe” UV with a longer wavelength than some of the UV in sunlight. This “safe” UV has enough photon energy to trigger the tanning mechanism. Is it likely to be able to cause cell damage and induce cancer with prolonged exposure?
5: Your pupils dilate when visible light intensity is reduced. Does wearing sunglasses that lack UV blockers increase or decrease the UV hazard to your eyes? Explain.
6: One could feel heat transfer in the form of infrared radiation from a large nuclear bomb detonated in the atmosphere 75 km from you. However, none of the profusely emitted x rays or rays reaches you. Explain.
7: Can a single microwave photon cause cell damage? Explain.
8: In an x-ray tube, the maximum photon energy is given by . Would it be technically more correct to say , where BE is the binding energy of electrons in the target anode? Why isn’t the energy stated the latter way?
Problems & Exercises
1: What is the energy in joules and eV of a photon in a radio wave from an AM station that has a 1530-kHz broadcast frequency?
2: (a) Find the energy in joules and eV of photons in radio waves from an FM station that has a 90.0-MHz broadcast frequency. (b) What does this imply about the number of photons per second that the radio station must broadcast?
3: Calculate the frequency in hertz of a 1.00-MeV -ray photon.
4: (a) What is the wavelength of a 1.00-eV photon? (b) Find its frequency in hertz. (c) Identify the type of EM radiation.
5: Do the unit conversions necessary to show that , as stated in the text.
6: Confirm the statement in the text that the range of photon energies for visible light is 1.63 to 3.26 eV, given that the range of visible wavelengths is 380 to 760 nm.
7: (a) Calculate the energy in eV of an IR photon of frequency . (b) How many of these photons would need to be absorbed simultaneously by a tightly bound molecule to break it apart? (c) What is the energy in eV of a ray of frequency ? (d) How many tightly bound molecules could a single such ray break apart?
8: Prove that, to three-digit accuracy, , as stated in the text.
9: (a) What is the maximum energy in eV of photons produced in a CRT using a 25.0-kV accelerating potential, such as a color TV? (b) What is their frequency?
10: What is the accelerating voltage of an x-ray tube that produces x rays with a shortest wavelength of 0.0103 nm?
11: (a) What is the ratio of power outputs by two microwave ovens having frequencies of 950 and 2560 MHz, if they emit the same number of photons per second? (b) What is the ratio of photons per second if they have the same power output?
12: How many photons per second are emitted by the antenna of a microwave oven, if its power output is 1.00 kW at a frequency of 2560 MHz?
13: Some satellites use nuclear power. (a) If such a satellite emits a 1.00-W flux of rays having an average energy of 0.500 MeV, how many are emitted per second? (b) These rays affect other satellites. How far away must another satellite be to only receive one ray per second per square meter?
14: (a) If the power output of a 650-kHz radio station is 50.0 kW, how many photons per second are produced? (b) If the radio waves are broadcast uniformly in all directions, find the number of photons per second per square meter at a distance of 100 km. Assume no reflection from the ground or absorption by the air.
15: How many x-ray photons per second are created by an x-ray tube that produces a flux of x rays having a power of 1.00 W? Assume the average energy per photon is 75.0 keV.
16: (a) How far away must you be from a 650-kHz radio station with power 50.0 kW for there to be only one photon per second per square meter? Assume no reflections or absorption, as if you were in deep outer space. (b) Discuss the implications for detecting intelligent life in other solar systems by detecting their radio broadcasts.
17: Assuming that 10.0% of a 100-W light bulb’s energy output is in the visible range (typical for incandescent bulbs) with an average wavelength of 580 nm, and that the photons spread out uniformly and are not absorbed by the atmosphere, how far away would you be if 500 photons per second enter the 3.00-mm diameter pupil of your eye? (This number easily stimulates the retina.)
18: Construct Your Own Problem
Consider a laser pen. Construct a problem in which you calculate the number of photons per second emitted by the pen. Among the things to be considered are the laser pen’s wavelength and power output. Your instructor may also wish for you to determine the minimum diffraction spreading in the beam and the number of photons per square centimeter the pen can project at some large distance. In this latter case, you will also need to consider the output size of the laser beam, the distance to the object being illuminated, and any absorption or scattering along the way.
Glossary
gamma ray
: also -ray; highest-energy photon in the EM spectrum
ionizing radiation
: radiation that ionizes materials that absorb it
x ray
: EM photon between -ray and UV in energy
bremsstrahlung
: German for braking radiation; produced when electrons are decelerated
characteristic x rays
: x rays whose energy depends on the material they were produced in
ultraviolet radiation
: UV; ionizing photons slightly more energetic than violet light
visible light
: the range of photon energies the human eye can detect
infrared radiation
: photons with energies slightly less than red light
microwaves
: photons with wavelengths on the order of a micron ()
Solutions
Problems & Exercises
1: ,
3:
5:
7: (a) 0.0829 eV
(b) 121
(c) 1.24 MeV
(d)
9: (a)
(b)
11: (a) 2.69
(b) 0.371
13: (a)
(b) 997 km
15:
17: 181 km |
189094 | https://docs.huihoo.com/godotengine/godot-docs/godot/tutorials/vector_math.html | Vector math — Godot Engine latest documentation
Godot Engine
latest
Tutorials
Learning step by step
Engine
2D tutorials
3D tutorials
Networking
Editor plugins
Miscellaneous
Math
Vector math
Introduction
Coordinate systems (2D)
Vectors in Godot
Perpendicular vectors
Unit vectors
Dot product
Siding with unit vectors
Planes
Area of a triangle
Plane of the triangle
Collision detection in 3D
Matrices and transforms
Shaders
Asset pipeline
Reference
Class reference
Languages
Cheat sheets
Compiling
Advanced
Community
Contributing
Godot Engine
Docs »
Miscellaneous »
Math »
Vector math
Edit on GitHub
Vector math¶
Introduction¶
This small tutorial aims to be a short and practical introduction to vector math, useful for 3D but also 2D games. Again, vector math is not only useful for 3D but also 2D games. It is an amazing tool once you get the grasp of it and makes programming of complex behaviors much simpler.
It often happens that young programmers rely too much on the incorrect math for solving a wide array of problems, for example using only trigonometry instead of vector of math for 2D games.
This tutorial will focus on practical usage, with immediate application to the art of game programming.
Coordinate systems (2D)¶
Typically, we define coordinates as an (x,y) pair, x representing the horizontal offset and y the vertical one. This makes sense given the screen is just a rectangle in two dimensions. As an example, here is a position in 2D space:
A position can be anywhere in space. The position (0,0) has a name, it’s called the origin. Remember this term well because it has more implicit uses later. The (0,0) of a n-dimensions coordinate system is the origin.
In vector math, coordinates have two different uses, both equally important. They are used to represent a position but also a vector. The same position as before, when imagined as a vector, has a different meaning.
When imagined as a vector, two properties can be inferred, the direction and the magnitude. Every position in space can be a vector, with the exception of the origin. This is because coordinates (0,0) can’t represent direction (magnitude 0).
Direction¶
Direction is simply towards where the vector points to. Imagine an arrow that starts at the origin and goes towards a [STRIKEOUT:position]. The tip of the arrow is in the position, so it always points outwards, away from the origin. Imagining vectors as arrows helps a lot.
Magnitude¶
Finally, the length of the vector is the distance from the origin to the position. Obtaining the length from a vector is easy, just use the Pythagorean Theorem.
var len = sqrt( xx + yy )
But... angles?¶
But why not using an angle? After all, we could also think of a vector as an angle and a magnitude, instead of a direction and a magnitude. Angles also are a more familiar concept.
To say truth, angles are not that useful in vector math, and most of the time they are not dealt with directly. Maybe they work in 2D, but in 3D a lot of what can usually be done with angles does not work anymore.
Still, using angles is still not an excuse, even for 2D. Most of what takes a lot of work with angles in 2D, is still much more natural easier to accomplish with vector math. In vector math, angles are useful only as measure, but take little part in the math. So, give up the trigonometry already, prepare to embrace vectors!
In any case, obtaining an angle from a vector is easy and can be accomplished with trig... er, what was that? I mean, the atan2() function.
Vectors in Godot¶
To make examples easier, it is worth explaining how vectors are implemented in GDScript. GDscript has both Vector2 and Vector3, for 2D and 3D math respectively. Godot uses Vector classes as both position and direction. They also contain x and y (for 2D) and x, y and z (for 3D) member variables.
create a vector with coordinates (2,5)
var a = Vector2(2,5)
create a vector and assign x and y manually
var b = Vector2()
b.x = 7
b.y = 8
When operating with vectors, it is not necessary to operate on the members directly (in fact this is much slower). Vectors support regular arithmetic operations:
add a and b
var c = a + b
will result in c vector, with value (9,13)
It is the same as doing:
var c = Vector2()
c.x = a.x + b.x
c.y = a.y + b.y
Except the former is way more efficient and readable.
Regular arithmetic operations such as addition, subtraction, multiplication and division are supported.
Vector multiplication and division can also be mixed with single-digit numbers, also named scalars.
multiplication of vector by scalar
var c = a2.0
will result in c vector, with value (4,10)
Which is the same as doing
var c = Vector2()
c.x = a.x2.0
c.y = a.y2.0
Except, again, the former is way more efficient and readable.
Perpendicular vectors¶
Rotating a 2D vector 90° degrees to either side, left or right, is really easy, just swap x and y, then negate either x or y (direction of rotation depends on which is negated).
Example:
var v = Vector2(0,1)
rotate right (clockwise)
var v_right = Vector2(-v.y, v.x)
rotate left (counter-clockwise)
var v_right = Vector2(v.y, -v.x)
This is a handy trick that is often of use. It is impossible to do with 3D vectors, because there are an infinite amount of perpendicular vectors.
Unit vectors¶
Ok, so we know what a vector is. It has a direction and a magnitude. We also know how to use them in Godot. The next step is learning about unit vectors. Any vector with magnitude of length 1 is considered a unit vector. In 2D, imagine drawing a circle of radius one. That circle contains all unit vectors in existence for 2 dimensions:
So, what is so special about unit vectors? Unit vectors are amazing. In other words, unit vectors have several, very useful properties.
Can’t wait to know more about the fantastic properties of unit vectors, but one step at a time. So, how is a unit vector created from a regular vector?
Normalization¶
Taking any vector and reducing it’s magnitude to 1.0 while keeping it’s direction is called normalization. Normalization is performed by dividing the x and y (and z in 3D) components of a vector by it’s magnitude:
var a = Vector2(2,4)
var m = sqrt(a.xa.x + a.ya.y)
a.x /= m
a.y /= m
As you might have guessed, if the vector has magnitude 0 (meaning, it’s not a vector but the origin also called null vector), a division by zero occurs and the universe goes through a second big bang, except in reverse polarity and then back. As a result, humanity is safe but Godot will print an error. Remember! Vector(0,0) can’t be normalized!.
Of course, Vector2 and Vector3 already provide a method to do this:
a = a.normalized()
Dot product¶
OK, the dot product is the most important part of vector math. Without the dot product, Quake would have never been made. This is the most important section of the tutorial, so make sure to grasp it properly. Most people trying to understand vector math give up here because, despite how simple it is, they can’t make head or tails from it. Why? Here’s why, it’s because...
The dot product takes two vectors and returns a scalar:
var s = a.xb.x + a.yb.y
Yes, pretty much that. Multiply x from vector a by x from vector b. Do the same with y and add it together. In 3D it’s pretty much the same:
var s = a.xb.x + a.yb.y + a.zb.z
I know, it’s totally meaningless! You can even do it with a built-in function:
var s = a.dot(b)
The order of two vectors does not matter, a.dot(b) returns the same value as b.dot(a).
This is where despair begins and books and tutorials show you this formula:
And you realize it’s time to give up making 3D games or complex 2D games. How can something so simple be so complex? Someone else will have to make the next Zelda or Call of Duty. Top down RPGs don’t look so bad after all. Yeah I hear someone did pretty will with one of those on Steam...
So this is your moment, this is your time to shine. DO NOT GIVE UP! At this point, this tutorial will take a sharp turn and focus on what makes the dot product useful. This is, why it is useful. We will focus one by one in the use cases for the dot product, with real-life applications. No more formulas that don’t make any sense. Formulas will make sense once you learn what they are useful for.
Siding¶
The first useful and most important property of the dot product is to check what side stuff is looking at. Let’s imagine we have any two vectors, a and b. Any direction or magnitude (neither origin). Does not matter what they are, but let’s imagine we compute the dot product between them.
var s = a.dot(b)
The operation will return a single floating point number (but since we are in vector world, we call them scalar, will keep using that term from now on). This number will tell us the following:
If the number is greater than zero, both are looking towards the same direction (the angle between them is < 90° degrees).
If the number is less than zero, both are looking towards opposite direction (the angle between them is > 90° degrees).
If the number is zero, vectors are shaped in L (the angle between them is 90° degrees).
So let’s think of a real use-case scenario. Imagine Snake is going through a forest, and then there is an enemy nearby. How can we quickly tell if the enemy has seen discovered Snake? In order to discover him, the enemy must be able to see Snake. Let’s say, then that:
Snake is in position A.
The enemy is in position B.
The enemy is facing towards direction vector F.
So, let’s create a new vector BA that goes from the guard (B) to Snake (A), by subtracting the two:
var BA = A - B
Ideally, if the guard was looking straight towards snake, to make eye to eye contact, it would do it in the same direction as vector BA.
If the dot product between F and BA is greater than 0, then Snake will be discovered. This happens because we will be able to tell that the guard is facing towards him:
if (BA.dot(F) > 0):
print("!")
Seems Snake is safe so far.
Siding with unit vectors¶
Ok, so now we know that dot product between two vectors will let us know if they are looking towards the same side, opposite sides or are just perpendicular to each other.
This works the same with all vectors, no matter the magnitude so unit vectors are not the exception. However, using the same property with unit vectors yields an even more interesting result, as an extra property is added:
If both vectors are facing towards the exact same direction (parallel to each other, angle between them is 0°), the resulting scalar is 1.
If both vectors are facing towards the exact opposite direction (parallel to each other, but angle between them is 180°), the resulting scalar is -1.
This means that dot product between unit vectors is always between the range of 1 and -1. So Again...
If their angle is 0° dot product is 1.
If their angle is 90°, then dot product is 0.
If their angle is 180°, then dot product is -1.
Uh.. this is oddly familiar... seen this before... where?
Let’s take two unit vectors. The first one is pointing up, the second too but we will rotate it all the way from up (0°) to down (180° degrees)...
While plotting the resulting scalar!
Aha! It all makes sense now, this is a Cosine function!
We can say that, then, as a rule...
The dot product between two unit vectors is the cosine of the angle between those two vectors. So, to obtain the angle between two vectors, we must do:
var angle_in_radians = acos( a.dot(b) )
What is this useful for? Well obtaining the angle directly is probably not as useful, but just being able to tell the angle is useful for reference. One example is in the Kinematic Character demo, when the character moves in a certain direction then we hit an object. How to tell if what we hit is the floor?
By comparing the normal of the collision point with a previously computed angle.
The beauty of this is that the same code works exactly the same and without modification in 3D. Vector math is, in a great deal, dimension-amount-independent, so adding or removing an axis only adds very little complexity.
Planes¶
The dot product has another interesting property with unit vectors. Imagine that perpendicular to that vector (and through the origin) passes a plane. Planes divide the entire space into positive (over the plane) and negative (under the plane), and (contrary to popular belief) you can also use their math in 2D:
Unit vectors that are perpendicular to a surface (so, they describe the orientation of the surface) are called unit normal vectors. Though, usually they are just abbreviated as normals. Normals appear in planes, 3D geometry (to determine where each face or vertex is siding), etc. A normalis a unit vector, but it’s called normal because of it’s usage. (Just like we call Origin to (0,0)!).
It’s as simple as it looks. The plane passes by the origin and the surface of it is perpendicular to the unit vector (or normal). The side towards the vector points to is the positive half-space, while the other side is the negative half-space. In 3D this is exactly the same, except that the plane is an infinite surface (imagine an infinite, flat sheet of paper that you can orient and is pinned to the origin) instead of a line.
Distance to plane¶
Now that it’s clear what a plane is, let’s go back to the dot product. The dot product between a unit vector and any point in space (yes, this time we do dot product between vector and position), returns the distance from the point to the plane:
var distance = normal.dot(point)
But not just the absolute distance, if the point is in the negative half space the distance will be negative, too:
This allows us to tell which side of the plane a point is.
Away from the origin¶
I know what you are thinking! So far this is nice, but real planes are everywhere in space, not only passing through the origin. You want real plane action and you want it now.
Remember that planes not only split space in two, but they also have polarity. This means that it is possible to have perfectly overlapping planes, but their negative and positive half-spaces are swapped.
With this in mind, let’s describe a full plane as a normalN and a distance from the origin scalar D. Thus, our plane is represented by N and D. For example:
For 3D math, Godot provides a Plane built-in type that handles this.
Basically, N and D can represent any plane in space, be it for 2D or 3D (depending on the amount of dimensions of N) and the math is the same for both. It’s the same as before, but D is the distance from the origin to the plane, travelling in N direction. As an example, imagine you want to reach a point in the plane, you will just do:
var point_in_plane = ND
This will stretch (resize) the normal vector and make it touch the plane. This math might seem confusing, but it’s actually much simpler than it seems. If we want to tell, again, the distance from the point to the plane, we do the same but adjusting for distance:
var distance = N.dot(point) - D
The same thing, using a built-in function:
var distance = plane.distance_to(point)
This will, again, return either a positive or negative distance.
Flipping the polarity of the plane is also very simple, just negate both N and D. This will result in a plane in the same position, but with inverted negative and positive half spaces:
N = -N
D = -D
Of course, Godot also implements this operator in Plane, so doing:
var inverted_plane = -plane
Will work as expected.
So, remember, a plane is just that and it’s main practical use is calculating the distance to it. So, why is it useful to calculate the distance from a point to a plane? It’s extremely useful! Let’s see some simple examples..
Constructing a plane in 2D¶
Planes clearly don’t come out of nowhere, so they must be built. Constructing them in 2D is easy, this can be done from either a normal (unit vector) and a point, or from two points in space.
In the case of a normal and a point, most of the work is done, as the normal is already computed, so just calculate D from the dot product of the normal and the point.
var N = normal
var D = normal.dot(point)
For two points in space, there are actually two planes that pass through them, sharing the same space but with normal pointing to the opposite directions. To compute the normal from the two points, the direction vector must be obtained first, and then it needs to be rotated 90° degrees to either side:
calculate vector from a to b
var dvec = (point_b - point_a).normalized()
rotate 90 degrees
var normal = Vector2(dvec.y,-dev.x)
or alternatively
var normal = Vector2(-dvec.y,dev.x)
depending the desired side of the normal
The rest is the same as the previous example, either point_a or point_b will work since they are in the same plane:
var N = normal
var D = normal.dot(point_a)
this works the same
var D = normal.dot(point_b)
Doing the same in 3D is a little more complex and will be explained further down.
Some examples of planes¶
Here is a simple example of what planes are useful for. Imagine you have a convex polygon. For example, a rectangle, a trapezoid, a triangle, or just any polygon where faces that don’t bend inwards.
For every segment of the polygon, we compute the plane that passes by that segment. Once we have the list of planes, we can do neat things, for example checking if a point is inside the polygon.
We go through all planes, if we can find a plane where the distance to the point is positive, then the point is outside the polygon. If we can’t, then the point is inside.
Code should be something like this:
var inside = true
for p in planes:
# check if distance to plane is positive
if (N.dot(point) - D > 0):
inside = false
break # with one that fails, it's enough
Pretty cool, huh? But this gets much better! With a little more effort, similar logic will let us know when two convex polygons are overlapping too. This is called the Separating Axis Theorem (or SAT) and most physics engines use this to detect collision.
The idea is really simple! With a point, just checking if a plane returns a positive distance is enough to tell if the point is outside. With another polygon, we must find a plane where _all the other_ polygon points return a positive distance to it. This check is performed with the planes of A against the points of B, and then with the planes of B against the points of A:
Code should be something like this:
var overlapping = true
for p in planes_of_A:
var all_out = true
for v in points_of_B:
if (p.distance_to(v) < 0):
all_out = false
break
if (all_out):
# a separating plane was found
# do not continue testing
overlapping = false
break
if (overlapping):
# only do this check if no separating plane
# was found in planes of A
for p in planes_of_B:
var all_out = true
for v in points_of_A:
if (p.distance_to(v) < 0):
all_out = false
break
if (all_out):
overlapping = false
break
if (overlapping):
print("Polygons Collided!")
As you can see, planes are quite useful, and this is the tip of the iceberg. You might be wondering what happens with non convex polygons. This is usually just handled by splitting the concave polygon into smaller convex polygons, or using a technique such as BSP (which is not used much nowadays).
Cross product¶
Quite a lot can be done with the dot product! But the party would not be complete without the cross product. Remember back at the beginning of this tutorial? Specifically how to obtain a perpendicular (rotated 90 degrees) vector by swapping x and y, then negating either of them for right (clockwise) or left (counter-clockwise) rotation? That ended up being useful for calculating a 2D plane normal from two points.
As mentioned before, no such thing exists in 3D because a 3D vector has infinite perpendicular vectors. It would also not make sense to obtain a 3D plane from 2 points, as 3 points are needed instead.
To aid in this kind stuff, the brightest minds of humanity’s top mathematicians brought us the cross product.
The cross product takes two vectors and returns another vector. The returned third vector is always perpendicular to the first two. The source vectors, of course, must not be the same, and must not be parallel or opposite, else the resulting vector will be (0,0,0):
The formula for the cross product is:
var c = Vector3()
c.x = (a.y + b.z) - (a.z + b.y)
c.y = (a.z + b.x) - (a.x + b.z)
c.z = (a.x + b.y) - (a.y + b.x)
This can be simplified, in Godot, to:
var c = a.cross(b)
However, unlike the dot product, doing a.cross(b) and b.cross(a) will yield different results. Specifically, the returned vector will be negated in the second case. As you might have realized, this coincides with creating perpendicular vectors in 2D. In 3D, there are also two possible perpendicular vectors to a pair of 2D vectors.
Also, the resulting cross product of two unit vectors is not a unit vector. Result will need to be renormalized.
Area of a triangle¶
Cross product can be used to obtain the surface area of a triangle in 3D. Given a triangle consisting of 3 points, A, B and C:
Take any of them as a pivot and compute the adjacent vectors to the other two points. As example, we will use B as a pivot:
var BA = A - B
var BC = C - B
Compute the cross product between BA and BC to obtain the perpendicular vector P:
var P = BA.cross(BC)
The length (magnitude) of P is the surface area of the parallelogram built by the two vectors BA and BC, therefore the surface area of the triangle is half of it.
var area = P.length()/2
Plane of the triangle¶
With P computed from the previous step, normalize it to get the normal of the plane.
var N = P.normalized()
And obtain the distance by doing the dot product of P with any of the 3 points of the ABC triangle:
var D = P.dot(A)
Fantastic! You computed the plane from a triangle!
Here’s some useful info (that you can find in Godot source code anyway). Computing a plane from a triangle can result in 2 planes, so a sort of convention needs to be set. This usually depends (in video games and 3D visualization) to use the front-facing side of the triangle.
In Godot, front-facing triangles are those that, when looking at the camera, are in clockwise order. Triangles that look Counter-clockwise when looking at the camera are not drawn (this helps to draw less, so the back-part of the objects is not drawn).
To make it a little clearer, in the image below, the triangle ABC appears clock-wise when looked at from the Front Camera, but to the Rear Camera it appears counter-clockwise so it will not be drawn.
Normals of triangles often are sided towards the direction they can be viewed from, so in this case, the normal of triangle ABC would point towards the front camera:
So, to obtain N, the correct formula is:
clockwise normal from triangle formula
var N = (A-C).cross(A-B).normalized()
for counter-clockwise:
var N = (A-B).cross(A-C).normalized()
var D = N.dot(A)
Collision detection in 3D¶
This is another bonus bit, a reward for being patient and keeping up with this long tutorial. Here is another piece of wisdom. This might not be something with a direct use case (Godot already does collision detection pretty well) but It’s a really cool algorithm to understand anyway, because it’s used by almost all physics engines and collision detection libraries :)
Remember that converting a convex shape in 2D to an array of 2D planes was useful for collision detection? You could detect if a point was inside any convex shape, or if two 2D convex shapes were overlapping.
Well, this works in 3D too, if two 3D polyhedral shapes are colliding, you won’t be able to find a separating plane. If a separating plane is found, then the shapes are definitely not colliding.
To refresh a bit a separating plane means that all vertices of polygon A are in one side of the plane, and all vertices of polygon B are in the other side. This plane is always one of the face-planes of either polygon A or polygon B.
In 3D though, there is a problem to this approach, because it is possible that, in some cases a separating plane can’t be found. This is an example of such situation:
To avoid it, some extra planes need to be tested as separators, these planes are the cross product between the edges of polygon A and the edges of polygon B
So the final algorithm is something like:
var overlapping = true
for p in planes_of_A:
var all_out = true
for v in points_of_B:
if (p.distance_to(v) < 0):
all_out = false
break
if (all_out):
# a separating plane was found
# do not continue testing
overlapping = false
break
if (overlapping):
# only do this check if no separating plane
# was found in planes of A
for p in planes_of_B:
var all_out = true
for v in points_of_A:
if (p.distance_to(v) < 0):
all_out = false
break
if (all_out):
overlapping = false
break
if (overlapping):
for ea in edges_of_A:
for eb in edges_of_B:
var n = ea.cross(eb)
if (n.length() == 0):
continue
var max_A = -1e20 # tiny number
var min_A = 1e20 # huge number
# we are using the dot product directly
# so we can map a maximum and minimum range
# for each polygon, then check if they
# overlap.
for v in points_of_A:
var d = n.dot(v)
if (d > max_A):
max_A = d
if (d < min_A):
min_A = d
var max_B = -1e20 # tiny number
var min_B = 1e20 # huge number
for v in points_of_B:
var d = n.dot(v)
if (d > max_B):
max_B = d
if (d < min_B):
min_B = d
if (min_A > max_B or min_B > max_A):
# not overlapping!
overlapping = false
break
if (not overlapping):
break
if (overlapping):
print("Polygons collided!")
This was all! Hope it was helpful, and please give feedback and let know if something in this tutorial is not clear! You should be now ready for the next challenge... Matrices and transforms!
NextPrevious
© Copyright 2014-2016, Juan Linietsky, Ariel Manzur and the Godot community (CC BY 3.0). Revision 577559cf.
Built with Sphinx using a theme provided by Read the Docs.
Read the Docs v: latest
VersionslatestDownloadspdfhtmlzipepubOn Read the DocsProject HomeBuilds
Free document hosting provided by Read the Docs. |
189095 | https://mathspace.co/textbooks/syllabuses/Syllabus-821/topics/Topic-18274/subtopics/Subtopic-249251/ | 9.07 Piecewise graphs
A piecewise linear function, as the name suggests, is made up of pieces or line segments, of various linear graphs. For example:
As we can see from the graph, we have three straight line pieces, which in this case are connected together, to create a complete graph over the domain $0\le x<10$0≤x<10. This makes this particular graph continuous over its range of $x$x values (though this isn't always the case).
Notice that at $x=0$x=0 there is a filled circle, indicating that at $x=0$x=0, we look at the first line segment to find the $y$y value. However, at $x=10$x=10 there is a hollow circle which means that at $x=10$x=10 there is no $y$y value that we can read from the graph, in this case, because this is where our graph stops.
Graphs of piecewise linear functions
Worked example
Example 1
Sketch a graph of the following piecewise linear function
$y$y | $=$= | | $-x+6$−x+6 | when$0\le x<5$0≤x<5
$1$1 | when$5\le x<8$5≤x<8
$0.5x-3$0.5x−3 | when$8\le x<10$8≤x<10
Solution:
We'll start with the first line segment from $x=0$x=0 to $x=5$x=5. We have many methods we can use to sketch graphs of straight lines and here we'll use a combination of strategies. First we'll plot the $y$y-intercept at $\left(0,6\right)$(0,6). Then we'll calculate the value of $y$y when $x=5$x=5, to find the point at which we want to end the drawing of this first line segment. This gives us $\left(5,1\right)$(5,1). As our equation showed $x<5$x<5, we'll need an open circle at the right hand end of the segment.
Next we'll draw the second linear function. We notice it's a horizontal line because we're drawing $y=1$y=1. This means that it will attach straight onto the endpoint of our last segment and we no longer need the open circle. And we'll draw this horizontal line from $x=5$x=5 to $x=8$x=8, with an open circle at the end.
Finally we'll sketch our last line segment. The best strategy here is to calculate the $y$y-coordinate at each $x$x-coordinate of our endpoints, plot the two coordinates and draw the line. So at $x=8$x=8 we have $y=1$y=1, which again means we can attach our final piece to the middle piece at $\left(8,1\right)$(8,1). At $x=10$x=10 we have $y=2$y=2. So at $\left(10,2\right)$(10,2) we'll draw an open circle to match the information given, and we'll draw the line between our two points.
Practice question
question 1
The function $f\left(x\right)$f(x) is defined as shown below. Sketch a graph of $f\left(x\right)$f(x).
$f\left(x\right)$f(x)
$=$=
$x+2$x+2
when $x\ge5$x≥5
$x-3$x−3
when $x<5$x<5
Loading Graph...Reveal Solution
The function $f\left(x\right)$f(x) is defined as shown below. Sketch a graph of $f\left(x\right)$f(x).
$f\left(x\right)$f(x) | $=$= | | $x+2$x+2 | when$x\ge5$x≥5
$x-3$x−3 | when$x<5$x<5
Finding the equation of a piecewise linear function
Recall we can find the equation of a linear graph by finding the $y$y-intercept and gradient. This is also the case for piecewise linear functions, where we find the equation of each piece and state the values of $x$x for that section covers.
Practice question
question 2
Define the function shown in the graph below.Loading Graph...
$y$y
$=$=
$\editable{}$
if $\editable{}$$\le$≤$x$x$\le$≤$\editable{}$
$\editable{}$
if $\editable{}$$<$<$x$x$\le$≤$\editable{}$
Reveal SolutionWatch video
Define the function shown in the graph below.
$y$y | $=$= | | $\editable{}$ | | | if$\editable{}$$\le$≤$x$x$\le$≤$\editable{}$
$\editable{}$ | | | if$\editable{}$$<$<$x$x$\le$≤$\editable{}$
Modelling with linear piecewise functions
Sometimes when attempting to create a linear model that describes a relationship between two variables, one linear model is not enough. As one (independent) variable changes, its relationship to the other (dependent) variable may also change. When this occurs, piecewise functions and step graphs can be used, so that multiple linear models can be applied to the one real-life scenario.
Worked example
Example 2
Sketch the piecewise linear function that describes the following:
Consider a runner who is running for exercise. Their exercise regime is to jog as a warm up, then start running, and then end their workout at rest. They begin jogging at a constant pace of $150$150 m/min for the first $2$2 minutes. After $2$2 minutes they start to run at $300$300 m/min. They continue at this speed for another $2$2 minutes before finally stopping and resting for the final $2$2 minutes of their workout.
Think: Let $y$y equal the distance covered by the runner, measured in metres and let $t$t equal the amount of time that the runner has been running, measured in minutes.
The person is moving at a constant speed in each part of their workout, so we can represent each section with a linear graph. Assess the information given for each section to determine the gradient and endpoints of the line segment.
Do: Assuming that no distance has been covered until the timer starts, this means that $y=0$y=0 when $t=0$t=0. So the first line will have a $y$y-axis intercept of $0$0 and a gradient of $150$150 metres per minute. The piecewise function follows this behaviour for $0\le t\le2$0≤t≤2.
At the $2$2 minute mark, the runner will have covered a distance of $2\times150=300$2×150=300 metres. We now have the end-points of the first section, $\left(0,0\right)$(0,0) and $\left(2,300\right)$(2,300), so we can plot and join the points to create the segment.
In the second section their speed now increases, which means the slope of the line also increases. The gradient of this second line would be $300$300 metres per minute, the speed at which the runner is now running. So the second line with this gradient can now be drawn between $2\le t\le4$2≤t≤4, connecting to the previous line when $t=2$t=2.
We can also find the distance covered at the $4$4 minute mark to plot the end-point. Since the runner's pace has been $300$300 metres per minute, they will have covered $2\times300=600$2×300=600 metres in the $2$2 minutes that they have been running at this pace, covering a total of $300+600=900$300+600=900 metres.
For the final $2$2 minutes of the workout, so for $4\le t\le6$4≤t≤6, the runner is resting and is therefore not covering any distance during this time. So a horizontal line can be drawn from the previous line at $t=4$t=4 until $t=6$t=6.
Reflect: The graph is now complete and represents the given scenario. The equations could also be defined to create the following piecewise function.
$y$y | $=$= | | $150t$150t | when$0\le t<2$0≤t<2
$300t-300$300t−300 | when$2\le t<4$2≤t<4
$900$900 | when$4\le t<6$4≤t<6
Practice questions
question 3
Ken starts travelling at $9$9 am from point $A$A and moves towards point $B$B in a straight line.
The distance between Ken and point $B$B at various times in his journey is shown on the graph below.
A line graph is displayed with the horizontal axis labeled "Time (hours)" marked from 9 to 4 in one-hour increments, and the vertical axis labeled "Distance (km)" marked from 0 to 120 in increments of 20. A line connects several points on the graph, starting at point A at $\left(9,120\right)$(9,120), to $\left(10,70\right)$(10,70), to $\left(12,100\right)$(12,100), to $\left(1.5,100\right)$(1.5,100), to $\left(2,40\right)$(2,40), to $\left(2.5,40\right)$(2.5,40), and ending at point B at $\left(4,0\right)$(4,0).
How far is point $B$B from point $A$A?How many hours was Ken's journey from point $A$A to point $B$B?
Between what times did Ken travel at the fastest speed?$2:30$2:30 pm $-$− $4$4 pmA$12$12 pm $-$− $1:30$1:30 pmB$9$9 am $-$− $10$10 amC$10$10 am $-$− $12$12 pmD$1:30$1:30 pm $-$− $2$2 pmEWhat was Ken's fastest speed in km/h?What was the distance in kilometres travelled by Ken between $1:30$1:30 pm and $4$4 pm?What is the total distance travelled by Ken from $9$9 am to $4$4 pm?Reveal SolutionWatch video
Ken starts travelling at $9$9 am from point $A$A and moves towards point $B$B in a straight line.
The distance between Ken and point $B$B at various times in his journey is shown on the graph below.
How far is point $B$B from point $A$A?
How many hours was Ken's journey from point $A$A to point $B$B?
Between what times did Ken travel at the fastest speed?
$2:30$2:30 pm $-$− $4$4 pm
$12$12 pm $-$− $1:30$1:30 pm
$9$9 am $-$− $10$10 am
$10$10 am $-$− $12$12 pm
$1:30$1:30 pm $-$− $2$2 pm
What was Ken's fastest speed in km/h?
What was the distance in kilometres travelled by Ken between $1:30$1:30 pm and $4$4 pm?
What is the total distance travelled by Ken from $9$9 am to $4$4 pm?
Question 4
A children's pool is being filled with water. The volume of water $V$V, in litres, after $t$t minutes is given by the piecewise graph below. The pool has a maximum capacity of $1500$1500 L.Loading Graph...Estimate the volume of water in litres after $45$45 minutes have elapsed, to the nearest $10$10 litresDetermine the equation that describes the volume of water $V$V in litres after $t$t minutes in the first $45$45 minutes.After $45$45 minutes, the rate at which the volume of water enters the pool is increased.
The piecewise relationship after $t=45$t=45 is given by $V=26t-970$V=26t−970.
Determine the value of $t$t (in minutes) until the pool is filled up.Reveal SolutionWatch video
A children's pool is being filled with water. The volume of water $V$V, in litres, after $t$t minutes is given by the piecewise graph below. The pool has a maximum capacity of $1500$1500 L.
Estimate the volume of water in litres after $45$45 minutes have elapsed, to the nearest $10$10 litres
Determine the equation that describes the volume of water $V$V in litres after $t$t minutes in the first $45$45 minutes.
After $45$45 minutes, the rate at which the volume of water enters the pool is increased.
The piecewise relationship after $t=45$t=45 is given by $V=26t-970$V=26t−970.
Determine the value of $t$t (in minutes) until the pool is filled up.
Piecewise linear functions with CAS
A CAS calculator can be used to generate graphs and evaluate piecewise functions. Select your brand of calculator below to view instructions to utilise these features.
How to use the CASIO Classpad to complete the following tasks regarding piecewise linear functions.
Consider the function $f\left(x\right)$f(x) is defined as:
$f\left(x\right)$f(x)
$=$=
$-4x+28$−4x+28
when $x\ge4$x≥4
$1.5x+6$1.5x+6
when $-2\le x<4$−2≤x<4
$3$3
when $x<-2$x<−2
Create a graph of the function $y=f(x)$y=f(x).Find the value of the function when $x=3.5$x=3.5.For what value(s) of $x$x is the function equal to $9$9.Reveal Solution
How to use the CASIO Classpad to complete the following tasks regarding piecewise linear functions.
Consider the function $f\left(x\right)$f(x) is defined as:
$f\left(x\right)$f(x) | $=$= | | $-4x+28$−4x+28 | when$x\ge4$x≥4
$1.5x+6$1.5x+6 | when$-2\le x<4$−2≤x<4
$3$3 | when$x<-2$x<−2
Create a graph of the function $y=f(x)$y=f(x).
Find the value of the function when $x=3.5$x=3.5.
For what value(s) of $x$x is the function equal to $9$9.
How to use the TI Nspire to complete the following tasks regarding piecewise linear functions.
Consider the function $f\left(x\right)$f(x) is defined as:
$f\left(x\right)$f(x)
$=$=
$-4x+28$−4x+28
when $x\ge4$x≥4
$1.5x+6$1.5x+6
when $-2\le x<4$−2≤x<4
$3$3
when $x<-2$x<−2
Create a graph of the function $y=f(x)$y=f(x).Find the value of the function when $x=3.5$x=3.5.For what value(s) of $x$x is the function equal to $9$9.Reveal Solution
How to use the TI Nspire to complete the following tasks regarding piecewise linear functions.
Consider the function $f\left(x\right)$f(x) is defined as:
$f\left(x\right)$f(x) | $=$= | | $-4x+28$−4x+28 | when$x\ge4$x≥4
$1.5x+6$1.5x+6 | when$-2\le x<4$−2≤x<4
$3$3 | when$x<-2$x<−2
Create a graph of the function $y=f(x)$y=f(x).
Find the value of the function when $x=3.5$x=3.5.
For what value(s) of $x$x is the function equal to $9$9.
Use technology to assist in graphing and analysing the piecewise function in the following practice question.
Practice question
Question 5
Harry goes out for a run. He accelerates from rest up to a desired speed and maintains that speed for some time. Feeling exhausted, his velocity drops until he's back at rest.
The speed $S$S in km/h after $t$t seconds is given by the following piecewise relationship.
$S$S
$=$=
$0.8t$0.8t
when $0\le t<15$0≤t<15
$12$12
when $15\le t<240$15≤t<240
$252-t$252−t
when $240\le t\le252$240≤t≤252
Find the speed in km/h after $10$10 seconds have elapsed.Find the speed in km/h after $50$50 seconds have elapsed.Find the speed in km/h after $150$150 seconds have elapsed.Find the speed in km/h after $250$250 seconds have elapsed.Which of the following graphs describes the piecewise relationship?Loading Graph...ALoading Graph...BLoading Graph...CLoading Graph...DReveal SolutionWatch video
Harry goes out for a run. He accelerates from rest up to a desired speed and maintains that speed for some time. Feeling exhausted, his velocity drops until he's back at rest.
The speed $S$S in km/h after $t$t seconds is given by the following piecewise relationship.
$S$S | $=$= | | $0.8t$0.8t | when$0\le t<15$0≤t<15
$12$12 | when$15\le t<240$15≤t<240
$252-t$252−t | when$240\le t\le252$240≤t≤252
Find the speed in km/h after $10$10 seconds have elapsed.
Find the speed in km/h after $50$50 seconds have elapsed.
Find the speed in km/h after $150$150 seconds have elapsed.
Find the speed in km/h after $250$250 seconds have elapsed.
Which of the following graphs describes the piecewise relationship?
Outcomes
2.3.9
sketch piece-wise linear graphs and step graphs, using technology when appropriate
2.3.10
interpret piece-wise linear and step graphs used to model practical situations; for example, the tax paid as income increases, the change in the level of water in a tank over time when water is drawn off at different intervals and for different periods of time, the charging scheme for sending parcels of different weights through the post
What is Mathspace
About Mathspace |
189096 | https://www.worldscientific.com/doi/pdf/10.1142/9789810248598_0013?srsltid=AfmBOopicqKss9rbXfVFosU_ym2_AtJzaCM0XSV511zpuSrLYsFX5urj | INTERACTIONS BETWEEN RADIATION AND MATTER (4062-4085) | Problems and Solutions on Atomic, Nuclear and Particle Physics
Login to your account
Email
Password
Forgot password?
Keep me logged in
[x]
New UserInstitutional Login
Change Password
Old Password
New Password
Too Short Weak Medium Strong Very Strong Too Long
Your password must have 8 characters or more and contain 3 of the following:
a lower case character,
an upper case character,
a special character
or a digit
Too Short
Password Changed Successfully
Your password has been changed
Create a new account
Email
Returning user
Can't sign in? Forgot your password?
Enter your email address below and we will send you the reset instructions
Email
Please check your inbox for the reset password link that is only valid for 24 hours.
Cancel
If the address matches an existing account you will receive an email with instructions to reset your password
Close
Request Username
Can't sign in? Forgot your username?
Enter your email address below and we will send you your username
Email
Close
If the address matches an existing account you will receive an email with instructions to retrieve your username
Search
This Book
This Book
Anywhere
Quick Search in Books
Enter words / phrases / DOI / ISBN / keywords / authors / etc
Search Search
Access type:- [x] Only show content I have full access to - [x] Only show Open Access
Quick Search anywhere
Enter words / phrases / DOI / ISBN / keywords / authors / etc
Search Search
Access type:- [x] Only show content I have full access to - [x] Only show Open Access
Advanced Search
0 My Cart
Sign in
Institutional Access
Skip main navigation
Open Drawer Menu Close Drawer MenuHome
Subject
All Subjects
Asian Studies
Business & Management
Chemistry
Children’s Books
Computer Science
Economics & Finance
Education
Engineering / Acoustics
Environmental Science
Life Sciences / Biology
Materials Science
Mathematics
Medicine
Nanotechnology & Nanoscience
Nonlinear Science, Chaos & Dynamical Systems
Physics & Astronomy
Popular & General Science
Social Sciences
华文书籍 (Chinese Titles)
Journals
Books
New Titles
New Reviews
Bestsellers
Major Reference Works
Browse by Subject
Resources for Partners
Publish with us
For Authors
For Booksellers
For Librarians
For Societies and Partners
For Individual Customers
Copyright & Permissions
Ethics and Integrity
Translation Rights
Open Access
About Us
About Us
World Scientific Connect
News
Press Releases
Contact Us
Privacy Policy
Sitemap
Help
Help
How to Order
Cookies Notification
We use cookies on this site to enhance your user experience. By continuing to browse the site, you consent to the use of our cookies. Learn More
×
System Upgrade on Monday, August 5th, 2025 at 2.30am (EDT)
During this period, the E-commerce and registration of new users may not available for up to 5 hours. For online Purchase, please visit us again. Contact us at customercare@wspc.com for any enquiries.
Problems and Solutions on Atomic, Nuclear and Particle PhysicsNo Access
INTERACTIONS BETWEEN RADIATION AND MATTER (4062-4085)
Doi:
PreviousNext
Article
Figures
References
Related
Details
Tools
Add to favorites
Download Citations
Track Citations
Recommend to Librarian
Share
Share on
Facebook
Twitter
Linked In
Reddit
Email
Abstract:
Please refer to full text.
Access content
To read the fulltext, please use one of the options below to sign in or purchase access.
Personal login
Institutional Login
Register for access
PurchaseSave for laterItem saved, go to cartISBN: 978-981-02-3917-6 (hardcover) USD 146.00 Add to cart ISBN: 978-981-02-3917-6 (hardcover) Checkout ISBN: 978-981-02-3918-3 (softcover) USD 82.00 Add to cart ISBN: 978-981-02-3918-3 (softcover) Checkout ISBN: 978-981-3102-81-1 (ebook) USD 66.00 Add to cart ISBN: 978-981-3102-81-1 (ebook) Checkout
Claim
Claim Number
Restore content access This functionality works only for purchases made as a guest
Figures
References
Related
Details
None
None
None
Problems and Solutions on Atomic, Nuclear and Particle Physics
MetricsDownloaded 7 times
History
Resources
For Authors
For Booksellers
For Librarians
Copyright & Permissions
Translation Rights
How to Order
Contact Us
Sitemap
About Us & Help
About Us
News
Author Services
Help
Links
World Scientific Europe
WS Education (K-12)
Global Publishing 八方文化
World Century
Privacy policy
© 2025 World Scientific Publishing Co Pte Ltd
Powered by Atypon® Literatum
✓
Thanks for sharing!
AddToAny
More…
Close crossmark popup
back |
189097 | https://arxiv.org/abs/0712.3640 | [0712.3640] Zeeman effect of the hyperfine structure levels in lithiumlike ions
Skip to main content
We gratefully acknowledge support from the Simons Foundation, member institutions, and all contributors.Donate
>physics> arXiv:0712.3640
Help | Advanced Search
Search
GO
quick links
Login
Help Pages
About
Physics > Atomic Physics
arXiv:0712.3640 (physics)
[Submitted on 21 Dec 2007]
Title:Zeeman effect of the hyperfine structure levels in lithiumlike ions
Authors:D.L. Moskovkin, V.M. Shabaev, W. Quint
View a PDF of the paper titled Zeeman effect of the hyperfine structure levels in lithiumlike ions, by D.L. Moskovkin and 2 other authors
View PDF
Abstract:The fully relativistic theory of the Zeeman splitting of the $(1s)^2 2s$ hyperfine-structure levels in lithiumlike ions with $Z=6 - 32$ is considered for the magnetic field magnitude in the range from 1 to 10 T. The second-order corrections to the Breit -- Rabi formula are calculated and discussed including the one-electron contributions as well as the interelectronic-interaction effects of order 1/Z. The 1/Z corrections are evaluated within a rigorous QED approach. These corrections are combined with other interelectronic-interaction, QED, nuclear recoil, and nuclear size corrections to obtain high-precision theoretical values for the Zeeman splitting in Li-like ions with nonzero nuclear spin. The results can be used for a precise determination of nuclear magnetic moments from $g$-factor experiments.
Comments:25 pages, 5 figures
Subjects:Atomic Physics (physics.atom-ph); Optics (physics.optics)
Cite as:arXiv:0712.3640 [physics.atom-ph]
(or arXiv:0712.3640v1 [physics.atom-ph] for this version)
Focus to learn more
arXiv-issued DOI via DataCite
Journal reference:Phys. Rev. A 77, 063421 (2008)
Related DOI:
Focus to learn more
DOI(s) linking to related resources
Submission history
From: Dmitriy Moskovkin [view email]
[v1] Fri, 21 Dec 2007 08:13:19 UTC (21 KB)
Full-text links:
Access Paper:
View a PDF of the paper titled Zeeman effect of the hyperfine structure levels in lithiumlike ions, by D.L. Moskovkin and 2 other authors
View PDF
TeX Source
Other Formats
view license
Current browse context:
physics.atom-ph
<prev | next>
new | recent | 2007-12
Change to browse by:
physics
physics.optics
References & Citations
NASA ADS
Google Scholar
Semantic Scholar
export BibTeX citation Loading...
BibTeX formatted citation
×
Data provided by:
Bookmark
Bibliographic Tools
Bibliographic and Citation Tools
[x] Bibliographic Explorer Toggle
Bibliographic Explorer (What is the Explorer?)
[x] Connected Papers Toggle
Connected Papers (What is Connected Papers?)
[x] Litmaps Toggle
Litmaps (What is Litmaps?)
[x] scite.ai Toggle
scite Smart Citations (What are Smart Citations?)
Code, Data, Media
Code, Data and Media Associated with this Article
[x] alphaXiv Toggle
alphaXiv (What is alphaXiv?)
[x] Links to Code Toggle
CatalyzeX Code Finder for Papers (What is CatalyzeX?)
[x] DagsHub Toggle
DagsHub (What is DagsHub?)
[x] GotitPub Toggle
Gotit.pub (What is GotitPub?)
[x] Huggingface Toggle
Hugging Face (What is Huggingface?)
[x] Links to Code Toggle
Papers with Code (What is Papers with Code?)
[x] ScienceCast Toggle
ScienceCast (What is ScienceCast?)
Demos
Demos
[x] Replicate Toggle
Replicate (What is Replicate?)
[x] Spaces Toggle
Hugging Face Spaces (What is Spaces?)
[x] Spaces Toggle
TXYZ.AI (What is TXYZ.AI?)
Related Papers
Recommenders and Search Tools
[x] Link to Influence Flower
Influence Flower (What are Influence Flowers?)
[x] Core recommender toggle
CORE Recommender (What is CORE?)
Author
Venue
Institution
Topic
About arXivLabs
arXivLabs: experimental projects with community collaborators
arXivLabs is a framework that allows collaborators to develop and share new arXiv features directly on our website.
Both individuals and organizations that work with arXivLabs have embraced and accepted our values of openness, community, excellence, and user data privacy. arXiv is committed to these values and only works with partners that adhere to them.
Have an idea for a project that will add value for arXiv's community? Learn more about arXivLabs.
Which authors of this paper are endorsers? | Disable MathJax (What is MathJax?)
About
Help
Contact
Subscribe
Copyright
Privacy Policy
Web Accessibility Assistance
arXiv Operational Status
Get status notifications via email or slack |
189098 | File:Acetone-structural.png - Wikimedia Commons
Jump to content
[x] Main menu
Main menu
move to sidebar hide
Navigate
Main page
Welcome
Community portal
Village pump
Help center
Language select
Participate
Upload file
Recent changes
Latest files
Random file
Contact us
Special pages
Search
Search
English
[x] Appearance
Appearance
move to sidebar hide
Text
Small Standard Large
This page always uses small font size
This page always uses small font size
Width
Standard Wide
The content is as wide as possible for your browser window.
Color (beta)
Automatic Light Dark
This page is always in light mode.
Donate
Create account
Log in
[x] Personal tools
Donate
Create account
Log in
Pages for logged out editors learn more
Contributions
Talk
[x] Toggle the table of contents
Contents
move to sidebar hide
Beginning
1 Summary
2 Licensing
File:Acetone-structural.png
File
Discussion
[x] English
Read
Edit
View history
[x] Tools
Tools
move to sidebar hide
Actions
Read
Edit
View history
General
What links here
Related changes
Permanent link
Page information
Cite this page
Get shortened URL
Download QR code
Concept URI
Expand all
Nominate for deletion
Print/export
Download as PDF
Printable version
In other projects
From Wikimedia Commons, the free media repository
File
File history
File usage on Commons
File usage on other wikis
Metadata
Download all sizesUse this file on the webUse this file on a wikiEmail a link to this fileInformation about reusing
Size of this preview: 800 × 516 pixels. Other resolutions: 320 × 206 pixels | 640 × 413 pixels | 1,024 × 660 pixels | 1,214 × 783 pixels.
Original file(1,214 × 783 pixels, file size: 14 KB, MIME type: image/png) Open in Media Viewer
File information
Structured data
Captions
Captions Edit
English
Add a one-line explanation of what this file represents
File:Aceton.svg is a vector version of this file. It should be used in place of this PNG file when not inferior.
File:Acetone-structural.png → File:Aceton.svg
For more information, see Help:SVG.
| ExpandIn other languages |
| Alemannisch∙العربية∙беларуская (тарашкевіца)∙български∙বাংলা∙català∙нохчийн∙čeština∙dansk∙Deutsch∙Ελληνικά∙English∙British English∙Esperanto∙español∙eesti∙euskara∙فارسی∙suomi∙français∙Frysk∙galego∙Alemannisch∙עברית∙हिन्दी∙hrvatski∙magyar∙հայերեն∙Bahasa Indonesia∙Ido∙italiano∙日本語∙ქართული∙한국어∙lietuvių∙македонски∙മലയാളം∙Bahasa Melayu∙မြန်မာဘာသာ∙norsk bokmål∙Plattdüütsch∙Nederlands∙norsk nynorsk∙norsk∙occitan∙polski∙prūsiskan∙português∙português do Brasil∙română∙русский∙sicilianu∙Scots∙slovenčina∙slovenščina∙српски / srpski∙svenska∙தமிழ்∙ไทย∙Türkçe∙татарча / tatarça∙українська∙vèneto∙Tiếng Việt∙中文∙中文（中国大陆）∙中文（简体）∙中文（繁體）∙中文（马来西亚）∙中文（新加坡）∙中文（臺灣）∙+/− |
Summary
[edit]
Description Acetone-structural.pngEnglish: Structural formula of acetone (CH 3 COCH 3).
Date March 2006
Source Own work
AuthorBen Mills
PNG development
InfoField
This structural formula was created with ChemDrawand Photoshop.
Licensing
[edit]
Public domain Public domain false false
This image of a simple structural formula is ineligible for copyright and therefore in the public domain, because it consists entirely of information that is common property and contains no original authorship.
File history
Click on a date/time to view the file as it appeared at that time.
| | Date/Time | Thumbnail | Dimensions | User | Comment |
--- --- --- |
| current | 14:08, 8 March 2006 | | 1,214 × 783 (14 KB) | Benjah-bmm27(talk | contribs) | '''Description''': Structural formula of acetone (CH3COCH3). '''Author, date of creation''': self made by Ben Mills at ~~~~~. '''Source''': user-created image '''Comments''': high-resolut |
You cannot overwrite this file.
File usage on Commons
There are no pages that use this file.
File usage on other wikis
The following other wikis use this file:
Usage on de.wikipedia.org
Diskussion:Aceton
Wikipedia:Auskunft/Archiv/2010/Woche 49
Wikipedia:Redaktion Chemie/Archiv/2016/Juli
Usage on eo.wikipedia.org
Izobuterata acido
Usage on es.wikiversity.org
Curso de Química:Grupos funcionales
Metadata
This file contains additional information such as Exif metadata which may have been added by the digital camera, scanner, or software program used to create or digitize it. If the file has been modified from its original state, some details such as the timestamp may not fully reflect those of the original file. The timestamp is only as accurate as the clock in the camera, and it may be completely wrong.
| Software used | Adobe ImageReady |
| Show extended details |
Structured data
Structured data labels link to their respective Wikidata pages and will open in a new window.Dismiss
Items portrayed in this file
depicts
Edit
from Wikidata
Custom value
Custom value
Some value
No value
inception
Edit
March 2006
Mark as prominent
source of file
Edit
original creation by uploader
Mark as prominent
media type
Edit
image/png
Mark as prominent
checksum
Edit
9c29b2ab719bd7232683feeee377357f18095d4c
Mark as prominent
determination method or standard: SHA-1
data size
Edit
14,310 byte
Mark as prominent
height
Edit
783 pixel
Mark as prominent
width
Edit
1,214 pixel
Mark as prominent
Add statement
Retrieved from "
Categories:
PNG created with ChemDraw
Chemical structures of acetone
Hidden categories:
Vector version available
PD chem
This page was last edited on 7 March 2025, at 10:56.
Files are available under licenses specified on their description page. All structured data from the file namespace is available under the Creative Commons CC0 License; all unstructured 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 the Privacy Policy.
Privacy policy
About Wikimedia Commons
Disclaimers
Code of Conduct
Developers
Statistics
Cookie statement
Mobile view
Search
Search
[x] Toggle the table of contents
File:Acetone-structural.png
Add topic
billion years
hundred million years
ten million years
million years
100,000 years
10,000 years
millennium
century
decade
year
month
day
hour
minute
second
Gregorian
Julian
Cancel Edit Delete Preview revert
Text of the note (may include Wiki markup)
Could not save your note (edit conflict or other problem). Please copy the text in the edit box below and insert it manually by editing this page.
Upon submitting the note will be published multi-licensed under the terms of the CC BY-SA 4.0 license and of the GFDL (unversioned, with no invariant sections, front-cover texts, or back-cover texts). See our terms of use for more details.
Add a note Draw a rectangle onto the image above (press the left mouse button, then drag and release).This file has annotations. Move the mouse pointer over the image to see them.To edit the notes, visit page X.Why do you want to remove this note?
SaveHelp about image annotations
To modify annotations, your browser needs to have the XMLHttpRequest object. Your browser does not have this object or does not allow it to be used (in Internet Explorer, it may be in a switched off ActiveX component), and thus you cannot modify annotations. We're sorry for the inconvenience.
$1$1$1 | |
189099 | https://artofproblemsolving.com/articles/files/MildorfInequalities.pdf?srsltid=AfmBOopgF1g1d1B6ktaaBzIwtK3QFlJAOkBY896nqm4glJHn-_FV3fWT | Olympiad Inequalities
Thomas J. Mildorf December 22, 2005
It is the purpose of this document to familiarize the reader with a wide range of theorems and techniques that can be used to solve inequalities of the variety typically appearing on mathematical olympiads or other elementary proof contests. The Standard Dozen is an exhibition of twelve famous inequalities which can be cited and applied without proof in a solution. It is expected that most problems will fall entirely within the span of these inequalities. The Examples section provides numerous complete solutions as well as remarks on inequality-solving intuition, all intended to increase the reader’s aptitude for the material covered here. It is organized in rough order of difficulty. Finally, the Problems section contains exercises without solutions, ranging from straightforward to quite difficult, for the purpose of practicing techniques contained in this document. I have compiled much of this from posts by my peers in a number of mathematical communities, particularly the Mathlinks-Art of Problem Solving forums, 1 as well as from various MOP lectures, 2 Kiran Kedlaya’s inequalities packet, 3 and John Scholes’ site. 4 I have tried to take note of original sources where possible. This work in progress is distributed for personal educational use only. In particular, any publication of all or part of this manuscript without prior consent of the author, as well as any original sources noted herein, is strictly prohibited. Please send comments - suggestions, corrections, missing information, 5 or other interesting problems - to the author at tmildorf@mit.edu .Without further delay...
1
and respectively, though they have merged into a single, very large and robust group. The forums there are also host to a considerable wealth of additional material outside of inequalities.
2
Math Olympiad Program. Although some people would try to convince me it is the Math Olympiad Summer Program and therefore is due the acronym MOSP, those who know acknowledge that the traditional “MOP” is the preferred appellation.
3
The particularly diligent student of inequalities would be interested in this document, which is available online at Further ma-terial is also available in the books Andreescu-Cartoaje-Dospinescu-Lascu, Old and New Inequalities , GIL Publishing House, and Hardy-Littlewood-P´ olya, Inequalities , Cambridge University Press. (The former is elementary and geared towards contests, the latter is more technical.)
4
where a seemingly inexhaustible supply of Olympiads is available.
5
Such as the source of the last problem in this document.
11 The Standard Dozen
Throughout this lecture, we refer to convex and concave functions. Write I and I′ for the intervals [ a, b ] and ( a, b ) respectively. A function f is said to be convex on I if and only if
λf (x) + (1 − λ)f (y) ≥ f (λx + (1 − λ)y) for all x, y ∈ I and 0 ≤ λ ≤ 1. Conversely, if the inequality always holds in the opposite direction, the function is said to be concave on the interval. A function f that is continuous on I and twice differentiable on I′ is convex on I
if and only if f ′′ (x) ≥ 0 for all x ∈ I (Concave if the inequality is flipped.) Let x1 ≥ x2 ≥ · · · ≥ xn; y1 ≥ y2 ≥ · · · ≥ yn be two sequences of real numbers. If
x1 + · · · + xk ≥ y1 + · · · + yk for k = 1 , 2, . . . , n with equality where k = n, then the sequence
{xi} is said to majorize the sequence {yi}. An equivalent criterion is that for all real numbers
t,
|t − x1| + |t − x2| + · · · + |t − xn| ≥ | t − y1| + |t − y2| + · · · + |t − yn|
We use these definitions to introduce some famous inequalities.
Theorem 1 (Jensen) Let f : I → R be a convex function. Then for any x1, . . . , x n ∈ I
and any nonnegative reals ω1, . . . , ω n,
ω1f (x1) + · · · + ωnf (xn) ≥ (ω1 + · · · + ωn) f
(ω1x1 + · · · + ωnxn
ω1 + · · · + ωn
)
If f is concave, then the inequality is flipped.
Theorem 2 (Weighted Power Mean) If x1, . . . , x n are nonnegative reals and ω1, . . . , ω n
are nonnegative reals with a postive sum, then
f (r) :=
(ω1xr
1
· · · + ωnxrn
ω1 + · · · + ωn
)1
r
is a non-decreasing function of r, with the convention that r = 0 is the weighted geometric mean. f is strictly increasing unless all the xi are equal except possibly for r ∈ (−∞ , 0] ,where if some xi is zero f is identically 0. In particular, f (1) ≥ f (0) ≥ f (−1) gives the AM-GM-HM inequality.
Theorem 3 (H¨ older) Let a1, . . . , a n; b1, . . . , b n; · · · ; z1, . . . , z n be sequences of nonnegative real numbers, and let λa, λ b, . . . , λ z positive reals which sum to 1. Then
(a1 + · · · + an)λa (b1 + · · · + bn)λb · · · (z1 + · · · + zn)λz ≥ aλa
1
bλb
1
· · · zλz
1
· · · + aλz
n
bλb
n
· · · zλz
n
This theorem is customarily identified as Cauchy when there are just two sequences.
Theorem 4 (Rearrangement) Let a1 ≤ a2 ≤ · · · ≤ an and b1 ≤ b2 ≤ · · · ≤ bn be two nondecreasing sequences of real numbers. Then, for any permutation π of {1, 2, . . . , n }, we have
a1b1 + a2b2 + · · · + anbn ≥ a1bπ(1) + a2bπ(2) + · · · + anbπ(n) ≥ a1bn + a2bn−1 + · · · + anb1
with equality on the left and right holding if and only if the sequence π(1) , . . . , π (n) is de-creasing and increasing respectively.
2Theorem 5 (Chebyshev) Let a1 ≤ a2 ≤ · · · ≤ an; b1 ≤ b2 ≤ · · · ≤ bn be two nondecreas-ing sequences of real numbers. Then
a1b1 + a2b2 + · · · + anbn
n ≥ a1 + a2 + · · · + an
n ·b1 + b2 + · · · + bn
n ≥ a1bn + a2bn−1 + · · · + anb1
n
Theorem 6 (Schur) Let a, b, c be nonnegative reals and r > 0. Then
ar(a − b)( a − c) + br(b − c)( b − a) + cr(c − a)( c − b) ≥ 0
with equality if and only if a = b = c or some two of a, b, c are equal and the other is 0.
Remark - This can be improved considerably. (See the problems section.) However, they are not as well known (as of now) as this form of Schur, and so should be proven whenever used on a contest.
Theorem 7 (Newton) Let x1, . . . , x n be nonnegative real numbers. Define the symmetric polynomials s0, s 1, . . . , s n by (x + x1)( x + x2) · · · (x + xn) = snxn + · · · + s1x + s0, and define the symmetric averages by di = si/(ni
). Then
d2
i
≥ di+1 di−1
Theorem 8 (Maclaurin) Let di be defined as above. Then
d1 ≥ √d2 ≥ 3
√d3 ≥ · · · ≥ n
√dn
Theorem 9 (Majorization) Let f : I → R be a convex on I and suppose that the sequence
x1, . . . , x n majorizes the sequence y1, . . . , y n, where xi, y i ∈ I. Then
f (x1) + · · · + f (xn) ≥ f (y1) + · · · + f (yn)
Theorem 10 (Popoviciu) Let f : I → R be convex on I, and let x, y, z ∈ I. Then for any positive reals p, q, r ,
pf (x) + qf (y) + rf (z) + (p + q + r)f
(px + qy + rz p + q + r
)
≥ (p + q)f
(px + qy p + q
)
( q + r)f
(qy + rz q + r
)
( r + p)f
(rz + px r + p
)
Theorem 11 (Bernoulli) For all r ≥ 1 and x ≥ − 1,
(1 + x)r ≥ 1 + xr
3Theorem 12 (Muirhead) Suppose the sequence a1, . . . , a n majorizes the sequence b1, . . . , b n.Then for any positive reals x1, . . . , x n,
∑
sym
xa1
1
xa2
2
· · · xan
n
≥ ∑
sym
xb1
1
xb2
2
· · · xbn
n
where the sums are taken over all permutations of n variables.
Remark - Although Muirhead’s theorem is a named theorem, it is generally not favor-ably regarded as part of a formal olympiad solution. Essentially, the majorization criterion guarantees that Muirhead’s inequality can be deduced from a suitable application of AM-GM. Hence, whenever possible, you should use Muirhead’s inequality only to deduce the correct relationship and then explicitly write all of the necessary applications of AM-GM. For a particular case this is a simple matter.
We now present an array of problems and solutions based primarily on these inequalities and ideas.
2 Examples
When solving any kind of problem, we should always look for a comparatively easy solu-tion first, and only later try medium or hard approaches. Although what constitutes this notoriously indeterminate “difficulty” varies widely from person to person, I usually con-sider “Dumbassing,” AM-GM (Power Mean), Cauchy, Chebyshev (Rearrangement), Jensen, H¨ older, in that order before moving to more clever techniques. (The first technique is de-scribed in remarks after example 1.) Weak inequalities will fall to AM-GM, which blatantly pins a sum to its smallest term. Weighted Jensen and H¨ older are “smarter” in that the effect of widely unequal terms does not cost a large degree of sharpness 6 - observe what happens when a weight of 0 appears. Especially sharp inequalities may be assailable only through clever algebra. Anyway, I have arranged the following with that in mind. 1. Show that for positive reals a, b, c
(a2b + b2c + c2a) ( ab 2 + bc 2 + ca 2) ≥ 9a2b2c2
Solution 1. Simply use AM-GM on the terms within each factor, obtaining
(a2b + b2c + c2a) ( ab 2 + bc 2 + ca 2) ≥
(
3 3
√a3b3c3
) (
3 3
√a3b3c3
)
= 9 a2b2c2
6The sharpness of an inequality generally refers to the extent to which the two sides mimic each other, particularly near equality cases.
4Solution 2. Rearrange the terms of each factor and apply Cauchy,
(a2b + b2c + c2a) ( bc 2 + ca 2 + ab 2) ≥
(√a3b3c3 + √a3b3c3 + √a3b3c3
)2
= 9 a2b2c2
Solution 3. Expand the left hand side, then apply AM-GM, obtaining
(a2b + b2c + c2a) ( ab 2 + bc 2 + ca 2) = a3b3 + a2b2c2 + a4bc
ab 4c + b3c3 + a2b2c2
a2b2c2 + abc 4 + a3c3
≥ 9 9
√a18 b18 c18 = 9 a2b2c2
We knew this solution existed by Muirhead, since (4 , 1, 1) , (3 , 3, 0), and (2 , 2, 2) all majorize (2 , 2, 2). The strategy of multiplying out all polynomial expressions and ap-plying AM-GM in conjunction with Schur is generally knowing as dumbassing because it requires only the calculational fortitude to compute polynomial products and no real ingenuity. As we shall see, dumbassing is a valuable technique. We also remark that the AM-GM combining all of the terms together was a particularly weak inequality, but the desired was a multiple of a2b2c2’s, the smallest 6th degree symmetric polynomial of three variables; such a singular AM-GM may not always suffice. 2. Let a, b, c be positive reals such that abc = 1. Prove that
a + b + c ≤ a2 + b2 + c2
Solution. First, we homogenize the inequality. that is, apply the constraint so as to make all terms of the same degree. Once an inequality is homogenous in degree
d, we may scale all of the variables by an arbitrary factor of k, causing both sides of the inequality to scale by the factor kd. This is valid in that it does not change the correctness of an inequality for any positive k, and if d is even, for any nonzero
k. Hence, we need consider a nonhomogenous constraint no futher. In this case, we multiply the left hand side by 3
√abc , obtaining
a43 b13 c13 + a13 b43 c13 + a13 b13 c43 ≤ a2 + b2 + c2
As abc = 1 is not homogenous, the above inequality must be true for all nonnegative
a, b, c . As (2 , 0, 0) majorizes (4 /3, 1/3, 1/3), we know it is true, and the necessary AM-GM is 2a2
3 + b2
6 + c2
6 = a2 + a2 + a2 + a2 + b2 + c2
6 ≥ 6
√a8b2c2 = a43 b13 c13
Let P (x) be a polynomial with positive coefficients. Prove that if
P
( 1
x
)
≥ 1
P (x)5holds for x = 1, then it holds for all x > 0.
Solution. Let P (x) = anxn + an−1xn−1 + · · · + a1x + a0. The first thing we notice is that the given is P (1) ≥ 1. Hence, the natural strategy is to combine P (x) and P ( 1
x
)
into P (1) in some fashion. The best way to accomplish this is Cauchy, which gives
P (x)P
( 1
x
)
= (anxn + · · · + a1x + a0)
(
an
1
xn + · · · + a1
1
x + a0
)
≥ (an + · · · + a1 + a0)2 = P (1) 2 ≥ 1as desired. This illustrates a useful means of eliminating denominators - by introducing similar factors weighted by reciprocals and applying Cauchy / H¨ older. 4. (USAMO 78/1) a, b, c, d, e are real numbers such that
a + b + c + d + e = 8
a2 + b2 + c2 + d2 + e2 = 16 What is the largest possible value of e?
Solution. Observe that the givens can be effectively combined by considering squares: (a − r)2 + ( b − r)2 + ( c − r)2 + ( d − r)2 + ( e − r)2 = (a2 + b2 + c2 + d2 + e2)
− 2r(a + b + c + d + e) + 5 r2
= 16 − 16 r + 5 r2
Since these squares are nonnegative, e ≤ √5r2 − 16 r + 16 + r = f (r) for all r. Since equality e = f (r) can be achieved when a = b = c = d = r, we need only compute the smallest value f (r). Since f grows large at either infinity, the minimum occurs when
f ′(r) = 1 + 10 r−16 2√5r2−16 r+16 = 0. The resultant quadratic is easily solved for r = 65 and
r = 2, with the latter being an extraneous root introduced by squaring. The largest possible e and greatest lower bound of f (r) is then f (6 /5) = 16 /5, which occurs when
a = b = c = d = 6 /5 and e = 16 /5. Alternatively, proceed as before except write
a = b = c = d = 8−e
4
since the maximum e must occur when the other four variables are equal. The second condition becomes a quadratic, and the largest solution is seen to be e = 16 5 .The notion of equating a, b, c, d is closely related to the idea of smoothing and Jensen’s inequality. If we are working with S1 = f (x1) + · · · + f (xn) under the constraint of a fixed sum x1 + · · · + xn, we can decrease S1 by moving several xi in the same interval
I together (that is, replacing xi1 < x i2 with x′
i1
= xi1 + ≤ < x i2 − ≤ = x′
i2
for any sufficiently small ≤) for any I where f is convex. S1 can also be decreased by spreading
xi in the same interval where f is concave. When seeking the maximum of S1, we proceed in the opposite fashion, pushing xi on the concave intervals of f together and moving xi on the convex intervals apart. 65. Show that for all positive reals a, b, c, d ,1
a + 1
b + 4
c + 16
d ≥ 64
a + b + c + d
Solution. Upon noticing that the numerators are all squares with √1 + √1 + √4 + √16 = √64, Cauchy should seem a natural choice. Indeed, multiplying through by
a + b + c + d and applying Cauchy, we have (a + b + c + d)
(12
a + 12
b + 22
c + 42
d
)
≥ (1 + 1 + 2 + 4) 2 = 64 as desired. 6. (USAMO 80/5) Show that for all non-negative reals a, b, c ≤ 1,
ab + c + 1 + bc + a + 1 + ca + b + 1 + (1 − a)(1 − b)(1 − c) ≤ 1
Solution. Let f (a, b, c ) denote the left hand side of the inequality. Since ∂2
∂a 2
f =
2b
(c+a+1) 3
2c
(a+b+1) 3
≥ 0, we have that f is convex in each of the three variables; hence, the maximum must occur where a, b, c ∈ { 0, 1}. Since f is 1 at each of these 8 points, the inequality follows. Second derivative testing for convexity/concavity is one of the few places where the use of Calculus is not seriously loathed by olympiad graders. It is one of the standard techniques in inequalities and deserves to be in any mental checklist of inequality solving. In this instance, it led to an easy solution. 7. (USAMO 77/5) If a, b, c, d, e are positive reals bounded by p and q with 0 < p ≤ q,prove that (a + b + c + d + e)
( 1
a + 1
b + 1
c + 1
d + 1
e
)
≤ 25 + 6
(√ pq −
√ qp
)2
and determine when equality holds.
Solution. As a function f of five variables, the left hand side is convex in each of
a, b, c, d, e ; hence, its maximum must occur when a, b, c, d, e ∈ { p, q }. When all five variables are p or all five are q, f is 25. If one is p and the other four are q, or vice versa, f becomes 17 + 4( pq + qp ), and when three are of one value and two of the other,
f = 13 + 6( pq + qp ). pq + qp ≥ 2, with equality if and only if p = q. Clearly, equality holds where p = q. Otherwise, the largest value assumed by f is 13 + 6( pq + qp ), which is obtained only when two of a, b, c, d, e are p and the other three are q, or vice versa. In such instances, f is identically the right hand side. This is a particular case of the Schweitzer inequality, which, in its weighted form, is sometimes known as the Kantorovich inequality. 78. a, b, c, are non-negative reals such that a + b + c = 1. Prove that
a3 + b3 + c3 + 6 abc ≥ 14
Solution. Multiplying by 4 and homogenizing, we seek 4a3 + 4 b3 + 4 c3 + 24 abc ≥ (a + b + c)3
= a3 + b3 + c3 + 3 (a2(b + c) + b2(c + a) + c2(a + b)) + 6 abc
⇐⇒ a3 + b3 + c3 + 6 abc ≥ a2(b + c) + b2(c + a) + c2(a + b)Recalling that Schur’s inequality gives a3 +b3 +c3 +3 abc ≥ a2(b+c)+ b2(c+a)+ c2(a+b), the inequality follows. In particular, equality necessitates that the extra 3 abc on the left is 0. Combined with the equality condition of Schur, we have equality where two of a, b, c are 12 and the third is 0. This is a typical dumbass solution.
Solution 2. Without loss of generality, take a ≥ b ≥ c. As a+b+c = 1, we have c ≤ 13
or 1 −3c ≥ 0. Write the left hand side as ( a+b)3 −3ab (a+b−2c) = ( a+b)3 −3ab (1 −3c). This is minimized for a fixed sum a + b where ab is made as large as possible. As by AM-GM ( a + b)2 ≥ 4ab , this minimum occurs if and only if a = b. Hence, we need only consider the one variable inequality 2 (1−c
2
)3 + c3 + 6 (1−c
2
)2 c = 14 · (9 c3 − 9c2 + 3 c + 1). Since c ≤ 13 , 3 c ≥ 9c2. Dropping this term and 9 c3, the inequality follows. Particularly, 9c3 = 0 if and only if c = 0, and the equality cases are when two variables are 12 and the third is 0. 9. (IMO 74/5) If a, b, c, d are positive reals, then determine the possible values of
aa + b + d + bb + c + a + cb + c + d + da + c + d
Solution. We can obtain any real value in (1 , 2). The lower bound is approached by
a → ∞ , b = d = √a, and c = 1. The upper bound is approached by a = c → ∞ ,
b = d = 1. As the expression is a continuous function of the variables, we can obtain all of the values in between these bounds. Finally, these bounds are strict because
aa + b + d + bb + c + a + cb + c + d + da + c + d >aa + b + c + d + ba + b + c + d + ca + b + c + d + da + b + c + d = 1 and
aa + b + d + bb + c + a + cb + c + d + da + c + d <aa + b + ba + b + cc + d + dc + d = 2 Whenever extrema occur for unusual parameterizations, we should expect the need for non-classical inequalities such as those of this problem where terms were completely dropped. 810. (IMO 95/2) a, b, c are positive reals with abc = 1. Prove that 1
a3(b + c) + 1
b3(c + a) + 1
c3(a + b) ≥ 32
Solution 1. Let x = 1
a
, y = 1
b
, and z = 1
c
. We perform this substitution to move terms out of the denominator. Since abc = xyz = 1, we have 1
a3(b + c) + 1
b3(c + a) + 1
c3(a + b) = x2
y + z + y2
x + z + z2
x + y
Now, multiplying through by ( x + y)( y + z)( z + x), we seek
x4 + y4 + z4 + x3y + x3z + y3z + xy 3 + xz 3 + yz 3 + x2yz + xy 2z + xyz 2 ≥
3
√xyz ·
(
3xyz + 32 · (x2y + x2z + y2x + xy 2 + xz 2 + yz 2))
which follows immediately by AM-GM, since x2yz +xy 2z+xyz 2 ≥ 3 3
√x4y4z4, x3y+xy 3+x3z
3
≥
3
√x7y4z and 7x4+4 y4+z4
12
≥ 3
√x7y4z - as guaranteed by Muirhead’s inequality.
Solution 2. Substitute x, y, z as before. Now, consider the convex function f (x) = x−1
for x > 0. ( f (x) = xc is convex for c < 0 and c ≥ 1, and concave for 0 < c ≤ 1, verify this with the second derivative test.) Now, by Jensen,
x2
y + z + y2
z + x + z2
x + y = xf
(y + zx
)
yf
(z + xy
)
zf
(x + yz
)
≥ (x + y + z)f
((y + z) + ( z + x) + ( x + y)
x + y + z
)
= x + y + z
2But x + y + z ≥ 3 3
√xyz = 3, as desired.
Solution 3. Perform the same substitution. Now, multiplying by ( x + y + z) and applying Cauchy, we have 12 (( y + z) + ( z + x) + ( x + y))
( x2
y + z + y2
z + x + z2
x + y
)
≥ 12(x + y + z)2
Upon recalling that x+y +z ≥ 3 we are done. Incidentally, the progress of this solution with Cauchy is very similar to the weighted Jensen solution shown above. This is no coincidence, it happens for many convex f (x) = xc.
Solution 4. Apply the same substitution, and put x ≥ y ≥ z. Simultaneously,
xy+z
≥ yz+x ≥ zx+y . Hence, by Chebyshev,
x ·
( xy + z
)
y ·
( yz + x
)
z ·
( zx + y
)
≥ x + y + z
3
( xy + z + yx + z + zx + y
)
Again, x + y + z ≥ 3. But now we have Nesbitt’s inequality, xy+z + yx+z + zx+y ≥ 32 . This follows immediately from AM-HM upon adding 1 to each term. 911. Let a, b, c be positive reals such that abc = 1. Show that 2(a + 1) 2 + b2 + 1 + 2(b + 1) 2 + c2 + 1 + 2(c + 1) 2 + a2 + 1 ≤ 1
Solution. The previous problem showed the substitution offers a way to rewrite an inequality in a more convenient form. Substitution can also be used to implicity use a given. First, expand the denominators and apply AM-GM, obtaining 2(a + 1) 2 + b2 + 1 = 2
a2 + b2 + 2 a + 2 ≤ 1
ab + a + 1 Now, write a = xy , b = yz , c = zx . We have 1
ab +a+1
= 1
xz+xy+1
= yz xy +yz +zx . It is now evident that the sum of the new fractions is 1. 12. (USAMO 98/3) Let a0, . . . , a n real numbers in the interval (0 , π
2
) such that tan
(
a0 − π
4
)
tan
(
a1 − π
4
)
· · · + tan
(
an − π
4
)
≥ n − 1Prove that tan( a0) tan( a1) · · · tan( an) ≥ nn+1
Solution 1. Let yi = tan (x − π
4
). We have tan( xi) = tan ((xi − π
4
) + π
4
) = yi+1 1−yi .Hence, given s = y0 + · · · + yn ≥ n − 1 we seek to prove ∏ni=0 1+ yi
1−yi
≥ nn+1 . Observe that for any a > b and fixed sum a + b, the expression
( 1 + a
1 − a
)
·
( 1 + b
1 − b
)
= 1 + 2( a + b)(1 − a)(1 − b)can be decreased by moving a and b any amount closer together. Hence, for any sequence y0, . . . , y n, we can replace any yi > sn+1 and yj < sn+1 with y′
i
= sn+1 and
y′
j
= yi + yj − sn+1 , decreasing the product. Now we have
n
∏
i=0
1 + yi
1 − yi
≥
( 1 + sn+1
1 − sn+1
)n+1
≥
( 2nn+1 2
n+1
)n+1
= nn+1
Where the last inequality follows from the fact that 1+ x
1−x
is an increasing function of x.
Solution 2. Perform the same substitution. The given can be written as 1 + yi ≥
∑
j6=i
(1 − yj ), which by AM-GM gives 1+ yn
n
≥ ∏
j6=i
(1 − yj ) 1
n
. Now we have
n
∏
i=0
1 + yi
n ≥
n
∏
i=0
∏
j6=i
(1 − yj ) 1
n
=
n
∏
i=0
(1 − yi)as desired. 10 13. Let a, b, c be positive reals. Prove that 1
a(1 + b) + 1
b(1 + c) + 1
c(1 + a) ≥ 31 + abc
with equality if and only if a = b = c = 1.
Solution. Multiply through by 1+ abc and add three to each side, on the left obtaining 1 + a + ab + abc a(1 + b) + 1 + b + bc + abc b(1 + c) + 1 + c + ac + abc c(1 + a)= (1 + a) + ab (1 + c)
a(1 + b) + (1 + b) + bc (1 + a)
b(1 + c) + (1 + c) + ac (1 + b)
c(1 + a)which is at least 6 by AM-GM, as desired. In particular, this AM-GM asserts the equivalence of (1+ a)
a(1+ b)
and a(1+ b)1+ a , or that they are both one. Likewise, all of the other terms must be 1. Now, (1 + a)2 = a2(1 + b)2 = a2b2(1 + c)2 = a2b2c2(1 + a)2, so the product abc = 1. Hence, 1+ aa(1+ b) = bc (1+ a)1+ b = bc (1+ a)
b(1+ c)
so that 1 + b = b + bc = b + 1
a
. It is now easy to see that equality holds if and only if a = b = c = 1. 14. (Romanian TST) Let a, b, x, y, z be positive reals. Show that
xay + bz + yaz + bx + zax + by ≥ 3
a + b
Solution. Note that ( a + b)( xy + yz + xz ) = ( x(ay + bz ) + y(az + bx ) + z(ax + by )). We introduce this factor in the inequality, obtaining (x(ay + bz ) + y(az + bx ) + z(ax + by ))
( xay + bz + yaz + bx + zax + by
)
≥
(x + y + z)2 ≥ 3( xy + yz + xz )Where the last inequality is simple AM-GM. The desired follows by simple algebra. Again we have used the idea of introducing a convenient factor to clear denominators with Cauchy. 15. The numbers x1, x 2, . . . , x n obey −1 ≤ x1, x 2, . . . , x n ≤ 1 and x 31 + x 32 + · · · + x 3
n
= 0. Prove that
x1 + x2 + · · · + xn ≤ n
3
Solution 1. Substitute yi = x3
i
so that y1 + · · · + yn = 0. In maximizing 3
√y1 +
· · · + 3
√yn, we note that f (y) = y 13 is concave over [0 , 1] and convex over [ −1, 0], with
|f ′(y1)| ≥ | f ′(y2)| ⇐⇒ 0 < |y1| ≤ | y2|. Hence, we may put y1 = · · · = yk = −1;
−1 ≤ yk+1 < 0, and yk+2 = · · · = yn = k−yk+1
n−k−1
. We first show that yk+1 leads to a maximal sum of 3
√yi if it is -1 or can be made positive. If |yk+1 | < |yk+2 |, we set 11 y′
k+1
= y′
k+2
= yk+1 +yk+2
2
, increasing the sum while making yk+1 positive. Otherwise, set y′
k+1
= −1 and y′
k+2
= 1 − yk+1 − yk+2 , again increasing the sum of the 3
√yi. Now we may apply Jensen to equate all positive variables, so that we need only show
k 3
√−1 + ( n − k) 3
√ kn − k ≤ n
3But we have (n + 3 k)3 − 27( n − k)2k = n3 − 18 n2k + 81 nk 2 = n(n − 9k)2 ≥ 0as desired. Particularly, as k is an integer, equality can hold only if 9 |n and then if and only if one ninth of the variables yi are -1 and the rest are 1/8.
Solution 2. Let xi = sin( αi), and write 0 = x31 + · · · + x3
n
= sin 3(α1) + · · · + sin 3(αn) =
14
((3 sin( α1) − sin(3 α1)) + · · · + (3 sin( αn) − sin(3 αn))). It follows that x1 + · · · + xn =sin( α1) + · · · + sin( αn) = sin(3 α1)+ ··· +sin(3 αn)3 ≤ n
3
. The only values of sin( α) which lead to sin(3 α) = 1 are 12 and -1. The condition for equality follows. 16. (Turkey) Let n ≥ 2 be an integer, and x1, x 2, . . . , x n positive reals such that x21 + x22 +
· · · + x2
n
= 1. Determine the smallest possible value of
x51
x2 + x3 + · · · + xn
x52
x3 + · · · + xn + x1
· · · + x5
n
x1 + · · · + xn−1
Solution. Observe that ∑ni=1 xi
∑
j6=i
xj ≤ n − 1, so that
( n∑
i=1
xi
(∑
j6=i
xj
)) ( n∑
i=1
x5
i
∑
j6=i
xi
)
≥ (x31 + · · · + x3
n
)2
= n2
(x31 + · · · + x3
n
n
)2
≥ n2
(x21 + · · · + x2
n
n
)3
= 1
n
Leads to n∑
i=1
x5
i
∑
j6=i
xi
≥ 1
n(n − 1) with equality if and only if x1 = · · · = xn = 1√n .17. (Poland 95) Let n be a positive integer. Compute the minimum value of the sum
x1 + x22
2 + x33
3 + · · · + xnn
n
12 where x1, x 2, . . . , x n are positive reals such that 1
x1
1
x2
· · · + 1
xn
= n
Solution. The given is that the harmonic mean of x1, . . . , x n is 1, which implies that the product x1x2 · · · xn is at least 1. Now, we apply weighted AM-GM
x1 + x22
2 + x33
3 + · · · + xnn
n ≥
(
1 + 12 + 13 + · · · + 1
n
)
1+ 12 +··· + 1
n
√x1x2 · · · xn
= 1 + 12 + 13 + · · · + 1
n
Prove that for all positive reals a, b, c, d ,
a4b + b4c + c4d + d4a ≥ abcd (a + b + c + d)
Solution. By AM-GM, 23 a4b + 7 b4c + 11 c4d + 10 ad 4
51 ≥ 51
√a102 b51 c51 d51 = a2bcd
from which the desired follows easily. Indeed, the most difficult part of this problem is determining suitable weights for the AM-GM. One way is to suppose arbitrary weights
x1, x 2, x 3, x 4 for a4b, b 4c, c 4d, ad 4 respectively, and solve the system
x1 + x2 + x3 + x4 = 14x1 + x2 = 24x2 + x3 = 14x3 + x4 = 119. (USAMO 01/3) Let a, b, c be nonnegative reals such that
a2 + b2 + c2 + abc = 4 Prove that 0 ≤ ab + bc + ca − abc ≤ 2
Solution [by Tony Zhang.] For the left hand side, note that we cannot have a, b, c >
Suppose WLOG that c ≤ 1. Then ab +bc +ca −abc ≥ ab +bc +ca −ab = c(a+b) ≥ 0. For the right, 4 = a2 + b2 + c2 + abc ≥ 4( abc )43 =⇒ abc ≤ 1. Since by the pigeon hole principle, among three numbers either two exceed 1 or two are at most 1. Hence, we assume WLOG that ( a − 1)( b − 1) ≥ 0, which gives ab + 1 ≥ a + b ⇐⇒ abc + c ≥ ac +
bc ⇐⇒ c ≥ ac + bc − abc . Now, we have ab + bc + ca − abc ≤ ab + c. Either we are done or ab +c > 2. But in the latter case, 4 = ( a2 +b2)+ c(c+2 ab ) > 2ab +2 c = 2( ab +c) > 4, a contradiction. 13 20. (Vietnam 98) Let x1, . . . , x n be positive reals such that 1
x1 + 1998 + 1
x2 + 1998 + · · · + 1
xn + 1998 = 11998 Prove that n
√x1x2 · · · xn
n − 1 ≥ 1998
Solution. Let yi = 1
xi+1998
so that y1 + · · · + yn = 11998 and xi = 1
yi
− 1998. Now
n
∏
i=1
xi =
n
∏
i=1
( 1
yi
− 1998
)
= e
Pni=1 ln
“1
yi−1998
”
Hence, to minimize the product of the xi, we equivalently minimize the sum of ln
( 1
yi
− 1998
)
.In particular,
ddy
(
ln
( 1
y − 1998
))
= 1
( 1
y
− 1998
)2 · −1
y2
= −1
y − 1998 y2
d2
dy 2
(
ln
( 1
y − 1998
))
= 1 − 3996 y
(y − 1998 y2)2
So ln
( 1
y
− 1998
)
is convex on [0 , 1/3996]. If we had 0 < y i ≤ 1/3996 for all i we could apply Jensen. Since yi + yj ≤ 1/1998 for all i, j , we consider
( 1
a − 1998
) ( 1
b − 1998
)
≥
( 2
a + b − 1998
)2
⇐⇒ 1
ab − 1998
( 1
a + 1
b
)
≥ 4(a + b)2 − 4 · 1998
a + b
⇐⇒ (a + b)2 − 1998( a + b)3 ≥ 4ab − 4ab (a + b) · 1998
⇐⇒ (a − b)2 ≥ 1998( a + b)( a − b)2
which incidentally holds for any a + b ≤ 11998 . Hence, any two yi and yj may be set to their average while decreasing the sum in question; hence, we may assume yi ∈ (0 , 13996 ]. Now Jensen’s inequality shows that the minimum occurs when yi = 11998 n for all i, or when xi = 1998( n − 1) for all i. It is easy to see that this yields equality. 21. (Romania 99) Show that for all positive reals x1, . . . , x n with x1x2 · · · xn = 1, we have 1
n − 1 + x1
· · · + 1
n − 1 + xn
≤ 114 Solution. First, we prove a lemma: the maximum of the sum occurs when n − 1 of the xi are equal. Consider f (y) = 1
k+ey
for an arbitrary nonnegative constant k. We have f ′(y) = −ey
(k+ey)2
and f ′′ (y) = ey (ey −k)(k+ey )3 . Evidently f ′′ (y) ≥ 0 ⇐⇒ ey ≥ k. Hence,
f (y) has a single inflexion point where y = ln( k), where f (y) is convex over the interval ((ln( k), ∞). Now, we employ the substitution yi = ln( xi) so that y1 + · · · + yn = 0 and
n
∑
i=1
1
n − 1 + xi
=
n
∑
i=1
f (yi)We take k = n − 1 and write k0 = ln( n − 1). Suppose that y1 ≥ · · · ≥ ym ≥ k0 ≥
ym+1 ≥ · · · xn for some positive m. Then by, Majorization,
f (y1) + · · · + f (ym) ≤ (m − 1) f (k0) + f (y1 + · · · + ym − (m − 1) k0)But then, also by Majorization, (m − 1) f (k0) + f (ym+1 ) + · · · + f (yn) ≤ (n − 1) f
((m − 1) k0 + ym+1 + · · · + yn
n − 1
)
Otherwise, all of the yi are less than k0. In that case we may directly apply Majorization to equate n − 1 of the yi whilst increasing the sum in question. Hence, the lemma is valid. 7 N
Applying the lemma, it would suffice to show
kk + x + 1
k + 1
xk
≤ 1Clearing the denominators,
(
k2 + kxk
)
( k + x) ≤ k2 + k
(
x + 1
xk
)
x1−k
−xk + x + k ≤ x1−k
But now this is evident. We have Bernoulli’s inequality, since x1−k = (1 + ( x − 1)) 1−k ≥
1 + ( x − 1)(1 − k) = x + k − xk . Equality holds only where x = 1 or n = 2. 22. (Darij Grinberg) Show that for all positive reals a, b, c ,
√b + ca +
√c + ab +
√a + bc ≥ 4( a + b + c)
√(a + b)( b + c)( c + a)
7This n−1 equal value principle is particularly useful. If a differentiable function has a single inflexion point and is evaluated at narbitrary reals with a fixed sum, any minimum or maximum must occur where some n−1 variables are equal.
15 Solution 1. By Cauchy, we have √(a + b)( a + c) ≥ a + √bc . Now,
∑
cyc
√b + ca ≥ 4( a + b + c)
√(a + b)( b + c)( c + a)
⇐⇒ ∑
cyc
b + ca
√(a + b)( a + c) ≥ 4( a + b + c)Substituting our result from Cauchy, it would suffice to show
∑
cyc
(b + c)
√bc a ≥ 2( a + b + c)WLOG a ≥ b ≥ c, implying b + c ≤ c + a ≤ a + b and √bc a ≤ √ca b ≤ √ab c . Hence, by Chebyshev and AM-GM,
∑
cyc
(b + c)
√bc a ≥
(2( a + b + c))
(√bc a + √ca b + √ab c
)
3
≥ 2( a + b + c)as desired.
Solution 2. Let x = √b + c, y = √c + a, z = √a + b. Then x, y, z are the sides of acute triangle XY Z (in the typical manner), since x2 + y2 = a + b + 2 c > a + b = z2.The inequality is equivalent to
∑
cyc
xy2 + z2 − x2 ≥ x2 + y2 + z2
xyz
Recalling that y2 + z2 − x2 = 2 yz cos( X), we reduce this to the equivalent
∑
cyc
x2
cos( X) ≥ 2( x2 + y2 + z2)WLOG, we have x ≥ y ≥ z, implying 1cos( X) ≥ 1cos( Y ) ≥ 1cos( Z) , so that applying Chebyshev to the left reduces the desired to proving that the sum of the reciprocals of the cosines is at least 6. By AM-HM, 1cos( X) + 1cos( Y ) + 1cos( Z) ≥ 9cos( X) + cos( Y ) + cos( Z)But recall from triangle geometry that cos( X) + cos( Y ) + cos( Z) = 1 + rR and R ≥ 2r.The desired is now evident. 16 23. Show that for all positive numbers x1, . . . , x n,
x31
x21 + x1x2 + x22
x32
x22 + x2x3 + x23
· · · + x3
n
x2
n
xnx1 + x21
≥ x1 + · · · + xn
3
Solution. Observe that 0 = ( x1 −x2)+( x2 −x3)+ · · · +( xn −x1) = ∑ni=1
x3
i−x3
i+1
x2
i+xixi+1 +x2
i+1
.Hence, (where xn+1 = x1)
n
∑
i=1
x3
i
x2
i
xixi+1 x2
i+1
= 12
n
∑
i=1
x3
i
x3
i+1
x2
i
xixi+1 + x2
i+1
But now a3 + b3 ≥ 13 a3 + 23 a2b + 23 ab 2 + 13 b3 = 13 (a + b)( a2 + ab + b2). Hence, 12
n
∑
i=1
x3
i
x3
i+1
x2
i
xixi+1 + x2
i+1
≥ 12
n
∑
i=1
xi + xi+3
3 = 13
n
∑
i=1
xi
as desired. 24. Let a, b, c be positive reals such that a + b ≥ c; b + c ≥ a; and c + a ≥ b, we have 2a2(b + c) + 2 b2(c + a) + 2 c2(a + b) ≥ a3 + b3 + c3 + 9 abc
Solution. After checking that equality holds for ( a, b, c ) = ( t, t, t ) and (2 t, t, t ), it is apparent that more than straight AM-GM will be required. To handle the condition, put a = y + z, b = z + x, c = x + y with x, y, z ≥ 0. Now, the left hand side becomes 4x3 + 4 y3 + 4 z3 + 10 x2(y + z) + 10 y2(z + x) + 10 z2(x + y) + 24 xyz while the right hand side becomes 2 x3 + 2 y3 + 2 z3 + 12 x2(y + z) + 12 y2(z + x) + 12 z2(x + y) + 18 xyz .The desired is seen to be equivalent to x3 + y3 + z3 + 3 xyz ≥ x2(y + z) + y2(z +
x) + z2(x + y), which is Schur’s inequality. Equality holds where x = y = z, which gives ( a, b, c ) = ( t, t, t ), or when two of x, y, z are equal and the third is 0, which gives (a, b, c ) ∈ { (2 t, t, t ), (t, 2t, t ), (t, t, 2t)}.25. Let a, b, c be the lengths of the sides of a triangle. Prove that
a
√2b2 + 2 c2 − a2 + b
√2c2 + 2 a2 − b2 + c
√2a2 + 2 b2 − c2 ≥ √3
Solution 1. Again write a = y + z, b = z + x, and c = x + y, noting that x, y, z are positive. (Triangles are generally taken to be non-degenerate when used in inequalities.) We have
∑
cyc
a
√2b2 + 2 c2 − a2 = ∑
cyc
y + z
√4x2 + 4 xy + 4 xz + y2 + z2 − 2yz
17 Consider the convex function f (x) = 1√x . (As we shall see, Jensen almost always provides a tractable means of eliminating radicals from inequalities.) Put x+y +z = 1. We have
∑
cyc
(y + z)f (4x2 + 4 xy + 4 xz + y2 + z2 − 2yz ) ≥
(( y + z) + ( z + x) + ( x + y)) f
(∑
cyc
(y + z) (4 x2 + 4 xy + 4 xz + y2 + z2 − 2yz )(y + z) + ( z + x) + ( x + y)
)
= 2√2
√∑
cyc
4x2(y + z) + (4 xy 2 + 4 xyz ) + (4 xyz + 4 xz 2) + y3 + z3 − y2z − yz 2
Noting that ∑
cyc
4x2(y + z) + (4 xy 2 + 4 xyz ) + (4 xyz + 4 xz 2) + y3 + z3 − y2z − yz 2 =
∑
cyc
2x3 + 7 x2(y + z) + 8 xyz ,8( x + y + z)3 ≥ 3 ∑
cyc
2x3 + 7 x2(y + z) + 8 xyz
⇐⇒ ∑
sym
4x3 + 24 x2y + 8 xyz ≥ ∑
sym
3x3 + 21 x2y + 12 xyz
⇐⇒ 2x3 + 2 y3 + 2 z3 + 3 (x2(y + z) + y2(z + x) + z2(x + y)) ≥ 24 xyz
which follows by AM-GM. As a follow up on an earlier mentioned connection, oberserve the similarity of the above application of Jensen and the following inequality (which follows by H¨ older’s inequality)
(∑
i
αiβi
) ( ∑
i
αi
1
√βi
)2
≥
(∑
i
αi
)3
Solution 2 [by Darij Grinberg.] Let ABC be a triangle of side lengths a, b, c
in the usual order. Denote by ma, m b, m c the lengths of the medians from A, B, C
respectively. Recall from triangle goemetry that ma = 12
√2b2 + 2 c2 − a2, so that we need only show ama + bmb + cmc ≥ 2√3. But a triangle with side lengths ma, m b, m c, in turn, has medians of length 3a
4
, 3b
4
, and 3c
4
. The desired inequality is therefore equivalent to 43 ma
a
43 mb
b
43 mc
c
≥ 2√3 where we refer to the new triangle ABC . Recalling that
23
ma = AG , where G is the centroid, the desired is seen to be equivalent to the geometric inequality AG a + BG b + CG c ≥ √3. But we are done as we recall from triangle geometry that AM a + BM b + CM c ≥ √3 holds for any point inside triangle ABC .8
8For a complete proof of this last inequality, see post #14.
18 26. (IMO 99/2) For n ≥ 2 a fixed positive integer, find the smallest constant C such that for all nonnegative reals x1, . . . , x n,
∑
1≤i<j ≤n
xixj (x2
i
x2
j
) ≤ C
( n∑
i=1
xi
)4
Solution. The answer is C = 18 , which is obtained when any two xi are non-zero and equal and the rest are 0. Observe that by AM-GM, (x1 + · · · + xn)4 =
( n∑
i=1
x2
i
2 ∑
1≤i<j ≤n
xixj
)2
≥ 4
( n∑
i=1
x2
i
) (
2 ∑
1≤i<j ≤n
xixj
)
= 8 ∑
1≤i<j ≤n
xixjn∑
k=1
x2
k
But x21 + · · · + x2
n
≥ x2
i
x2
j
with equality iff xk = 0 for all k 6 = i, j . It follows that (x1 + · · · + xn)4 ≥ 8 ∑
1≤i<j ≤n
xixj
(x2
i
x2
j
)
as desired. 27. Show that for nonnegative reals a, b, c ,2a6 + 2 b6 + 2 c6 + 16 a3b3 + 16 b3c3 + 16 c3a3 ≥ 9a4(b2 + c2) + 9 b4(c2 + a2) + 9 c4(a2 + b2)
Solution 1. Consider
∑
cyc
(a − b)6 = ∑
cyc
a6 − 6a5b + 15 a4b2 − 20 a3b3 + 15 a2b4 − 6ab 5 + b6 ≥ 0and ∑
cyc
ab (a − b)4 = ∑
cyc
a5b − 4a4b2 + 6 a3b3 − 4a2b4 + ab 5 ≥ 0Adding six times the latter to the former yields the desired result.
Solution 2. We shall prove a6 − 9a4b2 + 16 a3b3 − 9a2b4 + b6 ≥ 0. We have
a6 − 2a3b3 + b6 = (a3 − b3)2
= ((a − b)( a2 + ab + b2))2
≥ (a − b)2(3 ab )2 = 9 a4b2 − 18 a3b3 + 9 a2b4
19 As desired. The result now follows from adding this lemma cyclicly. The main difficulty with this problem is the absence of a5b terms on the right and also the presence of
a4b2 terms on the right - contrary to where Schur’s inequality would generate them. Evidently AM-GM is too weak to be applied directly, since a6 + 2 a3b3 ≥ 3a4b2 cannot be added symmetrically to deduce the problem. By introducing the factor ( a − b)2,however, we weight the AM-GM by a factor which we “know” will be zero at equality, thereby increasing its sharpness. 28. Let 0 ≤ a, b, c ≤ 12 be real numbers with a + b + c = 1. Show that
a3 + b3 + c3 + 4 abc ≤ 932
Solution. Let f (a, b, c ) = a3 + b3 + c3 + 4 abc and g(a, b, c ) = a + b + c = 1. Because
f and g are polynomials, they have continuous first partial derivatives. Moreover, the gradient of g is never zero. Hence, by the theorem of Lagrange Multipliers ,any extrema occur on the boundary or where ∇f = λ∇g for suitable scalars λ. As
∇f =< 3a2 + 4 bc, 3b2 + 4 ca, 3c2 + 4 ab > and ∇g =< 1, 1, 1 >, we have
λ = 3a2 + 4 bc
= 3b2 + 4 ca
= 3c2 + 4 ab g(a, b, c ) = a + b + c = 1 We have 3 a2 + 4 bc = 3 b2 + 4 ca or ( a − b)(3 a + 3 b − 4c) = ( a − b)(3 − 7c) = 0 for any permutation of a, b, c . Hence, without loss of generality, a = b. Now, 3 a2 + 4 ac =3c2 + 4 a2 and a2 − 4ac + 3 c2 = ( a − c)( a − 3c) = 0. The interior local extrema therefore occur when a = b = c or when two of {a, b, c } are three times as large as the third. Checking, we have f (13 , 13 , 13 ) = 7 /27 < 13 /49 = f (17 , 37 , 37 ). Recalling that f (a, b, c ) is symmetric in a, b, c , the only boundary check we need is f (12 , t, 12 −t) ≤ 932 for 0 ≤ t ≤ 12 .We solve
h(t) = f
(12, t, 12 − t
)
= 18 + t3 +
(12 − t
)3
2 t
(12 − t
)
= 14 + t
4 − t2
2
h(t) is 14 at either endpoint. Its derivative h′(t) = 14 − t is zero only at t = 14 . Checking,
h(14 ) = f (12 , 14 , 14 ) = 932 . Since h(t) has a continuous derivative, we are done. (As a further check, we could observe that h′′ (t) = −1 < 0, which guarantees that h(14 ) is a local minimum.) 20 Usage Note. The use of Lagrange Multipliers in any solution will almost certainly draw hostile review, in the sense that the tiniest of errors will be grounds for null marks. If you consider multipliers on Olympiads, be diligent and provide explicit, kosher remarks about the continuous first partial derivatives of both f (x1, . . . , x n) and the constraint g(x1, . . . , x n) = k, as well as ∇g 6 = 0, before proceeding to solve the system ∇f = λ∇g. The main reason this approach is so severely detested is that, given sufficient computational fortitude (if you are able to sort through the relevant algebra and Calculus), it can and will produce a complete solution. The example provided here is included for completeness of instruction; typical multipliers solutions will not be as clean or painless. 9
(Vascile Cartoaje) Let p ≥ 2 be a real number. Show that for all nonnegative reals
a, b, c ,
3
√
a3 + pabc
1 + p + 3
√
b3 + pabc
1 + p + 3
√
c3 + pabc
1 + p ≤ a + b + c
Solution. By H¨ older,
(∑
cyc
3
√
a3 + pabc
1 + p
)3
≤
(∑
cyc
11 + p
) ( ∑
cyc
a
) ( ∑
cyc
a2 + pbc
)
But a2 + b2 + c2 ≥ ab + bc + ca (proven by AM-GM, factoring, or a number of other methods) implies that
∑
cyc
a2 + pbc ≤ (p + 1) ∑
cyc
a2 + 2 bc
3 = p + 1 3 (a + b + c)2
From which we conclude
(∑
cyc
3
√
a3 + pabc
1 + p
)3
≤ (a + b + c)3
as desired. 30. Let a, b, c be real numbers such that abc = −1. Show that
a4 + b4 + c4 + 3( a + b + c) ≥ a2
b + a2
c + b2
c + b2
a + c2
a + c2
b
Solution. First we homogenize, obtaining a4 + b4 + c4 + a3(b + c) + b3(c + a) + c3(a +
b) − 3abc (a + b + c) ≥ 0. As this is homogenous in the fourth degree, we can scale a, b, c
9Just how painful can the calculations get? Most multipliers solutions will tend to look more like than this solution.
21 by any real k and hence may now ignore abc = −1. Equality holds at a = b = c = 1, but also at a = b = 1 , c = −2, a = 1 , b = 0 , c = −1, and a number of unusual locations with the commonality that a + b + c = 0. Indeed, c = −a − b is a parametric solution, and we discover the factorization ( a + b + c)2(a2 + b2 + c2 − ab − bc − ca ) ≥ 0. (We are motivated to work with factorizations because there are essentially no other inequalities with a + b + c = 0 as an equality condition.) 31. (MOP 2003) Show that for all nonnegative reals a, b, c ,
a4(b2 + c2) + b4(c2 + a2) + c4(a2 + b2) +2abc (a2b + a2c + b2c + b2a + c2a + c2b − a3 − b3 − c3 − 3abc )
≥ 2a3b3 + 2 b3c3 + 2 c3a3
Solution. As was suggested by the previous problem, checking for equality cases is important when deciding how to solve a problem. We see that setting a = b produces equality. As the expression is symmetric, this certainly implies that b = c and c = a are equality cases. Hence, if P (a, b, c ) is the difference LHS - RHS, then ( a − b)( b − c)( c −
a)|P (a, b, c ). Obviously, if the problem is going to be true, ( a−b) must be a double root of P , and accordingly we discover the factorization P (a, b, c ) = ( a − b)2(b − c)2(c − a)2.The result illustrated above was no accident. If ( x−y) divides a symmetric polynomial
P (x, y, z ), then ( x − y)2 divides the same polynomial. If we write P (x, y, z ) = ( x −
y)Q(x, y, z ), then ( x − y)Q(x, y, z ) = P (x, y, z ) = P (y, x, z ) = ( y − x)Q(y, x, z ), which gives Q(x, y, z ) = −Q(y, x, z ). Hence Q(x, x, z ) = 0, and ( x − y) also divides Q(x, y, z ). 32. (Cezar Lupu) Let a, b, c be positive reals such that a + b + c + abc = 4. Prove that
a
√b + c + b
√c + a + c
√a + b ≥√22 · (a + b + c)
Solution. By Cauchy
(∑
cyc
a√b + c
) ( ∑
cyc
a
√b + c
)
≥ (a + b + c)2
But, also by Cauchy,
√(a + b + c) ( a(b + c) + b(c + a) + c(a + b)) ≥ ∑
cyc
a√b + c
Hence, ∑
cyc
a
√b + c ≥√22 · (a + b + c) ·
√ a + b + cab + bc + ca
22 And we need only show a + b + c ≥ ab + bc + ca . Schur’s inequality for r = 1 can be expressed as 9abc a+b+c ≥ 4( ab + bc + ca ) − (a + b + c)2. Now, we suppose that
ab + bc + ca > a + b + c, and have 9abc a + b + c ≥ 4( ab + bc + ca ) − (a + b + c)2
(a + b + c) (4 − (a + b + c)) = abc (a + b + c)Hence, a + b + c < 3. But then abc < 1, which implies 4 = a + b + c + abc < 4. Contradiction, as desired. 33. (Iran 1996) Show that for all positive real numbers a, b, c ,(ab + bc + ca )
( 1(a + b)2 + 1(b + c)2 + 1(c + a)2
)
≥ 94
Solution. Fearless courage is the foundation of all success. 10 When everything else fails, return to the sure-fire strategy of clearing all denominators. In this case, we obtain 4( a + b)2(b + c)2(c + a)2(ab + bc + ca )
( 1(a + b)2 + 1(b + c)2 + 1(c + a)2
)
=
∑
sym
4a5b + 8 a4b2 + 10 a4bc + 6 a3b3 + 52 a3b2c + 16 a2b2c2
on the left, and on the right, 9( a + b)2(b + c)2(c + a)2 =
∑
sym
9a4b2 + 9 a4bc + 9 a3b3 + 54 a3b2c + 15 a2b2c2
Canceling like terms, we seek
∑
sym
4a5b − a4b2 + a4bc − 3a3b3 − 2a3b2c + a2b2c2
Sure enough, this is true, since 3a5b+ab 5
4
≥ a4b2 and a4b2+a2b4
2
≥ a3b3 by AM-GM, and
abc (a3 + b3 + c3 − a2(b + c) + b2(c + a) + c2(a + b) + 3 abc ) ≥ 0 by Schur. 34. (Japan 1997) Show that for all positive reals a, b, c ,(a + b − c)2
(a + b)2 + c2 + (b + c − a)2
(b + c)2 + a2 + (c + a − b)2
(c + a)2 + b2 ≥ 35
10 Found on a fortune cookie by Po-Ru Loh while grading an inequality on 2005 Mock IMO Day 2 that was solved by brutal force.
23 Solution. Put a + b + c = 3 so that equality will hold at a = b = c = 1 and suppose that there exists some k for which (b + c − a)2
(b + c)2 + a2 = (3 − 2a)2
(3 − a)2 + a2 ≥ 15 + ka − k
for all positive a, b, c ; such an inequality would allow us to add cyclicly to deduce the desired inequality. As the inequality is parametrically contrived to yield equality where
a = 1, we need to find k such that a = 1 is a double root. At a = 1, the derivative on the left is (2(3 −2a)·− 2)((3 −a)2+a2)−((3 −2a)2)(2(3 −a)·− 1+2 a)((3 −a)2+a2)2 = −18 25 . The derivative on the right is k, so we set k = −18 25 . But for this k we find (3 − 2a)2 −
(15 + ka − k
) ((3 − a)2 + a2) = 18 25 − 54 a2
25 + 36 a3
25 = 18 25 (a − 1) 2(2 a + 1) ≥ 0as desired. Alternatively, we could have used AM-GM to show a3 + a3 + 1 ≥ 3a2. As hinted at by a previous problem, inequalities are closely linked to polynomials with roots of even multiplicity. The isolated manipulation idea used in this solution offers a completely different approach to the inequalities which work with every term. 35. (MOP 02) Let a, b, c be positive reals. Prove that
( 2ab + c
)23
+
( 2bc + a
)23
+
( 2ca + b
)23
≥ 3
Solution. Suppose that there exists some r such that
( 2ab + c
)23
≥ 3ar
ar + br + cr
We could sum the inequality cyclicly to deduce what we want. Since equality holds at
a = b = c = 1, we use derivatives to find a suitable r. At the said equality case, on the left, the partial derivative with respect to a is 23 , while the same derivative on the right is 23 r. Equating the two we have r = 1. (This is necessary since otherwise the inequality will not hold for either a = 1 + ≤ or a = 1 − ≤.) 11 Now, 3aa + b + c ≤ 3a
3 3
√
a · (b+c
2
)2
11 Actually, even this is a special case of the general sense that the convexity of one side must exceed the convexity of the other. More precisely, we have the following result: Let fand gfunctions over the domain Dwith continuous partial derivatives. If f(ν)≥g(ν) for all ν∈D, then at every equality case ν0,
∇(f−g)( ν0) = 0and every component of ∇2(f−g) ( ν0) is nonnegative.
24 = a23
(b+c
2
)23
=
( 2ab + c
)23
by AM-GM, as desired. 36. (Mildorf) Let n ≥ 2 be an integer. Prove that for all reals a1, a 2, . . . , a n > 0 and reals
p, k ≥ 1, ( a1 + a2 + · · · + an
ap
1
ap
2
· · · + apn
)k
≥ ak
1
ak
2
· · · + akn
apk
1
apk
2
· · · + apk n
where inequality holds iff p = 1 or k = 1 or a1 = a2 = · · · = an, flips if instead 0 < p < 1, and flips (possibly again) if instead 0 < k < 1.
Solution. Taking the kth root of both sides, we see that the inequality is equivalent to n∑
i=1
k
√
aki
ak
1
ak
2
· · · + akn
≥
n
∑
i=1
k
√
apk i
apk
1
apk
2
· · · apk n
WLOG, suppose that a1 ≥ a2 ≥ · · · ≥ an. We prove a lemma. Let Si = api
ap
1+··· +apn
and
Ti = aqi
aq
1+··· +aqn
for i = 1 , 2, . . . , n where 0 < q < p . Then the sequence S1, S 2, . . . , S n
majorizes the sequence T1, T 2, . . . , T n.To prove the claim, we note that S1 ≥ · · · ≥ Sn and T1 ≥ · · · ≥ Tn and have, for
m ≤ n,
m
∑
i=1
Si ≥
m
∑
i=1
Ti
⇐⇒ (ap
1
· · · + apm) ( aq
1
· · · + aqn) ≥ (aq
1
· · · + aqm) ( ap
1
· · · + apn)
⇐⇒ (ap
1
· · · + apm) (aqm+1 + · · · + aqn
) ≥ (aq
1
· · · + aqm) (apm+1 + · · · + apn
)
⇐⇒ ∑
(i,j )| { 1≤i≤m<j ≤n}
api aqj − aqi apj ≥ 0Which is obvious. In particular, m = n is the equality case, and the claim is established. But now the desired is a direct consequence of the Majorization inequality applied to the sequences in question and the function f (x) = k
√x.37. (Vascile Cartoaje) Show that for all real numbers a, b, c,(a2 + b2 + c2)2 ≥ 3 (a3b + b3c + c3a)
25 Solution. We will be content to give the identity (a2 + b2 + c2)2 − 3( a3b + b3c + c3a) = 12
∑
cyc
(a2 − 2ab + bc − c2 + ca )2
Any Olympiad partipant should be comfortable constructing various inequalities through well-chosen squares. Here, we could certainly have figured we were summing the square of a quadratic that is 0 when a = b = c such that no term a2bc is left uncancelled. A good exercise is to show that equality actually holds iff a = b = c or, for some cyclic permutation, a : b : c ≡ sin 2 (4π
7
) : sin 2 (2π
7
) : sin 2 (π
7
).38. (Anh-Cuong) Show that for all nonnegative reals a, b, c ,
a3 + b3 + c3 + 3 abc ≥ ab √2a2 + 2 b2 + bc √2b2 + 2 c2 + ca √2c2 + 2 a2
Solution. Upon observing that this inequality is stronger than Schur’s inequality for
r = 1, we are inspired to prove a sharp lemma to eliminate the radical. Knowing that √2x2 + 2 y2 ≥ x + y ≥ 2xy x+y , we seek a combination of the latter two that exceeds the former. We find 3x2 + 2 xy + 3 y2
2( x + y) ≥ √2x2 + 2 y2
This follows from algebra, since (3 x2 + 2 xy + 3 y2)2 = 9 x4 + 12 x3y + 22 x2y2 + 12 xy 3 +9y4 ≥ 8x4 + 16 x3y + 16 x2y2 + 16 xy 3 + 8 y4 = 4( x + y)2(2 x2 + 2 y2), so that (3 x2 + 2 xy +3y2)2 − 4( x + y)2(2 x2 + 2 y2) = x4 − 4x3y + 6 x2y2 − 4xy 3 + y4 = ( x − y)4 ≥ 0. Now,
∑
cyc
ab √2a2 + 2 b2 ≤ ∑
cyc
(3 a2 + 2 ab + 3 b2)ab
2( a + b)So it would suffice to show
∑
cyc
a(a − b)( a − c) = ∑
cyc
(a3 + abc − ab (a + b))
≥ ∑
cyc
(3 a2 + 2 ab + 3 b2)ab
2( a + b) − ab (a + b)= ∑
cyc
3a3b + 2 a2b2 + 3 ab 3 − 2a3b − 4a2c2 − 2ab 3
2( a + b)= ∑
cyc
ab (a − b)2
2( a + b)But ∑
cyc
(b + c − a)( b − c)2 = 2 ∑
cyc
a(a − b)( a − c)26 so that the desired is
∑
cyc
(
b + c − a − bc b + c
)
(b − c)2 ≥ 0which is evident, since without loss of generality we may assume a ≥ b ≥ c and find
(
a + b − c − ab a + b
)
(a − b)2 ≥ 0
(
c + a − b − ac a + c
) ((a − c)2 − (b − c)2) ≥ 0
(
b + c − a − bc b + c
)
(b − c)2 +
(
c + a − b − ac a + c
)
(b − c)2 ≥ 0The key to this solution was the sharp upper bound on the root-mean-square. At first glance our lemma seems rather arbitrary and contrived. Actually, it is a special case of a very sharp bound on the two variable power mean that I have conjectured and proved.
Mildorf’s Lemma 1 Let k ≥ − 1 be an integer. Then for all positive reals a and b,
(1 + k)( a − b)2 + 8 ab
4( a + b) ≥ k
√ak + bk
2
with equality if and only if a = b or k = ±1, where the power mean k = 0 is interpreted to be the geometric mean √ab . Moreover, if k < −1, then the inequality holds in the reverse direction, with equality if and only if a = b.
Usage Note. As of early November 2005, I have proven an extension of this lemma to additional values of k.12 Thus, you may rest assured that the result stated above is true. I was unable to get this result published, so I have instead posted the proof here as “ASharpBound.pdf.” However, the proof is rather difficult (or at least so I think, being as though it took me nearly half a year) and the lemma is far from mainstream. Thus, should you require it on an Olympiad, you should prove it for whatever particular value of k you are invoking. This is not terribly difficult if k is a small integer. One simply takes the kth power of both sides and factors the difference of the two sides as
(a − b)4 · P (a, b ), etc.
For x ≥ y ≥ 1, prove that
x
√x + y + y
√y + 1 + 1
√x + 1 ≥ y
√x + y + x
√x + 1 + 1
√y + 1
12 In particular, the inequality holds for all kin ( −∞ ,−1) ,{− 1,0,1},(1 ,3/2] ,[2 ,∞) with the signs ≤,≥,≤
,≥respectively, with equality iff a=bor k=±1.
27 Solution. By observation, equality holds when y = 1 and when x = y. Combining this with the restriction, it makes sense to write x = y + a and y = 1 + b where a, b ≥ 0. Now we can write
x − y
√x + y + y − 1
√y + 1 + 1 − x
√1 + x ≥ 0
⇐⇒ a
√2 + a + 2 b + b
√2 + b ≥ a + b
√2 + a + b
But this is evident by Jensen’s inequality applied to the convex function f (x) = 1√x ,since
af (2 + a + 2 b) + bf (2 + b) ≥ (a + b)f
(a(2 + a + 2 b) + b(2 + b)
a + b
)
= (a + b)f
((a + b)2 + 2( a + b)
a + b
)
= a + b
√2 + a + b
as desired. 40. (MOP) For n ≥ 2 a fixed positive integer, let x1, . . . , x n be positive reals such that
x1 + x2 + · · · + xn = 1
x1
1
x2
· · · + 1
xn
Prove that 1
n − 1 + x1
1
n − 1 + x2
· · · + 1
n − 1 + xn
≤ 1
Solution. We will prove the contrapositive. (We are motivated to do this for two good reasons: 1) it is usually difficult the show that the sum of some reciprocals is bounded above, and 2) the given relation in its current form is an abomination.) Take
yi = 1
n−1+ xi
, and for the sake of contradiction assume y1 + · · · + yn > 1. Since the yi
are too large, the xi are too small and we shall prove 1
x1
· · · + 1
xn
x 1 + · · · + xn.Since xiyi = 1 − (n − 1) yi, we have (n − 1) yi > (n − 1)
(
yi + 1 −
n
∑
j=1
yj
)
= (n − 1) yi − 1 +
n
∑
j=1
(1 − (n − 1) yj )= −xiyi +
n
∑
j=1
xj yj (∗)=⇒ n − 1
xi
−1 +
n
∑
j=1
xj yj
xiyi
(∗∗ )28 Summing () over i,(n − 1)
( 1
x1
· · · + 1
xn
)
n
∑
i=1
xiyi
(( n∑
j=1
1
xj yj
)
− 1
xiyi
)
But by Cauchy and (), we have
( n∑
j=1
1
xj yj
)
− 1
xiyi
≥ (n − 1) 2
(∑nj=1 xj yj
)
− xiyi
(n − 1) 2
(n − 1) yi
= n − 1
yi
Hence, (n − 1)
( 1
x1
· · · + 1
xn
)
n
∑
i=1
xiyi
(n − 1
yi
)
= ( n − 1)( x1 + · · · + xn)as desired. 41. (Vascile Cartoaje) Show that for positive reals a, b, c ,14a2 − ab + 4 b2 + 14b2 − bc + 4 c2 + 14c2 − ca + 4 a2 ≥ 97( a2 + b2 + c2)
Solution. Upon expansion, we see that it is equivalent to
∑
sym
56 a6 − 28 a5b + 128 a4b2 + 44 a3b3 + 95 2 a4bc + 31 a3b2c − 45 2 a2b2c2 ≥ 0We conjure up the following inequalities:
∑
sym
a6 − 2a5b + a4bc ≥ 0 (1)
∑
sym
a5b − 4a4b2 + 3 a3b3 ≥ 0 (2)
∑
sym
a4b2 − a4bc − a3b3 + 2 a3b2c − a2b2c2 ≥ 0 (3)
∑
sym
a4bc − 2a3b2c + a2b2c2 ≥ 0 (4) (1) and (4) follow from Schur’s inequality for r = 4 and r = 1 (multiplied by abc )respectively. (2) is the result of expanding ∑
cyc
ab (a − b)4 ≥ 0, and (3) is the expanded form of the famous ( a − b)2(b − c)2(c − a)2 ≥ 0. The desired now follows by subtracting 56 times (1), 84 times (2), 208 times (3), 399 2 times (4), and then simple AM-GM to clear the remaining a2b2c2.29 This is about as difficult as a dumbass solution can get. A good general strategy is to work with the sharpest inequalities you can find until you reduce a problem to something obvious, starting with the most powerful (most bunched, in this case ∑
sym
a6) term and work your way down to the weak terms while keeping the most powerful term’s coefficient positive. My solution to this problem starts with (1), Schur with r = 4 (Schur is stronger for larger r), which is almost certainly sharper than the inequality in question. Next, inequality (2) is a sharp cyclic sum to use the a5b terms. In particular, it relates terms involving only two of the three variables. Most of the time, the only inequality that can “pull up” symmetric sums involving three variables to stronger ones involving just two is Schur, although it does so at the expense of a very strong term with only one variable. Hence, we made a logical choice. Inequality (3) is extremely sharp, and allowed us to obtain more a4bc and a3b3 terms simultaneously. In particular, it was necessary to cancel the a3b3 terms. I’ll note that this inequality is peculiar to sixth degree symmetry in three variables - it does not belong to a family of similar, nice inequalities. Finally, inequality (4), which is a handy corollary to (3), is another Schur. Every inequality we have used so far is quite sharp, and so it is no surprise that the leftovers are the comparatively weak AM-GM. 42. (Reid Barton, IMO Shortlist 03/A6.) Let n ≥ 2 be a positive integer and x1, x 2, . . . , x n,y1, y 2, . . . , y n a sequence of 2 n positive reals. Suppose z2, z 3, . . . , z 2n is such that z2
i+j
≥
xiyj for all i, j ∈ { 1, . . . , n }. Let M = max {z2, z 3, . . . , z 2n}. Prove that
(M + z2 + z3 + · · · + z2n
2n
)2
≥
(x1 + · · · + xn
n
) ( y1 + · · · + yn
n
)
Reid’s official solution. Let max( x1, . . . , x n) = max( y1, . . . , y n) = 1. (We can do this by factoring X from every xi, Y from every yj , and √XY from every zi+j
without changing the sign of the inequality.) We will prove M + z2 + · · · + z2n ≥
x1 + x2 + · · · + xn + y1 + y2 + · · · + yn, after which the desired follows by AM-GM. We will show that the number of terms on the left which are greater than r is at least as large as the number of terms on the right which are greater than r, for all r ≥ 0. For r ≥ 1, the claim is obvious, since all terms on the right are at most 1. Now take
r < 1. Let A and B denote the set of i for which xi > r and the set of j for which
yj > r respectively, and write a = |A|, b = |B|. Evidently, from our scaling, a, b ≥ 1. Now, xi > r and yj > r implies zi+j ≥ √xiyj ≥ r. Hence, if C is the set of k for which zk > r , we have |C| ≥ | A + B|, where the set addition is defined by the set of possible values if we take an element of A and add it to an element of B. How-ever, |A + B| ≥ | A| + |B| − 1, since if A and B consist of the values p1 < · · · < p a and
q1 < · · · < q b respectively we have all of the values p1 +q1 < . . . < p a +q1 < · · · < p a +qb
in A + B. Hence, |C| ≥ a + b − 1. Since |C| ≥ 1, there is some zk > r , and hence,
M > r . Therefore, the left side of the inequality in question has at least a + b terms which exceed r, as desired. •
30 The preponderance of difficulty here stemmed from dealing with the superabundance of givens, especially the mysterious M . Scaling allowed us to introduce some degree of control and, with marked audacity, a profoundly clever idea. As it turned out, the in-equality was no sharper than simple AM-GM! It is my opinion that it is highly unlikely that a problem as staggeringly pernicious as this one will appear on an Olympiad - at least in the foreseeable future. Nevertheless, I have included it here for the purpose of illustrating just how unusual and creative a solution can be.
3 Problems
(MOP 04) Show that for all positive reals a, b, c ,
( a + 2 ba + 2 c
)3
+
( b + 2 cb + 2 a
)3
+
(c + 2 ac + 2 b
)3
≥ 32. (MOP) Show that if k is a positive integer and a1, a 2, . . . , a n are positive reals which sum to 1, then n∏
i=1
1 − aki
aki
≥ (nk − 1)n
Let a1, a 2, . . . , a n be nonnegative reals with a sum of 1. Prove that
a1a2 + a2a3 + · · · + an−1an ≤ 144. (Ukraine 01) Let a, b, c, x, y, z be nonnegative reals such that x + y + z = 1. Show that
ax + by + cz + 2 √(ab + bc + ca )( xy + yz + zx ) ≤ a + b + c
Let n > 1 be a positive integer and a1, a 2, . . . , a n positive reals such that a1a2 . . . a n = 1. Show that 11 + a1
· · · + 11 + an
≤ a1 + · · · + an + n
(Aaron Pixton) Let a, b, c be positive reals with product 1. Show that 5 + ab + bc + ca ≥ (1 + a)(1 + b)(1 + c)7. (Valentin Vornicu 13 ) Let a, b, c, x, y, z be arbitrary reals such that a ≥ b ≥ c and either
x ≥ y ≥ z or x ≤ y ≤ z. Let f : R → R+0 be either monotonic or convex, and let k be a positive integer. Prove that
f (x)( a − b)k(a − c)k + f (y)( b − c)k(b − a)k + f (z)( c − a)k(c − b)k ≥ 0
13 This improvement is more widely known than the other one in this packet, and is published in his book,
Olimpiada de Matematica... de la provocare la experienta , GIL Publishing House, Zalau, Romania. (In English, “The Math Olympiad... from challenge to experience.”)
31 8. (IMO 01/2) Let a, b, c be positive reals. Prove that
a
√a2 + 8 bc + b
√b2 + 8 ca + c
√c2 + 8 ab ≥ 19. (USAMO 04/5) Let a, b, c be positive reals. Prove that
(a5 − a2 + 3 ) ( b5 − b2 + 3 ) ( c5 − c2 + 3 ) ≥ (a + b + c)3
(Titu Andreescu) Show that for all nonzero reals a, b, c ,
a2
b2 + b2
c2 + c2
a2 ≥ ac + cb + ba
(IMO 96 Shortlist) Let a, b, c be positive reals with abc = 1. Show that
ab a5 + b5 + ab + bc b5 + c5 + bc + ca c5 + a5 + ca ≤ 112. Let a, b, c be positive reals such that a + b + c = 1. Prove that
√ab + c + √bc + a + √ca + b ≥ 1 + √ab + √bc + √ca
(APMO 2005/2) Let a, b, c be positive reals with abc = 8. Prove that
a2
√(a3 + 1) ( b3 + 1) + b2
√(b3 + 1) ( c3 + 1) + c2
√(c3 + 1) ( a3 + 1) ≥ 4314. Show that for all positive reals a, b, c ,
a3
b2 − bc + c2 + b3
c2 − ca + a2 + c3
a2 − ab + b2 ≥ a + b + c
(USAMO 97/5) Prove that for all positive reals a, b, c ,1
a3 + b3 + abc + 1
b3 + c3 + abc + 1
c3 + a3 + abc ≤ 1
abc
(Mathlinks Lore) Show that for all positive reals a, b, c, d with abcd = 1, and k ≥ 2, 1(1 + a)k + 1(1 + b)k + 1(1 + c)k + 1(1 + d)k ≥ 22−k
(IMO 05/3) Prove that for all positive a, b, c with product at least 1,
a5 − a2
a5 + b2 + c2 + b5 − b2
b5 + c2 + a2 + c5 − c2
c5 + a2 + b2 ≥ 032 18. (Mildorf) Let a, b, c, k be positive reals. Determine a simple, necessary and sufficient condition for the following inequality to hold: (a + b + c)k (akbk + bkck + ckak) ≤ (ab + bc + ca )k(ak + bk + ck)19. Let a, b, c be reals with a + b + c = 1 and a, b, c ≥ − 34 . Prove that
aa2 + 1 + bb2 + 1 + cc2 + 1 ≤ 910 20. (Mildorf) Show that for all positive reals a, b, c ,
3
√4a3 + 4 b3 + 3
√4b3 + 4 c3 + 3
√4c3 + 4 a3 ≤ 4a2
a + b + 4b2
b + c + 4c2
c + a
Let a, b, c, x, y, z be real numbers such that (a + b + c)( x + y + z) = 3 , (a2 + b2 + c2)( x2 + y2 + z2) = 4 Prove that
ax + by + cz ≥ 022. (Po-Ru Loh) Let a, b, c be reals with a, b, c > 1 such that 1
a2 − 1 + 1
b2 − 1 + 1
c2 − 1 = 1 Prove that 1
a + 1 + 1
b + 1 + 1
c + 1 ≤ 123. (Weighao Wu) Prove that (sin x)sin x < (cos x)cos x
for all real numbers 0 < x < π
4
.24. (Mock IMO 05/2) Let a, b, c be positive reals. Show that 1 < a
√a2 + b2 + b
√b2 + c2 + c
√c2 + a2 ≤ 3√2225. (Gabriel Dospinescu) Let n ≥ 2 be a positive integer. Show that for all positive reals
a1, a 2, . . . , a n with a1a2 . . . a n = 1,
√a21 + 1 2 + · · · +
√a2
n
1 2 ≤ a1 + · · · + an
33 26. Let n ≥ 2 be a positive integer, and let k ≥ n−1
n
be a real number. Show that for all positive reals a1, a 2, . . . , a n,
( (n − 1) a1
a2 + · · · + an
)k
+
( (n − 1) a2
a3 + · · · + an + a1
)k
· · · +
( (n − 1) an
a1 + · · · + an−1
)k
≥ n
(Mildorf) Let a, b, c be arbitrary reals such that a ≥ b ≥ c, and let x, y, z be nonnegative reals with x + z ≥ y. Prove that
x2(a − b)( a − c) + y2(b − c)( b − a) + z2(c − a)( c − b) ≥ 0and determine where equality holds. 28. (USAMO 00/6) Let n ≥ 2 be an integer and S = {1, 2, . . . , n }. Show that for all nonnegative reals a1, a 2, . . . , a n, b 1, b 2, . . . , b n,
∑
i,j ∈S
min {aiaj , b ibj } ≤ ∑
i,j ∈S
min {aibj , a j bi}
(Kiran Kedlaya) Show that for all nonnegative a1, a 2, . . . , a n,
a1 + √a1a2 + · · · + n
√a1 · · · an
n ≤ n
√
a1 · a1 + a2
2 · · · a1 + · · · + an
n
(Vascile Cartoaje) Prove that for all positive reals a, b, c such that a + b + c = 3,
aab + 1 + bbc + 1 + cca + 1 ≥ 3231. (Gabriel Dospinescu) Prove that ∀a, b, c, x, y, z ∈ R+| xy + yz + zx = 3,
a(y + z)
b + c + b(z + x)
c + a + c(x + y)
a + b ≥ 332. (Mildorf) Let a, b, c be non-negative reals. Show that for all real k,
∑
cyc
max( ak, b k)( a − b)2
2 ≥ ∑
cyc
ak(a − b)( a − c) ≥ ∑
cyc
min( ak, b k)( a − b)2
2(where a, b, c 6 = 0 if k ≤ 0) and determine where equality holds for k > 0, k = 0, and
k < 0 respectively. 33. (Vascile Cartoaje) Let a, b, c, k be positive reals. Prove that
ab + ( k − 3) bc + ca
(b − c)2 + kbc + bc + ( k − 3) ca + ab
(c − a)2 + kca + ca + ( k − 3) ab + bc
(a − b)2 + kab ≥ 3( k − 1)
k
(Taiwan? 02) Show that for all positive a, b, c, d ≤ k, we have
abcd
(2 k − a)(2 k − b)(2 k − c)(2 k − d) ≤ a4 + b4 + c4 + d4
(2 k − a)4 + (2 k − b)4 + (2 k − c)4 + (2 k − d)4
34 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.